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Plant cells (possessing cellulose) and fungal cells (possessing chitin) both have a rigid cell wall that provides structural support and protection. Animal cells lack a cell wall entirely. Chloroplasts are unique to photosynthesising plant cells and are not found in fungi or animals. Large, permanent vacuoles are typical of plant cells, and mitochondria are found in all three eukaryotic types.

1 mark for identifying cell wall (A).

When the concentration of mineral ions in the soil is lower than inside the root hair cell, movement must occur against a concentration gradient. This requires active transport, an active process that uses energy released from aerobic respiration in the form of ATP.

1 mark for identifying active transport requiring respiratory energy (A).

Wind-pollinated flowers have adapted to catch pollen blowing in the wind. Feathery, exposed stigmas hang outside the flower to provide a large surface area to trap drifting pollen grains. Bright petals, sticky pollen, and nectaries are all features characteristic of insect-pollinated flowers to attract animal vectors.

1 mark for identifying the correct adaptation of wind-pollinated flowers (A).

In a spinal reflex arc, the stimulus is detected by a receptor. The electrical impulse travels along the sensory neurone to the central nervous system, where it passes across a synapse to a relay neurone, and then across another synapse to a motor neurone. Finally, the impulse reaches the effector (muscle or gland) to produce a rapid response.

1 mark for the correct sequence of neurones (A).

Arteries must withstand and maintain high blood pressure from the heart, so they have thick, muscular, and elastic walls. Veins carry blood at low pressure back to the heart, have a wider lumen, and contain valves to prevent backflow of blood.

1 mark for identifying the correct structural difference (A).

The heterozygous black guinea pig has the genotype Bb. The brown guinea pig, being recessive, must have the genotype bb. A genetic cross (Bb \(\times\) bb) produces offspring genotypes in a 1:1 ratio: \(50\%\) Bb (heterozygous black) and \(50\%\) bb (homozygous recessive brown). Thus, the probability of brown fur is 0.50.

1 mark for calculating the probability as 0.50 (A).

Fertiliser runoff causes nutrient enrichment leading to an algal bloom. The thick layer of algae blocks light from reaching submerged plants, which subsequently die. Aerobic bacteria decompose the dead plants, rapidly consuming dissolved oxygen. This oxygen depletion causes aquatic animals, such as fish, to suffocate and die.

1 mark for the correct chronological sequence of eutrophication events (A).

Sweating causes water loss, making the blood plasma more concentrated (low water potential). The hypothalamus detects this change and stimulates the pituitary gland to release more ADH. ADH increases the permeability of the collecting ducts in the kidney, so more water is reabsorbed back into the blood, resulting in a smaller volume of highly concentrated urine.

1 mark for identifying that ADH levels increase and urine volume decreases (A).

When plant cells are placed in a concentrated sucrose solution, the solution outside has a lower water potential than the cell sap inside the vacuole. Therefore, water moves out of the vacuole and cytoplasm down a water potential gradient by osmosis. This loss of water causes the cell membrane to pull away from the cell wall, leaving the cells plasmolysed.

A is the correct option (1 mark) because water leaves the cell by osmosis causing plasmolysis. B is incorrect as water would enter if the outside water potential was higher. C and D are incorrect because sucrose does not actively transport across the membrane to cause this immediate physical change.

To focus on a near object, light rays must be refracted (bent) more sharply. The ciliary muscles contract, which reduces the tension on the suspensory ligaments, causing them to slacken (loosen). This allows the lens to return to its natural, more rounded and thicker shape, increasing its refractive power.

A is the correct option (1 mark). B describes the process of accommodation for a distant object. C and D represent incorrect physiological combinations of muscle contraction and ligament tension.

A mature red blood cell is specialized to transport oxygen, containing no mitochondria so it does not consume this oxygen. Instead, it respires anaerobically. In contrast, skeletal muscle cells require a continuous and rapid supply of ATP energy for active contraction. They contain many mitochondria to carry out aerobic respiration to meet this high energy demand.

1 mark: Explanation of why red blood cells do not have mitochondria (e.g., they respire anaerobically or to maximize space for haemoglobin/oxygen transport). 1 mark: Explanation of why muscle cells have many mitochondria (to release energy/ATP via aerobic respiration). 0.5 marks: Linking muscle energy requirement to contraction or physical movement.

When red onion cells are placed in a concentrated (hypertonic) sodium chloride solution, the water potential outside the cell is lower than inside. Water moves out of the cell's vacuole and cytoplasm across the partially permeable cell membrane by osmosis. This loss of water causes the vacuole to shrink, pulling the cell membrane away from the cell wall, leaving the cell plasmolysed.

1 mark: Correct description of plasmolysis (cell membrane/cytoplasm pulling away from the cell wall). 1 mark: Explanation that water moves out of the cell by osmosis down a water potential gradient (from high to low water potential). 0.5 marks: Identifying that this movement occurs across a partially/selectively permeable membrane.

During ventilation, the lungs continuously expand and contract within the thoracic cavity. The pleural membranes produce pleural fluid, which fills the pleural cavity. This fluid acts as a physical lubricant, reducing friction between the outer surface of the lungs and the inner rib cage. This prevents physical wear and damage to the delicate lung tissues during breathing.

1 mark: Stating that pleural fluid acts as a lubricant. 1 mark: Explaining that this reduces friction between the lungs and the rib cage / chest wall during expansion and deflation. 0.5 marks: Correctly naming the pleural membranes/cavity as the site of this fluid.

When a plant shoot is exposed to light from one direction (unidirectional light), the hormone auxin, which is synthesized in the shoot tip, diffuses downwards. It accumulates in greater concentrations on the shaded side of the shoot. Higher concentrations of auxin stimulate cell elongation on this shaded side, causing it to grow faster than the light side, resulting in the shoot bending towards the light.

1 mark: Mentioning that auxin is produced at the tip and accumulates on the shaded/darker side of the shoot. 1 mark: Explaining that auxin stimulates cell elongation on this shaded side. 0.5 marks: Concluding that uneven growth/elongation causes the shoot to bend towards the light source.

Xylem and phloem are the vascular tissues of plants. Structurally, xylem vessels are made of dead cells with no end walls, reinforced with waterproof lignin, whereas phloem sieve tubes consist of living cells with perforated end plates (sieve plates) supported by companion cells. Xylem transports water and dissolved mineral ions upward from the roots, whereas phloem transports organic nutrients like sucrose and amino acids bidirectionally.

1 mark: Stating at least one structural difference (e.g., dead vs living cells, lignin present in xylem vs sieve plates in phloem). 1 mark: Identifying the substances transported by both (xylem transports water/minerals AND phloem transports sucrose/amino acids). 0.5 marks: Correctly linking both tissues to their respective functions without confusion.

Ultrafiltration occurs in the nephron under high hydrostatic pressure. The afferent arteriole entering the glomerulus is wider in diameter than the efferent arteriole leaving it, which creates a high-pressure filter system. This pressure forces small molecules (such as water, glucose, urea, and salts) out of the blood capillaries, through the tiny pores in the capillary walls and the basement membrane, and into the Bowman's capsule as glomerular filtrate. Large proteins and blood cells are too large to pass through the filter.

1 mark: Explaining how high pressure is created (afferent arteriole wider than efferent arteriole). 1 mark: Describing how small molecules (glucose, urea, water, ions) are forced through the capillary walls/basement membrane. 0.5 marks: Stating that large components (proteins, red blood cells) cannot pass through due to size.

The ABO blood group system is determined by three alleles: \(I^A\), \(I^B\), and \(I^O\). The alleles \(I^A\) and \(I^B\) are codominant, while \(I^O\) is recessive. For a child to have blood group O, they must be homozygous recessive with the genotype \(I^O I^O\). This means they must inherit one \(I^O\) allele from each parent. Therefore, the parent with blood group A must be heterozygous with genotype \(I^A I^O\), and the parent with blood group B must be heterozygous with genotype \(I^B I^O\). The cross yields a 25% probability of producing an \(I^O I^O\) offspring.

1 mark: Stating that both parents must be heterozygous (genotypes \(I^A I^O\) and \(I^B I^O\)). 1 mark: Explaining that the child must inherit one recessive \(I^O\) allele from each parent to have genotype \(I^O I^O\). 0.5 marks: Stating that the probability of this occurrence is 25% or 1 in 4.

To maximize the rate of photosynthesis, a grower must overcome limiting factors. First, they can use heaters to maintain a temperature that matches the optimum for photosynthetic enzymes. Second, they can burn fuels like paraffin, which releases carbon dioxide gas, providing more reactant for the light-independent stage of photosynthesis. Third, they can install artificial grow lights to maintain high light intensity and prolong day length, ensuring continuous light absorption by chlorophyll.

1 mark: Explaining how temperature control (heaters) optimizes enzyme activity for photosynthesis. 1 mark: Explaining how increasing carbon dioxide concentration (burning paraffin) provides more reactant for the process. 0.5 marks: Explaining how artificial lighting maintains high light intensity or extends photoperiod.

Root hair cells are specialized cells found in plant roots. Their primary adaptation is a long, finger-like projection which significantly increases the surface area to volume ratio of the cell, allowing more water to be absorbed via osmosis and more mineral ions via active transport. Additionally, they have a thin cell wall, which shortens the distance that water and minerals must travel. They also contain a high density of mitochondria to release ATP through aerobic respiration, which powers the active transport of mineral ions against their concentration gradient.

1 mark: Mentioning the long, hair-like projection increases the surface area for osmosis or active transport. 1 mark: Explaining that the thin cell wall provides a short diffusion distance. 0.5 marks: Stating that they contain many mitochondria to provide ATP or energy for the active transport of mineral ions.

Increasing the temperature increases the kinetic energy of both the particles and the enzymes involved in aerobic respiration. This leads to an increased rate of respiration, which produces more ATP (energy). Since active transport is an active process requiring energy from ATP, more active transport can occur, increasing its rate. (At temperatures above 40 degrees, the carrier proteins and respiratory enzymes would denature, stopping the process, but within the 20 to 40 degree range, the rate increases).

1 mark: Stating that the rate of active transport increases because respiration rate increases. 1 mark: Explaining that higher temperature increases kinetic energy and produces more ATP. 0.5 marks: Identifying that active transport requires energy or ATP from respiration to move ions against a concentration gradient.

Alveoli have several key adaptations: 1. They provide a massive total surface area because there are millions of them in each lung. 2. Their walls are extremely thin (only one cell thick, made of squamous epithelium), and the adjacent capillary walls are also only one cell thick, minimizing the diffusion distance for oxygen and carbon dioxide. 3. A dense network of capillaries ensures a continuous flow of blood, maintaining a steep concentration gradient for diffusion.

1 mark: Mentioning the thin walls (one cell thick) which provide a short diffusion path. 1 mark: Mentioning the large total surface area provided by millions of alveoli. 0.5 marks: Mentioning the rich capillary blood supply or constant ventilation that maintains a steep concentration gradient.

When the hand touches a hot object, thermoreceptors in the skin detect the heat. They initiate an electrical impulse that travels along the sensory neurone to the spinal cord (CNS). In the spinal cord, the impulse crosses a synapse to a relay neurone, and then across another synapse to a motor neurone. The motor neurone carries the impulse to the effector (biceps muscle), causing it to contract and withdraw the hand. This pathway is fast because it bypasses the conscious areas of the brain, involving a minimal number of synapses.

1 mark: Describing the pathway of the impulse (sensory neurone to relay neurone in spinal cord, then to motor neurone). 1 mark: Identifying the effector as a muscle that contracts to cause the withdrawal movement. 0.5 marks: Explaining that bypassing the conscious brain or having few synapses makes the response extremely rapid and protective.

Capillaries are adapted for exchange in several ways. Firstly, their walls are composed of a single layer of endothelial cells (one cell thick), which provides an extremely short diffusion distance for oxygen, glucose, and carbon dioxide. Secondly, the capillary lumen is very narrow (about the width of a single red blood cell), which slows down blood flow and forces red blood cells close to the wall, reducing diffusion distance and allowing more time for exchange. Finally, capillaries have tiny gaps or pores between endothelial cells, making them highly permeable to small soluble molecules.

1 mark: Stating that walls are one cell thick to provide a short diffusion distance. 1 mark: Explaining that a narrow lumen slows blood flow or pushes red blood cells close to the vessel wall to facilitate diffusion. 0.5 marks: Mentioning that capillary walls are permeable or have pores to allow small molecules to pass through.

When a person is dehydrated, osmoreceptors in the hypothalamus detect the low water potential of the blood. This stimulates the pituitary gland to release more Antidiuretic Hormone (ADH) into the bloodstream. ADH travels to the kidneys, where it increases the permeability of the collecting ducts to water. As a result, more water is reabsorbed by osmosis back into the blood capillaries. This restores the blood water potential towards normal and produces a small, highly concentrated volume of urine.

1 mark: Stating that the pituitary gland releases more ADH in response to dehydration. 1 mark: Explaining that ADH increases the permeability of the kidney collecting ducts, leading to more water reabsorption. 0.5 marks: Stating that this results in a smaller, more concentrated volume of urine.

Biological control involves introducing a natural predator, parasite, or pathogen to target a specific pest species. This is more sustainable than chemical pesticides because: 1. Pests do not develop resistance to biological control agents, unlike chemical pesticides where resistance rapidly evolves. 2. Biological controls do not cause environmental pollution, bioaccumulation, or biomagnification up the food chain. 3. It is highly specific, meaning non-target beneficial insects (like bees) are not harmed. 4. Once established, the biological control agent can self-sustain, whereas pesticides require expensive, repeated chemical applications.

1 mark: Explaining that pests cannot develop resistance to predators/pathogens. 1 mark: Explaining that there is no chemical pollution or risk of bioaccumulation/biomagnification in the food chain. 0.5 marks: Mentioning that biological control is highly specific and does not harm non-target species, or that it is self-sustaining.

Untreated sewage contains organic waste and high concentrations of minerals such as nitrates and phosphates. When discharged into a river, it provides a rich food source for aerobic decomposers (bacteria). These bacteria multiply rapidly and decompose the organic matter. In doing so, they carry out aerobic respiration, consuming vast amounts of dissolved oxygen from the water. This rapid depletion of oxygen levels creates an anoxic environment, meaning fish and other aquatic animals cannot respire and consequently suffocate and die.

1 mark: Stating that decomposers (bacteria) feed on the organic matter in sewage and multiply rapidly. 1 mark: Explaining that these bacteria use up dissolved oxygen during aerobic respiration. 0.5 marks: Linking the resulting low oxygen levels to the suffocation/death of fish and other aquatic organisms.

Mature red blood cells are biconcave discs that lack a nucleus. This absence of a nucleus allows more space inside the cell for haemoglobin, which binds to oxygen and increases the oxygen-carrying capacity of the blood. However, because the nucleus contains the cell's DNA, its absence means the cell lacks the genetic instructions and structures necessary to replicate its organelles and genetic material to undergo mitosis.

1 mark for stating that the lack of nucleus allows more space for haemoglobin or oxygen transport. 1 mark for explaining that the lack of nucleus means there is no DNA or genetic material. 0.5 marks for concluding that without DNA, the cell cannot undergo mitosis or cell division.

Both cells are in a hypertonic environment, meaning the water potential outside is lower than inside. Water leaves both cells by osmosis down a water potential gradient. In the plant cell, the vacuole shrinks and the cell membrane pulls away from the cell wall, becoming plasmolysed, but the rigid cellulose cell wall prevents it from collapsing completely. In the red blood cell, which lacks a cell wall, the loss of water causes the entire cell membrane to shrivel and crenate.

1 mark for identifying that water leaves both cells by osmosis. 1 mark for explaining that the plant cell becomes plasmolysed but its cell wall maintains structural support. 0.5 marks for explaining that the red blood cell shrivels or crenates because it lacks a cell wall.

As temperature increases beyond the optimum, the increased kinetic energy causes the enzyme molecule to vibrate excessively. This breaks the hydrogen and ionic bonds holding the tertiary structure of the protein together. Consequently, the shape of the active site changes permanently (denaturation), meaning the substrate is no longer complementary and cannot bind to form enzyme-substrate complexes.

1 mark for stating that bonds within the enzyme are broken by high kinetic energy or vibrations. 1 mark for explaining that the active site changes shape or denatures. 0.5 marks for stating that the substrate can no longer fit or bind to form enzyme-substrate complexes.

Alveolar surfactant is a phospholipid mixture that lowers the surface tension of the moisture layer inside the alveoli, preventing them from sticking together and collapsing during exhalation. The alveolar wall is extremely thin, consisting of a single layer of squamous epithelial cells, which minimizes the physical distance oxygen and carbon dioxide must diffuse, significantly speeding up the rate of gas exchange.

1 mark for explaining that surfactant reduces surface tension to prevent alveolar collapse or keeps them open. 1 mark for stating the alveolar wall is one cell thick or extremely thin. 0.5 marks for linking the thinness of the wall to a short diffusion distance that increases the rate of diffusion.

Transport in the phloem (translocation) is active because companion cells must use energy in the form of ATP to actively pump sucrose into sieve tube elements against a concentration gradient. In contrast, movement in the xylem is passive because it is driven by transpiration: the evaporation of water from leaves creates a tension (negative pressure) that pulls water molecules up the stem through cohesive and adhesive forces, requiring no metabolic energy from the plant.

1 mark for stating phloem transport involves active loading of sucrose using energy or ATP. 1 mark for explaining xylem transport is driven by transpiration pull or evaporation from leaves. 0.5 marks for clarifying that xylem transport does not require metabolic energy (ATP) from the plant.

The afferent arteriole, which brings blood into the glomerulus, has a wider lumen diameter than the efferent arteriole, which carries blood away. This bottleneck effect creates high hydrostatic pressure within the glomerular capillaries. This high pressure is essential because it acts as the driving force to push water, glucose, amino acids, urea, and inorganic ions across the basement membrane and into the Bowman's capsule, while larger proteins and blood cells remain behind.

1 mark for stating that the afferent arteriole has a wider diameter than the efferent arteriole. 1 mark for explaining that this difference in diameter creates a high hydrostatic pressure. 0.5 marks for stating that this pressure is needed to force water and small solutes through the basement membrane into the Bowman's capsule.

During bread production, yeast is mixed with flour and water, where it undergoes anaerobic respiration (fermentation) due to low oxygen levels inside the dough. This process produces carbon dioxide gas and ethanol. The carbon dioxide gas gets trapped as bubbles within the gluten network of the dough, causing it to expand and rise. During the baking process, the high temperatures of the oven cause any ethanol (alcohol) that was produced to evaporate, leaving the baked bread alcohol-free.

1 mark for explaining that yeast respires anaerobically to produce carbon dioxide which forms gas bubbles. 1 mark for stating that these carbon dioxide bubbles expand the dough. 0.5 marks for explaining that the ethanol produced evaporates during the baking process.

When nitrate fertilisers leach into a pond, they act as a rich nutrient source, causing rapid growth of algae on the water surface (an algal bloom). This thick layer of algae blocks sunlight from reaching aquatic plants deeper in the water, preventing photosynthesis and causing them to die. Decomposing bacteria then multiply rapidly as they feed on the dead plant matter. These bacteria respire aerobically, consuming vast amounts of dissolved oxygen, which severely lowers the oxygen concentration of the pond.

1 mark for explaining that nitrates cause an algal bloom which blocks sunlight and kills submerged plants. 1 mark for explaining that decomposers or bacteria break down the dead plants. 0.5 marks for stating that these bacteria consume oxygen through aerobic respiration, lowering dissolved oxygen levels.

Root hair cells are found in soil/underground where light cannot penetrate. Since chloroplasts absorb light energy for photosynthesis, they are not needed in these cells. The vacuole contains concentrated cell sap, which maintains a lower water potential than the surrounding soil water, enabling water absorption by osmosis and maintaining cell turgor.

1 mark for explaining that underground cells receive no light and thus cannot photosynthesise. 0.5 mark for identifying that chloroplasts are therefore redundant. 1 mark for vacuole function (0.5 mark for maintaining low water potential / osmosis and 0.5 mark for cell turgidity).

The pleural membranes form a double envelope surrounding the lungs. The space between them (the pleural cavity) contains pleural fluid. This fluid acts as a lubricant, allowing the lungs to slide smoothly over the chest wall during inhalation and exhalation, preventing friction and potential tearing of the lung tissues.

1 mark for mentioning the presence of pleural fluid within the pleural cavity. 1 mark for identifying its role as a lubricant that reduces friction. 0.5 mark for connecting this to the protection of lung tissues during chest expansion/movement.

When humidity is low, the concentration of water vapor in the atmosphere is much lower than inside the air spaces of the leaf, establishing a steep concentration gradient that speeds up diffusion out through the stomata. At the same time, higher temperatures increase the kinetic energy of water molecules, which increases the rate of evaporation from the wet cell walls of the spongy mesophyll.

1 mark for stating that lower humidity increases/maintains the water vapor concentration gradient. 1 mark for stating that higher temperature increases the kinetic energy of water molecules. 0.5 mark for linking this kinetic energy to faster evaporation from mesophyll cell surfaces.

Codominance is a genetic scenario where neither of the two alleles is dominant over the other, so both contribute to the phenotype of a heterozygote. In the ABO system, alleles \(I^A\) and \(I^B\) are codominant. A person inheriting one of each has the genotype \(I^AI^B\) and the phenotype blood group AB, displaying both antigen A and antigen B on their red blood cells.

1 mark for the definition of codominance (both alleles expressed in the phenotype). 0.5 mark for identifying that the alleles \(I^A\) and \(I^B\) are codominant. 1 mark for explaining that genotype \(I^AI^B\) leads to phenotype AB with both antigens present on the red blood cells.

Consuming a high salt diet decreases the water potential of the blood plasma. Osmoreceptors in the hypothalamus detect this change and stimulate the pituitary gland to secrete more ADH into the bloodstream. The increased ADH concentration makes the walls of the kidney collecting ducts more permeable to water. Consequently, more water is reabsorbed back into the blood, leading to the production of a smaller volume of highly concentrated urine.

0.5 mark for noting that a high salt diet lowers the water potential of the blood. 1 mark for stating that the pituitary gland releases more ADH in response. 0.5 mark for explaining that this increases the permeability of the collecting ducts to water, increasing reabsorption. 0.5 mark for stating that this produces a smaller volume of urine.

Sewage contains high concentrations of organic waste and mineral ions. This provides a source of nutrients for decomposers such as bacteria, causing their population to increase rapidly. As these bacteria break down the organic matter, they respire aerobically, using up large quantities of the dissolved oxygen in the river water.

1 mark for explaining that sewage provides organic waste/nutrients for bacteria/decomposers. 0.5 mark for stating that this leads to rapid bacterial reproduction/growth. 1 mark for explaining that the bacteria use up dissolved oxygen during aerobic respiration.

To find the actual size from magnification, use the formula: \(\text{Actual Size} = \frac{\text{Image Size}}{\text{Magnification}}\). First, convert the image size from millimetres (\(\text{mm}\)) to micrometres (\(\mu\text{m}\)): \(45\text{ mm} \times 1000 = 45,000\\ \mu\text{m}\). Next, divide by the magnification: \(\text{Actual Size} = \frac{45,000\\ \mu\text{m}}{15,000} = 3\\ \mu\text{m}\).

- 1 mark for converting mm to \(\mu\text{m}\) (\(45 \times 1000 = 45,000\)) OR for dividing image size by magnification first to get \(0.003\text{ mm}\) - 1 mark for dividing \(45,000\) by \(15,000\) OR for converting \(0.003\text{ mm}\) to \(3\\ \mu\text{m}\) - 0.5 marks for correct final answer: 3

First, calculate the volume of water taken up by multiplying the distance the bubble moved by the cross-sectional area of the tube: \(60\text{ mm} \times 0.8\text{ mm}^2 = 48\text{ mm}^3\). Next, calculate the rate per hour. Since there are \(15\text{ minutes}\) in the measurement period, and \(60\text{ minutes}\) in an hour, the scaling factor is \(60 / 15 = 4\). Therefore, the rate of water uptake is \(48\text{ mm}^3 \times 4 = 192\text{ mm}^3\text{ per hour}\).

- 1 mark for calculating the volume of water taken up: \(60 \times 0.8 = 48\text{ mm}^3\) - 1 mark for dividing by time and converting to hours: \((48 / 15) \times 60 = 192\) - 0.5 marks for the correct final answer: 192

To calculate the percentage efficiency of energy transfer, divide the energy in the secondary consumers by the energy in the primary consumers and multiply by 100: \(\text{Efficiency} = (1,920 / 24,000) \times 100 = 0.08 \times 100 = 8\\%\).

- 1 mark for the correct division setup: \(1,920 / 24,000\) - 1 mark for multiplying the fraction by 100 to find the percentage - 0.5 marks for the correct final answer: 8

First, calculate the total area sampled by the quadrats: \(10 \times 0.25\text{ m}^2 = 2.5\text{ m}^2\). Next, determine the multiplier to scale up to the entire field: \(800\text{ m}^2 / 2.5\text{ m}^2 = 320\). Finally, multiply the total number of dandelions counted by this multiplier: \(35 \times 320 = 11,200\) dandelions. Alternatively, find the mean density per square metre: \(35 / 2.5 = 14\) dandelions per \(\text{m}^2\), then multiply by the total field area: \(14 \times 800 = 11,200\).

- 1 mark for calculating the total area sampled: \(10 \times 0.25 = 2.5\text{ m}^2\) OR for calculating the mean number of dandelions per quadrat: \(3.5\) - 1 mark for multiplying the mean density by the total area: \((35 / 2.5) \times 800\) OR \((3.5 / 0.25) \times 800\) - 0.5 marks for the correct final answer: 11,200

To investigate the effect of glucose concentration on the rate of anaerobic respiration in yeast, a CORMS approach should be used:

- **C (Change):** Use at least five different concentrations of glucose solution (e.g., 0%, 2%, 4%, 6%, and 8%).
- **O (Organism):** Use the same species of yeast (e.g., *Saccharomyces cerevisiae*) and ensure the yeast suspension is of the same starting concentration and age.
- **R (Repeat):** Repeat the experiment at least three times for each glucose concentration to calculate a mean and identify any anomalous results.
- **M1 (Measurement 1):** Measure the volume of carbon dioxide gas produced (using a gas syringe or an inverted measuring cylinder over water) OR count the number of bubbles produced.
- **M2 (Measurement 2):** Measure this volume or count the bubbles over a fixed time period (e.g., 10 minutes or per minute).
- **S (Same / Standardisation):** Control other variables that could affect the rate of reaction:
- Keep the temperature constant using a thermostatically controlled water bath (e.g., at 35°C).
- Keep the volume of the yeast suspension constant.
- Ensure anaerobic conditions are maintained, for example, by adding a layer of liquid paraffin (oil) on top of the mixture to prevent oxygen from entering.

One mark for each point, up to a maximum of 6 marks:

- **C (Change):** Identify at least 5 different concentrations of glucose used (e.g. 0% to 8% / a range of concentrations) (1)
- **O (Organism):** Use the same species / concentration / volume of yeast suspension (1)
- **R (Repeat):** Repeat the investigation at least three times at each concentration to calculate a mean / identify anomalies (1)
- **M1 (Measure 1):** Measure the volume of carbon dioxide produced / count the number of bubbles (1)
- **M2 (Measure 2):** State a specific time period for the measurement (e.g. in 5 minutes / per minute) (1)
- **S1 & S2 (Same):** Control two of the following variables (1 mark for each, max 2):
- Maintain a constant temperature using a water bath (1)
- Keep the pH constant using a buffer (1)
- Keep the volume of yeast/glucose mixture constant (1)
- Maintain anaerobic conditions using a layer of paraffin/oil on top (1)

(a)
1. First, calculate the change in mass:
\(\text{Change in mass} = \text{Final mass} - \text{Initial mass} = 2.96\text{ g} - 3.20\text{ g} = -0.24\text{ g}\)
2. Next, calculate the percentage change:
\(\text{Percentage change} = \frac{-0.24}{3.20} \times 100 = -7.5\%\) (or a decrease of \(7.5\%\)).

(b)
- The water potential of the surrounding solution (pure water / 0.0 mol/dm³) is higher than the water potential inside the vacuole/cytoplasm of the potato cells.
- Water enters the potato cells by osmosis.
- Osmosis is the net movement of water from a region of higher water potential to a region of lower water potential across a partially permeable membrane.
- The entry of water increases the volume and mass of the potato cylinder.

(a) [2 marks total]
- Award 1 mark for correct working showing calculation of change in mass and division by initial mass: \(\frac{2.96 - 3.20}{3.20}\) or \(\frac{-0.24}{3.20}\) or \(\frac{0.24}{3.20}\)
- Award 1 mark for correct final answer: -7.5% / 7.5% decrease (accept 7.5% with working showing decrease, reject 7.5% without indicating it is a loss/negative unless formula is shown)

(b) [3 marks total]
- Award 1 mark for stating that the water potential of the solution (0.0 mol/dm³) is higher than the water potential inside the potato cells.
- Award 1 mark for stating that water moves into the potato cells by osmosis / down the water potential gradient.
- Award 1 mark for mentioning the partially permeable membrane.

(a)
- As light intensity increases from 5 to 25 arbitrary units, the volume of oxygen produced per minute (rate of photosynthesis) increases from 12 to 48 \(\text{mm}^3/\text{minute}\).
- Above 25 arbitrary units, further increases in light intensity (to 30 arbitrary units) do not change the rate of photosynthesis, which remains constant at 48 \(\text{mm}^3/\text{minute}\).

(b)
- Below 25 arbitrary units, light intensity is the limiting factor for photosynthesis.
- Above 25 arbitrary units, light intensity is no longer the limiting factor.
- This is because some other factor is now limiting the rate of photosynthesis.
- Examples of other limiting factors include temperature or carbon dioxide concentration.
- Since these factors are at a constant, sub-optimal level, the rate of photosynthesis cannot increase any further even if light intensity continues to rise.

(a) [2 marks total]
- Award 1 mark for stating that the rate of photosynthesis / volume of oxygen increases as light intensity increases up to 25 arbitrary units.
- Award 1 mark for stating that the rate levels off / becomes constant / remains at 48 \(\text{mm}^3/\text{minute}\) at or after 25 arbitrary units.

(b) [3 marks total]
- Award 1 mark for identifying that light intensity is no longer the limiting factor.
- Award 1 mark for stating that another factor is now limiting the rate of photosynthesis.
- Award 1 mark for providing a correct example of another limiting factor, such as carbon dioxide concentration or temperature.

Sterile air is bubbled into the fermenter to provide oxygen, which is essential for the aerobic respiration of the Penicillium fungus. The air is sterilized (filtered) to prevent contamination by other unwanted microorganisms that might compete for nutrients or spoil the product.

1 mark for the correct answer (B). Reject all other options.

When focusing on a near object (accommodation), the ciliary muscles contract. This causes the suspensory ligaments to slacken (become loose). Consequently, the elastic lens becomes rounder and fatter (more convex), refracting light rays more strongly to focus the image on the retina.

1 mark for the correct answer (A). Reject all other options.

Drinking a large volume of water raises the water potential of the blood. The pituitary gland releases less ADH into the bloodstream. A lower concentration of ADH makes the collecting ducts of the nephrons less permeable to water, meaning less water is reabsorbed. This results in a large volume of dilute urine being produced.

1 mark for the correct answer (A). Reject all other options.

First, biological control agents (like wrasse) are specific to the pest (sea lice) and do not harm other non-target marine species, whereas chemical pesticides can toxic to other aquatic life. Second, there is no risk of bioaccumulation or biomagnification up the marine food chain, which can occur with persistent chemical pesticides. Third, the pest cannot develop genetic resistance to being eaten by a predator, whereas they can rapidly evolve resistance to chemical pesticides.

1 mark for identifying that biological control does not cause bioaccumulation/environmental persistence. 1 mark for explaining that biological control is highly specific and does not harm non-target marine species. 1 mark for stating that pests cannot develop resistance to being eaten by a predator.

High pressure is created in the glomerulus because the afferent arteriole entering the glomerulus is wider than the efferent arteriole leaving it. This high pressure forces small molecules (such as water, glucose, urea, amino acids, and salts) out of the capillary blood. The basement membrane acts as a selective filter, allowing these small molecules to pass into the Bowman's capsule while preventing large proteins and red blood cells from leaving the capillary.

1 mark for explaining that high pressure is generated because the afferent arteriole is wider than the efferent arteriole. 1 mark for stating that small molecules (water, glucose, urea, etc.) are forced out of the blood. 1 mark for describing the role of the basement membrane as a filter that prevents large molecules (proteins/blood cells) from passing.

To focus on a near object, the ciliary muscles in the eye contract. This contraction causes the suspensory ligaments to slacken (become loose). Consequently, the tension on the elastic lens is released, allowing it to become more rounded, convex, and thick. This rounded lens refracts (bends) the incoming light rays more strongly, bringing them to a sharp focus on the retina.

1 mark for stating that ciliary muscles contract. 1 mark for stating that suspensory ligaments slacken (become loose). 1 mark for stating that the lens becomes more rounded/convex/thick to refract light more.

Water travels from the root cortex cells and enters the xylem vessels. Inside the xylem, water molecules stick together because of cohesive forces (cohesion) between water molecules. As water evaporates and transpires from the stomata in the leaves, it creates a transpirational pull (tension) that draws water upwards. This creates a continuous, uninterrupted column of water moving from the roots up to the leaves.

1 mark for describing water entering the xylem vessels. 1 mark for explaining cohesion (water molecules sticking together). 1 mark for explaining that transpiration/evaporation at the leaves creates a pull/tension that draws the column upwards.

For a child to have blood group O, they must have the homozygous recessive genotype \(I^O I^O\). This means the child must inherit one recessive \(I^O\) allele from the mother and one from the father. For this to happen, both parents must be heterozygous. The mother with blood group A must have the genotype \(I^A I^O\), and the father with blood group B must have the genotype \(I^B I^O\). A genetic cross shows there is a 25% (1 in 4) chance of producing an \(I^O I^O\) child.

1 mark for identifying that blood group O requires the homozygous recessive genotype \(I^O I^O\). 1 mark for stating that both parents must be heterozygous (mother genotype \(I^A I^O\) and father genotype \(I^B I^O\)). 1 mark for explaining that the child inherits one recessive \(I^O\) allele from each parent.

The pleural membranes form a double-layered barrier surrounding the lungs and lining the inside of the chest cavity (thorax). The space between these membranes contains pleural fluid, which is secreted by the membranes. This fluid acts as a lubricant to reduce friction between the expanding lungs and the ribs during ventilation. It also creates a strong surface tension (cohesion) that sticks the lungs to the thoracic wall, ensuring that when the rib cage expands, the lungs are pulled outward to draw air in.

1 mark for explaining that pleural membranes line the lungs and thorax to secrete pleural fluid. 1 mark for stating that pleural fluid acts as a lubricant to reduce friction during breathing. 1 mark for explaining that surface tension/suction forces the lungs to expand along with the chest wall.

The runoff of fertilizers containing nitrates and phosphates causes rapid growth of algae on the water surface, known as an algal bloom. This layer of algae blocks sunlight from reaching submerged aquatic plants, preventing them from photosynthesizing, which causes them to die. Aerobic bacteria (decomposers) multiply rapidly as they feed on the abundant dead plant matter. These bacteria respire aerobically, depleting the dissolved oxygen levels in the water, which leads to the suffocation and death of fish and other active aquatic organisms.

1 mark for stating that fertilizers cause an algal bloom which blocks sunlight, leading to the death of submerged plants. 1 mark for stating that decomposers (aerobic bacteria) multiply rapidly as they break down dead plant tissue. 1 mark for explaining that bacteria consume dissolved oxygen during aerobic respiration, leading to the suffocation of fish.

Restriction enzymes are used to cut the human insulin gene out of human DNA and also to cut open the bacterial plasmid (the vector). These enzymes cut the DNA at specific base sequences, leaving single-stranded overhangs called complementary sticky ends. DNA ligase is then used to join the sticky ends of the human insulin gene with those of the opened bacterial plasmid by forming covalent phosphodiester bonds. This molecular gluing action successfully creates a recombinant plasmid containing the human gene.

1 mark for stating that restriction enzymes cut the human gene and the plasmid at specific sequences to produce complementary sticky ends. 1 mark for stating that DNA ligase is used to join/glue the sticky ends of the gene and plasmid together. 1 mark for explaining that this process results in the formation of recombinant DNA/plasmid.

During exercise, increased rate of aerobic respiration in muscle cells produces more \(CO_2\) as a waste product. This \(CO_2\) dissolves in blood plasma, forming carbonic acid and lowering the pH of the blood. The decrease in pH is detected by chemoreceptors in the medulla oblongata, aorta, and carotid arteries. Nerve impulses are sent via the phrenic and intercostal nerves to the diaphragm and intercostal muscles, causing them to contract and relax faster and more forcefully, thereby increasing the rate of ventilation to expel excess \(CO_2\) and absorb more \(O_2\).

1. Exercise increases respiration rate, releasing more \(CO_2\) into the blood (1 mark). 2. Increased \(CO_2\) lowers blood pH / makes blood more acidic (1 mark). 3. Change in pH is detected by chemoreceptors / medulla (1 mark). 4. Nerve impulses sent to diaphragm / intercostal muscles to contract faster or deeper (1 mark). [Maximum 3.2 marks]

Ultrafiltration is driven by high blood pressure in the glomerulus, which is generated because the afferent arteriole (entering) is wider than the efferent arteriole (leaving). The endothelial cells of the glomerular capillaries have tiny pores. Beneath these is the basement membrane, which acts as a physical mesh/filter. This selective barrier permits small molecules (water, salts, glucose, amino acids, urea) to pass into the capsule cavity, forming the glomerular filtrate, while larger structures like red blood cells and plasma proteins are retained in the capillary.

1. Afferent arteriole is wider than efferent arteriole, creating high hydrostatic pressure (1 mark). 2. Glomerular capillaries have tiny pores/fenestrations (1 mark). 3. Basement membrane acts as a selective filter/mesh (1 mark). 4. Small molecules (water, urea, glucose, ions) pass through while large proteins or blood cells cannot (1 mark). [Maximum 3.2 marks]

When fish are crowded, physical contact increases, facilitating the spread of infectious agents (bacteria, viruses, parasites like sea lice). Excess faeces and uneaten food accumulate, leading to eutrophication within cages, lower dissolved oxygen, and higher ammonia levels. This environment stresses fish, compromising their immune defense. Non-chemical control methods include introducing wrasse or lumpsuckers (cleaner fish) to feed on sea lice, using underwater bubble curtains or skirts to block parasite larvae, and periodically leaving cage sites empty (fallowing) to break pathogen life cycles.

1. Close proximity of fish allows pathogens/parasites to spread rapidly by direct contact (1 mark). 2. High organic waste lowers water quality and stresses fish, weakening their immune systems (1 mark). 3. Use of biological control, such as cleaner fish (e.g., wrasse), to consume parasites (1 mark). 4. Physical methods such as fallowing (rotating cages) or physical barriers (protective skirts) (1 mark). [Maximum 3.2 marks]

Nitrate ions act as limiting nutrients for aquatic primary producers. An excess of nitrates (often from fertilizer runoff) triggers an algal bloom. The thick layer of algae on the surface blocks light penetration to the bottom of the pond. Submerged macrophytes cannot photosynthesise and consequently die. Aerobic decomposers (bacteria) break down this abundant dead organic matter. Due to the high food supply, the bacterial population grows exponentially, and their cellular respiration consumes the dissolved oxygen in the water faster than it can be replenished.

1. Excess nitrates cause rapid growth of surface algae / algal bloom (1 mark). 2. Algal bloom blocks sunlight, stopping photosynthesis and causing deeper plants to die (1 mark). 3. Aerobic bacteria/decomposers multiply to decay the dead plant matter (1 mark). 4. Decomposers respire aerobically, depleting the dissolved oxygen in the pond (1 mark). [Maximum 3.2 marks]

Auxins are plant growth regulators synthesised in the apical meristem of the shoot. In response to unidirectional light, lateral transport of auxin occurs from the illuminated side to the shaded side of the stem. The higher concentration of auxin on the shaded side promotes active hydrogen ion pumping into cell walls, loosening them and allowing water uptake. This results in greater cell elongation on the shaded side compared to the light side, resulting in an asymmetric growth rate that causes the shoot to bend towards the light source (positive phototropism).

1. Auxin is produced in the stem tip and moves downwards (1 mark). 2. Auxin accumulates on the shaded side of the shoot (1 mark). 3. High auxin concentration stimulates cell elongation on the shaded side (1 mark). 4. Asymmetrical growth/elongation causes the shoot to bend towards the light (1 mark). [Maximum 3.2 marks]

Transpiration depends on the rate of water vapour diffusion out of stomata, which is governed by the water vapour concentration gradient. When wind speed decreases, there is no air movement to sweep away the escaping water molecules. Consequently, a saturated boundary layer of water vapour forms immediately outside the stomatal pores. This narrows the concentration gradient between the internal airspace of the leaf (which is highly humid) and the immediate external environment, leading to a reduced rate of diffusion and thus a lower rate of transpiration.

1. Transpiration rate decreases (1 mark). 2. Still air allows a layer of water vapour / high humidity to build up around the stomata (1 mark). 3. This reduces the concentration gradient of water vapour between the inside and outside of the leaf (1 mark). 4. The rate of diffusion of water vapour out of the stomata is reduced (1 mark). [Maximum 3.2 marks]

Bile, produced by the liver and stored in the gallbladder, acts as an emulsifier in the duodenum. Bile salts have both hydrophobic and hydrophilic regions, allowing them to disperse large fat globules into micro-droplets. This massive increase in surface-area-to-volume ratio allows lipase enzymes to hydrolyse the lipid molecules into fatty acids and glycerol much faster. The process is entirely physical because it does not alter the molecular structure of the lipids (no ester bonds are cleaved) and is a physical change of state/dispersion rather than a chemical reaction.

1. Bile emulsifies large lipid droplets into smaller droplets (1 mark). 2. This increases the total surface area available for lipase to act on (1 mark). 3. It is a physical process because no chemical/covalent/ester bonds are broken (1 mark). 4. No new chemical substances are formed / lipids remain chemically unchanged (1 mark). [Maximum 3.2 marks]

To calculate the percentage efficiency of energy transfer, use the formula: \(\text{Efficiency} = \frac{\text{Energy in consumer biomass}}{\text{Energy in plant biomass}} \times 100\). Substituting the values: \(\text{Efficiency} = \frac{1440}{12000} \times 100 = 12\%\).

1 mark for showing correct working: \(\frac{1440}{12000} \times 100\) (or equivalent). 1 mark for the correct final answer of \(12\%\). Accept 12.

First, find the radius \(r\) of the capillary tube: \(r = 1.0\text{ mm} / 2 = 0.5\text{ mm}\). Next, calculate the volume of water taken up using the cylinder volume formula: \(\text{Volume} = 3.14 \times (0.5)^2 \times 60 = 3.14 \times 0.25 \times 60 = 47.1\text{ mm}^3\). Then, calculate the rate of water uptake per minute: \(\text{Rate} = \frac{47.1\text{ mm}^3}{15\text{ minutes}} = 3.14\text{ mm}^3\text{ min}^{-1}\).

1 mark for calculating the volume of water taken up as \(47.1\text{ mm}^3\) (or showing correct substitution into the volume formula). 1 mark for the correct final rate of \(3.14\) (accept \(3.14\text{ mm}^3\text{ min}^{-1}\)).

First, convert the image size from millimetres to micrometres: \(12\text{ mm} \times 1000 = 12,000\ \mu\text{m}\). Next, use the magnification formula: \(\text{Actual Size} = \frac{\text{Image Size}}{\text{Magnification}}\). Therefore, \(\text{Actual Size} = \frac{12,000\ \mu\text{m}}{8,000} = 1.5\ \mu\text{m}\).

1 mark for converting \(12\text{ mm}\) to \(12,000\ \mu\text{m}\) or dividing \(12\) by \(8,000\) to get \(0.0015\text{ mm}\). 1 mark for the correct final answer of \(1.5\) (accept \(1.5\ \mu\text{m}\)).

First, determine the rate per minute: \(\frac{18\text{ cm}^3}{12\text{ min}} = 1.5\text{ cm}^3\text{ min}^{-1}\). Since there are 60 minutes in an hour, multiply this rate by 60 to find the hourly rate: \(1.5 \times 60 = 90\text{ cm}^3\text{ hr}^{-1}\). Alternatively, convert 12 minutes to hours: \(\frac{12}{60} = 0.2\text{ hours}\). Then divide the volume by the time in hours: \(\frac{18\text{ cm}^3}{0.2\text{ hours}} = 90\text{ cm}^3\text{ hr}^{-1}\).

1 mark for calculating the rate per minute as \(1.5\text{ cm}^3\text{ min}^{-1}\) or converting 12 minutes to \(0.2\text{ hours}\). 1 mark for the correct final answer of \(90\) (accept \(90\text{ cm}^3\text{ hr}^{-1}\)).

To calculate the percentage increase in the control pond: Day 1 = 0.5 mg/dm³, Day 5 = 4.2 mg/dm³. Increase = 4.2 - 0.5 = 3.7 mg/dm³. Percentage increase = (3.7 / 0.5) * 100 = 740%. For the explanation: bacteria convert toxic ammonia into nitrite and then nitrate (nitrification) or assimilate the nitrogen directly into bacterial proteins/biomass, reducing free ammonia in the water.

Calculation (2 marks): 1 mark for calculating correct increase of 3.7 mg/dm³; 1 mark for correct percentage of 740%. Explanation (1.7 marks): 1 mark for mentioning nitrification/conversion to nitrates; 0.7 marks for mentioning assimilation into microbial biomass/proteins.

Calculation: Change in calcification rate = 15.0 - 6.3 = 8.7 mg CaCO₃ per gram per day. Percentage decrease = (8.7 / 15.0) * 100 = 58%. Ecological consequence: Reduced calcification weakens the structural integrity of coral reefs. This leads to habitat destruction and a loss of shelter/breeding grounds for many marine organisms, which can collapse local food webs and reduce biodiversity.

Calculation (2 marks): 1 mark for showing change of 8.7; 1 mark for correct percentage calculation of 58%. Ecological consequence (1.7 marks): 1 mark for identifying loss of habitat/shelter; 0.7 marks for linking this to reduced biodiversity or food web collapse.

First, find the radius from the diameter: r = 0.8 mm / 2 = 0.4 mm. The height (h) is the distance the bubble moved in one minute, which is 8.4 mm. Apply the formula: Volume = 3.142 * (0.4)^2 * 8.4 = 3.142 * 0.16 * 8.4 = 4.222848. Rounding to 2 decimal places gives 4.22. The units are cubic millimetres (mm³).

Radius calculation (0.7 marks): 0.7 marks for correctly identifying radius as 0.4 mm. Substitution (1 mark): 1 mark for substituting the correct values into the cylinder formula (3.142 * 0.4² * 8.4). Correct value and units (2 marks): 1 mark for 4.22; 1 mark for correct unit of mm³.