2024 Cambridge IAL Biology (9700) 模擬試題連答案詳解

A competitive inhibitor binds to the active site of the enzyme, increasing the apparent \(K_m\) because a higher substrate concentration is required to reach half \(V_{max}\). However, \(V_{max}\) remains unchanged because high substrate concentrations can overcome the inhibition. A non-competitive inhibitor binds to an allosteric site, decreasing \(V_{max}\) because it reduces the concentration of active enzyme molecules, but the \(K_m\) remains unchanged because the affinity of the remaining active enzymes for the substrate is unaffected.

1 mark for identifying the correct effects of competitive and non-competitive inhibitors on both parameters (A).

Both areas have exactly 4 species, so their species richness is identical (A is incorrect). Area Y has a much more balanced distribution of individuals among the species (10, 12, 13, 15) compared to Area X, which is heavily dominated by Species 1 (45 individuals). This means Area Y has greater species evenness (B is correct). A higher species evenness results in a higher Simpson's Index of Diversity value, so Area Y will have a higher \(D\) value than Area X (C is incorrect). The index can be calculated regardless of whether sample sizes are identical (D is incorrect).

1 mark for the correct deduction regarding species evenness in Area Y (B).

At high light intensities, light is no longer the limiting factor. Since the concentration of carbon dioxide is kept constant and high, temperature becomes the primary limiting factor. The reactions of the light-independent stage (Calvin cycle) are catalyzed by enzymes such as RuBisCO. An increase in temperature from \(15\ ^\circ\text{C}\) to \(25\ ^\circ\text{C}\) increases the kinetic energy of these enzymes and substrates, increasing the rate of reaction and therefore the overall rate of photosynthesis.

1 mark for explaining that temperature is the limiting factor at high light intensity and affects enzyme-controlled reactions (A).

Sieve tube elements have peripheral cytoplasm and lack a nucleus and vacuole (1) to provide an unobstructed pathway, reducing resistance to the mass flow of organic solutes. They also have sieve plates with pores (3) to allow continuous flow from one element to the next. Lignified cell walls (2) are characteristic of xylem vessels, not sieve tube elements, which have non-lignified cellulose walls.

1 mark for identifying statements 1 and 3 as correct structural adaptations of sieve tube elements (A).

According to the acid growth hypothesis, auxin (IAA) stimulates proton pumps in the cell membrane to actively transport hydrogen ions (\(\text{H}^+\)) into the cell wall. This lowers the pH of the cell wall, activating expansins (proteins that break hydrogen bonds and cross-links between cellulose microfibrils). This relaxes the wall's structure, allowing the cell to expand as water enters the vacuole and increases turgor pressure.

1 mark for the correct description of the acid-growth hypothesis involving proton pumps, low pH, and expansins (A).

Monoclonal antibodies are identical antibodies produced from cloned hybridoma cells (formed by fusing a plasma B-lymphocyte with a myeloma cell, so A is incorrect). Because they are identical, they all have the same antigen-binding site complementary to one specific epitope (B is correct). Like all IgG antibodies, they consist of two heavy and two light polypeptide chains (C is incorrect). Administered monoclonal antibodies provide passive immunity, which is temporary because no memory cells are produced in the recipient (D is incorrect).

1 mark for identifying that monoclonal antibodies are identical and complementary to a single specific epitope (B).

Mitochondria have an inner membrane folded into cristae to increase surface area, and chloroplasts contain grana which are stacks of thylakoid membranes (Row A is correct). Both mitochondria and chloroplasts contain 70S ribosomes, not 80S (Row B is incorrect). Mitochondria are the site of oxidative phosphorylation, while chloroplasts are the site of photophosphorylation (Row C is incorrect). Both organelles are surrounded by a double membrane envelope (Row D is incorrect).

1 mark for identifying the correct structural features of mitochondria and chloroplasts in Row A.

During lactate fermentation in mammalian muscle cells, pyruvate acts as the hydrogen acceptor. It is reduced to lactate by accepting hydrogen atoms from reduced \(\text{NAD}\) (\(\text{NADH} + \text{H}^+\)), a reaction catalysed by lactate dehydrogenase. This regenerates oxidized \(\text{NAD}\), which is essential for glycolysis to continue. Option C describes ethanol fermentation, which occurs in yeast and plants, not mammals.

1 mark for identifying that pyruvate is reduced to lactate to regenerate NAD (A).

An increase in \(K_m\) indicates a lower apparent affinity of the enzyme for its substrate, as more substrate is required to reach half of the maximum velocity (\(\frac{1}{2}V_{max}\)). Since the \(V_{max}\) remains unchanged, the inhibitor must be competitive. Competitive inhibitors bind reversibly to the active site and can be outcompeted at very high substrate concentrations, allowing the original maximum velocity to be achieved.

Award 1 mark for identifying the inhibitor as competitive and explaining that an increased \(K_m\) corresponds to decreased apparent affinity of the enzyme for the substrate.

Simpson's Index of Diversity (\(D\)) ranges between 0 and 1, where a value closer to 1 indicates high biodiversity. High biodiversity is characterized by both high species richness (the number of different species present) and high species evenness (the relative abundance of each species). Such ecosystems are stable and resilient, meaning they can better withstand environmental disturbances.

Award 1 mark for explaining that a high value of \(D\) (close to 1) represents high species richness and evenness, resulting in a more resilient ecosystem.

Sodium hydrogencarbonate dissociates in water to release carbon dioxide (\(CO_2\)), which ensures that carbon dioxide concentration is not a limiting factor for photosynthesis. The gas released as a product of the light-dependent stage of photosynthesis is oxygen (\(O_2\)), which is collected and measured in the capillary tube.

Award 1 mark for identifying the role of sodium hydrogencarbonate as a carbon dioxide source and the collected gas as oxygen.

Xylem vessel elements have highly lignified cell walls to withstand tension and provide mechanical support, and their end walls are completely absent to allow a continuous column of water to flow. Phloem sieve tube elements have non-lignified cellulose walls and retaining end walls modified into sieve plates with sieve pores to allow the flow of organic solutes.

Award 1 mark for the correct combination of cell wall lignification and end wall structures in both transport tissues.

According to the acid growth hypothesis, auxin binds to receptors, stimulating proton pumps in the cell membrane to pump \(H^+\) ions into the cell wall, decreasing the pH. The low pH activates expansin proteins, which break hydrogen bonds between cellulose microfibrils. The cell wall becomes loose, allowing water to enter the vacuole by osmosis, causing the cell to expand.

Award 1 mark for the correct sequence of events from proton pumping to cell wall loosening and osmotic water entry.

Artificial active immunity occurs when an antigen is intentionally introduced into the body (artificial), such as via a vaccine containing weakened (attenuated) pathogens, triggering an immune response where the body produces its own antibodies and memory cells (active).

Award 1 mark for identifying vaccination with a weakened pathogen as artificial active immunity.

Resolution is limited by the wavelength of the radiation used to image the specimen. The wavelength of an electron beam is much shorter than the wavelength of visible light, which allows the TEM to distinguish between two points that are much closer together, achieving a significantly higher resolution.

Award 1 mark for identifying that the shorter wavelength of the electron beam gives the TEM a higher resolution than the light microscope.

When blood water potential falls (e.g., dehydration), water leaves osmoreceptors in the hypothalamus by osmosis, causing them to shrink. This stimulates neurosecretory cells to release ADH from the posterior pituitary gland. ADH travels in the blood and binds to receptors on the collecting duct walls, causing vesicle fusion to add aquaporins to the luminal membranes, increasing water reabsorption.

Award 1 mark for the correct sequence: osmoreceptors shrink, ADH is released from the posterior pituitary, and aquaporins are inserted into the collecting duct membranes.

An enzyme catalyzing a reaction lowers the activation energy required to initiate the reaction. However, it does not alter the free energy or overall energy change of the reaction (the difference in energy levels between the reactants and products remains constant).

Correct answer is A (1 mark). B is incorrect as overall energy change is unaffected. C is incorrect as activation energy must decrease. D is incorrect as activation energy decreases, not increases.

To calculate Simpson's Index of Diversity (D) for Field 1: total individuals (N) = 100. sum of (n/N)^2 = (50/100)^2 + (30/100)^2 + (20/100)^2 = 0.25 + 0.09 + 0.04 = 0.38. D = 1 - 0.38 = 0.62. For Field 2: total individuals (N) = 100. sum of (n/N)^2 = (80/100)^2 + (15/100)^2 + (5/100)^2 = 0.64 + 0.0225 + 0.0025 = 0.665. D = 1 - 0.665 = 0.335 (rounds to 0.34). Field 1 has a higher D value because the individuals are more evenly distributed among the species (greater species evenness), whereas Field 2 is heavily dominated by one species.

Correct answer is A (1 mark) for correct calculation of D values and linking higher biodiversity to greater species evenness.

Since the rate of photosynthesis remains constant at 45 cubic millimetres per minute when the distance is changed from 20 cm to 10 cm, increasing the light intensity does not increase the rate of photosynthesis. Therefore, light intensity is not the limiting factor in this range; another factor such as temperature or carbon dioxide concentration must be limiting.

Correct answer is B (1 mark). A is incorrect because rate does not change. C is incorrect because positive net oxygen production shows photosynthesis exceeds respiration. D is incorrect because the relationship is not linear.

Mature xylem vessel elements are dead cells with cell walls reinforced with lignin and completely lack cytoplasm. In contrast, mature phloem sieve tube elements are living cells with cellulose cell walls and contain a thin layer of peripheral cytoplasm (though they lack a nucleus, ribosomes, and vacuole).

Correct answer is A (1 mark). B is incorrect as xylem vessels do not have a plasma membrane and sieve tube elements do. C is incorrect as xylem vessel walls contain lignin, not cellulose only. D is incorrect as xylem vessels are perforated and sieve tube elements do have a plasma membrane.

Seed germination begins with the absorption of water (imbibition), which stimulates the embryo to synthesize and release gibberellin. Gibberellin diffuses to the surrounding aleurone layer, where it activates the transcription of the gene for amylase. The synthesized amylase is then secreted into the endosperm, where it hydrolyses stored starch into maltose and glucose.

Correct answer is A (1 mark). B, C, and D describe incorrect sites of synthesis or incorrect sequence of physiological steps.

Phagocytes (such as macrophages) process and present foreign antigens on their cell surface bound to MHC class II proteins. Helper T-lymphocytes recognize these antigens and secrete cytokines (e.g., interleukins) to coordinate the immune response. B-lymphocytes, upon activation, proliferate and differentiate into plasma cells that secrete specific antibodies.

Correct answer is A (1 mark). Other options represent incorrect cellular functions or anatomical locations.

A non-competitive inhibitor binds to an alternative site on the enzyme (the allosteric site) rather than the active site. This binding alters the tertiary structure of the enzyme, changing the shape of its active site so that the substrate can no longer bind or the reaction can no longer be catalysed. Because it does not compete with the substrate for the active site, increasing substrate concentration cannot overcome this inhibition, resulting in a decreased Vmax.

Correct answer is B (1 mark). A describes competitive inhibition. C and D describe incorrect or physically impossible mechanisms.

ADH is a peptide hormone and binds to G-protein coupled receptors on the basolateral membrane of collecting duct cells. This activates adenylyl cyclase, raising intracellular cAMP (second messenger) levels. This triggers a signaling cascade that causes vesicles containing aquaporins to fuse with the luminal (apical) membrane. The increased water permeability allows water to exit the collecting duct lumen down its water potential gradient via osmosis.

Correct answer is A (1 mark). B is incorrect as ADH is peptide-based and does not cross the membrane. C and D are incorrect as water movement through aquaporins is passive (osmosis), and ADH does not act via active transport of sodium ions to drive this process.

A competitive inhibitor binds to the active site of the enzyme, competing directly with the substrate. This increases the apparent Michaelis constant (\(K_m\)) because a higher concentration of substrate is required to achieve half the maximum velocity (\(V_{\text{max}}\)). Since the inhibition can be overcome at very high substrate concentrations, the \(V_{\text{max}}\) remains unchanged. An increase in \(K_m\) corresponds to a decrease in the affinity of the enzyme for its substrate.

1 mark for selecting B. Reject other options: A and D are incorrect because non-competitive inhibitors decrease \(V_{\text{max}}\); C is incorrect because an increased \(K_m\) represents a decreased affinity, not an increased affinity.

A high Simpson’s Index of Diversity (close to 1) indicates high species richness and evenness, meaning the community is stable and not dominated by just one or two species. In such a diverse community (Area X), the ecological niches are complex, and the loss of a single species due to a disease is less likely to cause the collapse of the entire community compared to Area Y, which has low diversity and is highly unstable.

1 mark for identifying C as the correct option. Reject A because high diversity means it is not dominated by one or two species; reject B because Area Y has lower diversity; reject D because high-diversity areas are more resilient to small environmental changes.

Under Condition 1 (\(15^\circ\text{C}\) and \(0.04\%\text{ CO}_2\)) at a light intensity of \(30\text{ au}\), increasing the concentration of carbon dioxide to \(0.12\%\) (Condition 2) while keeping temperature and light intensity constant resulted in a doubling of the rate of photosynthesis. This shows that the rate was limited by the concentration of carbon dioxide in Condition 1.

1 mark for B. Reject A because light intensity is already at \(30\text{ au}\) and changing \(CO_2\) increased the rate; reject C because temperature was kept constant at \(15^\circ\text{C}\) in both conditions and yet the rate increased; reject D because water is rarely a limiting factor in aquatic plants like *Elodea*.

A mature phloem sieve tube element is a living cell. Although it has lost many organelles, it retains a thin layer of cytoplasm (2), a cell surface membrane (1), and is connected to adjacent companion cells via plasmodesmata (3). A mature xylem vessel element is a dead, hollow tube with no cytoplasm, cell membrane, or plasmodesmata. Both types of cells lack a nucleus at maturity, so feature 4 (absence of a nucleus) does not distinguish between them.

1 mark for selecting A. Reject B because feature 3 (plasmodesmata) is also a distinguishing feature. Reject C and D because feature 4 does not distinguish them as both lack a nucleus.

The acid growth hypothesis states that auxin (IAA) stimulates proton pumps (\(\text{H}^+\)-ATPases) in the cell surface membrane. These pumps move protons (\(\text{H}^+\) ions) from the cytoplasm into the cell wall, causing a drop in cell wall pH. This acidic environment activates expansin proteins, which disrupt the hydrogen bonds holding the cellulose microfibrils to other hemicelluloses, loosening the cell wall and allowing water uptake to cause cell elongation.

1 mark for A. Reject B because expansins break hydrogen bonds, not covalent glycosidic bonds; reject C because \(\text{H}^+\) ions are pumped *into* the cell wall, not the cytoplasm, and this decreases pH, not increases it; reject D because IAA does not denature proteins to loosen the cell wall.

Clonal selection is the process by which a specific antigen binds to the complementary antibody receptor on the surface of a unique B-lymphocyte, selecting that particular cell to be activated. Clonal expansion (option A) is the subsequent mitotic division of these selected cells. Differentiation (option C) occurs after expansion.

1 mark for B. Reject A (describes clonal expansion); reject C (describes differentiation); reject D (describes T-helper cell action, which is not B-cell clonal selection).

The temperature coefficient (\(Q_{10}\)) represents the factor by which the rate of reaction increases for every \(10^\circ\text{C}\) rise in temperature. The temperature increases from \(15^\circ\text{C}\) to \(35^\circ\text{C}\), which is an increase of \(20^\circ\text{C}\) (two \(10^\circ\text{C}\) intervals).

* Rate at \(25^\circ\text{C} = 1.2 \times 2.0 = 2.4\text{ mmol dm}^{-3}\text{ s}^{-1}\)
* Rate at \(35^\circ\text{C} = 2.4 \times 2.0 = 4.8\text{ mmol dm}^{-3}\text{ s}^{-1}\)

1 mark for selecting C. Option A represents only a \(10^\circ\text{C}\) rise. Option B is an incorrect linear addition. Option D represents a three-fold doubling (\(30^\circ\text{C}\) rise).

Substance W is present in blood plasma but absent in both glomerular filtrate and urine, identifying it as plasma protein (which is too large to pass through the basement membrane during ultrafiltration). Substance X is filtered into the nephron (same concentration as plasma) but completely reabsorbed in the proximal convoluted tubule, identifying it as glucose. Substance Y is filtered and highly concentrated in the urine, identifying it as urea. Substance Z is filtered and partially reabsorbed but concentrated due to water reabsorption, identifying it as sodium ions.

1 mark for selecting A. Reject other options based on the incorrect placement of protein (cannot be filtered) and glucose (should be fully reabsorbed under normal conditions).

(a) A non-competitive inhibitor binds to an allosteric site (a site other than the active site) of the enzyme. This binding alters the overall tertiary structure (three-dimensional shape) of the enzyme, causing a conformational change in the active site. Consequently, the active site is no longer complementary to the substrate, preventing the formation of enzyme-substrate complexes.

(b) In the absence of lead ions, the rate of reaction increases rapidly at low substrate concentrations as the substrate is limiting. As substrate concentration increases further, the curve plateaus (reaches a maximum) because the enzyme concentration becomes the limiting factor (all active sites are fully saturated with substrate). In the presence of lead ions, the rate of reaction is significantly lower at all substrate concentrations. The curve plateaus at a much lower maximum rate because the inhibitor inactivates a portion of the enzyme molecules, and increasing the substrate concentration does not displace the non-competitive inhibitor or overcome the inhibition.

(c) (i) \(V_{\text{max}}\) is decreased because the number of functional enzymes is reduced. (ii) \(K_{\text{m}}\) remains unchanged because the affinity of the remaining active enzymes for the substrate is unaffected by a non-competitive inhibitor.

(a) Max 3 marks:
- Inhibitor binds to a site other than the active site / allosteric site [1]
- Alters the tertiary structure / 3D shape of the enzyme [1]
- Changes the conformation / shape of the active site [1]
- Substrate is no longer complementary / cannot bind / fewer enzyme-substrate complexes form [1]

(b) Max 5 marks:
- Without lead ions: rate increases as substrate concentration increases and then plateaus [1]
- Explanation: at low substrate concentrations, substrate is limiting; at high substrate concentrations, enzyme active sites are saturated / enzyme concentration is limiting [1]
- With lead ions: rate is lower at all substrate concentrations [1]
- The curve plateaus at a lower maximum rate / maximum rate is never reached [1]
- Explanation: increasing substrate concentration does not overcome the inhibition / does not displace the inhibitor from the enzyme [1]

(c) 2 marks:
- (i) \(V_{\text{max}}\) decreases [1]
- (ii) \(K_{\text{m}}\) remains unchanged [1]

(a) The formula for Simpson's Index of Diversity is:
\(D = 1 - \sum \left(\frac{n}{N}\right)^2\)
where:
- \(n\) = number of individuals of a particular species
- \(N\) = total number of individuals of all species
- \(\sum\) = sum of
A high value of \(D\) (close to 1) indicates high species richness and high species evenness. This suggests a highly stable ecosystem that is resilient to environmental changes.

(b) Working for Fragment A:
- Species 1: \(n/N = 50/100 = 0.5\), \((n/N)^2 = 0.25\)
- Species 2: \(n/N = 30/100 = 0.3\), \((n/N)^2 = 0.09\)
- Species 3: \(n/N = 20/100 = 0.2\), \((n/N)^2 = 0.04\)
- Sum of \((n/N)^2 = 0.25 + 0.09 + 0.04 = 0.38\)
- \(D = 1 - 0.38 = 0.62\)

Working for Fragment B:
- Species 1: \(n/N = 90/100 = 0.9\), \((n/N)^2 = 0.81\)
- Species 2: \(n/N = 8/100 = 0.08\), \((n/N)^2 = 0.0064\)
- Species 3: \(n/N = 2/100 = 0.02\), \((n/N)^2 = 0.0004\)
- Sum of \((n/N)^2 = 0.81 + 0.0064 + 0.0004 = 0.8168\)
- \(D = 1 - 0.8168 = 0.1832\) (or \(0.18\))

(c) Habitat fragmentation breaks up continuous forests into smaller, isolated patches, which reduces gene flow between populations. This leads to inbreeding depression, loss of genetic variation, and increases vulnerability to local extinction. It also increases edge effects, exposing species to harsher microclimates and invasive species. To mitigate these effects, conservationists can establish wildlife corridors (or habitat corridors) to link isolated patches, allowing animals to travel, forage, and mate safely.

(a) Max 3 marks:
- Correct formula: \(D = 1 - \sum (n/N)^2\) [1]
- Identifies \(n\) as number of individuals of a species, and \(N\) as total number of individuals of all species [1]
- Explains that a high \(D\) value indicates high species diversity / high evenness / stable community [1]

(b) Max 4 marks:
- Correct working of \(\sum(n/N)^2\) for Fragment A (\(0.38\)) [1]
- Correct calculation of \(D\) for Fragment A (\(0.62\)) [1]
- Correct working of \(\sum(n/N)^2\) for Fragment B (\(0.8168\)) [1]
- Correct calculation of \(D\) for Fragment B (\(0.18\) or \(0.183\)) [1]

(c) Max 3 marks (at least 1 for effects, 1 for mitigation):
- Effects: smaller/isolated populations are susceptible to genetic drift / inbreeding depression [1]
- Effects: edge effects increase / harsher microclimate at borders [1]
- Mitigation: wildlife corridors / habitat corridors to connect fragments [1]
- Mitigation: designation of national parks / nature reserves with strict protection [1]

(a) A limiting factor is an environmental factor that is in the shortest supply and directly determines and limits the rate of a physiological process. If its value is increased, the rate of the process will increase.

(b) To vary light intensity, the student can position the light source (lamp) at varying, precisely measured distances from the photosynthetic apparatus. The relative light intensity can be calculated using the inverse-square law, \(I = 1/d^2\). To prevent temperature from confounding the results, the student should place a thick, clear glass water bath or heat sink between the lamp and the boiling tube containing the plant. This water bath absorbs the thermal energy (heat) emitted by the lamp while allowing light to pass through. Alternatively, an LED light source can be used as it produces very little heat.

(c) (i) At \(10\text{ }^\circ\text{C}\), the rate of photosynthesis increases linearly with light intensity at first, and then plateaus at a relatively low maximum rate.
(ii) At \(25\text{ }^\circ\text{C}\), the rate also increases linearly at low light levels but plateaus at a much higher maximum rate.
The difference in the plateau region is because at high light intensities, light is no longer the limiting factor. At \(10\text{ }^\circ\text{C}\), temperature is the limiting factor. The enzymes involved in the light-independent stage of photosynthesis, such as Rubisco, have low kinetic energy at \(10\text{ }^\circ\text{C}\), resulting in fewer successful collisions and a slower rate of carbon fixation. At \(25\text{ }^\circ\text{C}\), the kinetic energy is higher, and enzymes operate much closer to their optimum rate.

(a) Max 2 marks:
- Factor that is in shortest supply / nearest its minimum value [1]
- Which directly determines / limits the rate of photosynthesis when increased [1]

(b) Max 4 marks:
- Move the lamp to different, measured distances from the plant [1]
- Explain relationship between distance and light intensity (\(I \propto 1/d^2\)) [1]
- Use a water bath / glass shield / beaker of water between the light source and plant [1]
- To absorb heat / maintain constant temperature around the plant [1]
- (Accept) Use an LED bulb that emits very little heat [1]

(c) Max 4 marks:
- At both temperatures, the rate increases linearly at low light intensities / light is limiting [1]
- Both curves plateau / level off at high light intensities [1]
- The plateau for \(10\text{ }^\circ\text{C}\) is lower than for \(25\text{ }^\circ\text{C}\) [1]
- Explanation: at high light intensity, temperature is limiting; at \(10\text{ }^\circ\text{C}\), enzymes (e.g., Rubisco) have lower kinetic energy / fewer successful collisions [1]

(a) Xylem vessel elements are highly adapted for transport:
- They are dead cells containing no cytoplasm, nuclei, or organelles, which leaves a completely empty, hollow lumen. This drastically reduces the resistance to water movement.
- The end walls of the individual elements are lost completely, forming continuous, open tubes that allow uninterrupted water columns to flow upwards.
- The cell walls are thickened with lignin, which provides high mechanical strength to prevent the vessel from collapsing inwards under the negative pressure (tension) generated by transpiration.
- Unlignified areas, called pits, are present in the cell walls, allowing lateral movement of water and dissolved minerals to adjacent vessels or cells.

(b) Sieve tube elements and companion cells work together as a functional unit:
- Sieve tube elements have an extremely simplified cytoplasm (lacking a nucleus, vacuole, and ribosomes, and with only a thin peripheral layer of cytoplasm). This reduces resistance to the mass flow of phloem sap.
- Sieve plates with sieve pores exist at the junctions between elements, permitting the continuous passage of assimilates under hydrostatic pressure.
- Sieve tube elements have tough, reinforced cellulose walls to withstand high hydrostatic pressure without bursting.
- Sieve tube elements are connected to companion cells via abundant plasmodesmata, which allows the direct passage of organic solutes and proteins.
- Companion cells contain a large number of mitochondria, which produce ATP for the active transport of hydrogen ions out of the companion cell cytoplasm into the cell wall.
- Companion cells also contain proton pumps and cotransporter proteins in their cell membranes, enabling the secondary active loading of sucrose against its concentration gradient.

(a) Max 4 marks:
- Lacking cytoplasm / nucleus / organelles to create a hollow lumen / reduce resistance to water flow [1]
- Loss of end walls / continuous tubes to allow uninterrupted water columns [1]
- Lignification of walls to prevent collapse under tension / negative pressure [1]
- Presence of pits to allow lateral movement of water [1]

(b) Max 6 marks:
- Sieve tube elements have minimal organelles / no nucleus / thin peripheral cytoplasm to reduce resistance to mass flow [1]
- Sieve plates have pores to allow mass flow of phloem sap [1]
- Strong cellulose walls to withstand high hydrostatic pressure [1]
- Plasmodesmata connect companion cells and sieve tube elements for exchange of substances [1]
- Companion cells have numerous mitochondria to produce ATP for active loading [1]
- Companion cells have proton pumps to pump \(\text{H}^+\) out into the wall, and cotransporter proteins to transport sucrose into the cell [1]

(a) According to the acid-growth hypothesis, auxin binds to cell surface receptors on target cells. This stimulates proton pumps (\(\text{H}^+\)-ATPases) in the cell membrane to actively pump hydrogen ions (\(\text{H}^+\)) from the cytoplasm into the cell wall space. The resulting decrease in pH activates expansins, which are cell-wall-loosening proteins. Expansins break the hydrogen bonds between cellulose microfibrils and other matrix polysaccharides. This relaxes and loosens the cell wall, making it highly extensible. Water then moves into the vacuole of the cell down a water potential gradient by osmosis. This builds up high turgor pressure inside the cell, stretching the weakened cell wall and causing the cell to elongate.

(b) When a shoot tip is exposed to unilateral light (light from one direction), auxin is transported laterally away from the illuminated side to the shaded side of the stem. This results in a higher concentration of auxin on the shaded side. This high concentration of auxin stimulates rapid cell elongation on the shaded side, while the cells on the illuminated side elongate much less. This differential growth causes the stem to bend towards the light source.

(c) During germination, water is absorbed (imbibition) by the seed, which triggers the embryo to synthesize and release gibberellin. The gibberellin diffuses to the aleurone layer surrounding the endosperm. Here, it stimulates the transcription and synthesis of hydrolytic enzymes, specifically \(\alpha\)-amylase. Amylase then diffuses into the endosperm and hydrolyses stored starch into maltose and glucose, which are transported to the growing embryo to provide energy for respiration and materials for growth.

(a) Max 5 marks:
- Auxin binds to cell membrane receptors [1]
- Stimulates proton pumps to actively pump \(\text{H}^+\) into the cell wall [1]
- Lowers the pH of the cell wall [1]
- Activates expansins [1]
- Expansins disrupt/break hydrogen bonds between cellulose microfibrils [1]
- Cell wall becomes more elastic/extensible [1]
- Water enters by osmosis, and turgor pressure causes cell expansion [1]

(b) Max 3 marks:
- Unilateral light causes lateral movement of auxin to the shaded side [1]
- Accumulation of auxin on the shaded side [1]
- Stimulates greater cell elongation on the shaded side relative to the illuminated side [1]
- Causes the shoot to bend towards the light [1]

(c) Max 2 marks:
- Water absorption triggers release of gibberellin from the embryo [1]
- Gibberellin diffuses to the aleurone layer [1]
- Stimulates production of amylase [1]
- Amylase breaks down starch to maltose / glucose for respiration/growth [1]

(a) The hybridoma method begins by injecting a specific antigen into an experimental animal (e.g., a mouse) to trigger an immune response. After some time, plasma B-cells producing the complementary antibody are harvested from the animal's spleen. Since these plasma cells have a limited lifespan and cannot divide, they are fused with immortal cancerous plasma cells (myeloma cells) using a chemical fusing agent such as polyethylene glycol (PEG) or through electrofusion. This forms hybridoma cells. These hybridoma cells are cultured on selective media (such as HAT medium) to select for successful fusions. Individual hybridoma cells are then screened to identify the clone producing the desired monoclonal antibody, which is then cloned in large bioreactors for industrial-scale antibody harvesting.

(b) Active immunity and passive immunity differ in several fundamental ways:
1. Active immunity involves the production of antibodies by the individual's own plasma cells, whereas passive immunity involves receiving pre-formed antibodies from an external source (such as maternal transfer or injection).
2. Active immunity is stimulated by direct contact with an antigen, whereas passive immunity does not require direct antigen contact.
3. Active immunity results in the production of memory cells and provides long-lasting protection, whereas passive immunity produces no memory cells and provides only immediate, temporary protection.

(c) Passive immunity is short-lived because the introduced antibodies are proteins that are eventually recognized as foreign and broken down (metabolised) and excreted by the recipient's body. Furthermore, because passive immunity does not involve clonal selection and expansion of the recipient's own lymphocytes, no memory B-cells or memory T-cells are produced. Consequently, the recipient's immune system cannot manufacture new antibodies when the introduced ones disappear.

(a) Max 5 marks:
- Inject specific antigen into a mammal / mouse [1]
- Extract plasma B-cells from the spleen of the mouse [1]
- Fuse plasma cells with myeloma / cancer cells [1]
- Use PEG (polyethylene glycol) / electrofusion [1]
- Culture on selective media to select hybridoma cells [1]
- Screen and clone hybridomas that produce the desired antibody [1]

(b) Max 3 marks:
- Active involves individual's own plasma cells producing antibodies, passive involves getting pre-made antibodies [1]
- Active is triggered by exposure to antigen, passive does not require exposure to antigen [1]
- Active produces memory cells, passive does not produce memory cells [1]
- Active provides long-term protection, passive provides short-term protection [1]
- Active has a lag phase before protection, passive provides immediate protection [1]

(c) Max 2 marks:
- Foreign antibodies are broken down / degraded by the body [1]
- No memory cells are formed, so the body cannot produce more antibodies on its own [1]

### Solution

**(a) (i) Preparation of Serial Dilution (Table 1.1)**
- **0.50%**: Take 5.0 cm³ of 1.00% amylase stock solution and add 5.0 cm³ of distilled water, **W**.
- **0.25%**: Take 5.0 cm³ of 0.50% amylase solution and add 5.0 cm³ of distilled water, **W**.
- **0.125%**: Take 5.0 cm³ of 0.25% amylase solution and add 5.0 cm³ of distilled water, **W**.
- **0.0625%**: Take 5.0 cm³ of 0.125% amylase solution and add 5.0 cm³ of distilled water, **W**.

**(a) (ii) Identification of Variables**
- **Independent variable**: Concentration of amylase solution / %.
- **Dependent variable**: Time taken for complete starch hydrolysis / s (or rate of reaction / s⁻¹).

**(a) (iii) Table of Results Design**
A suitable table must have a fully bordered grid with correct column headings and units. Amylase concentrations should be listed in descending or ascending order:

| Amylase concentration / % | Time taken for starch hydrolysis, \(t\) / s | Rate of reaction, \(\frac{1}{t}\) / s⁻¹ |
| :--- | :--- | :--- |
| 1.00 | [insert raw data] | [calculated value] |
| 0.50 | [insert raw data] | [calculated value] |
| 0.25 | [insert raw data] | [calculated value] |
| 0.125 | [insert raw data] | [calculated value] |
| 0.0625 | [insert raw data] | [calculated value] |

**(b) (i) Control Experiment**
- Keep all other variables (temperature, volume of starch, volume of enzyme, pH) identical.
- Boil the amylase solution, **A**, and allow it to cool before mixing with the starch solution.
- Observe that no hydrolysis occurs (the mixture remains blue-black with iodine throughout the investigation).

**(b) (ii) Graph Plotting Steps**
1. **Axes**: Plot rate of reaction / s⁻¹ on the y-axis and temperature / °C on the x-axis. Both axes must be fully labeled with units.
2. **Scale**: Choose a linear scale where the plotted points occupy more than half of the grid in both directions (e.g., x-axis: 0 to 70 °C with 10 °C intervals; y-axis: 0 to 0.020 s⁻¹ with 0.002 s⁻¹ intervals).
3. **Plotting**: Plot points using small, neat crosses (x) or circled dots. All points from Table 1.2 must be accurately positioned.
4. **Line**: Join points directly using ruled straight lines from point to point, or draw a smooth, appropriate curve of best fit through all points.

**(b) (iii) Molecular Changes (30 °C vs 60 °C)**
- **At 30 °C**: The enzyme and starch molecules have moderate kinetic energy. They collide successfully, forming active enzyme-substrate complexes (ESCs) at a steady rate. The tertiary structure is stable because weak hydrogen and ionic bonds are intact.
- **At 60 °C**: The high temperature provides excessive thermal and kinetic energy, causing the atoms in the enzyme molecule to vibrate violently.
- This breaks the hydrogen and ionic bonds that maintain the specific tertiary structure of the amylase active site.
- The active site undergoes a permanent conformational change (denaturation) and is no longer complementary to the substrate (starch).
- Therefore, no enzyme-substrate complexes can form, which explains the extremely low rate of reaction (0.0021 s⁻¹ compared to 0.0083 s⁻¹ at 30 °C).

**(a) (i) Dilution Table [3 marks total]**
- Award 1 mark for correct, decreasing sequence of amylase concentrations (0.50, 0.25, 0.125, 0.0625).
- Award 1 mark for keeping the total volume constant at 10.0 cm³ for all steps.
- Award 1 mark for correct volumes of amylase transferred (5.0 cm³ each time) and distilled water added (5.0 cm³ each time).

**(a) (ii) Variables [2 marks total]**
- Award 1 mark for identifying the independent variable as the amylase concentration.
- Award 1 mark for identifying the dependent variable as the time taken for complete hydrolysis / rate of reaction.

**(a) (iii) Table of Results [4 marks total]**
- Award 1 mark for complete table with ruled grid lines and clear column headers.
- Award 1 mark for correct headings with units separated by a slash (e.g., `Amylase concentration / %`, `Time / s`, `Rate / s⁻¹`).
- Award 1 mark for listing amylase concentrations in logical order (ascending or descending).
- Award 1 mark for indicating a calculated rate column with values calculated to a consistent number of significant figures.

**(b) (i) Control Experiment [2 marks total]**
- Award 1 mark for stating that amylase must be boiled (and cooled) to denature the enzyme, or replacing amylase with an equal volume of distilled water.
- Award 1 mark for stating that all other variables (volumes, temperature, pH) must remain constant and that starch remains unhydrolysed (iodine remains blue-black).

**(b) (ii) Graph Plotting [4 marks total]**
- Award 1 mark for correct axes labels with units: x-axis = `Temperature / °C`, y-axis = `Rate of reaction / s⁻¹`.
- Award 1 mark for appropriate scale selection where plotted data occupies more than half of the grid.
- Award 1 mark for precise plotting of all 5 data points from Table 1.2 (within half a small square).
- Award 1 mark for joining the plotted points with a series of straight, ruled lines or a smooth curve.

**(b) (iii) Molecular Explanation [5 marks total]**
- Award 1 mark for stating that at 30 °C, molecules have kinetic energy, leading to successful collisions and enzyme-substrate complex (ESC) formation.
- Award 1 mark for stating that 60 °C is above the optimum temperature, providing excessive kinetic energy that causes bonds to vibrate and break.
- Award 1 mark for specifying that hydrogen bonds / ionic bonds / hydrophobic interactions within the enzyme are broken.
- Award 1 mark for stating that this causes a change in the tertiary structure and denaturation of the active site.
- Award 1 mark for stating that the substrate (starch) can no longer fit into the active site / no ESCs form, reducing the rate of reaction.

### Solution

**(a) Low-Power Plan Diagram of Leaf Section**
- **Quality**: Drawn with a sharp HB pencil, single unbroken lines, no shading or cross-hatching.
- **Size**: Large, filling at least half of the available page space.
- **Shape**: Shows a hollow circle/oval with an opening, representing the rolled leaf. The inner (adaxial) surface should show ridges (pleats) and furrows (crypts). The outer (abaxial) surface should be smooth and have a thick cuticle layer indicated by double lines.
- **Tissues**: Show the positions of vascular bundles located inside the ridges of the leaf. The bundle sheath and vascular tissue (xylem on the inside, phloem on the outside) should be outlined. No individual cells should be drawn.
- **Labels**: Two clear label lines drawn with a ruler pointing precisely to the:
- **Adaxial epidermis** (inner, folded surface of the rolled leaf).
- **Vascular bundle** (found within the ridges of the mesophyll tissue).

**(b) High-Power Drawing of Parenchyma Cells**
- **Quality**: Drawn with clear, sharp lines. No sketchy lines or shading.
- **Detail**: Three adjacent cells showing cell walls drawn as double lines, with the middle lamella represented where the walls touch.
- **Shape**: Polyhedral, rounded or slightly oval shapes, showing typical thin-walled parenchyma characteristics.
- **Proportions**: Cell sizes and shapes are drawn realistically in relation to each other.

**(c) (i) Calibration Calculation**
- 1 division of stage micrometer = 0.1 mm.
- 6 divisions of stage micrometer = \(6 \times 0.1\text{ mm} = 0.6\text{ mm}\).
- Convert mm to micrometers (\(\mu\text{m}\)):
$$0.6\text{ mm} \times 1000 = 600\ \mu\text{m}$$
- 40 eyepiece graticule units = \(600\ \mu\text{m}\).
- 1 eyepiece graticule unit = \(\frac{600}{40} = 15\ \mu\text{m}\).

**(c) (ii) Xylem Vessel Diameter Calculation**
- Diameter = 8 eyepiece graticule units.
- 1 eyepiece unit = \(15\ \mu\text{m}\).
- Actual diameter = \(8 \times 15\ \mu\text{m} = 120\ \mu\text{m}\).

**(d) Comparison Table**

| Feature | Xylem vessel elements | Phloem sieve tube elements |
| :--- | :--- | :--- |
| **Lignification** | Lignified cell walls present (stain red/blue depending on dye) | Non-lignified cellulose cell walls |
| **End walls** | End walls are completely lost (open, continuous tubes) | Sieve plates with sieve pores are present |
| **Cell lumen** | Wide, completely hollow lumen | Narrower lumen, containing peripheral cytoplasm |
| **Cell viability** | Dead cells (no cytoplasm, organelles, or nucleus) | Living cells (contain thin cytoplasm, but lack nucleus and key organelles) |
| **Companion cells** | Absent | Present, closely associated with each sieve tube element |
| **Wall thickness** | Thick, reinforced walls (spiral, annular, or pitted patterns) | Thin, unreinforced walls |

**(a) Plan Diagram [5 marks total]**
- Award 1 mark for clear, thin, unbroken lines with no sketching and absolutely no shading.
- Award 1 mark for a large drawing that occupies at least 50% of the available space.
- Award 1 mark for drawing the correct rolled shape showing the inner surface folded into ridges and furrows.
- Award 1 mark for showing vascular bundles correctly distributed in the ridges and not drawing individual cells.
- Award 1 mark for correct labels of `adaxial epidermis` (on the folded inner surface) and `vascular bundle` with ruled label lines.

**(b) High-Power Drawing [4 marks total]**
- Award 1 mark for sharp, single-line boundaries representing cell shapes without sketching.
- Award 1 mark for drawing only three adjacent cells where all three touch each other.
- Award 1 mark for representing cell walls as double lines with appropriate thickness.
- Award 1 mark for drawing cells of polygonal or oval shape typical of parenchyma cells, with no organelles drawn inside.

**(c) (i) Calibration [3 marks total]**
- Award 1 mark for converting stage micrometer divisions to millimeters or micrometers (\(6 \times 0.1 = 0.6\text{ mm}\) or \(600\ \mu\text{m}\)).
- Award 1 mark for showing the correct division of the distance by the number of eyepiece units (\(600 / 40\)).
- Award 1 mark for the correct final answer of \(15\ \mu\text{m}\) with working.

**(c) (ii) Measurement [2 marks total]**
- Award 1 mark for multiplying the calibrated value by 8 (\(8 \times 15\)).
- Award 1 mark for the correct final answer of \(120\ \mu\text{m}\) (allow ecf if calibration in (c)(i) was incorrect).

**(d) Comparison Table [6 marks total]**
- Award 1 mark for drawing a organized, ruled table comparing features side-by-side.
- Award 1 mark for contrasting lignin: xylem is lignified vs phloem is non-lignified / cellulose-only.
- Award 1 mark for contrasting end walls: xylem has no end walls / open vessels vs phloem has sieve plates.
- Award 1 mark for contrasting cell contents: xylem is empty / dead vs phloem contains cytoplasm / living.
- Award 1 mark for contrasting companion cells: xylem has no companion cells vs phloem is associated with companion cells.
- Award 1 mark for contrasting wall thickness: xylem has thick walls vs phloem has thin walls.

(a)(i) Heating to \(95^\circ\text{C}\) breaks the hydrogen bonds between complementary base pairs of the double-stranded DNA. This denatures the DNA, separating it into two single strands to act as templates.
(ii) Cooling to \(55^\circ\text{C}\) allows short, single-stranded DNA primers to bind (anneal) via hydrogen bonding to complementary sequences at the \(3'\) ends of the target DNA strands.

(b) DNA fragments have a net negative charge due to the phosphate groups in their sugar-phosphate backbone. When placed in an agarose gel and an electric current is applied, they migrate towards the positive electrode (anode). The agarose gel acts as a molecular sieve, resisting the movement of larger DNA fragments. Consequently, smaller DNA fragments move faster and further through the gel pores than larger fragments, separating them purely by molecular size (length in base pairs).

(c) Comparing the DNA bands (microsatellite profiles) allows conservationists to assess the level of genetic diversity within and between populations. Populations with high heterozygosity (many different bands) are genetically diverse and should be prioritized for protection. It helps prevent inbreeding depression by identifying genetically distinct individuals from different populations to cross-breed, maximizing genetic variation and evolutionary potential in offspring.

**Part (a)(i)** [Max 2 marks]
* M1: heating to \(95^\circ\text{C}\) denatures DNA / breaks hydrogen bonds (between complementary bases); [1]
* M2: separating double-stranded DNA into single strands / exposing bases to act as templates; [1]
* *Reject: breaking covalent/phosphodiester bonds.*

**Part (a)(ii)** [Max 2 marks]
* M3: cooling to \(55^\circ\text{C}\) allows primers to anneal / bind / hybridize; [1]
* M4: to complementary sequences on single-stranded template DNA / at the \(3'\) end of target sequences; [1]

**Part (b)** [Max 3 marks]
* M5: DNA is negatively charged (due to phosphate groups); [1]
* M6: DNA fragments migrate towards the positive electrode / anode (when an electric field is applied); [1]
* M7: agarose gel acts as a mesh / sieve / matrix; [1]
* M8: smaller fragments move faster / further through the gel than larger fragments (or vice versa); [1]

**Part (c)** [Max 3 marks]
* M9: (banding patterns show) genetic diversity / heterozygosity / unique alleles; [1]
* M10: choose genetically distinct individuals to mate / cross-breed; [1]
* M11: to minimize inbreeding / prevent inbreeding depression / maintain or increase gene pool; [1]
* M12: identify distinct sub-populations for separate conservation management / prioritize genetically diverse populations; [1]

(a) NAD acts as a coenzyme that accepts hydrogen atoms (along with their associated electrons and protons) during dehydrogenation reactions in glycolysis, the link reaction, and the Krebs cycle, becoming reduced NAD (NADH). It then transports these high-energy electrons and protons to the inner mitochondrial membrane (cristae) to enter the electron transport chain.

(b) High-energy electrons from NADH and \(\text{FADH}_2\) pass along carriers in the electron transport chain, releasing energy. This energy is used by the electron carriers (proton pumps) to actively pump protons (\(\text{H}^+\)) from the matrix across the inner membrane into the intermembrane space. This establishes a high concentration of protons in the intermembrane space relative to the matrix, forming an electrochemical proton gradient. Protons then diffuse down their gradient back into the matrix via the enzyme ATP synthase (chemiosmosis). This flow of protons through ATP synthase causes a conformational change that drives the phosphorylation of ADP and \(\text{P}_i\) to synthesize ATP.

(c) Oligomycin blocks the proton channel in ATP synthase, meaning protons can no longer flow back into the mitochondrial matrix. As a result, the proton gradient across the inner membrane becomes extremely steep, and the energy required to pump additional protons into the intermembrane space exceeds the energy released by the transfer of electrons along the ETC. Electron transport halts, meaning electrons can no longer be passed down the chain to oxygen, the terminal electron acceptor. Consequently, the reduction of oxygen to water stops, and oxygen consumption decreases significantly.

**Part (a)** [Max 2 marks]
* M1: acts as a hydrogen carrier / coenzyme; [1]
* M2: undergoes reduction / becomes reduced NAD / accepts electrons and protons during dehydrogenation; [1]
* M3: carries / transfers hydrogen / electrons to the electron transport chain (on the cristae); [1]

**Part (b)** [Max 4 marks]
* M4: energy from electrons passing along the ETC is used to pump protons (\(\text{H}^+\)) from matrix into intermembrane space; [1]
* M5: creates a high concentration of protons / electrochemical gradient / proton motive force in intermembrane space; [1]
* M6: protons diffuse / flow back into the matrix down their concentration/electrochemical gradient; [1]
* M7: through ATP synthase; [1]
* M8: (flow of protons) drives the phosphorylation of ADP and \(\text{P}_i\) to form ATP (chemiosmosis); [1]

**Part (c)** [Max 4 marks]
* M9: oligomycin blocks ATP synthase / prevents protons returning to the matrix; [1]
* M10: proton concentration in intermembrane space remains high / gradient becomes too steep; [1]
* M11: proton pumping by ETC stops because it cannot pump against such a high gradient; [1]
* M12: electron transport chain stops / carrier proteins remain reduced / cannot accept more electrons; [1]
* M13: oxygen is the terminal/final electron acceptor; [1]
* M14: since electrons cannot flow down the chain, oxygen is not reduced to water, so oxygen consumption drops; [1]

(a) The light-independent stage occurs in the stroma of the chloroplast.

(b)(i) Enzyme X is ribulose bisphosphate carboxylase-oxygenase (RuBisCO).
(ii) GP is converted into TP in a reduction reaction. This reaction requires ATP (which provides phosphate and energy) and reduced NADP (NADPH, which provides hydrogen atoms/electrons), both of which are products of the light-dependent stage. This produces triose phosphate (TP) and releases ADP, inorganic phosphate, and oxidized NADP.
Five out of every six molecules of TP are then used to regenerate ribulose bisphosphate (RuBP). This regeneration process is a multi-step pathway that requires energy, which is supplied by the hydrolysis of ATP (converting ATP to ADP and \(\text{P}_i\)).

(c) In continuous darkness, the light-dependent stage stops, meaning no ATP or reduced NADP are synthesized. Consequently, GP cannot be reduced to TP. This causes GP to accumulate in the stroma, while the concentration of TP drops. Because TP is needed to regenerate RuBP, the concentration of RuBP also declines rapidly. Eventually, the Calvin cycle stops entirely due to the depletion of RuBP and the lack of ATP and reduced NADP.

**Part (a)** [1 mark]
* M1: stroma (of chloroplast); [1]

**Part (b)(i)** [1 mark]
* M2: RuBisCO / ribulose bisphosphate carboxylase-oxygenase; [1]

**Part (b)(ii)** [Max 5 marks]
* M3: GP is reduced to TP; [1]
* M4: using reduced NADP / NADPH (to supply hydrogens/electrons); [1]
* M5: using ATP (to supply energy / phosphate); [1]
* M6: ADP and oxidized NADP are regenerated; [1]
* M7: most TP / \(\frac{5}{6}\) of TP is used to regenerate RuBP; [1]
* M8: \(\frac{1}{6}\) of TP is used to make hexose sugars / glucose / organic substances; [1]
* M9: regeneration of RuBP requires ATP; [1]

**Part (c)** [Max 3 marks]
* M10: light-dependent stage stops so no ATP or reduced NADP is produced; [1]
* M11: GP cannot be converted / reduced to TP; [1]
* M12: GP level increases / accumulates; [1]
* M13: TP and RuBP levels decrease / are depleted; [1]
* M14: Calvin cycle stops; [1]

(a) Osmoreceptors in the hypothalamus detect a decrease in the water potential of the blood (dehydration, high solute concentration). They stimulate neurosecretory cells, which transmit action potentials to the posterior pituitary gland, triggering the release of ADH into the blood.

(b) ADH travels in the bloodstream to the kidneys, where it binds to specific cell surface receptors on the basolateral membrane of the collecting duct epithelial cells. This binding activates a G-protein, which in turn activates the enzyme adenylyl cyclase. Adenylyl cyclase catalyzes the conversion of ATP to cyclic AMP (cAMP), which acts as a second messenger. cAMP activates a cascade of protein kinases, causing vesicles containing water channel proteins (aquaporins) to move towards and fuse with the luminal (apical) membrane of the collecting duct cells. This increases the permeability of the luminal membrane to water. Water then moves out of the collecting duct lumen by osmosis down a water potential gradient, into the highly concentrated tissue fluid of the renal medulla, and eventually back into the blood capillaries.

(c) A mutation in the ADH receptor gene results in an altered receptor protein with a different tertiary structure, preventing ADH from binding. This stops the intracellular signaling cascade, so aquaporin-containing vesicles do not fuse with the luminal membrane. The collecting duct remains impermeable to water, meaning water cannot be reabsorbed. Consequently, the patient excretes a very large volume of dilute urine (polyuria) and suffers from extreme thirst (polydipsia).

**Part (a)** [2 marks]
* M1: low water potential of blood / decreased blood volume / dehydration; [1]
* M2: posterior pituitary gland; [1]

**Part (b)** [Max 5 marks]
* M3: ADH binds to receptors on cell surface / basolateral membrane of collecting duct (epithelial) cells; [1]
* M4: activates G-protein which activates adenylyl cyclase (to produce cyclic AMP / cAMP); [1]
* M5: cAMP acts as a second messenger / activates a kinase cascade; [1]
* M6: causes vesicles containing aquaporins (water channels) to move towards and fuse with the luminal / apical membrane; [1]
* M7: increases the water permeability of the luminal membrane; [1]
* M8: water moves out of the collecting duct lumen by osmosis down a water potential gradient (into the hypertonic medulla / blood); [1]

**Part (c)** [Max 3 marks]
* M9: mutation changes tertiary structure of ADH receptor so ADH cannot bind / signal cascade not activated; [1]
* M10: aquaporins are not inserted into the luminal membrane / collecting duct remains impermeable to water; [1]
* M11: large volume of urine produced; [1]
* M12: urine is very dilute / has a high water potential / low concentration; [1]

(a)(i) Schwann cells produce the myelin sheath in the peripheral nervous system. They wrap themselves repeatedly around the axon, leaving concentric layers of cell membrane containing myelin, which is lipid-rich and acts as an electrical insulator.
(ii) The gaps in the myelin sheath are called the Nodes of Ranvier.

(b) Myelin acts as an electrical insulator, preventing the movement of ions (\(\text{Na}^+\) and \(\text{K}^+\)) across the axon membrane. Consequently, depolarization and action potentials cannot occur along the myelinated regions. Voltage-gated sodium ion channels are highly concentrated only at the unmyelinated Nodes of Ranvier. When an action potential occurs at one node, the local circuits of ionic current flow internally from the active node to the next adjacent node. This depolarizes the membrane at the next node to threshold, triggering an action potential there. This causes the action potential to 'jump' from one node of Ranvier to the next, a process called saltatory conduction, which greatly increases the speed of transmission.

(c) During an action potential, voltage-gated sodium channels open and then rapidly close and become inactivated (the refractory period). During this brief recovery time, the membrane cannot undergo depolarization again, regardless of the strength of the stimulus. Since the region of the axon membrane behind the advancing action potential is in its refractory period, the local circuits can only depolarize the resting membrane ahead of the action potential, ensuring one-way propagation along the neurone.

**Part (a)(i)** [2 marks]
* M1: Schwann cells; [1]
* M2: wrap repeatedly around the axon to form concentric layers of membrane / lipid-rich / myelin layer; [1]

**Part (a)(ii)** [1 mark]
* M3: Node(s) of Ranvier; [1]

**Part (b)** [Max 4 marks]
* M4: myelin is an electrical insulator / prevents ion movement across the axon membrane; [1]
* M5: action potentials / depolarization can only occur at the Nodes of Ranvier; [1]
* M6: voltage-gated sodium channels are concentrated at the nodes; [1]
* M7: local currents / ionic currents flow from one node to the next; [1]
* M8: action potential 'jumps' from node to node / saltatory conduction; [1]

**Part (c)** [Max 3 marks]
* M9: reference to refractory period; [1]
* M10: voltage-gated sodium channels are inactivated / closed and cannot open; [1]
* M11: resting potential must be restored before another action potential can occur / membrane hyperpolarized; [1]
* M12: prevents action potentials from moving backwards / local currents can only depolarize the membrane in front / ahead of the action potential; [1]

(a) Originally, the pocket mouse population had predominantly light-colored coats, which provided camouflage against granite rocks. A random mutation occurred in the gene controlling coat color, producing a dominant allele for dark-colored coats.
In the volcanic basalt environment, dark-colored mice are well camouflaged from visual predators, such as owls, whereas light-colored mice are highly visible. As a result, predators selectively prey upon the light-colored mice. The dark-colored mice have a higher chance of survival and reproduction (selective advantage). They pass on the allele for dark coat color to their offspring. Over many generations, the frequency of the dark coat allele increased significantly in the basalt population.
In the granite environment, the reverse selection pressure occurs: light-colored mice are camouflaged, while dark-colored mice are preyed upon, maintaining a high frequency of the light coat allele there.

(b) Allopatric speciation would occur. This is because the two populations are geographically isolated by a physical barrier (the wide strip of grassland they rarely cross), which prevents gene flow between the two gene pools.

(c) The biological species concept states that a species is a group of organisms that can interbreed to produce fertile offspring. To determine if they are separate species, researchers should bring individuals from both populations together under controlled laboratory conditions and attempt to mate them. If they either do not mate, fail to produce offspring, or produce offspring that are sterile (infertile), then they have evolved reproductive isolation and are classified as two distinct species.

**Part (a)** [Max 5 marks]
* M1: mutation occurs producing allele for dark coat color; [1]
* M2: variation in coat color exists in the population; [1]
* M3: predators (e.g. owls) act as selection pressure; [1]
* M4: on dark basalt, dark mice are camouflaged / less visible to predators AND on granite, light mice are camouflaged; [1]
* M5: camouflaged mice survive and reproduce / have selective advantage / differential survival; [1]
* M6: pass on their advantageous alleles to offspring; [1]
* M7: allele frequency for dark coat increases in dark environment / allele frequency for light coat increases in light environment; [1]

**Part (b)** [2 marks]
* M8: allopatric speciation; [1]
* M9: because there is geographical isolation / physical barrier / no gene flow between populations; [1]

**Part (c)** [Max 3 marks]
* M10: attempt to interbreed / mate individuals from both populations; [1]
* M11: observe if they produce offspring; [1]
* M12: check if the offspring are fertile; [1]
* M13: if they cannot interbreed / cannot produce fertile offspring, they are separate species; [1]

(a) Epistasis is the interaction between two non-allelic genes, where the expression of one gene (the epistatic gene) masks, suppresses, or modifies the phenotypic expression of another gene (the hypostatic gene). In this example, gene **W** is the epistatic gene because the dominant allele **W** masks the phenotypic expression of the **Y**/**y** gene.

(b) The cross is **WwYy** \(\times\) **WwYy**. Because the two genes assort independently, we expect a typical 9:3:3:1 ratio of genotypes in the \(F_2\) generation out of 16 combinations:
- \(9\) **W_Y_**
- \(3\) **W_yy**
- \(3\) **wwY_**
- \(1\) **wwyy**

Now we apply the epistatic rules to determine phenotypes:
1. Any genotype with a dominant **W** allele (**W_Y_** and **W_yy**) will be white.
- Total white = \(9 \text{ (W_Y_)} + 3 \text{ (W_yy)} = 12\).
2. Genotypes with **ww** and at least one dominant **Y** (**wwY_**) will be yellow.
- Total yellow = \(3\).
3. Genotypes with **wwyy** will be green.
- Total green = \(1\).

Thus, the expected phenotypic ratio is **12 white : 3 yellow : 1 green**.

(c) Metabolic pathways consist of a sequence of chemical reactions, each catalyzed by a specific enzyme. These enzymes are coded for by genes.
The dominant allele **W** likely codes for an active repressor protein or inhibitor enzyme that blocks the pathway converting the white precursor molecule into a colored intermediate. Thus, in **W_** plants, the pathway is blocked, and the fruit remains white. The recessive allele **w** is a mutated, non-functional version that does not produce the active inhibitor. Therefore, in **ww** plants, the pathway is not blocked, allowing the white precursor to be converted.
The **Y** gene codes for an enzyme that catalyzes the next step in the pathway (e.g., converting yellow pigment to green, or vice-versa). If the plant is **ww**, the pathway can proceed, and the specific allele of gene **Y** determines whether yellow or green pigment is produced.

**Part (a)** [2 marks]
* M1: epistasis is the interaction of genes where one gene masks / suppresses / affects the expression of another gene; [1]
* M2: gene **W** is epistatic (over gene **Y**); [1]

**Part (b)** [Max 5 marks]
* M3: correct identification of genotypes for white phenotype: **W_Y_** and **W_yy** (or listing all 12 white genotypes); [1]
* M4: correct identification of genotypes for yellow phenotype: **wwY_** (or **wwYY** and **wwYy**); [1]
* M5: correct identification of genotype for green phenotype: **wwyy**; [1]
* M6: correct determination of individual ratios: 12 white, 3 yellow, 1 green (or equivalent fractions out of 16); [1]
* M7: correct final simplified ratio written clearly: **12 white : 3 yellow : 1 green**; [1]

**Part (c)** [Max 3 marks]
* M8: genes code for enzymes / proteins that catalyze steps in a metabolic pathway; [1]
* M9: allele **W** codes for an inhibitor / repressor protein that blocks the conversion of white precursor to color; [1]
* M10: allele **w** is a mutated/non-functional allele that does not block the pathway, allowing color development; [1]
* M11: gene **Y** codes for an enzyme that converts the colored intermediate to yellow/green; [1]

(a) *In situ* conservation is the conservation of ecosystems and natural habitats and the maintenance and recovery of viable populations of species in their natural surroundings. Examples include national parks, nature reserves, and marine protected areas.
*Ex situ* conservation is the conservation of components of biological diversity outside their natural habitats. Examples include seed banks, botanic gardens, zoos, and captive breeding programs.

(b)(i) A high value of \(D\) (close to 1) indicates high species diversity, meaning there is both high species richness (many different species) and high species evenness (similar numbers of individuals of each species). In Area A, the ecosystem is stable because food webs are complex; if one species population declines, predators have alternative food sources, and the community is resilient.
A low value of \(D\) in Area B indicates low species diversity, with the community dominated by only one or a few species. This ecosystem is unstable; a small environmental change (like a disease or climate change) that affects the dominant species can easily disrupt the entire ecosystem.

(ii) Simpson's Index of Diversity is a better measure because it takes into account species evenness (the relative abundance of each species) in addition to species richness. Simply counting the number of species does not reveal if one species is highly dominant while all others are extremely rare.

(c) In seed banks, seeds are first cleaned and checked for viability. They are then dehydrated (dried) to very low moisture content (around \(5\%\) to \(8\%\)) and stored at low temperatures (typically \(-20^\circ\text{C}\)). This dry, cold environment significantly reduces the rate of metabolic processes (respiration) in the seeds, preventing germination and protecting them from decay by bacteria or fungi, ensuring they remain viable for decades.

**Part (a)** [4 marks]
* M1: *In situ*: conservation of species within their natural / native habitat; [1]
* M2: example of *in situ*: national parks / nature reserves / marine reserves / protecting habitats; [1]
* M3: *Ex situ*: conservation of species outside of their natural / native habitat; [1]
* M4: example of *ex situ*: seed banks / botanic gardens / zoos / captive breeding programs; [1]

**Part (b)(i)** [Max 3 marks]
* M5: high \(D\) value indicates high species diversity / high richness and high evenness; [1]
* M6: high \(D\) indicates stable ecosystem / complex food webs / community less affected by changes in one species; [1]
* M7: low \(D\) indicates low diversity / dominated by one or a few species; [1]
* M8: low \(D\) indicates unstable ecosystem / vulnerable to environmental change / pests / diseases; [1]

**Part (b)(ii)** [1 mark]
* M9: takes into account evenness / relative abundance / relative numbers of each species (not just richness); [1]

**Part (c)** [Max 2 marks]
* M10: seeds are cleaned (and sorted/checked); [1]
* M11: dried / dehydrated (to low moisture content); [1]
* M12: kept at very low temperatures / frozen / stored at \(-20^\circ\text{C}\); [1]
* M13: (dry and cold conditions) reduce metabolic rate / prevent fungal growth / prevent germination; [1]

(a) Reduced NAD (NADH) is produced during glycolysis, the link reaction, and the Krebs cycle. During oxidative phosphorylation, NADH binds to Complex I of the electron transport chain (ETC) in the inner mitochondrial membrane. It undergoes oxidation, releasing protons (\(H^+\)) and high-energy electrons. The electrons are transferred down the electron transport chain, releasing energy as they move from one carrier to the next. This energy is harnessed to pump protons across the membrane. The final electron acceptor is oxygen.

(b) As electrons pass along the electron carriers of the ETC, they lose energy. This energy is used by membrane proteins (complexes I, III, and IV) to actively pump protons (\(H^+\)) from the mitochondrial matrix, across the inner membrane, and into the intermembrane space. Because the inner mitochondrial membrane is impermeable to protons, they accumulate in the intermembrane space, establishing both a concentration and charge gradient (an electrochemical gradient) relative to the matrix.

(c)(i) The rate of oxygen consumption increases or remains high. Because DNP constantly dissipates the proton gradient, the electron transport chain continues to transport electrons rapidly to oxygen (the terminal electron acceptor) in an ongoing attempt to restore the electrochemical gradient.

(c)(ii) ATP production decreases significantly because protons bypass ATP synthase, meaning there is insufficient proton flow through the channel to drive the phosphorylation of ADP. Instead, the potential energy stored in the gradient is released as heat energy.

(a) Max 3 marks:
1. Reduced NAD releases hydrogen atoms / splits to release protons (\(H^+\)) and electrons (\(e^-\));
2. Electrons are passed to the electron transport chain / electron carriers;
3. Energy is released as electrons flow along the electron transport chain;
4. Protons are left in the matrix (to be pumped) / electrons are accepted by oxygen at the end of the chain;
[Reject: NADH enters the Krebs cycle]

(b) Max 3 marks:
1. Energy released from electron transport chain is used to pump protons (\(H^+\));
2. Protons are pumped from the mitochondrial matrix into the intermembrane space;
3. Through transmembrane proteins / Complexes I, III, IV;
4. Inner mitochondrial membrane is impermeable to protons, preventing them diffusing back easily / resulting in high concentration in intermembrane space;
[Accept: reference to electrochemical gradient / proton motive force]

(c)(i) Max 2 marks:
1. Rate of oxygen consumption increases / remains high;
2. Because electron transport chain continues / operates faster to try to rebuild the gradient;
3. Oxygen is still required as the final electron acceptor (to accept electrons and protons to form water);

(c)(ii) Max 2 marks:
1. ATP production decreases / ceases because protons bypass ATP synthase / no proton flow through ATP synthase / no proton motive force;
2. Energy from the gradient / electron transport is released as heat instead of being stored as ATP;

(a) According to the acid growth hypothesis, auxin (IAA) binds to specific receptor proteins (such as Auxin Binding Protein 1, ABP1) on the cell surface membrane of target shoot cells. This binding stimulates proton pumps (\(H^+\)-ATPases) located in the cell surface membrane to actively transport protons (\(H^+\)) from the cytoplasm into the cell wall space (apoplast). This active transport lowers the pH of the cell wall, making it acidic. The acidic environment activates wall-loosening proteins called expansins, which disrupt the non-covalent hydrogen bonds between cellulose microfibrils and matrix hemicelluloses.

(b) Expansins break the hydrogen bonds holding the cellulose microfibrils together, allowing them to slide past one another. Because the cell wall structure is loosened, it becomes more elastic. As water enters the cell's central vacuole by osmosis, it generates turgor pressure. This internal pressure pushes against the weakened, plastic cell wall, stretching it and causing the cell to elongate.

(c) ABA binds to receptors on the cell surface membrane of guard cells. This triggers a signaling cascade that inhibits the proton pumps in the guard cell membrane and opens calcium channel proteins, leading to an influx of calcium ions (\(Ca^{2+}\)) into the cytoplasm. The rise in intracellular calcium acts as a second messenger, triggering the opening of anion and potassium (\(K^+\)) efflux channels. Potassium and malate ions leave the guard cells, raising the water potential inside the cells. Water then leaves the guard cells by osmosis down a water potential gradient. The guard cells lose turgidity and become flaccid, closing the stomatal pore.

(a) Max 5 marks:
1. Auxin / IAA binds to (cell surface) receptors / ABP1;
2. Stimulates proton pumps / \(H^+\)-ATPase in the cell surface membrane;
3. Protons (\(H^+\)) are actively pumped from cytoplasm into the cell wall / apoplast;
4. This lowers the pH / acidifies the cell wall;
5. The low pH activates expansin proteins;
6. Expansins break hydrogen bonds / non-covalent bonds between cellulose microfibrils and other cell wall components (hemicellulose);

(b) Max 2 marks:
1. Expansins loosen / disrupt the rigid arrangement of cellulose microfibrils / cell wall matrix;
2. This allows the cell wall to expand / stretch / yield in response to turgor pressure (generated by water entry);

(c) Max 3 marks:
1. ABA binds to receptors on guard cell membranes;
2. Promotes calcium ion (\(Ca^{2+}\)) entry into cytoplasm / inhibits proton pumps;
3. Calcium acts as a second messenger, opening efflux channels for potassium ions (\(K^+\)) / malate / anions;
4. \(K^+\) / solutes leave the guard cells;
5. Water potential inside guard cells increases, so water leaves by osmosis;
6. Guard cells lose turgor / become flaccid, causing the stomatal pore to close;

(a) To prepare five concentrations of copper sulfate (e.g., 0.10, 0.08, 0.06, 0.04, and 0.02 mol dm\(^{-3}\)) each of volume 10 cm\(^{3}\), the student can mix the following volumes of 0.10 mol dm\(^{-3}\) stock solution and distilled water: For 0.10 mol dm\(^{-3}\): 10.0 cm\(^{3}\) stock + 0.0 cm\(^{3}\) water; For 0.08 mol dm\(^{-3}\): 8.0 cm\(^{3}\) stock + 2.0 cm\(^{3}\) water; For 0.06 mol dm\(^{-3}\): 6.0 cm\(^{3}\) stock + 4.0 cm\(^{3}\) water; For 0.04 mol dm\(^{-3}\): 4.0 cm\(^{3}\) stock + 6.0 cm\(^{3}\) water; For 0.02 mol dm\(^{-3}\): 2.0 cm\(^{3}\) stock + 8.0 cm\(^{3}\) water. (b) Method details: 1. Set up a water bath at a constant temperature (e.g., 35-40 deg C) using a thermostatically controlled water bath. 2. Measure equal volumes of starch solution (e.g., 5 cm\(^{3}\) of 1.0%) into 5 different tubes, and equal volumes of amylase (e.g., 2 cm\(^{3}\) of 1.0%) mixed with 1 cm\(^{3}\) of each prepared copper sulfate concentration into another set of 5 tubes. 3. Place all tubes in the water bath for 5 minutes to equilibrate. 4. Put drops of iodine solution in the wells of a spotting tile. 5. Mix the enzyme-inhibitor mixture with the starch substrate, start a stopwatch, and return the tube to the water bath. 6. Take a sample every 30 seconds and add it to a drop of iodine solution. 7. Record the time taken for the blue-black color to stop appearing (color remains yellow/brown). Calculate the rate as 1 / time. 8. Repeat the entire procedure at least 3 times for each concentration and calculate the mean rate. 9. Safety: Wear goggles and gloves as copper sulfate is toxic and irritating. (c) Modify the experiment by keeping the inhibitor concentration constant (e.g., at 0.04 mol dm\(^{-3}\)) while varying the starch (substrate) concentration (e.g., from 0.2% to 1.0%). Compare the maximum rate (Vmax) achieved to a control setup without inhibitor. If competitive, Vmax will eventually reach the control value at high starch concentrations. If non-competitive, Vmax will remain significantly lower. (d)(i) Null hypothesis: There is no significant difference between the mean rate of starch hydrolysis in the presence of 0.00 mol dm\(^{-3}\) and 0.04 mol dm\(^{-3}\) copper sulfate. (ii) Test: Student's t-test (or unpaired t-test). Reason: It is used to compare the means of two independent groups of continuous, normally distributed data.

Part (a) [3 marks]:
- 1 mark for calculating correct volumes of stock and water to make at least 4 non-zero concentrations.
- 1 mark for specifying a total volume of 10 cm\(^{3}\) for each concentration.
- 1 mark for identifying that 0.00 mol dm\(^{-3}\) (distilled water only) acts as the control.

Part (b) [6 marks]:
- 1 mark for identifying the independent variable (at least 5 concentrations of copper sulfate) and dependent variable (time taken for starch to be completely hydrolysed/no blue-black color with iodine).
- 1 mark for describing how rate is determined (e.g., 1/time or colorimeter absorbance over time).
- 1 mark for controlling temperature using a thermostatically controlled water bath and equilibrating solutions before mixing.
- 1 mark for keeping starch volume and concentration, and amylase volume and concentration constant.
- 1 mark for repeating the experiment at least 3 times for each concentration to calculate a mean / identify anomalies.
- 1 mark for identifying a valid safety precaution (e.g., wearing gloves/goggles because copper sulfate is harmful/irritant).

Part (c) [3 marks]:
- 1 mark for stating that substrate (starch) concentration must be varied while inhibitor concentration is kept constant.
- 1 mark for stating that if competitive, the maximum rate (Vmax) can still be reached / high substrate concentration overcomes inhibition.
- 1 mark for stating that if non-competitive, the maximum rate (Vmax) is reduced / high substrate concentration does not overcome inhibition.

Part (d) [3 marks]:
- 1 mark for a correct null hypothesis stating 'no significant difference' between the two means.
- 1 mark for identifying the Student's t-test (or unpaired t-test).
- 1 mark for explaining that it compares the means of two independent, continuous data sets.

(a)(i) Multiplying by 1000 avoids having very small decimal values (e.g., 0.0004 vs 0.4), making the numbers easier to plot on a graph and work with. (a)(ii) Variable 1: Carbon dioxide concentration. Control: Add a fixed volume and concentration of sodium hydrogencarbonate solution (e.g., 50 cm\(^{3}\) of 0.1 mol dm\(^{-3}\)) to the beaker. Variable 2: Biomass of Chlorella. Control: Use an equal number of algal beads of the same size/diameter in each trial. (b) Description: For both temperatures, as the light intensity index increases (distance decreases), the mean rate of oxygen production increases. At all light intensities, the rate is higher at 25 deg C than at 15 deg C. Explanation: At low light intensity index (0.4 to 1.1), light intensity is the limiting factor for both temperatures because increasing light energy increases the rate of photolysis / light-dependent reaction. At higher light intensity index (2.5 to 10.0), the rate of oxygen production at 15 deg C starts to plateau (e.g., only increasing from 68.2 to 72.4), indicating that light is no longer the limiting factor; temperature has become the limiting factor (enzyme activity of Rubisco in the light-independent stage is limited). (c) Calculation: Rate at 15 deg C = 48.0, Rate at 25 deg C = 62.1. Increase = 62.1 - 48.0 = 14.1. Percentage increase = (14.1 / 48.0) * 100 = 29.375% (or 29.4%). (d)(i) Spearman's rank correlation coefficient (or Pearson's linear correlation coefficient). (ii) 1. Find the critical value in a statistical table for Spearman's rank at the p = 0.05 level of significance for the number of pairs of data (n = 5). 2. Compare the calculated value of rs with the critical value. 3. If the calculated rs is greater than or equal to the critical value, reject the null hypothesis and conclude there is a statistically significant correlation (not due to chance).

Part (a)(i) [1 mark]:
- 1 mark for stating that it avoids very small decimals / makes values easier to plot or read.

Part (a)(ii) [4 marks]:
- 1 mark for identifying carbon dioxide concentration AND 1 mark for describing the use of a fixed concentration/volume of sodium hydrogencarbonate solution.
- 1 mark for identifying the concentration/mass/volume of algae/beads AND 1 mark for describing using the same number or size of algal beads in each tube.

Part (b) [4 marks]:
- 1 mark for describing that as light intensity index increases, the rate of oxygen production increases at both temperatures.
- 1 mark for stating that the rate of oxygen production is higher at 25 deg C than at 15 deg C at all light intensities.
- 1 mark for explaining that at low light intensity (0.4 to 1.1), light is the limiting factor because it limits photolysis/light-dependent stage.
- 1 mark for explaining that at high light intensity (2.5 to 10.0), the rate at 15 deg C plateaus/temperature is limiting enzyme-controlled reactions (e.g., Rubisco/light-independent stage).

Part (c) [2 marks]:
- 1 mark for showing correct working: ((62.1 - 48.0) / 48.0) * 100.
- 1 mark for the correct answer of 29.4% or 29.38% (accept 29.375%).

Part (d) [4 marks]:
- 1 mark for naming Spearman's rank correlation coefficient (or Pearson's correlation).
- 1 mark for stating that the degrees of freedom / number of pairs (n = 5) is determined.
- 1 mark for stating that the calculated value is compared with a critical value from tables at p = 0.05 (5% probability).
- 1 mark for explaining that if the calculated value is greater than the critical value, the null hypothesis is rejected / the correlation is significant.