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When a plant cell is placed in a concentrated sucrose solution, which has a lower water potential than the cell cytoplasm, water moves out of the cell down the water potential gradient by osmosis. This causes the cell membrane to pull away from the cell wall, making the cell plasmolysed.

1 mark for selecting C. Award 1 mark for identifying both the correct cell state (plasmolysed) and the correct net direction of water movement (out of the cell).

Biuret reagent is used to test for proteins, turning from blue to lilac-purple if protein is present. The ethanol emulsion test is used to detect lipids, forming a cloudy white emulsion if lipids are present.

1 mark for selecting B. Award 1 mark for correctly matching the positive test result for protein (lilac-purple) with the positive test result for lipid (cloudy white emulsion).

Eutrophication begins with nutrient runoff leading to rapid algal growth (algal bloom). This blocks light from reaching plants lower down in the water column, causing them to die. Aerobic bacteria decompose the dead plant material, rapidly consuming dissolved oxygen, which causes fish to suffocate and die.

1 mark for selecting A. Award 1 mark for the correct chronological sequence of ecological changes leading to fish suffocation.

Arteries carry blood away from the heart under high pressure. They have a narrow lumen (small) to maintain pressure, a thick wall containing many muscle and elastic fibres to withstand and smooth out the high pressure, and no valves (unlike veins).

1 mark for selecting B. Award 1 mark for identifying the correct combination of a small lumen, thick wall with many muscle/elastic fibres, and lack of valves.

In a reflex arc, the stimulus is detected by a receptor, which sends an electrical impulse along a sensory neurone to the central nervous system (spinal cord). Inside the spinal cord, it synapses with a relay neurone, which then synapses with a motor neurone. The motor neurone carries the impulse to the effector (muscle) to produce a response.

1 mark for selecting C. Award 1 mark for identifying the correct order: receptor, sensory neurone, relay neurone, motor neurone, effector.

To find magnification, use the formula: \(\text{Magnification} = \frac{\text{Image size}}{\text{Actual size}}\). Substituting the given values: \(\text{Magnification} = \frac{2.4\text{ mm}}{0.06\text{ mm}} = 40\). Therefore, the magnification is \(\times 40\).

1. Shows correct working or substitution: \(\frac{2.4}{0.06}\) (1 mark)
2. Correct final answer of \(40\) or \(\times 40\) (1 mark)

Osmosis causes water to enter both cells down a water potential gradient. The red blood cell has no cell wall, so the cell membrane cannot withstand the increased internal pressure, causing it to burst (lysis). The plant cell has a strong, rigid cell wall made of cellulose, which resists the turgor pressure and prevents bursting.

1. Water enters the cells by osmosis / down a water potential gradient (1 mark)
2. Plant cell has a rigid cell wall that prevents bursting OR red blood cell lacks a cell wall and bursts (1 mark)

To test for glucose, mix the food sample with Benedict's reagent in a test tube. Heat the tube in a hot water bath (at about 80 °C) for 5 minutes. If glucose is present, the mixture will change colour from blue to green, yellow, orange, or brick-red, depending on the concentration.

1. Add Benedict's reagent/solution and heat in a water bath (1 mark)
2. Correct colour change observed: from blue to green / yellow / orange / brick-red (1 mark)

The ileum is highly adapted for absorption: 1. It is very long and its inner lining is folded into millions of tiny finger-like projections called villi (which themselves have microvilli), greatly increasing the surface area for diffusion and active transport. 2. Each villus has an extensive network of blood capillaries, which rapidly carry away absorbed food to maintain a steep concentration gradient.

Any two from:
- Highly folded / long / contains villi / microvilli to increase surface area (1 mark)
- Wall of villus is only one cell thick to provide a short diffusion pathway (1 mark)
- Excellent blood supply / rich capillary network to maintain a concentration gradient (1 mark)
- Presence of lacteals to absorb fats/lipids (1 mark)

Capillaries are adapted for exchange because their walls are extremely thin (one cell thick), reducing the diffusion distance for gases and nutrients. Their walls are also permeable, allowing small molecules like oxygen and glucose to pass out into the surrounding tissue fluid.

1. Wall is one cell thick / very thin to provide a short diffusion distance (1 mark)
2. Walls are permeable / have pores to allow substances to pass through (1 mark)

The myelin sheath is a fatty layer that surrounds the axon of a neurone. It acts as an electrical insulator, preventing the loss of the electrical signal. It also speeds up the transmission of the nerve impulse by forcing the impulse to jump from one node (Node of Ranvier) to the next (saltatory conduction).

1. Insulates the axon / prevents electrical loss / prevents dissipation of the signal (1 mark)
2. Speeds up transmission / allows the impulse to jump between nodes (1 mark)

Selective reabsorption of glucose takes place in the proximal convoluted tubule (PCT). All of the glucose is actively transported back into the blood capillaries against its concentration gradient, requiring energy (ATP) from active respiration.

1. Reabsorption occurs in the proximal convoluted tubule / PCT (1 mark)
2. By active transport / using energy / ATP (1 mark)

When untreated sewage enters a river, decomposer bacteria feed on the organic matter and rapidly multiply. These bacteria respire aerobically, consuming vast amounts of dissolved oxygen from the water. The resulting oxygen depletion (anoxia) means fish and other aquatic organisms cannot get enough oxygen for their own respiration, leading to their death.

1. Bacteria/decomposers multiply/feed on organic matter and respire aerobically / use up dissolved oxygen (1 mark)
2. Depletion of oxygen causes fish to suffocate/die as they cannot respire (1 mark)

First, calculate the increase in carbon dioxide concentration: \( 412 - 317 = 95 \text{ ppm} \). Next, calculate the percentage increase using the 1960 concentration as the base: \( \frac{95}{317} \times 100 = 29.968... \% \). Rounded to one decimal place, this is \( 30.0\% \).

Award 1 mark for showing correct working: \( \frac{412 - 317}{317} \times 100 \) or \( \frac{95}{317} \times 100 \). Award 1 mark for correct final answer of \( 30.0\% \) (accept 30% or 30.0).

First, calculate the temperature rise of the water: \( 28.5 - 18.5 = 10.0 \text{ °C} \). Then, substitute the values into the given formula: \( \text{Energy released} = 20.0 \times 10.0 \times 4.2 = 840 \text{ J} \).

Award 1 mark for calculating the correct temperature rise of 10.0 °C or showing correct substitution. Award 1 mark for the correct final answer of 840 (J).

First, calculate the change in mass: \( 3.74 - 4.25 = -0.51 \text{ g} \). Next, calculate the percentage change relative to the initial mass: \( \frac{-0.51}{4.25} \times 100 = -12\% \).

Award 1 mark for calculating the correct mass change of -0.51 g or showing the correct fraction: \( \frac{3.74 - 4.25}{4.25} \times 100 \). Award 1 mark for the correct final answer of -12% (accept -12 or 12% loss).

First, calculate the cardiac output in \( \text{cm}^3 \) per minute: \( 140 \times 85 = 11900 \text{ cm}^3/\text{min} \). Then, convert this volume to \( \text{dm}^3 \) by dividing by 1000: \( 11900 \div 1000 = 11.9 \text{ dm}^3/\text{min} \).

Award 1 mark for the correct calculation of cardiac output in \( \text{cm}^3 \) as 11900 or showing correct step. Award 1 mark for the correct final answer of 11.9.

First, calculate the total area sampled: \( 10 \times 0.25 = 2.5 \text{ m}^2 \). Next, calculate the mean density of dandelions: \( 15 \div 2.5 = 6 \text{ dandelions per m}^2 \). Finally, multiply this density by the total field area: \( 6 \times 800 = 4800 \text{ dandelions} \).

Award 1 mark for calculating the mean density of 6 per \( \text{m}^2 \) or showing the correct ratio \( \frac{15}{2.5} \). Award 1 mark for the correct final answer of 4800.

First, calculate the actual size in mm: \( \text{Actual size} = \frac{\text{Image size}}{\text{Magnification}} = \frac{45}{1500} = 0.03 \text{ mm} \). Then, convert mm to micrometres: \( 0.03 \times 1000 = 30 \mu\text{m} \).

Award 1 mark for dividing image size by magnification or converting units correctly. Award 1 mark for the correct final answer of 30.

First, calculate the volume of filtrate reabsorbed: \( 180 - 1.5 = 178.5 \text{ dm}^3 \). Next, calculate this as a percentage of the total filtrate: \( \frac{178.5}{180} \times 100 = 99.166... \% \), which rounds to \( 99.2\% \).

Award 1 mark for calculating the volume reabsorbed as 178.5 or setting up the correct fraction. Award 1 mark for the correct final answer of 99.2% (accept 99% or 99.17%).

First, calculate the energy lost by subtracting stored energy from the total input: \( 25000 - 3750 = 21250 \text{ kJ} \). Next, calculate the percentage lost: \( \frac{21250}{25000} \times 100 = 85\% \).

Award 1 mark for calculating the energy lost as 21250 kJ or the percentage stored as 15%. Award 1 mark for the correct final answer of 85%.

1. First, calculate the change in mass:
\(\text{Change in mass} = \text{Final mass} - \text{Initial mass} = 2.15 - 2.50 = -0.35\text{ g}\)

2. Next, calculate the percentage change relative to the initial mass:
\(\text{Percentage change} = \left(\frac{-0.35}{2.50}\right) \times 100 = -14\%\)

Therefore, the percentage change is \(-14\%\) (which represents a \(14\%\) decrease).

Award 1 mark for correct working showing the change in mass divided by the initial mass:
\(\frac{-0.35}{2.50} \times 100\) or \(\frac{0.35}{2.50} \times 100\)

Award 1 mark for the correct final answer of \(-14\%\) or a \(14\%\) decrease.
[Accept: 14% if it is clearly indicated as a decrease, loss, or negative change. Award 1 mark max for a final answer of 14% without stating it is a decrease if no working is shown.]

When fossil fuels are burned, sulfur dioxide gas is released. This gas dissolves in water vapor in the atmosphere to form acidic solutions (acid rain). When this acid rain falls into freshwater lakes, it lowers the pH of the water. The increased acidity directly damages fish gills and disrupts their biological processes, and it can also cause toxic heavy metals like aluminium to leach from the surrounding soil into the lake water, killing the fish.

Award 1 mark for explaining that sulfur dioxide dissolves in water vapor / rain to form acid rain (or lowers the pH of rainwater).
Award 1 mark for explaining that the acidic water in lakes damages fish gills / disrupts enzymes / causes toxic aluminium ions to leach from soil into the water, leading to death.

The process is known as eutrophication:
1. The leaching of nitrate fertilizers causes an 'algal bloom' where algae on the surface grow rapidly.
2. This layer of algae blocks sunlight from reaching aquatic plants deeper in the stream.
3. These plants cannot photosynthesise and therefore die.
4. Decomposers, such as aerobic bacteria, decompose the dead plant material and multiply rapidly.
5. The large population of bacteria uses up the dissolved oxygen in the water during aerobic respiration.
6. The resulting low oxygen conditions mean that fish cannot get enough oxygen for their respiration, leading to their death.

Max 4.6 marks:
- Algal bloom / rapid growth of algae on surface (1 mark)
- Blocks light so plants below cannot photosynthesise (1 mark)
- Plants die and are decomposed by bacteria / decomposers (1 mark)
- Bacteria reproduce / increase in number and respire aerobically (1 mark)
- Depletes dissolved oxygen in water so fish suffocate / cannot respire (1 mark)
Accept: microbes/fungi for bacteria.
Reject: plants die due to lack of oxygen.

Part (a):
Magnification = Image size / Actual size
Image size = \( 45\text{ mm} \)
Actual size = \( 0.09\text{ mm} \)
Magnification = \( \frac{45}{0.09} = 500 \) times (\( \times 500 \)).

Part (b):
Chloroplasts are the site of photosynthesis. A high density of chloroplasts ensures maximum absorption of light energy to produce glucose.
Mitochondria are the sites of aerobic respiration. They release energy in the form of ATP, which is required for active metabolic processes within the cell, such as the active transport of mineral ions or protein synthesis.

Max 4.6 marks:
- Correct calculation of magnification: \( \times 500 \) (2 marks: 1 mark for correct formula/working with units matched, 1 mark for correct final answer)
- High density of chloroplasts for absorbing light / photosynthesising to make glucose (1 mark)
- Mitochondria for aerobic respiration to release energy / produce ATP (1 mark)
- ATP/energy used for active transport / cellular processes (1 mark)
Accept: alternative correct workings.
Reject: mitochondria 'create' or 'produce' energy.

1. When an electrical impulse (action potential) reaches the end of the pre-synaptic neurone, it causes vesicles containing neurotransmitters to fuse with the pre-synaptic membrane.
2. The neurotransmitter molecules are released into the synaptic cleft by exocytosis.
3. The neurotransmitters diffuse across the synaptic gap down a concentration gradient.
4. They bind to specific complementary receptor proteins on the post-synaptic membrane.
5. This binding triggers a new electrical impulse in the post-synaptic neurone.
6. A drug that blocks these receptors prevents the neurotransmitters from binding. As a result, sodium channels do not open, no depolarization occurs, and the electrical impulse cannot be transmitted across the synapse.

Max 4.6 marks:
- Neurotransmitters released from vesicles in pre-synaptic membrane (1 mark)
- Diffuse across the synaptic cleft / gap (1 mark)
- Bind to specific receptors on the post-synaptic membrane (1 mark)
- Triggers an electrical impulse in the post-synaptic neurone (1 mark)
- Blocking drug prevents binding, so no electrical impulse is generated (1 mark)
Accept: named neurotransmitters like acetylcholine.
Reject: electrical impulse jumping across the synapse.

Part (a):
Volume of water taken up = Distance moved by bubble \( \times \) Cross-sectional area of capillary tube
Volume = \( 60\text{ mm} \times 0.8\text{ mm}^2 = 48\text{ mm}^3 \)
Time = \( 15\text{ minutes} \)
Rate of water uptake = \( \frac{\text{Volume}}{\text{Time}} = \frac{48\text{ mm}^3}{15\text{ minutes}} = 3.2\text{ mm}^3\text{ min}^{-1} \).

Part (b):
An increase in wind speed moves humid air (water vapour) away from the surface of the leaf. This maintains a steep water vapour concentration gradient between the air spaces inside the leaf and the atmosphere outside. Therefore, the rate of diffusion of water vapour out through the stomata (transpiration) increases.

Max 4.6 marks:
- Correct calculation of volume: \( 48\text{ mm}^3 \) (1 mark)
- Correct calculation of rate: \( 3.2\text{ mm}^3\text{ min}^{-1} \) (1 mark)
- Wind moves water vapour away from leaf surface / stomata (1 mark)
- This maintains/increases the concentration gradient of water vapour (1 mark)
- Leading to faster diffusion / evaporation of water out of stomata (1 mark)
Accept: alternative correct methods of calculating rate, e.g., \( \frac{60}{15} = 4\text{ mm/min} \), then \( 4 \times 0.8 = 3.2 \).
Reject: water 'active transport' or incorrect unit conversions.

Part (a):
Parental Phenotypes: Pink x Pink
Parental Genotypes: \( C^R C^W \times C^R C^W \)
Gametes: \( C^R \) and \( C^W \) from each parent.
Punnett Square:
- Gametes: \( C^R \) and \( C^W \) crossed with \( C^R \) and \( C^W \)
- Offspring Genotypes: \( C^R C^R \) (Red), \( C^R C^W \) (Pink), \( C^R C^W \) (Pink), \( C^W C^W \) (White)
Offspring Phenotypes and Ratio: 1 Red : 2 Pink : 1 White.

Part (b):
The probability of obtaining a red-flowered plant (\( C^R C^R \)) is \( \frac{1}{4} \) or \( 0.25 \) or \( 25\% \).

Max 4.6 marks:
- Correct parental genotypes and gametes identified (1 mark)
- Correct Punnett square/genetic cross showing offspring genotypes (1 mark)
- Correct phenotype ratio of offspring: 1 Red : 2 Pink : 1 White (1 mark)
- Correct probability calculated as \( 0.25 \) or \( \frac{1}{4} \) or \( 25\% \) (1 mark)
- Clear explanation of codominance where both alleles are expressed in the heterozygote (1 mark)
Accept: equivalent phenotypic descriptions.
Reject: dominance-recessive relationship representations.

1. Ultrafiltration: High blood pressure is created in the glomerulus because the afferent arteriole is wider than the efferent arteriole. This pressure forces water, ions, glucose, amino acids, and urea out of the blood through the capillary walls and the basement membrane (acting as a filter) into the Bowman's capsule. Large proteins and blood cells are too big to pass through and remain in the blood.
2. Selective Reabsorption: This takes place in the proximal convoluted tubule (PCT). All glucose and amino acids, and some water and salts, are reabsorbed back into the blood capillaries.
3. Adaptations of the PCT: The cells lining the PCT have microvilli to increase the surface area for diffusion and active transport. They also contain a high number of mitochondria to provide ATP for the active transport of glucose against its concentration gradient.

Max 4.6 marks:
- High pressure in glomerulus forces small molecules/filtrate into Bowman's capsule (1 mark)
- Basement membrane acts as a filter preventing large proteins/blood cells from passing (1 mark)
- Selective reabsorption of all glucose/amino acids in the PCT (1 mark)
- PCT has microvilli to increase surface area for absorption (1 mark)
- PCT cells have many mitochondria to supply ATP/energy for active transport (1 mark)
Accept: active uptake for active transport.
Reject: glucose reabsorbed by simple diffusion alone.

1. The human gene responsible for producing insulin is identified and cut out of human DNA using a specific restriction enzyme. This enzyme cuts the DNA leaving 'sticky ends' (short sections of single-stranded DNA).
2. A plasmid (a circular piece of DNA) is isolated from a bacterium and cut open using the same restriction enzyme, ensuring it has complementary sticky ends.
3. The human insulin gene and the cut plasmid are mixed together. The enzyme DNA ligase is used to join the sticky ends together, forming a recombinant plasmid.
4. The recombinant plasmid is inserted back into a bacterial cell (often E. coli). This bacterium is now transgenic.
5. The transgenic bacteria are grown in an industrial fermenter under controlled conditions (correct temperature, pH, oxygen) where they reproduce rapidly and express the gene, producing human insulin which can then be extracted and purified.

Max 4.6 marks:
- Cut human insulin gene using restriction enzyme (1 mark)
- Cut plasmid with the SAME restriction enzyme to produce complementary sticky ends (1 mark)
- Use DNA ligase to join gene and plasmid to make recombinant plasmid (1 mark)
- Insert plasmid into bacterium / transformation (1 mark)
- Grow bacteria in a fermenter to produce insulin at scale (1 mark)
Accept: vector for plasmid.
Reject: incorrect enzymes like amylase or protease instead of restriction/ligase.

Part (a):
At \( 20^\circ\text{C} \):
Time = \( 12\text{ minutes} = 12 \times 60 = 720\text{ seconds} \)
Rate = \( \frac{1}{720} \approx 0.00139\text{ s}^{-1} \) (or \( 1.39 \times 10^{-3}\text{ s}^{-1} \)).

At \( 40^\circ\text{C} \):
Time = \( 3\text{ minutes} = 3 \times 60 = 180\text{ seconds} \)
Rate = \( \frac{1}{180} \approx 0.00556\text{ s}^{-1} \) (or \( 5.56 \times 10^{-3}\text{ s}^{-1} \)).

Part (b):
At \( 40^\circ\text{C} \), the enzymes and starch (substrate) molecules have more kinetic energy than at \( 20^\circ\text{C} \). This causes them to move faster, leading to more frequent collisions per unit time. As a result, there are more successful collisions between the substrate molecules and the active site of the amylase enzyme, forming more enzyme-substrate complexes and increasing the rate of reaction.

Max 4.6 marks:
- Correct conversion of minutes to seconds for both temperatures: \( 720\text{ s} \) and \( 180\text{ s} \) (1 mark)
- Correct calculation of rates: \( 0.00139\text{ s}^{-1} \) and \( 0.00556\text{ s}^{-1} \) (accept rounded values like 0.0014 and 0.0056) (1 mark)
- Higher temperature increases kinetic energy of molecules (1 mark)
- Molecules move faster, leading to more frequent collisions (1 mark)
- Increased probability of successful collisions / forming enzyme-substrate complexes (1 mark)
Accept: reciprocal of minutes if specified, but formula requires 'seconds' so stick to seconds.
Reject: enzymes 'die' or 'are killed' at high temperatures.

1. Water moves into both types of cells by osmosis, down a water potential gradient (from a higher water potential in the \(0.1\%\) salt solution to a lower water potential inside the cells) across a selectively permeable membrane.
2. In red blood cells, this influx of water causes the volume to increase. Since red blood cells lack a cell wall, the cell membrane cannot withstand the increased internal osmotic pressure, causing them to swell and burst (lysis).
3. In plant palisade cells, water also enters the vacuole, causing the protoplast to swell. However, plant cells possess a strong, rigid cell wall made of cellulose that exerts turgor pressure, opposing further water entry and preventing the cell from bursting. Instead, the plant cells become turgid.

Maximum 4.6 marks:
- Water enters both cells by osmosis / down a water potential gradient (1 mark)
- Red blood cells swell and burst / undergo lysis (1 mark)
- Explanation that red blood cells lack a cell wall to resist pressure (1 mark)
- Plant cells become turgid / do not burst (1 mark)
- Explanation that plant cells have a rigid cellulose cell wall that prevents lysis (1 mark)

1. Leaching/runoff of nitrates into the pond causes rapid growth of algae (algal bloom) at the surface.
2. This layer of algae blocks sunlight, preventing submerged aquatic plants from photosynthesising, which eventually leads to their death.
3. Decomposing bacteria multiply rapidly as they feed on the large amount of dead plant material.
4. These bacteria respire aerobically, using up the dissolved oxygen in the water.
5. The resulting lack of oxygen prevents fish and other aquatic animals from respiring, leading to their death.

Maximum 4.6 marks:
- Nitrates leach/run off into pond causing rapid algal growth/algal bloom (1 mark)
- Algal bloom blocks sunlight, preventing photosynthesis of deeper plants, leading to their death (1 mark)
- Bacteria/decomposers multiply and decompose the dead plant matter (1 mark)
- Bacteria respire aerobically, consuming dissolved oxygen (1 mark)
- Oxygen depletion leads to suffocation and death of fish (1 mark)

1. Auxin is a plant hormone synthesized in the shoot tip that stimulates cell elongation.
2. In Group A (untreated), unilateral light causes auxin to migrate to and accumulate on the shaded side of the shoot. As auxin diffuses downwards, it causes cells on the shaded side to elongate more rapidly than cells on the illuminated side, bending the shoot toward the light source (positive phototropism).
3. In Group B, the removal of the tip means there is no source of auxin, so little to no growth or bending will occur.
4. In Group C, the opaque cap blocks the unilateral light stimulus from reaching the photoreceptors in the tip. Consequently, auxin is distributed symmetrically down all sides of the shoot, resulting in uniform cell elongation and straight vertical growth.

Maximum 4.6 marks:
- Auxin is synthesized in the tip and stimulates cell elongation (1 mark)
- Group A: Auxin moves to/accumulates on the shaded side, causing unilateral cell elongation and bending towards the light (1 mark)
- Group B: No growth or bending because the source of auxin has been removed (1 mark)
- Group C: Grows straight up (no bending) because the tip cannot perceive the unilateral light direction, leading to equal auxin distribution (1.6 marks)

1. Restriction enzymes are sequence-specific endonucleases. They are used to cut out the target human insulin gene from human DNA.
2. The same restriction enzymes are used to cut open the bacterial plasmid vector. Using the same enzyme ensures that both the gene and the plasmid have complementary 'sticky ends' (short, single-stranded DNA overhangs).
3. These sticky ends can associate through hydrogen bonding of complementary base pairs.
4. DNA ligase is then used to join the sugar-phosphate backbones of the insulin gene and the plasmid, forming covalent phosphodiester bonds. This permanently seals the gene into the vector, creating a recombinant plasmid.

Maximum 4.6 marks:
- Restriction enzymes cut out the insulin gene and cut open the plasmid (1 mark)
- Using the same restriction enzyme creates complementary sticky ends (1.6 marks)
- Sticky ends contain exposed bases that can pair together (1 mark)
- DNA ligase joins the gene and the plasmid backbones together (1 mark)
- This produces a recombinant plasmid (1 mark)

1. Amylase is a protein enzyme with a specific three-dimensional tertiary structure containing a uniquely shaped active site.
2. pH 7 represents the optimum pH for amylase. At this pH, the shape of the active site is completely complementary to the substrate, starch, permitting maximum formation of enzyme-substrate complexes and the fastest rate of reaction.
3. At pH 4, the high concentration of hydrogen ions (acidic conditions) disrupts the ionic and hydrogen bonds holding the tertiary structure of the amylase enzyme.
4. This causes the enzyme to denature, permanently altering the shape of the active site.
5. The starch substrate can no longer bind to the altered active site, meaning no enzyme-substrate complexes are formed, and starch is not digested.

Maximum 4.6 marks:
- pH 7 is the optimum pH where the active site is complementary to starch (1 mark)
- At optimum pH, enzyme-substrate complexes form at the highest rate (1 mark)
- At pH 4, high hydrogen ion concentration denatures the enzyme (1 mark)
- Denaturation is caused by the disruption of hydrogen and ionic bonds (1 mark)
- This permanently changes the shape of the active site so that starch can no longer bind (0.6 marks)

1. During ultrafiltration in the glomerulus, glucose is small enough to pass into the Bowman's capsule as part of the glomerular filtrate.
2. In a healthy individual, all of this glucose is selectively reabsorbed back into the blood from the proximal convoluted tubule (PCT) via active transport, using specific carrier proteins, meaning zero glucose is excreted in the urine.
3. In a person with untreated diabetes, blood glucose levels are extremely elevated, leading to a much higher concentration of glucose in the glomerular filtrate.
4. The rate of glucose filtration exceeds the maximum rate of reabsorption (transport maximum) because the carrier proteins in the PCT become fully saturated.
5. The excess unreabsorbed glucose continues through the nephron and is excreted in the urine.

Maximum 4.6 marks:
- Glucose is filtered into the Bowman's capsule during ultrafiltration (1 mark)
- In a healthy individual, all glucose is selectively reabsorbed in the proximal convoluted tubule (1 mark)
- Reabsorption occurs via active transport / using carrier proteins (1 mark)
- In diabetics, blood glucose levels are too high, meaning the quantity of filtered glucose exceeds reabsorption capacity (1 mark)
- Carrier proteins become saturated, allowing excess glucose to escape in the urine (0.6 marks)

1. Methods to maximize yield:
- Controlled Feeding: Farmers provide high-protein, nutrient-rich pellets in small, frequent portions. This maximizes growth rates and minimizes uneaten food decomposing on the seabed.
- Separation of Fish: Fish are separated by age, size, and species (intraspecific and interspecific separation). This reduces competition for food and prevents larger fish from preying on smaller ones.
- Disease Control: High densities encourage rapid spread of disease. Farmers use antibiotics, chemical treatments, or biological controls (such as cleaner fish) to keep fish healthy.
2. Environmental Disadvantage:
- Eutrophication: Large amounts of organic waste (faeces and uneaten food) fall beneath the cages. Decomposers break down this waste, using up dissolved oxygen and creating dead zones.
- Spread of Pathogens: Parasites (e.g., sea lice) multiply rapidly on the farm and can easily spread to wild fish populations migrating past the cages.
- Bioaccumulation of Chemicals: Antibiotics and pesticides can leak into surrounding marine ecosystems, harming non-target organisms.

Maximum 4.6 marks:
- Describe and explain Method 1 (e.g., feeding high-protein diets in small portions to maximize growth and reduce decomposition) (1.5 marks)
- Describe and explain Method 2 (e.g., separating fish by size to prevent intraspecific predation / using biological control for parasites) (1.5 marks)
- State one environmental disadvantage (e.g., eutrophication, escape of parasites/disease, or chemical leakage) (1 mark)
- Explain the consequence of this disadvantage on the local ecosystem (0.6 marks)

When water potential of the blood decreases, the pituitary gland is stimulated to release more ADH. ADH increases the permeability of the collecting duct walls to water. This allows more water to be reabsorbed back into the blood by osmosis, resulting in a smaller, more concentrated volume of urine.

1 mark for choosing option A, which correctly identifies increased ADH secretion and increased permeability of the collecting duct walls leading to higher water reabsorption.

Fertiliser runoff leads to an algal bloom on the water's surface, which blocks sunlight from reaching underwater plants. These plants die as they cannot photosynthesise. Decomposer bacteria feed on the dead plant matter, multiplying rapidly and using up dissolved oxygen during aerobic respiration, which causes fish to suffocate.

1 mark for choosing option A, which lists the correct biological sequence of events during eutrophication.

In micropropagation, tissue explants are grown in nutrient-rich agar media which would also support rapid growth of unwanted bacteria and fungi. To prevent contamination, both the explants and the agar medium must be sterilised (using disinfectants and autoclaving respectively) before cultivation.

1 mark for choosing option B, which identifies the correct sterilisation procedure to prevent microbial contamination.

When an impulse reaches the end of the pre-synaptic neurone, neurotransmitter molecules are released from synaptic vesicles. These molecules diffuse across the synaptic gap down a concentration gradient and bind to complementary receptor proteins on the post-synaptic membrane, initiating a new electrical impulse.

1 mark for choosing option C, which correctly states that neurotransmitters diffuse across the gap and bind to receptors on the post-synaptic membrane.

The amount of glucose reabsorbed is the difference between the starting concentration and the ending concentration: \(5.5 \text{ mmol/dm}^3 - 0.0 \text{ mmol/dm}^3 = 5.5 \text{ mmol/dm}^3\). To find the percentage of glucose reabsorbed, divide the amount reabsorbed by the initial concentration and multiply by 100: \((5.5 / 5.5) \times 100 = 100\%\).

1 mark for calculating the amount of glucose reabsorbed (5.5 mmol/dm³) or showing the percentage division setup. 1.1 marks for the correct final answer of 100%.

First, calculate the absolute increase in nitrate concentration: \(15.6 \text{ mg/dm}^3 - 1.2 \text{ mg/dm}^3 = 14.4 \text{ mg/dm}^3\). Next, divide this increase by the original upstream concentration and multiply by 100 to get the percentage increase: \((14.4 / 1.2) \times 100 = 1200\%\).

1 mark for calculating the absolute increase of 14.4 mg/dm³ or showing the correct formula for percentage increase. 1.1 marks for the correct final percentage of 1200%.

Since the plasmid is circular, cutting it at two sites will produce exactly two fragments. The sum of the lengths of these two fragments must equal the total size of the original plasmid. Therefore, the length of the second fragment is: \(4500 \text{ bp} - 1800 \text{ bp} = 2700 \text{ bp}\).

1 mark for recognizing that the sum of the fragments must equal 4500 bp or showing the subtraction calculation. 1.1 marks for the correct final answer of 2700 bp.

The period of 9 weeks represents 3 cycles of subculturing (since \(9 / 3 = 3\)). After the first 3 weeks (cycle 1): \(5 \times 4 = 20\) explants. After 6 weeks (cycle 2): \(20 \times 4 = 80\) explants. After 9 weeks (cycle 3): \(80 \times 4 = 320\) explants. This can also be calculated as \(5 \times 4^3 = 5 \times 64 = 320\).

1 mark for identifying that 3 cycles of division occur or calculating 80 explants at 6 weeks. 1.1 marks for the correct final answer of 320.

Cardiac output is calculated by multiplying heart rate by stroke volume: \(60 \text{ beats/min} \times 80 \text{ cm}^3/\text{beat} = 4800 \text{ cm}^3/\text{min}\). To convert this volume into dm³, divide by 1000 (since \(1 \text{ dm}^3 = 1000 \text{ cm}^3\)): \(4800 / 1000 = 4.8 \text{ dm}^3/\text{min}\).

1 mark for calculating the cardiac output in cm³ (4800 cm³/min) or showing the division by 1000. 1.1 marks for the correct final answer of 4.8 dm³/min with correct units.

A cross between two heterozygotes (Aa x Aa) yields offspring with genotypes in the ratio 1 AA : 2 Aa : 1 aa. The homozygous individuals are AA (homozygous dominant) and aa (homozygous recessive). The probability of being AA is 0.25 (25%) and the probability of being aa is 0.25 (25%). The total probability of being homozygous is \(0.25 + 0.25 = 0.50\), which is \(50\%\).

1 mark for constructing a genetic diagram / Punnett square or showing the probability of AA (25%) and aa (25%). 1.1 marks for the correct final percentage of 50%.

Speed is defined as distance divided by time. Using the values provided: \(\text{Speed} = 1.2 \text{ m} / 0.015 \text{ s} = 80 \text{ m/s}\).

1 mark for showing the correct speed formula with values substituted: \(1.2 / 0.015\). 1.1 marks for the correct final speed of 80 m/s.

First, calculate the total energy absorbed by the water using the formula: \(\text{Energy} = \text{mass of water} \times \text{specific heat capacity} \times \text{temperature rise} = 20 \text{ g} \times 4.2 \text{ J/g}^\circ\text{C} \times 18 ^\circ\text{C} = 1512 \text{ J}\). Next, calculate the energy released per gram of food by dividing the total energy by the mass of the food sample: \(1512 \text{ J} / 1.5 \text{ g} = 1008 \text{ J/g}\).

1 mark for calculating the total energy absorbed by the water (1512 J). 1.1 marks for the correct final answer of 1008 J/g.

1. ADH increases the permeability of the collecting duct to water. 2. This causes more water to be reabsorbed back into the bloodstream by osmosis. 3. Consequently, the volume of urine decreases and its concentration increases.

Award 1 mark for explaining that ADH increases the permeability of the collecting duct to water. Award 1 mark for explaining that more water is reabsorbed into the blood by osmosis, resulting in a lower volume of more concentrated urine.

1. Denitrifying bacteria convert soluble nitrates in the soil into nitrogen gas, which is lost to the atmosphere, thereby reducing soil fertility. 2. These bacteria are anaerobic organisms, meaning they are most active in conditions lacking oxygen, such as waterlogged or highly compacted soils.

Award 1 mark for stating that they convert nitrates to nitrogen gas / decrease nitrate concentration in soil. Award 1 mark for identifying anaerobic conditions / waterlogged soil / lack of oxygen.

1. Micropropagation produces clones, ensuring all offspring retain the exact desirable traits of the parent plant. 2. It allows for the rapid production of large numbers of plants at any time of year in a controlled environment, free from pathogens.

Accept any two of the following for 1 mark each: - Produces genetically identical plants / clones (retains desired traits); - Allows rapid production of a large number of plants; - Can be done at any time of the year / not dependent on seasons; - Produces disease-free / sterile plants.

1. Follicle-stimulating hormone (FSH) is secreted by the pituitary gland and travels to the ovaries where it stimulates the maturation of an egg follicle. 2. Luteinising hormone (LH) is also released by the pituitary gland and causes the mature follicle to rupture, releasing the egg into the oviduct (ovulation).

Award 1 mark for describing the role of FSH: stimulates follicle growth / development / maturation of the egg. Award 1 mark for describing the role of LH: triggers ovulation / release of the egg.

Part (a):
- The manure contains organic matter / sewage / nutrients that enter the water.
- This organic waste acts as a food source for decomposers, causing a rapid increase in the population of bacteria.
- These bacteria respire aerobically, extracting dissolved oxygen from the water faster than it can be replenished, causing oxygen levels to drop.

Part (b):
- Dissolved oxygen at 0 m = \(9.2 \text{ mg/dm}^3\)
- Dissolved oxygen at 150 m = \(1.8 \text{ mg/dm}^3\)
- Decrease in oxygen = \(9.2 - 1.8 = 7.4 \text{ mg/dm}^3\)
- Percentage decrease = \(\frac{7.4}{9.2} \times 100 = 80.4347...\%\)
- Rounded to 3 significant figures = \(80.4\%\)

Part (a) [Max 3 marks]:
- 1 mark for stating that manure contains organic nutrients / acts as food for decomposers / bacteria.
- 1 mark for stating that the population of decomposers / bacteria increases rapidly.
- 1 mark for explaining that these bacteria respire aerobically and use up the dissolved oxygen.

Part (b) [1.5 marks]:
- 0.5 marks for showing the correct calculation method: \(\frac{9.2 - 1.8}{9.2} \times 100\) or \(\frac{7.4}{9.2} \times 100\).
- 1.0 mark for the correct final percentage: \(80.4\%\) (accept range \(80\%\) to \(80.43\%\)).

Part (a):
- The internal diameter of the tube is \(1.0 \text{ mm}\), so the radius \(r = 0.5 \text{ mm}\).
- The distance the bubble moved under control conditions is \(h = 48 \text{ mm}\).
- The volume of water taken up is \(V = 3.142 \times (0.5)^2 \times 48\)
- \(V = 3.142 \times 0.25 \times 48 = 37.704 \text{ mm}^3\).
- Time = \(15 \text{ minutes}\).
- Rate of water uptake = \(\frac{37.704}{15} = 2.5136 \text{ mm}^3/\text{min}\).
- Rounded to 2 decimal places = \(2.51 \text{ mm}^3/\text{min}\).

Part (b):
- Wind moves water molecules away from the surface of the leaf / stomatal pores.
- This prevents the accumulation of water vapour outside the stomata (reduces the boundary layer).
- This maintains a steep water potential (or water vapour concentration) gradient between the air spaces inside the leaf and the atmosphere.
- Thus, the rate of diffusion / evaporation of water out of the stomata increases.

Part (a) [2.5 marks]:
- 0.5 marks for identifying the radius is 0.5 mm (or converting diameter to radius).
- 1.0 mark for the correct volume calculation: \(37.7 \text{ mm}^3\) (accept \(37.70\) to \(37.704\)).
- 1.0 mark for dividing the volume by time to get the correct rate: \(2.51 \text{ mm}^3/\text{min}\) (accept range \(2.51\) to \(2.514\)).

Part (b) [2 marks]:
- 1 mark for stating that wind blows away water vapour / prevents accumulation near the stomata / boundary layer.
- 1 mark for explaining that this maintains or steepens the water potential / concentration gradient, leading to faster diffusion / evaporation.

Part (a):
- Auxin is produced in the growing tip of the shoot and diffuses down the stem.
- When the shoot is horizontal, gravity causes auxin to accumulate in higher concentrations on the lower side of the shoot.
- In stems/shoots, auxin promotes / stimulates cell elongation.
- The cells on the lower side elongate more than the cells on the upper side, resulting in unequal growth and causing the shoot to bend upwards.

Part (b):
- Initial length = \(4.2 \text{ cm} = 42 \text{ mm}\)
- Final length = \(5.4 \text{ cm} = 54 \text{ mm}\)
- Growth in length = \(54 \text{ mm} - 42 \text{ mm} = 12 \text{ mm}\)
- Duration = \(48 \text{ hours}\)
- Rate of growth = \(\frac{12 \text{ mm}}{48 \text{ hours}} = 0.25 \text{ mm/hour}\)

Part (a) [Max 3 marks]:
- 1 mark for stating that gravity causes auxin to accumulate on the lower side of the shoot.
- 1 mark for stating that auxins stimulate cell elongation in shoots.
- 1 mark for explaining that unequal elongation (more elongation on the lower side than the upper side) causes the upward bending.

Part (b) [1.5 marks]:
- 0.5 marks for calculating the growth in mm (\(12 \text{ mm}\)) or showing correct conversion from cm to mm.
- 1.0 mark for dividing the growth by 48 to get the correct rate of \(0.25 \text{ mm/hour}\).

Part (a):
- Drinking 1.5 litres of water increases the water potential of the blood, which is detected by osmoreceptors in the hypothalamus.
- This leads to a decrease in the secretion of ADH from the pituitary gland.
- With less ADH in the blood, the walls of the collecting ducts in the kidney nephrons become less permeable to water.
- Consequently, less water is reabsorbed from the filtrate back into the surrounding blood capillaries, leaving more water in the urine (producing a large volume of dilute urine).

Part (b):
- New urine output = \(3.2 \text{ cm}^3/\text{min}\)
- Normal urine output = \(0.8 \text{ cm}^3/\text{min}\)
- Ratio (New : Normal) = \(3.2 : 0.8\)
- Simplifying by dividing both numbers by 0.8:
\(\frac{3.2}{0.8} = 4\)
\(\frac{0.8}{0.8} = 1\)
- Simple whole-number ratio = \(4 : 1\)

Part (a) [Max 3 marks]:
- 1 mark for stating that the pituitary gland releases less ADH / ADH secretion decreases.
- 1 mark for stating that the walls of the collecting ducts become less permeable to water.
- 1 mark for explaining that less water is reabsorbed back into the blood / more water remains in the urine.

Part (b) [1.5 marks]:
- 0.5 marks for setting up the ratio \(3.2 : 0.8\) (or equivalent division).
- 1.0 mark for simplifying to the correct whole-number ratio \(4 : 1\) (reject \(1 : 4\)).

Part (a):
- The human gene for insulin is isolated and cut out of human DNA using a specific restriction enzyme.
- A bacterial plasmid (which acts as the vector) is cut open using the same restriction enzyme, leaving complementary sticky ends.
- The cut insulin gene and the open plasmid are mixed together and joined using the enzyme DNA ligase to produce a recombinant plasmid.
- This recombinant plasmid is then inserted back into a host bacterium (transformation).

Part (b):
- Volume of fermenter = \(15,000 \text{ litres}\)
- Production rate = \(0.12 \text{ g/litre/hour}\)
- Mass of insulin produced per hour = \(15,000 \times 0.12 = 1,800 \text{ g/hour}\)
- Mass of insulin produced in 24 hours = \(1,800 \times 24 = 43,200 \text{ g}\)
- Converting to kilograms (optional but acceptable): \(\frac{43,200}{1000} = 43.2 \text{ kg}\)

Part (a) [Max 3 marks]:
- 1 mark for stating that a restriction enzyme is used to cut out the insulin gene and open the plasmid vector, creating complementary sticky ends.
- 1 mark for stating that DNA ligase is used to join the insulin gene and the plasmid vector together to form a recombinant plasmid.
- 1 mark for explaining that the recombinant plasmid vector is inserted into a bacterium.

Part (b) [1.5 marks]:
- 0.5 marks for showing correct working (multiplying volume, hourly rate, and 24 hours): \(15,000 \times 0.12 \times 24\).
- 1.0 mark for the correct numerical value and correct unit: \(43,200 \text{ g}\) or \(43.2 \text{ kg}\).

Part (a):
- Small pieces of plant tissue (explants) are cut from the growing tip or stem of the desirable parent plant.
- The explants are sterilized using a disinfectant (like bleach or alcohol) to kill any micro-organisms (bacteria/fungi).
- The sterile explants are placed onto a sterile agar medium containing nutrients (glucose, amino acids, mineral ions) and plant growth regulators (auxins and cytokinins).
- The cells divide by mitosis to form a callus, which then differentiates into tiny plantlets with roots and shoots under controlled light and temperature.

Part (b):
- Success rate = \(85\% = 0.85\)
- Desired number of plantlets = \(1,190\)
- Let \(x\) be the number of explants started with.
- \(0.85 \times x = 1190\)
- \(x = \frac{1190}{0.85} = 1400\)
- Therefore, the grower must start with a minimum of 1,400 explants.

Part (a) [Max 3 marks]:
- 1 mark for stating that explants are cut from the parent plant and sterilized (to prevent contamination by microbes).
- 1 mark for stating that they are placed on sterile agar containing nutrients (glucose/minerals) and hormones.
- 1 mark for explaining that the cells divide by mitosis to form plantlets.

Part (b) [1.5 marks]:
- 0.5 marks for showing the correct algebraic arrangement: \(\frac{1190}{0.85}\).
- 1.0 mark for the correct answer of 1400.

Part (a):
- Methods include using antibiotics/fungicides in the food or water to treat bacterial/fungal infections.
- Biological controls can be used, such as introducing cleaner fish (e.g., wrasse) to eat sea lice.
- Farmers can minimize stocking density so that the fish are less crowded, which slows down the transmission of infectious agents.
- Regular removal of dead or visibly sick fish prevents pathogens from spreading to healthy fish.

Part (b):
- Increase in fish mass = \(2,700 \text{ kg} - 1,200 \text{ kg} = 1,500 \text{ kg}\)
- Mass of feed consumed = \(3,750 \text{ kg}\)
- \(\text{FCR} = \frac{3750}{1500} = 2.5\)
- The FCR is 2.5 (meaning 2.5 kg of feed is needed to produce 1 kg of salmon mass increase).

Part (a) [2 marks]:
- 1 mark for each described method up to a maximum of 2. Acceptable methods include: antibiotics/pesticides/chemicals, separating different age groups, biological control (e.g., cleaner fish), reducing stocking density, removal of dead fish, or keeping water clean/filtered.

Part (b) [2.5 marks]:
- 0.5 marks for calculating the correct mass increase: \(2700 - 1200 = 1500 \text{ kg}\).
- 1.0 mark for substituting correct values into the formula: \(\frac{3750}{1500}\).
- 1.0 mark for the correct final answer of 2.5 (no units needed, accept 2.5 : 1).

Part (a):
- Secure the slide on the stage using stage clips.
- Rotate the nosepiece to select the lowest power objective lens.
- Look from the side and use the coarse focus knob to move the stage up / lens down until it is close to the slide, then look through the eyepiece and turn the coarse focus knob in the opposite direction until the cells are in focus.
- Switch to the high-power objective lens.
- Adjust the fine focus knob only to sharpen and obtain a clear, detailed image.

Part (b):
- Formula: \(\text{Magnification} = \frac{\text{Image size}}{\text{Actual size}}\)
- Convert units to be consistent: Image size = \(45 \text{ mm} = 45 \times 1000 = 45,000 \text{ µm}\) (or actual size = \(30 \text{ µm} = 0.03 \text{ mm}\)).
- \(\text{Magnification} = \frac{45,000 \text{ µm}}{30 \text{ µm}} = 1500\)
- Therefore, the magnification is \(\times 1500\).

Part (a) [Max 3 marks]:
- 1 mark for starting with the lowest power objective lens and using the coarse focus knob to locate the cells.
- 1 mark for switching to the high-power objective lens.
- 1 mark for using the fine focus knob only at high power to sharpen the image.

Part (b) [1.5 marks]:
- 0.5 marks for converting units correctly (either \(45 \text{ mm}\) to \(45,000 \text{ µm}\) or \(30 \text{ µm}\) to \(0.03 \text{ mm}\)).
- 1.0 mark for dividing the image size by the actual size to obtain \(1500\) or \(\times 1500\).

The dialysis machine separates the patient's blood from the dialysis fluid using a partially permeable membrane. 1. Proteins: Although blood plasma contains \(70.0\text{ g/dm}^{3}\) of protein and the dialysis fluid contains none, proteins are large macromolecules. They cannot pass through the pores of the partially permeable membrane and therefore remain in the blood. 2. Glucose: The dialysis fluid is formulated to have a glucose concentration of \(1.0\text{ g/dm}^{3}\), which is identical to the normal glucose concentration in the blood. Because there is no concentration gradient, there is no net diffusion of glucose out of the blood. 3. Urea: The dialysis fluid has a urea concentration of \(0.0\text{ g/dm}^{3}\), whereas the blood contains \(1.8\text{ g/dm}^{3}\). This creates a steep concentration gradient, allowing urea to rapidly diffuse out of the blood and into the dialysis fluid. The dialysis fluid is constantly refreshed to maintain this gradient.

1. Membrane: State that blood and dialysis fluid are separated by a selectively / partially permeable membrane (1 mark). 2. Protein size: Explain that proteins are too large to pass through the membrane pores (1 mark). 3. Glucose concentration: Reference the equal glucose concentrations of \(1.0\text{ g/dm}^{3}\) in both blood and fluid, meaning there is no concentration gradient / no net movement of glucose (1 mark). 4. Urea gradient: Reference the concentration gradient for urea (\(1.8\text{ g/dm}^{3}\) in blood vs \(0.0\text{ g/dm}^{3}\) in fluid) (1 mark). 5. Urea diffusion: Explain that urea moves out of the blood by diffusion down this concentration gradient (1 mark).