2024 Edexcel A-Level Biology A (Salters-Nuffield) (9BN0) 模擬試題連答案詳解

During non-cyclic photophosphorylation, photolysis of water splits molecules to release electrons, protons, and oxygen. These electrons replace the excited electrons lost from Photosystem II (PSII). The excited electrons from PSII are passed down an electron transport chain to Photosystem I (PSI). Finally, electrons from PSI are used to reduce NADP+ to form reduced NADP (NADPH). Therefore, the correct pathway of electron flow is: H2O -> Photosystem II -> Electron Transport Chain -> Photosystem I -> NADP+.

[1 mark] A is the correct answer. Accept 'a' or 'A'. Reject B, C, and D as they do not represent the correct sequence of electron transfer or involve the incorrect coenzyme.

Ecological succession occurs on a corpse over time. The first colonisers are typically blowflies (Calliphoridae), which lay eggs that hatch into larvae within hours. Rigor mortis occurs within 2 to 36 hours after death. The core temperature typically cools to match the ambient temperature within 24 hours. Predatory beetles and parasitic wasps arrive later in the succession sequence (after several days or weeks) to feed on the decomposing tissues and on the larvae of earlier colonisers. Therefore, their presence indicates the longest period of time has elapsed since death.

[1 mark] C is the correct answer. Accept 'c' or 'C'. Reject A, B, and D because blowfly eggs/first-instar larvae (A), elevated body temperature/early cooling stages (B), and rigor mortis (D) are all characteristic of early post-mortem intervals.

Histone acetylation involves adding acetyl groups to lysine residues on histones, which neutralises their positive charge. This weakens the attraction between histones and the negatively charged DNA, causing the chromatin to decondense into a more open structure (euchromatin). Transcription factors and RNA polymerase can access the DNA more easily, thereby increasing transcription. In contrast, DNA methylation and histone deacetylation lead to tighter packing (heterochromatin) and transcriptional silencing.

[1 mark] B is the correct answer. Accept 'b' or 'B'. Reject A, C, and D as these changes promote a closed chromatin structure (heterochromatin) and inhibit/silence gene transcription.

When mucus in the airways is too dry, chloride ions are actively pumped into epithelial cells across the basal membrane and then diffuse out of the cells into the mucus layer via open CFTR channel proteins on the apical membrane. This accumulation of negative charge in the mucus draws sodium ions out of the cells and tissue fluid down an electrical gradient. The resulting high concentration of solute in the mucus lowers its water potential, causing water to move out of the epithelial cells into the mucus by osmosis, which thins the mucus.

[1 mark] A is the correct answer. Accept 'a' or 'A'. Reject B, C, and D because CFTR channels transport chloride ions out of cells, not into them (B); sodium ions move passively following the electrical gradient established by chloride ions (C); and water cannot be actively pumped (D).

Molecular phylogeny is the study of evolutionary relationships based on comparing molecular differences, specifically DNA/RNA nucleotide sequences or amino acid sequences of conserved proteins (such as cytochrome c). Homologous structures, behavioral adaptations, and ecological niches are anatomical, behavioral, and ecological observations rather than molecular phylogenetic evidence.

[1 mark] B is the correct answer. Accept 'b' or 'B'. Reject A, C, and D as they describe anatomical, behavioral, or ecological characteristics, not molecular phylogeny.

When a person dies, metabolic reactions stop and the body cools until it matches the temperature of the surroundings. The rate of this cooling depends directly on the temperature gradient between the body and the environment. A higher environmental temperature reduces this gradient, slowing down heat loss. If this factor is not accounted for, the pathologist may underestimate the post-mortem interval, wrongly concluding that the individual died more recently than they actually did.

1. Correctly identifies that environmental/ambient temperature determines the temperature gradient between the body and its surroundings (1 mark). 2. Explains that a smaller gradient (higher ambient temperature) reduces the rate of heat loss/cooling (or vice versa) (1 mark). 3. Explains the impact on estimating time of death (e.g., if ambient temperature is high, cooling is slower, which would lead to an underestimation of the time of death if not corrected) (1 mark).

First, state the standard ecological formula: \(NPP = GPP - R\). Next, calculate the energy lost to respiration: \(R = 0.65 \times 2.4 \times 10^4 = 1.56 \times 10^4\text{ kJ m}^{-2}\text{ year}^{-1}\). Finally, subtract this respiration value from the GPP: \(NPP = 2.4 \times 10^4 - 1.56 \times 10^4 = 0.84 \times 10^4\text{ kJ m}^{-2}\text{ year}^{-1}\) (which can also be written as \(8400\text{ kJ m}^{-2}\text{ year}^{-1}\)).

1. States the correct formula: \(NPP = GPP - R\) (1 mark). 2. Shows correct working for calculating respiration losses: \(1.56 \times 10^4\) or shows expression \(2.4 \times 10^4 \times (1 - 0.65)\) (1 mark). 3. Provides correct final answer: \(8400\) or \(8.4 \times 10^3\) or \(0.84 \times 10^4\) with correct units \(\text{kJ m}^{-2}\text{ year}^{-1}\) (1 mark). [Accept equivalent units like \(\text{kJ/m}^{2}\text{/year}\). Reject answers without units or with incorrect units.]

Increased methylation involves the addition of methyl groups to the cytosine bases (often at CpG sites) within the promoter region of a gene. This physical modification, along with the recruitment of proteins that condense chromatin, prevents transcription factors and RNA polymerase from binding to the promoter. Consequently, transcription of the gene is blocked, no mRNA is produced, and the gene is not expressed.

1. State that methyl groups are added to the cytosine bases / CpG islands in the promoter region of the gene (1 mark). 2. Explain that this prevents the binding of transcription factors / RNA polymerase to the DNA (1 mark). 3. Conclude that transcription is prevented, meaning no mRNA is produced (1 mark). [Accept: prevents translation of the protein because no mRNA is transcribed.]

During a primary immune response, a B cell binds to its complementary antigen, engulfs it, and presents it on its MHC class II molecules. A mature, active T helper cell with a complementary CD4 receptor binds to this antigen-MHC complex. Once bound, the T helper cell secretes cytokines (such as interleukins). These chemical signals trigger the B cell to undergo clonal expansion (rapid division by mitosis) and differentiate into antibody-secreting plasma cells and memory B cells.

1. Identifies that CD4 / T helper receptors bind to the complementary antigen presented on the MHC (class II) of a B cell (1 mark). 2. Explains that the activated T helper cell secretes cytokines / interleukins (1 mark). 3. Explains that these cytokines stimulate the B cell to undergo clonal expansion / mitosis / divide to produce plasma cells / memory cells (1 mark).

Xylem vessels must transport water under negative pressure (tension) without collapsing. They are adapted for this by having cell walls heavily impregnated with lignin, a polymer that provides high compressive strength and waterproofing. Furthermore, the cellulose microfibrils in the primary cell walls are arranged in a specific helical or net-like arrangement, providing excellent tensile strength to resist pulling forces.

1. State that xylem vessel walls are thickened / reinforced with lignin (1 mark). 2. Explain that lignin provides strength / rigidity to prevent the walls collapsing inward under tension (1 mark). 3. Mention that cellulose microfibrils are arranged in spirals / meshes / layers which provides high tensile strength (1 mark).

Each year, a tree grows a new layer of xylem vessels, which appears as a distinct ring in a cross-section of the trunk. The width of this ring is directly affected by the climate conditions during that growing season; warmer temperatures and higher water availability lead to more rapid photosynthesis and growth, producing a wider ring. By taking core samples from living trees and older preserved wood, scientists can find overlapping ring patterns to build up a continuous historical record of climate variations over thousands of years.

1. Explains that each ring represents one year of growth, and the width of the ring depends on the environmental conditions / temperature / moisture during that year (1 mark). 2. Identifies that wider rings indicate warmer / wetter conditions (more photosynthesis/growth) and narrower rings indicate colder / drier conditions (1 mark). 3. Explains that overlapping ring patterns from living and dead / preserved wood allow researchers to build a chronological timeline of past climate (1 mark).

Under normal conditions, the functional CFTR protein pumps chloride ions out of the epithelial cells into the mucus, creating a low water potential that draws water out of the cells by osmosis to keep the mucus runny. When the CFTR protein is non-functional or absent, chloride ions are not transported out. Consequently, sodium ions are actively reabsorbed into the cells, carrying water with them by osmosis. This deprives the mucus of water, making it dehydrated, thick, and highly viscous.

1. Explains that chloride ions cannot be transported out of the epithelial cells / across the apical membrane (into the mucus) (1 mark). 2. Explains that sodium ions (and water) are actively reabsorbed into the epithelial cells (1 mark). 3. States that water does not move out of the cells / moves into the cells by osmosis, leaving the mucus dehydrated / thick / sticky (1 mark).

The formation of the river acts as a geographical barrier, preventing gene flow between the two snail populations (geographical isolation). Over time, the environments on either side of the river may differ, resulting in different selection pressures. Beneficial mutations will arise independently in each population, and natural selection will change the allele frequencies in different ways. Eventually, the genetic differences become so significant that individuals from the two populations can no longer interbreed to produce fertile offspring, meaning speciation has occurred.

1. Identifies that the river acts as a geographical barrier that prevents gene flow / interbreeding between the two populations (1 mark). 2. Explains that different selection pressures / environmental conditions (or genetic drift) lead to natural selection of different alleles in each population (1 mark). 3. Explains that over time, genetic/phenotypic differences accumulate until the populations can no longer interbreed to produce fertile offspring / are reproductively isolated (1 mark).

1. Net Primary Productivity (\(NPP\)) is calculated using the equation: \(NPP = GPP - R\).
2. Calculate the energy lost through respiration (\(R\)): \(R = 0.55 \times 24400 = 13420\text{ kJ m}^{-2}\text{ yr}^{-1}\).
3. Subtract \(R\) from \(GPP\) to find \(NPP\): \(NPP = 24400 - 13420 = 10980\text{ kJ m}^{-2}\text{ yr}^{-1}\).

Alternative Method:
Calculate the percentage of energy left as \(NPP\): \(100\% - 55\% = 45\%\).
Calculate \(45\%\) of GPP: \(0.45 \times 24400 = 10980\text{ kJ m}^{-2}\text{ yr}^{-1}\).

- **MP1 (Method)**: Correctly identifies the formula \(NPP = GPP - R\) or calculates \(45\%\) of \(24400\) (1 mark).
- **MP2 (Method)**: Correct calculation of \(R\) as \(13420\text{ kJ m}^{-2}\text{ yr}^{-1}\) OR sets up the equation \(0.45 \times 24400\) (1 mark).
- **MP3 (Accuracy)**: Correct final value of \(10980\) with correct units (\(\text{kJ m}^{-2}\text{ yr}^{-1}\) or equivalent) (1 mark).

*Note: Correct final answer with correct units and no working scores 3 marks. Correct numerical answer with no/incorrect units scores 2 marks.*

During the humoral response, a B cell binds to a complementary antigen, processes it, and presents it on its MHC class II molecules on its surface, becoming an antigen-presenting cell (APC). An activated T helper cell with a complementary receptor binds to the presented antigen. This binding triggers the T helper cell to release cytokines, such as interleukins. These cytokines stimulate the specific B cell to rapidly divide by mitosis (clonal expansion) and differentiate into antibody-secreting plasma cells and memory B cells.

- **MP1**: Reference to the T helper cell binding to the antigen presented on the MHC class II complex of the B cell / APC (using its complementary T-cell receptor/CD4 receptor) (1 mark).
- **MP2**: Active T helper cell releases cytokines / interleukins (1 mark).
- **MP3**: Cytokines stimulate the B cell to undergo clonal expansion (mitosis / cell division) OR to differentiate into plasma cells / memory cells (1 mark).

Pluripotent stem cells have the capacity to express all genes. Transcription factors act as activators or repressors by binding to specific promoter/regulatory regions of DNA. This controls which genes are transcribed into mRNA (selective gene expression). Only the transcribed mRNA is translated into proteins. These specific proteins permanently change the cell's structure and control its chemical activities, resulting in a specialized cell.

- **MP1**: Transcription factors bind to specific regions of DNA (such as promoters / enhancers / regulatory sequences) (1 mark).
- **MP2**: This stimulates (activates) or inhibits (represses) the transcription of specific genes / synthesis of mRNA (1 mark).
- **MP3**: Only active genes are translated to produce specific proteins, which permanently alter the structure and function of the cell (leading to differentiation) (1 mark).

1. The CFTR gene mutation alters the primary structure of the CFTR protein, resulting in a misfolded, non-functional, or absent chloride channel in the apical membrane of epithelial cells.
2. Because of this, chloride ions (\(\text{Cl}^-\)) cannot be actively transported out of the epithelial cells into the mucus.
3. Sodium ions (\(\text{Na}^+\)) are excessively reabsorbed into the cells, creating a high solute concentration inside. Consequently, water is drawn out of the mucus into the cells by osmosis, leaving the mucus highly dehydrated, thick, and sticky.

- **MP1**: The mutation changes the structure of the CFTR protein, resulting in absent or non-functional chloride (\(\text{Cl}^-\)) channels (1 mark).
- **MP2**: Chloride ions are not transported out of the epithelial cells (into the mucus) OR sodium ions/water are excessively reabsorbed into the epithelial cells (1 mark).
- **MP3**: No water potential gradient is established to draw water out / water does not move out of the cells by osmosis into the mucus, leaving the mucus dehydrated, thick, and sticky (1 mark).

1. The formula for the heterozygosity index is: \(H = \frac{\text{number of heterozygous loci}}{\text{total number of loci examined}}\).
2. Substitute the given values: \(H = \frac{72}{450} = 0.16\).
3. A heterozygosity index closer to 1 means that a high proportion of gene loci in the population are heterozygous, which indicates high genetic diversity within the population.

- **MP1 (Method)**: Correctly sets up the division: \(\frac{72}{450}\) (1 mark).
- **MP2 (Accuracy)**: Correct calculation of \(0.16\) (accept \(\frac{4}{25}\) or \(16\%\)) (1 mark).
- **MP3 (Analysis)**: States that a value closer to 1 indicates higher/greater genetic diversity / a wider gene pool (1 mark).

Every year, trees produce a new layer of xylem vessels under the bark, visible as an annual growth ring. The width of these rings depends on the climate conditions during that growing season: warmer temperatures and higher rainfall lead to more rapid photosynthesis and growth, resulting in wider rings. By taking core samples from living trees and preserved old wood, scientists can match ring patterns (cross-dating) to construct a timeline going back thousands of years. This timeline provides detailed historical data on climate variation, showing trends such as global warming.

- **MP1**: Trees produce one growth ring per year, and the width of this ring reflects the growth rate/environmental conditions of that year (1 mark).
- **MP2**: Wider rings indicate warmer temperatures / more rainfall / more rapid growth/photosynthesis (or narrower rings indicate colder/dryer conditions) (1 mark).
- **MP3**: Comparing/correlating ring patterns from living and preserved wood (cross-dating) allows scientists to reconstruct past temperatures/climate trends over thousands of years (1 mark).

In PCR, primers are short, single-stranded sequences of DNA that are complementary to the start of the target DNA sequences. They bind (anneal) to the single-stranded DNA after denaturation, defining the region to be amplified and providing a double-stranded section that DNA polymerase requires to begin synthesis. DNA polymerase (typically heat-stable Taq polymerase) then binds to the primers and synthesizes the complementary strand of DNA by aligning and joining free nucleotides (dNTPs) via phosphodiester bonds.

- **MP1**: Primers bind (anneal) to complementary sequences on single-stranded DNA (at the start of the target region) (1 mark).
- **MP2**: Primers identify/mark the target region to be amplified OR provide a double-stranded attachment site for DNA polymerase to begin synthesis (1 mark).
- **MP3**: DNA polymerase synthesizes the complementary DNA strand by aligning and joining free nucleotides (via phosphodiester bonds) (1 mark). [Accept reference to Taq polymerase being thermostable / not denaturing at high temperatures].

Step 1: Calculate NPP for 2000 using the formula \(NPP = GPP - R\), which gives \(1250 - 750 = 500\text{ g m}^{-2}\text{ yr}^{-1}\). Step 2: Calculate NPP for 2020: \(1420 - 980 = 440\text{ g m}^{-2}\text{ yr}^{-1}\). Step 3: Calculate the percentage change: \(((440 - 500) / 500) \times 100 = -12\%\) (a decrease of 12%). Explanation: Higher temperatures increase the kinetic energy of enzymes and substrates involved in respiration (e.g., decarboxylases in the Link reaction and Krebs cycle), significantly raising respiration rates. If the rate of respiration increases more than the rate of photosynthesis (GPP) under global warming, less glucose is accumulated as biomass, resulting in a net decrease in NPP.

1. Calculates NPP for both years: 500 in 2000 and 440 in 2020 (1 mark). 2. Calculates percentage change as -12% or a 12% decrease (1 mark). 3. Explains that higher temperatures increase the rate of enzyme-controlled respiration more than photosynthesis/GPP (1 mark). 4. Concludes that more organic molecules are oxidized for ATP production, leaving less stored biomass/NPP (1 mark).

Step 1: Calculate larval growth rate between 48 and 72 hours: \((14.0\text{ mm} - 8.5\text{ mm}) / (72\text{ hours} - 48\text{ hours}) = 5.5\text{ mm} / 24\text{ hours} = 0.229\text{ mm hr}^{-1}\). Step 2: Determine growth from the 48-hour baseline: \(11.25\text{ mm} - 8.5\text{ mm} = 2.75\text{ mm}\). Step 3: Calculate the time taken to grow this amount: \(2.75\text{ mm} / 0.229\text{ mm hr}^{-1} = 12\text{ hours}\). Step 4: Add to the baseline hours: \(48 + 12 = 60\text{ hours}\). Factors other than temperature affecting growth include: 1. Presence of drugs or toxins in the tissues (e.g., cocaine speeds up development, while toxins like amitriptyline delay it). 2. Species competition or food availability on the carcass.

1. Correct calculation of linear growth rate as 0.229 mm/hr (or equivalent fraction) (1 mark). 2. Correct age of 60 hours calculated (1 mark). 3. Identifies one non-temperature factor (e.g., presence of drugs/toxins, competition, humidity/moisture) and states its directional effect on growth (1 mark). 4. Identifies a second valid non-temperature factor and states its directional effect on growth (1 mark).

Step 1: Calculate the overall increase in methylation: \(78\% - 15\% = 63\%\). Step 2: Calculate the days elapsed: \(9 - 3 = 6\text{ days}\). Step 3: Calculate the daily rate: \(63\% / 6 = 10.5\%\text{ per day}\). Biological explanation: DNA methylation adds methyl groups to cytosine bases in CpG islands located in the promoter region of the OCT4 gene. This physical modification prevents the binding of transcription factors and RNA polymerase. Consequently, transcription of OCT4 is prevented, blocking the synthesis of the OCT4 protein. Since the OCT4 protein maintains pluripotency, its silencing allows differentiation pathways to proceed.

1. Correct calculation of the rate as 10.5% per day (must include units) (1 mark). 2. Explains that methylation involves attaching methyl groups to CpG islands / cytosine bases in the promoter region (1 mark). 3. States that methylation physically blocks transcription factors / RNA polymerase from binding to the promoter (1 mark). 4. Concludes that transcription is silenced, preventing OCT4 protein expression and allowing the cell to lose pluripotency and differentiate (1 mark).

Step 1: Calculate the heterozygosity index for Population A: \(H_A = 48 / 150 = 0.32\). Step 2: Calculate the heterozygosity index for Population B: \(H_B = 32 / 80 = 0.40\). Step 3: Conclude that Population A is at greater risk because its index (0.32) is lower than Population B's (0.40). Explanation: A lower heterozygosity index represents lower genetic variation and a restricted gene pool. This reduces the likelihood that individuals within Population A will possess alleles enabling them to adapt to environmental changes, such as new diseases or shifting climates. It also increases the frequency of inbreeding, which can lead to inbreeding depression and the accumulation of homozygous recessive deleterious alleles.

1. Correct calculation of heterozygosity indices: HA = 0.32 and HB = 0.40 (1 mark). 2. Identifies Population A as having the lower genetic diversity and being at higher risk of extinction (1 mark). 3. Explains that a low heterozygosity index reflects a small gene pool with fewer distinct alleles (1 mark). 4. Explains that a smaller gene pool reduces adaptive capacity to environmental change OR increases the risk of inbreeding depression/homozygous recessive diseases (1 mark).

1. Calculate Gross Primary Productivity (GPP):
\(GPP = 3.20 \times 10^6 \times 0.0150 = 48,000 \text{ kJ m}^{-2}\text{ yr}^{-1}\) (or \(4.80 \times 10^4\))

2. Calculate Net Primary Productivity (NPP):
\(NPP = GPP - R\)
Since \(58.0\%\) of GPP is lost as respiration, NPP is \(42.0\%\) of GPP:
\(NPP = 48,000 \times (1 - 0.580) = 48,000 \times 0.420 = 20,160 \text{ kJ m}^{-2}\text{ yr}^{-1}\)

3. Convert to standard form and round to two significant figures:
\(2.0 \times 10^4 \text{ kJ m}^{-2}\text{ yr}^{-1}\)

1 mark: Correct calculation of GPP as \(48,000\text{ kJ m}^{-2}\text{ yr}^{-1}\) (or \(4.80 \times 10^4\)).
1 mark: Correct calculation of NPP before rounding as \(20,160\text{ kJ m}^{-2}\text{ yr}^{-1}\).
1 mark: Correct final answer in standard form to 2 significant figures, with correct units: \(2.0 \times 10^4 \text{ kJ m}^{-2}\text{ yr}^{-1}\) (accept \(\text{y}^{-1}\) or \(\text{year}^{-1}\)). Allow error carried forward (ECF) for rounding and standard form from incorrect previous steps.

1. Identify \(q^2\), the frequency of homozygous recessive individuals:
\(q^2 = \frac{1}{10,000} = 0.0001\)

2. Calculate the recessive allele frequency (\(q\)):
\(q = \sqrt{0.0001} = 0.01\)

3. Calculate the dominant allele frequency (\(p\)):
\(p = 1 - q = 1 - 0.01 = 0.99\)

4. Calculate the frequency of heterozygotes (\(2pq\)):
\(2pq = 2 \times 0.99 \times 0.01 = 0.0198\)

5. Convert to percentage:
\(0.0198 \times 100\% = 1.98\%\)

1 mark: Correct calculation of \(q = 0.01\).
1 mark: Correct substitution into the Hardy-Weinberg formula to find heterozygote frequency: \(2 \times 0.99 \times 0.01 = 0.0198\).
1 mark: Correct percentage to 3 significant figures: \(1.98\%\) (accept \(1.98\) if percentage sign is already understood, reject \(2\%\) or \(2.0\%\) unless correct ECF is applied).

1. Calculate the rate of oxygen production at \(12^\circ\text{C}\):
\(\text{Rate}_{12} = \frac{4.5 \text{ cm}^3}{30 \text{ min}} = 0.15 \text{ cm}^3\text{ min}^{-1}\) (or \(9.0 \text{ cm}^3\text{ h}^{-1}\))

2. Calculate the rate of oxygen production at \(22^\circ\text{C}\):
\(\text{Rate}_{22} = \frac{11.7 \text{ cm}^3}{45 \text{ min}} = 0.26 \text{ cm}^3\text{ min}^{-1}\) (or \(15.6 \text{ cm}^3\text{ h}^{-1}\))

3. Calculate \(Q_{10}\) using the rate values:
\(Q_{10} = \frac{\text{Rate at } 22^\circ\text{C}}{\text{Rate at } 12^\circ\text{C}} = \frac{0.26}{0.15} \approx 1.7333...\)

4. Round to 2 decimal places:
\(1.73\)

1 mark: Correct calculation of rate at \(12^\circ\text{C}\) (\(0.15\)) and rate at \(22^\circ\text{C}\) (\(0.26\)) in consistent units.
1 mark: Correct substitution into the \(Q_{10}\) formula: \(\frac{0.26}{0.15}\) (or \(\frac{15.6}{9.0}\)).
1 mark: Correct final value of \(1.73\) (accept ECF if minor calculation errors occurred in rates, provided final value is correctly rounded to 2 decimal places).

Climate change alters key abiotic factors, specifically decreasing water availability and increasing temperatures, which act as selection pressures on species within a community. In a deciduous woodland, succession leads to a stable climax community dominated by large tree species. Under current conditions, beech and sessile oak coexist, with beech often being dominant. However, severe summer drought stress will cause stomatal closure in beech trees to conserve water, significantly reducing their rate of photosynthesis, growth, and overall survival. In contrast, sessile oak trees have physiological adaptations that make them more drought-tolerant, enabling them to maintain higher rates of photosynthesis and biomass production under water-stressed conditions. This shifts the competitive balance: sessile oaks will outcompete beech trees for resources such as light, soil water, and mineral ions. Over time, the composition of the climax community will change, with sessile oak becoming the dominant species, while beech populations decline and their distribution shifts away from drought-prone areas. Furthermore, the rapid death of beech trees may open up canopy gaps, initiating secondary succession where pioneer species colonise the newly lit forest floor before oak seedlings eventually establish dominance. This change in the dominant tree species will also alter the microclimate and food webs, affecting the niches and distribution of associated insects, fungi, and birds.

Level 1 (1-2 marks): The candidate identifies that climate change reduces water availability and states that this will impact beech trees more than oak trees. There is a simple reference to succession or dominant species. Level 2 (3-4 marks): The candidate explains how drought tolerance allows sessile oak to outcompete beech, linking this to a shift in the dominant species of the climax community. Level 3 (5-6 marks): The candidate provides a detailed and balanced discussion linking the change in abiotic factors (water availability) to physiological impacts (photosynthesis), competitive exclusion, succession stages (including potential secondary succession in gaps), and the wider ecological consequences on niches and other species. Indicative content: 1. Water availability and temperature act as abiotic factors/selection pressures. 2. Beech trees suffer reduced photosynthesis and growth due to drought-induced stomatal closure. 3. Sessile oak is drought-tolerant and maintains photosynthesis, giving it a competitive advantage. 4. The dominant species in the climax community shifts from beech to oak. 5. Altered canopy structure/beech dieback initiates secondary succession in gaps. 6. Shifts in tree species distribution alter niches, affecting wider woodland biodiversity (insects, birds, decomposers).

To estimate the time of death of a mammal, forensic scientists combine multiple independent methods. 1. Body Temperature (Algor Mortis): Post-mortem, metabolic heat production ceases, and the core body temperature decreases until it equilibrates with the ambient temperature, following a predictable sigmoid cooling curve. The scientist can measure the rectal or liver temperature to estimate how long cooling has been occurring. The main limitation is that the cooling rate is highly dependent on external factors such as ambient temperature, body mass, clothing, humidity, and wind speed, making it reliable only within the first 24 hours. 2. State of Decomposition: This includes rigor mortis (temporary muscle stiffening caused by the depletion of ATP, preventing actin and myosin filaments from separating, typically lasting 12 to 36 hours) and subsequent autolysis and putrefaction by bacteria. The state of decay can indicate whether death occurred days or weeks ago. The limitation is that decomposition rate is extremely temperature-dependent (occurring much faster in warm conditions) and is affected by the accessibility of the body to decomposers. 3. Forensic Entomology: Insects colonise a corpse in a predictable sequence (ecological succession), starting with blowflies (Calliphoridae). By collecting larvae and pupae, identifying the species, and measuring their length or developmental stage, scientists can use known growth rates at specific temperatures to calculate the minimum time since death. The limitation is that insect development rates are highly sensitive to temperature fluctuations, and colonization can be delayed by physical barriers, burial, or the presence of drugs/toxins in the tissues.

Level 1 (1-2 marks): The candidate identifies at least two methods (e.g., body temperature, insects, or decomposition) and gives a basic description of how they change over time. Level 2 (3-4 marks): The candidate explains how at least two of the methods are used to estimate the time of death, linking them to underlying biological processes (e.g., cooling curve, rigor mortis/ATP, or insect life cycles) and outlines at least one limitation. Level 3 (5-6 marks): The candidate provides a detailed and comprehensive explanation of all three methods (body temperature, decomposition, and entomology), clearly explaining the biological mechanisms involved and discussing major environmental limitations for each method. Indicative content: 1. Core body temperature cools along a sigmoid curve (algor mortis); limited by ambient temperature, body size, and clothing (mostly useful <24 hours). 2. Rigor mortis is muscle stiffening due to lack of ATP preventing muscle relaxation; limited by temperature and pre-death activity. 3. General decomposition (autolysis/bacterial decay) occurs in stages; limited by temperature, moisture, and decomposer access. 4. Forensic entomology uses the predictable succession of insect species colonising the body. 5. Analysis of fly life cycle stages (eggs, larvae, pupae) and larval length determines age; limited by temperature, season, and presence of toxins.

During anaerobic respiration, oxygen is unavailable to act as the terminal electron acceptor in the electron transport chain. Therefore, reduced NAD cannot be reoxidized by the mitochondria. To allow glycolysis (which yields a net 2 ATP per glucose) to continue, pyruvate is reduced to lactate. This reaction is catalyzed by lactate dehydrogenase and oxidizes NADH back to \(\text{NAD}^+\), regenerating the coenzyme needed for the triose phosphate dehydrogenase step of glycolysis.

1 mark for option B. [Reject: other options]

MDMA target cells are primarily serotonergic neurons. MDMA binds to and reverses the action of serotonin reuptake transporters (SERT). Instead of transporting serotonin back into the pre-synaptic terminal, it pumps serotonin out into the synaptic cleft, significantly increasing serotonin concentration and stimulating post-synaptic receptors.

1 mark for option C. [Reject: other options]

First, calculate the time interval between consecutive R waves (representing one heartbeat): \(\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{20\text{ mm}}{25\text{ mm s}^{-1}} = 0.8\text{ s}\). Next, calculate the heart rate per minute: \(\text{Heart rate} = \frac{60\text{ s}}{0.8\text{ s beat}^{-1}} = 75\text{ bpm}\).

1 mark for option B (correct calculations and final value of 75 bpm). [Reject: other options]

Transcription factors are proteins that bind to specific promoter regions of DNA. Activators assist the binding of RNA polymerase to initiate transcription, while repressors prevent RNA polymerase from binding, thereby regulating transcription.

1 mark for option B. [Reject: other options]

In the light, light energy causes rhodopsin to break down (bleach) into opsin and retinal. This activates a cascade involving transducin and phosphodiesterase, which breaks down cGMP to GMP. The reduction in cGMP concentration causes sodium channels in the outer segment to close. Since sodium ions continue to be actively pumped out of the inner segment, the cell becomes hyperpolarized and neurotransmitter (glutamate) release is inhibited.

1 mark for option B. [Reject: other options]

Cyanide binds to the active site of cytochrome c oxidase, inhibiting its activity. This prevents the transfer of electrons to oxygen, meaning oxygen cannot act as the terminal electron acceptor. Consequently, the electron transport chain stops operating. This prevents the active transport of protons (\(H^+\) ions) from the mitochondrial matrix into the intermembrane space. Without this proton movement, the electrochemical proton gradient is lost, meaning protons cannot flow back down their concentration gradient through ATP synthase (chemiosmosis), preventing the phosphorylation of ADP to ATP.

1. MP1: Cytochrome c oxidase cannot transfer electrons to oxygen / oxygen cannot act as the terminal electron acceptor (1 mark).
2. MP2: This stops the pumping of protons (\(H^+\) ions) into the intermembrane space, preventing the formation of an electrochemical proton gradient (1 mark).
3. MP3: No protons flow down their concentration gradient through ATP synthase, so ADP cannot be phosphorylated to ATP (1 mark).

When an action potential depolarises the sarcolemma, calcium ions are released from the sarcoplasmic reticulum into the sarcoplasm. These calcium ions bind to the protein troponin, causing it to undergo a conformational change. This change pulls the attached tropomyosin molecule, moving it away from the myosin-binding sites on the actin filaments. With these sites exposed, myosin heads can bind to the actin filaments to form actin-myosin cross-bridges, which are required for myofibril contraction via the power stroke.

1. MP1: Calcium ions bind to troponin, causing a conformational change (1 mark).
2. MP2: This causes tropomyosin to displace/move, exposing the myosin-binding sites on the actin filament (1 mark).
3. MP3: Allows myosin heads to bind to actin to form actin-myosin cross-bridges (1 mark).

In a healthy, myelinated neurone, myelin acts as an electrical insulator. This prevents depolarisation from occurring along the myelinated sections of the axon, restricting ion exchange to the nodes of Ranvier. Consequently, the action potential "jumps" from one node to the next, a process called saltatory conduction. When the myelin sheath is destroyed in MS patients, depolarisation must occur continuously along the entire length of the axon membrane. This is a significantly slower process, which reduces the overall speed of conduction of the nerve impulse.

1. MP1: Myelin normally acts as an electrical insulator, forcing depolarisation to occur only at the nodes of Ranvier (1 mark).
2. MP2: In healthy neurones, action potentials jump from node to node via saltatory conduction (1 mark).
3. MP3: Without myelin, depolarisation must occur continuously along the entire length of the axon membrane, significantly reducing impulse conduction speed (1 mark).

MRI and fMRI differ in what they measure and display. MRI provides high-resolution, static structural images of the brain's anatomy, which is useful for identifying physical degeneration in structures like the substantia nigra. In contrast, fMRI provides dynamic images showing changes in brain activity over time by measuring blood oxygenation levels (the BOLD signal). This allows researchers to observe brain function in real-time while the patient performs motor tasks, helping to identify functional deficits in motor pathway activation.

1. MP1: MRI provides static structural images of the brain's anatomy, whereas fMRI provides dynamic images showing changes in brain activity over time (1 mark).
2. MP2: fMRI measures changes in blood oxygenation / blood flow (the BOLD signal) in active areas of the brain, whereas MRI does not measure blood flow (1 mark).
3. MP3: fMRI can be used to observe brain activity during active motor tasks, whereas MRI is used to identify anatomical degeneration / structural damage in the substantia nigra (1 mark).

During habituation, repeated stimulation of the sensory neurone causes voltage-gated calcium ion (\(Ca^{2+}\)) channels in the presynaptic membrane to become less responsive and open less frequently. As a result, fewer calcium ions enter the presynaptic knob. This reduction in calcium influx prevents synaptic vesicles from fusing with the presynaptic membrane, leading to less neurotransmitter being released by exocytosis into the synaptic cleft. Consequently, fewer sodium ion channels on the postsynaptic membrane open, preventing the threshold potential from being reached and reducing the generation of action potentials in the postsynaptic neurone.

1. MP1: Repeated stimulation causes calcium ion (\(Ca^{2+}\)) channels in the presynaptic membrane to become less responsive / open less (1 mark).
2. MP2: Fewer calcium ions enter the presynaptic terminal (1 mark).
3. MP3: Fewer vesicles of neurotransmitter undergo exocytosis / less neurotransmitter is released into the synaptic cleft, meaning fewer action potentials are generated in the postsynaptic neurone (1 mark).

To calculate the cardiac output, use the formula: \(\text{Cardiac output} = \text{Heart rate} \times \text{Stroke volume}\).
Substitute the given values: \(\text{Cardiac output} = 145\ \text{bpm} \times 110\ \text{cm}^3 = 15,950\ \text{cm}^3\ \text{min}^{-1}\).
Convert the volume from \(cm^3\) to \(dm^3\) by dividing by 1,000:
\(15,950 / 1,000 = 15.95\ \text{dm}^3\ \text{min}^{-1}\).

1. MP1: Uses the correct formula or substitution: \(145 \times 110\) (1 mark).
2. MP2: Correct calculation in \(cm^3\): \(15,950\ \text{cm}^3\ \text{min}^{-1}\) (1 mark).
3. MP3: Correct final answer with units: \(15.95\ \text{dm}^3\ \text{min}^{-1}\) (Accept \(16\ \text{dm}^3\ \text{min}^{-1}\) if correctly rounded) (1 mark).
[Note: Award 3 marks for correct final answer with units even if no working is shown.]

Transcription factors are proteins that control the rate of transcription of genetic information from DNA to messenger RNA. They do this by binding to specific promoter regions located near target genes. In the process of differentiation, specific transcription factors are activated. These either promote (by helping RNA polymerase bind) or repress the transcription of specific genes. Consequently, only genes relevant to cardiac muscle cells are transcribed into mRNA and translated into proteins (such as actin and myosin), establishing the cell's specialized structure and function.

1. MP1: Transcription factors bind to specific promoter regions / DNA sequence near target genes (1 mark).
2. MP2: They stimulate (or inhibit) the binding of RNA polymerase to the DNA to control transcription (1 mark).
3. MP3: Only genes specific to cardiac muscle cells are transcribed into mRNA (and translated), producing the proteins needed for the cell's specialized structure/function (1 mark).

Liposomes are small, artificial spheres composed of a phospholipid bilayer surrounding an aqueous core containing the therapeutic DNA (functional CFTR gene). Because their membrane structure is highly similar to the cell surface membrane of target epithelial cells, they can easily fuse with them to deliver the gene inside the cell. Furthermore, the lipid bilayer protects the DNA from being degraded by extracellular nucleases. Finally, unlike viral vectors, liposomes are non-pathogenic and do not typically elicit a significant immune response, making them a safer alternative for repeated clinical use.

1. MP1: Liposomes are made of a phospholipid bilayer, allowing them to easily fuse with the cell surface membranes of the target epithelial cells (1 mark).
2. MP2: They protect the therapeutic DNA/CFTR gene from degradation by extracellular enzymes (1 mark).
3. MP3: They are non-pathogenic / do not trigger an immune response (unlike viral vectors), making them safer for repeated clinical treatments (1 mark).

To allow glycolysis to continue under anaerobic conditions, the cell must regenerate oxidized NAD. During the conversion of pyruvate to lactate, reduced NAD (NADH) is oxidized, donating its hydrogens to pyruvate. The regenerated NAD can then be used again in glycolysis to oxidize triose phosphate, which produces a net yield of 2 ATP molecules per glucose molecule.

1. Pyruvate is reduced / accepts hydrogen from reduced NAD (NADH) to form lactate (1 mark). 2. This oxidizes NADH / regenerates oxidized NAD (1 mark). 3. Oxidized NAD is required in glycolysis to accept hydrogens from triose phosphate, allowing ATP production to continue (1 mark).

MDMA targets the serotonin reuptake proteins (transporters) on the presynaptic membrane. By binding to these transporters, it blocks the reuptake of serotonin from the synaptic cleft. Additionally, MDMA can reverse the action of these transporters, pumping serotonin out of the presynaptic neuron. This leads to a high concentration of serotonin remaining in the synaptic cleft, resulting in continuous activation of the postsynaptic receptors.

1. MDMA binds to and blocks the serotonin reuptake transporter proteins on the presynaptic membrane (1 mark). 2. This prevents serotonin from being reabsorbed / removed from the synaptic cleft, or causes transporters to run in reverse (1 mark). 3. Consequently, there is an increased concentration of serotonin in the synaptic cleft, leading to continuous/increased stimulation of the postsynaptic receptors (1 mark).

First, convert the units so they are consistent. Since \(1\text{ dm}^3 = 1000\text{ cm}^3\), the cardiac output of \(6.3\text{ dm}^3\text{ min}^{-1}\) is equal to \(6300\text{ cm}^3\text{ min}^{-1}\). Alternatively, convert the stroke volume of \(75\text{ cm}^3\) to \(0.075\text{ dm}^3\). The formula is: \(\text{Cardiac Output} = \text{Stroke Volume} \times \text{Heart Rate}\). Rearranging the formula gives: \(\text{Heart Rate} = \frac{\text{Cardiac Output}}{\text{Stroke Volume}}\). Therefore, \(\text{Heart Rate} = \frac{6300}{75} = 84\text{ bpm}\).

1. Converts units correctly: \(6.3\text{ dm}^3\) to \(6300\text{ cm}^3\) OR \(75\text{ cm}^3\) to \(0.075\text{ dm}^3\) (1 mark). 2. Correctly rearranges formula: \(\text{Heart Rate} = \frac{\text{Cardiac Output}}{\text{Stroke Volume}}\) (1 mark). 3. Correct calculation of 84 (bpm) (1 mark). Note: Accept correct answer with no working shown for 3 marks.

During habituation, repeated stimulation of the sensory neurone leads to a decrease in the responsiveness of calcium ion channels in the presynaptic membrane. As a result, fewer calcium ions (\(\text{Ca}^{2+}\)) enter the presynaptic neurone when an action potential arrives. This reduces the number of neurotransmitter vesicles fusing with the presynaptic membrane, leading to less neurotransmitter being released into the synaptic cleft. Consequently, fewer ligand-gated sodium channels open on the postsynaptic membrane, and the depolarisation is insufficient to reach the threshold required to trigger an action potential in the postsynaptic neurone.

1. Calcium ion channels on the presynaptic membrane become less responsive, so fewer calcium ions enter the presynaptic neurone (1 mark). 2. Fewer synaptic vesicles fuse with the presynaptic membrane, releasing less neurotransmitter into the synaptic cleft (1 mark). 3. Fewer sodium channels open on the postsynaptic membrane, so there is insufficient depolarisation to reach the threshold for an action potential (1 mark).

Antioxidants work by donating electrons to free radicals, thereby neutralizing them and preventing them from reacting with other cellular components. Free radicals cause oxidative damage to LDL cholesterol and to the endothelial cells lining the arteries. By neutralizing free radicals, antioxidants minimize damage to the endothelium. This prevents the inflammatory response that leads to the accumulation of white blood cells and the deposition of lipids, which reduces plaque (atheroma) formation.

1. Antioxidants neutralize free radicals by donating electrons (1 mark). 2. This prevents oxidative damage to endothelial cell linings of arteries / to LDL cholesterol (1 mark). 3. This reduces endothelial inflammation/damage, preventing the cascade of events that leads to the accumulation of lipids and atheroma/plaque formation (1 mark).

DNA methylation involves the addition of methyl groups (\(-\text{CH}_3\)) to cytosine bases, typically where they are adjacent to guanine bases (CpG islands) in the promoter region of a gene. This modification alters the physical shape of the DNA and recruits proteins that condense chromatin. This prevents transcription factors and RNA polymerase from binding to the promoter region, so the gene cannot be transcribed into mRNA, effectively silencing the gene.

1. Methyl groups are added to cytosine bases / CpG islands within the promoter region of the gene (1 mark). 2. This blocks the binding of transcription factors / RNA polymerase to the promoter (1 mark). 3. This prevents transcription from occurring, so no mRNA is produced, silencing/preventing gene expression (1 mark).

1. **Calculate the absolute reduction in blood lactate at 14 \(\text{km h}^{-1}\):**
\(3.5 - 1.9 = 1.6 \text{ mmol dm}^{-3}\)

2. **Calculate the percentage reduction:**
\(\frac{1.6}{3.5} \times 100 = 45.71\%\) (or \(45.7\%\) rounded to 3 significant figures).

3. **Biological Explanation:**
- High-intensity training increases aerobic capacity (e.g., through increased density of mitochondria, increased capillary density, or stroke volume).
- This allows runners to perform at higher speeds (such as 14 \(\text{km h}^{-1}\)) using aerobic respiration, delaying the onset of anaerobic respiration and lactate production.
- Reduced lactate accumulation prevents the accumulation of hydrogen ions (\(\text{H}^+\)), which lowers muscle pH and causes fatigue/inhibits muscle contraction enzymes. Runners can therefore maintain a faster pace for longer.

Award up to 4 marks:
- **Mark 1 (Method):** Correct calculation of absolute decrease: \(3.5 - 1.9 = 1.6\) AND correct formula setup: \(\frac{1.6}{3.5} \times 100\). [1 mark]
- **Mark 2 (Accuracy):** Correct percentage reduction of \(45.7\%\) (accept \(46\%\) or \(45.71\%\)). [1 mark]
- **Mark 3 (Analysis):** Lower blood lactate levels show that training increased aerobic capacity / delayed the threshold for anaerobic respiration at higher speeds. [1 mark]
- **Mark 4 (Explanation):** Reduced lactate accumulation means less acidosis (lower accumulation of \(\text{H}^+\) ions), which reduces muscle fatigue and allows the runner to sustain a high speed for a longer duration. [1 mark]

1. **Calculate the rate of increase in conduction velocity for myelinated neurons:**
\(\frac{105.0 - 45.0}{40 - 20} = \frac{60}{20} = 3.0 \text{ m s}^{-1}\ ^{\circ}\text{C}^{-1}\)

2. **Calculate the rate of increase in conduction velocity for unmyelinated neurons:**
\(\frac{2.4 - 1.2}{40 - 20} = \frac{1.2}{20} = 0.06 \text{ m s}^{-1}\ ^{\circ}\text{C}^{-1}\)

3. **Calculate the difference in rates:**
\(3.0 - 0.06 = 2.94 \text{ m s}^{-1}\ ^{\circ}\text{C}^{-1}\)

4. **Comparison/Biological analysis:**
- Conduction velocity in myelinated neurons increases at a much faster rate with temperature than in unmyelinated neurons.
- Both show an increase because higher temperature increases the kinetic energy of sodium (\(\text{Na}^+\)) and potassium (\(\text{K}^+\)) ions, speeding up facilitated diffusion during action potential propagation.
- Myelinated neurons are much faster overall because they undergo saltatory conduction, where depolarization only occurs at the nodes of Ranvier, whereas in unmyelinated neurons, depolarization must occur along the entire length of the axon.

Award up to 4 marks:
- **Mark 1 (Method):** Clear attempt to calculate rate of change per degree for both types of neurons (e.g., finding the gradient of conduction velocity over temperature: \(3.0\) and \(0.06\)). [1 mark]
- **Mark 2 (Accuracy):** Correct calculation of the difference in the rate of increase: \(2.94 \text{ m s}^{-1}\ ^{\circ}\text{C}^{-1}\) (must include correct unit). [1 mark]
- **Mark 3 (Analysis):** Conduction velocity is higher in myelinated neurons at all temperatures AND both types show a positive linear correlation with temperature. [1 mark]
- **Mark 4 (Explanation):** Higher temperatures increase ion kinetic energy / rate of diffusion. Myelination enables saltatory conduction (action potential jumps between nodes of Ranvier) which dramatically amplifies this velocity advantage compared to continuous conduction in unmyelinated fibers. [1 mark]

1. **Calculate the percentage change for the experimental group:**
- Change = \(3.6 - 4.3 = -0.7 \text{ mmol L}^{-1}\)
- Percentage Change = \(\frac{-0.7}{4.3} \times 100 = -16.279\%\) (or a \(16.3\%\) decrease).

2. **Explain the biological pathway to atherosclerosis:**
- LDLs carry cholesterol in the blood. If blood LDL is high, excess LDL is more likely to accumulate under damaged endothelial linings of arteries.
- Reducing LDL levels reduces lipid/cholesterol deposition in the artery walls.
- This prevents the inflammatory response where white blood cells (macrophages) engulf cholesterol to form foam cells.
- Fewer foam cells and less lipid accumulation prevent the development of fatty streaks into an atheroma (fibrous plaque), keeping the artery lumen wide and maintaining healthy blood flow.

Award up to 4 marks:
- **Mark 1 (Method):** Correct calculation of absolute change (\(-0.7\)) and division by baseline value (\(4.3\)). [1 mark]
- **Mark 2 (Accuracy):** Correct percentage change of \(-16.3\%\) or stated as a \(16.3\%\) decrease (accept \(16.28\%\) or \(16\%\); reject positive \(16.3\%\) without the word 'decrease' or negative sign). [1 mark]
- **Mark 3 (Biological link):** Explains that lower LDL levels reduce the amount of cholesterol/lipids depositing inside the endothelial lining / wall of damaged arteries. [1 mark]
- **Mark 4 (Explanation):** This reduces the inflammatory response / macrophage infiltration / foam cell formation, thereby preventing the formation of an atheroma (plaque) that narrows the lumen. [1 mark]

1. **Calculate the fold change:**
- Fold change = \(\frac{\text{Value at Day 6}}{\text{Value at Day 0}}\)
- Fold change = \(\frac{12.8}{0.1} = 128\) (or a 128-fold increase).

2. **Explain the biological role of transcription factors:**
- Transcription factors are proteins that bind to specific promoter or enhancer regions of target genes in DNA.
- They stimulate (or inhibit) the transcription of these genes by facilitating the binding of RNA polymerase.
- *MYOD1* upregulation leads to the transcription of muscle-specific genes (producing mRNAs for proteins like actin and myosin).
- These specific proteins determine the structure and specialized function of the mature skeletal muscle cell, leading to successful differentiation.

Award up to 4 marks:
- **Mark 1 (Method):** Clear attempt to calculate fold change by dividing the Day 6 value by the Day 0 value (\(\frac{12.8}{0.1}\)). [1 mark]
- **Mark 2 (Accuracy):** Correct calculation of fold change as \(128\) (or 128-fold / 128\(\times\)). [1 mark]
- **Mark 3 (Explanation):** Transcription factors bind to specific promoter / regulatory regions of DNA to switch genes on/off (or initiate/regulate transcription of mRNA). [1 mark]
- **Mark 4 (Explanation):** This ensures only muscle-specific genes are expressed, leading to the synthesis of specialized proteins (e.g., actin/myosin) that structurally and functionally define the differentiated cell. [1 mark]

1. Convert the exercise cardiac output from dm³ min⁻¹ to cm³ min⁻¹: \(24.3 \times 1000 = 24300 \text{ cm}^3 \text{ min}^{-1}\). 2. Calculate the exercise stroke volume: \(\text{Stroke volume} = \frac{\text{Cardiac output}}{\text{Heart rate}} = \frac{24300}{180} = 135 \text{ cm}^3\). 3. Calculate the percentage increase in stroke volume from the resting stroke volume of 75 cm³: \(\frac{135 - 75}{75} \times 100 = \frac{60}{75} \times 100 = 80\%\).

1. Convert cardiac output to cm³ min⁻¹: 24,300 cm³ min⁻¹ (1 mark).
2. Calculate exercise stroke volume: 135 cm³ (1 mark).
3. Correct percentage increase: 80% (1 mark). Accept correct calculation if resting stroke volume is converted to dm³ instead.

1. Calculate conduction time in the neurones: \(\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{1.45 \text{ m}}{75 \text{ m s}^{-1}} = 0.01933 \text{ s}\). Convert this time to milliseconds: \(0.01933 \times 1000 = 19.33 \text{ ms}\). 2. Calculate the total synaptic delay from the two synapses: \(2 \times 1.5 \text{ ms} = 3.0 \text{ ms}\). 3. Calculate total time: \(19.33 \text{ ms} + 3.0 \text{ ms} = 22.33 \text{ ms}\). Rounded to 3 significant figures, this is 22.3 ms.

1. Calculate conduction time in neurones as 19.3 ms or 0.0193 s (1 mark).
2. Show addition of total synaptic delay of 3.0 ms to neurone conduction time (1 mark).
3. Correct final answer of 22.3 ms (or 22.3) rounded to 3 significant figures (1 mark).

1. Calculate the total energy stored in the 32 moles of ATP produced: \(32 \times 30.5 \text{ kJ} = 976 \text{ kJ}\). 2. Express this stored energy as a percentage of the total energy available in one mole of glucose: \(\frac{976}{2870} \times 100 = 34.007\%\). 3. Round to 3 significant figures: 34.0%.

1. Calculate energy stored in ATP: 976 kJ (1 mark).
2. Express as a percentage of total energy from glucose: \(\frac{976}{2870} \times 100\) (1 mark).
3. Correct final answer of 34.0% (or 34.0) rounded to 3 significant figures (1 mark).

Indicative content:
- Rotenone inhibits Complex I, stopping the transfer of electrons from NADH into the electron transport chain (ETC).
- This stops the movement of electrons down the ETC to the final electron acceptor, oxygen.
- As a result, active transport of protons \(H^+\) from the mitochondrial matrix into the intermembrane space is reduced or halted.
- The proton / electrochemical gradient across the inner mitochondrial membrane is not established or maintained.
- Therefore, there is no or significantly reduced flow of protons through ATP synthase (chemiosmosis), leading to a drastic decrease in ATP synthesis.
- Active transport of calcium ions \(Ca^{2+}\) back into the sarcoplasmic reticulum cannot occur without ATP.
- ATP binding is required to break the actin-myosin cross-bridge; without ATP, myosin heads remain bound to actin filaments.
- ATP hydrolysis is also required to reset (cock) the myosin head for the next power stroke.
- This leads to muscle stiffness, sustained contraction (rigidity), and eventual failure of muscle contraction (paralysis).

Level 1 (1-2 marks): Describes basic processes.
- Mentions that ATP production decreases because the electron transport chain or aerobic respiration is blocked.
- Mentions that ATP is needed for muscle contraction or that muscles will stop contracting/stiffen.

Level 2 (3-4 marks): Links the inhibition of the ETC to a specific cellular consequence, or links ATP lack to a specific stage of muscle contraction.
- Explains how blocking Complex I stops electron flow, preventing proton pumping and thus reducing the proton gradient or ATP synthase activity.
- OR explains how a lack of ATP prevents the detachment of myosin heads from actin filaments or prevents calcium reuptake.

Level 3 (5-6 marks): Provides a fully integrated and logical explanation linking the mitochondrial defect to the muscular failure.
- Clearly details how the loss of the proton gradient across the inner mitochondrial membrane prevents chemiosmosis and ATP synthesis.
- AND explains how the absence of ATP prevents both the breaking of cross-bridges (causing rigidity) and the active transport of calcium back into the sarcoplasmic reticulum.

Indicative content:
- Anatoxin-a binds to nicotinic acetylcholine receptors on the post-synaptic membrane (sarcolemma) of the neuromuscular junction.
- This binding opens ligand-gated sodium ion channels, leading to a rapid influx of sodium ions \(Na^+\) into the muscle cell sarcoplasm.
- This causes depolarization of the post-synaptic membrane and generates action potentials.
- Because acetylcholinesterase cannot break down anatoxin-a, the toxin remains bound, and the receptors/channels remain continuously open.
- This leads to persistent, continuous depolarization of the post-synaptic membrane.
- Initially, this continuous depolarization triggers continuous release of calcium ions \(Ca^{2+}\) from the sarcoplasmic reticulum, causing rapid, uncontrolled muscle contractions (convulsions/spasms).
- Over time, persistent depolarization prevents the resetting of voltage-gated sodium channels (depolarizing block), preventing further action potentials from being generated.
- This ultimately results in flaccid paralysis and can lead to death by respiratory failure (as the diaphragm cannot contract).

Level 1 (1-2 marks): Describes basic synaptic action or muscle consequence.
- Identifies that anatoxin-a binds to acetylcholine receptors or causes continuous stimulation.
- Mentions that the muscle will continuously contract or become paralyzed.

Level 2 (3-4 marks): Explains the cellular mechanisms of continuous depolarization or initial contraction.
- Explains that binding of anatoxin-a causes sodium channels to open, leading to depolarization because it cannot be broken down.
- Describes how continuous action potentials or calcium release lead to muscle spasms/continuous contraction.

Level 3 (5-6 marks): Provides a comprehensive, logical explanation connecting the lack of degradation to the dual phase of muscle response (spasm followed by paralysis).
- Detailed explanation of persistent depolarization due to lack of acetylcholinesterase action.
- Explains how this initially causes spasms due to continuous calcium release, but eventually leads to paralysis because voltage-gated sodium channels cannot reset to propagate new action potentials.

To understand the distribution of the plant species, other abiotic factors that affect plant growth must be measured. These include soil water content (soil moisture), soil pH, temperature of the soil or air, wind speed, salinity of the soil water, and soil mineral nutrient concentration.

Award 1 mark for each valid abiotic factor named, up to a maximum of 2 marks. Acceptable answers include: soil moisture / water content, soil pH, soil / air temperature, salinity, soil mineral concentration, wind speed, slope. Reject: light intensity (stated in question), any biotic factors (e.g., grazing, competition).

During respiration, germinating seeds absorb oxygen and release carbon dioxide. Potassium hydroxide (KOH) solution absorbs the carbon dioxide gas produced. Therefore, any reduction in gas volume inside the respirometer tube is due solely to the uptake of oxygen, allowing the rate of oxygen consumption to be measured by the movement of the liquid in the capillary tube.

MP1: Absorbs the carbon dioxide (CO2) produced by the respiring seeds [1 mark]. MP2: Ensures that any change in gas volume / pressure (which causes the liquid to move) is due solely to the uptake of oxygen (O2) [1 mark].

The distance travelled by DNA fragments in gel electrophoresis depends on the size (molecular weight or length in base pairs) of the DNA fragments because smaller fragments can navigate through the gel matrix faster. It also depends on the concentration of the agarose gel (denser gels slow down movement), the voltage applied across the gel, and the run time.

Award 1 mark for each correct factor listed, up to a maximum of 2 marks. Accept: size / length / molecular weight / number of base pairs of the DNA fragments; concentration of the agarose gel; voltage / electric current applied; composition/pH of the buffer solution.

To ensure a valid investigation of habituation, several key variables must be controlled. These include the force or strength of the mechanical stimulus (to ensure it is consistent and doesn't cause damage or fail to elicit a response), the time interval between consecutive stimuli (as varying intervals affect habituation rate), the background noise and light levels, and the temperature of the environment.

Award 1 mark for each valid controlled variable, up to a maximum of 2 marks. Accept: strength / force of the stimulus / tap (e.g. dropping a cotton bud from a fixed height); time interval between consecutive taps; temperature of the testing environment; background noise / distraction level; same species / age / size of snail; snail must be allowed to fully extend before starting.

To ensure a fair comparison of tensile strength between stinging nettle and flax fibres, physical dimensions of the fibres must be controlled. Fibres must have the same length and the same diameter / cross-sectional area, as thicker or shorter fibres are inherently stronger. Additionally, they should be prepared using the same extraction (retting) method and tested under the same environmental conditions (temperature and humidity).

MP1: Use fibres of the same length and same diameter / thickness / cross-sectional area [1 mark]. MP2: Ensure environmental conditions (e.g. temperature / humidity) are kept constant OR ensure the fibres are extracted / prepared using the same method (e.g. same retting time) [1 mark].

Cutting the beetroot into discs damages cell membranes and tonoplasts, releasing betalain pigment onto the surface of the discs. If not washed away, this pigment would dissolve into the experimental solutions and cause an overestimation of membrane leakage. Thorough washing in distilled water removes this pre-existing pigment so that any leakage measured is due solely to the temperature treatments.

MP1: To wash away pigment released from cell membranes / tonoplasts damaged during cutting of the beetroot discs [1 mark]. MP2: To ensure that any pigment measured during the experiment is due only to the effect of the temperature treatment on membrane permeability (ensuring validity) [1 mark].

Incubating bacterial cultures at or near human body temperature (37 degrees Celsius) encourages the rapid growth of potential human pathogens (disease-causing microorganisms). In school laboratories, incubation temperatures are restricted to 25 degrees Celsius or below (never above 30 degrees Celsius) to prevent this hazard and ensure safe handling of bacterial plates.

Award 1 mark for each valid reason, up to a maximum of 2 marks. MP1: To prevent the growth of potential human pathogens / harmful bacteria (that thrive at body temperature) [1 mark]. MP2: Health and safety precaution / risk assessment guidelines for school microbiology [1 mark].

Hydrochloric acid breaks down the pectins that form the middle lamella holding the plant cell walls together. This macerates the tissue, allowing the cells of the meristem to separate easily. When light pressure is applied to the coverslip, the cells can be spread out into a single layer (monolayer), allowing light to pass through and making individual chromosomes and mitotic stages visible.

MP1: To break down / digest the middle lamella / pectin that holds the cell walls together [1 mark]. MP2: To allow the cells to separate and be squashed into a single / thin layer of cells (to allow light to pass through under the microscope) [1 mark].

During the light-dependent stage of photosynthesis, water undergoes photolysis, releasing electrons. These electrons are accepted by the blue dye DCPIP instead of NADP. As the rate of the Hill reaction increases, DCPIP is reduced more rapidly. Since reduced DCPIP is colourless, the overall absorbance of the solution decreases.

1. Reference to DCPIP acting as an electron acceptor and being reduced by electrons from photolysis or light-dependent reactions (1 mark). 2. Reference to reduced DCPIP changing from blue to colourless, which causes the absorbance to decrease (1 mark).

Potassium hydroxide is normally used to absorb the carbon dioxide released during respiration. Without it, the carbon dioxide remains in the tube. Since the volume of oxygen absorbed is equal to the volume of carbon dioxide released (assuming a respiratory quotient of 1.0), the total volume and pressure of gas inside the tube do not change. Therefore, there is no pressure difference to move the liquid.

1. Description: Liquid in the capillary tube does not move / remains stationary (1 mark). 2. Explanation: Carbon dioxide produced is not absorbed, so the volume of oxygen absorbed equals the volume of carbon dioxide released, resulting in no net change in gas volume or pressure (1 mark).

First, convert the percentage concentration of the standard solution to a mass concentration: \(0.1\%\text{ w/v}\) is equivalent to \(1\text{ mg cm}^{-3}\). Second, calculate the mass of vitamin C in \(1.50\text{ cm}^3\) of the standard solution that reacts with the DCPIP: \(1.50\text{ cm}^3 \times 1\text{ mg cm}^{-3} = 1.50\text{ mg}\). Third, since \(2.40\text{ cm}^3\) of orange juice contains this same mass of vitamin C, calculate its concentration: \(1.50\text{ mg} / 2.40\text{ cm}^3 = 0.625\text{ mg cm}^{-3}\).

1. Correct calculation of the mass of vitamin C needed to decolourise the DCPIP as \(1.50\text{ mg}\) (or conversion of \(0.1\%\) to \(1\text{ mg cm}^{-3}\)) (1 mark). 2. Correct calculation of the vitamin C concentration in orange juice as \(0.625\text{ mg cm}^{-3}\) or \(0.63\text{ mg cm}^{-3}\) (1 mark).

Magnesium is a key component in the chemical structure of chlorophyll. A deficiency in magnesium ions prevents the plant from producing enough chlorophyll, reducing its ability to absorb light energy. This decreases the rate of photosynthesis, meaning less glucose and other organic molecules (such as cellulose and proteins) are made, leading to a reduction in dry mass.

1. Reference to magnesium ions being required for chlorophyll synthesis / production of light-absorbing pigments (1 mark). 2. Reference to reduced light absorption / photosynthesis leading to less production of organic molecules / glucose / biomass (1 mark).

Step 1: Calculate the total number of organisms \(N\). \(N = 12 + 28 + 10 = 50\). Step 2: Calculate \(N(N-1)\). \(N(N-1) = 50 \times 49 = 2450\). Step 3: Calculate \(\sum n(n-1)\) for all species. For Species A: \(12 \times 11 = 132\). For Species B: \(28 \times 27 = 756\). For Species C: \(10 \times 9 = 90\). \(\sum n(n-1) = 132 + 756 + 90 = 978\). Step 4: Calculate \(D\). \(D = 2450 / 978 = 2.5051\). Rounding to 2 decimal places gives 2.51.

M1: Correct calculation of \(N(N-1)\) as 2450. M2: Correct calculation of \(\sum n(n-1)\) as 978. A1: Correct index of diversity of 2.51 (Accept 2.5 or 2.51).

Step 1: Calculate resting cardiac output. \\text{Resting Cardiac Output} = 60 \\times 75 = 4500\\text{ cm}^3\\text{ min}^{-1}\. Step 2: Calculate stroke volume during exercise. \(75 \times 1.40 = 105\text{ cm}^3\). Step 3: Calculate cardiac output during exercise. \\text{Exercise Cardiac Output} = 165 \\times 105 = 17325\\text{ cm}^3\\text{ min}^{-1}\. Step 4: Calculate percentage increase. \\frac{17325 - 4500}{4500} \\times 100 = 285\\%\.

M1: Correct calculation of resting cardiac output (4500) OR exercise stroke volume (105). M2: Correct calculation of exercise cardiac output (17325). A1: Correct percentage increase of 285% (Accept 285).

Step 1: Determine the frequency of the homozygous recessive genotype \(q^2\). \(q^2 = 1 / 2500 = 0.0004\). Step 2: Calculate the frequency of the recessive allele \(q\). \(q = \sqrt{0.0004} = 0.02\). Step 3: Calculate the frequency of the dominant allele \(p\). \(p = 1 - 0.02 = 0.98\). Step 4: Calculate the frequency of heterozygous carriers \(2pq\). \(2pq = 2 \times 0.98 \times 0.02 = 0.0392\). Step 5: Calculate the total number of carriers. \(120,000 \times 0.0392 = 4704\).

M1: Correct calculation of allele frequencies (q = 0.02 and p = 0.98). M2: Correct calculation of carrier frequency (2pq = 0.0392). A1: Correct number of carriers of 4704.

Step 1: Calculate the volume of oxygen consumed. \(V = \pi \times (0.5)^2 \times 42 = 10.5\pi \approx 32.987\text{ mm}^3\). Step 2: Calculate the rate of oxygen consumption. \\text{Rate} = 32.987 / 15 \\approx 2.1991\\text{ mm}^3\\text{ min}^{-1}\. Step 3: Round to 3 significant figures, which yields 2.20.

M1: Correct calculation of total volume of oxygen consumed (32.99 or 32.987 or 10.5pi). M2: Division of volume by 15. A1: Correct rate of oxygen consumption of 2.20 (Accept 2.2).

1. Sea ice provides a substrate and habitat for the growth and attachment of sea ice algae. 2. Loss of sea ice reduces the available surface area, leading to a smaller population of algae. 3. Reduced photosynthesis means lower Gross Primary Productivity (GPP). 4. Since Net Primary Productivity (NPP) = GPP - Respiration, a decrease in GPP results in a decrease in NPP.

Mark 1: Loss of habitat/substrate for attachment/growth of sea ice algae. Mark 2: Reduced rate of photosynthesis (accept less light energy converted to chemical energy). Mark 3: Reference to NPP = GPP - R, so a decrease in GPP (or increase in respiratory loss relative to photosynthesis) decreases NPP.

1. Anaerobic respiration releases lactate and hydrogen ions (H+), which lowers the pH of muscle cells (sarcoplasm). 2. Low pH denatures or reduces the activity of enzymes involved in muscle contraction (e.g., ATPase) or glycolysis. 3. Low pH/H+ ions compete with calcium ions for binding sites on troponin. 4. This prevents tropomyosin from moving, reducing actin-myosin cross-bridge formation, leading to weaker muscle contractions (muscle fatigue).

Mark 1: Lactate/H+ accumulation causes a decrease in pH within muscle cells/sarcoplasm. Mark 2: Decreased pH affects enzyme activity/denatures enzymes (e.g., ATPase / respiratory enzymes). Mark 3: Low pH disrupts calcium ion binding to troponin, preventing actin-myosin cross-bridge formation.

1. Methyl groups are added to cytosine bases (CpG islands) in the promoter region of the Hsp70 gene. 2. This modifies the structure of the DNA, preventing the binding of transcription factors/RNA polymerase. 3. Consequently, transcription of the Hsp70 gene into mRNA is inhibited/prevented.

Mark 1: Methyl groups attach to cytosine bases / CpG islands in the promoter. Mark 2: Prevents transcription factors / RNA polymerase from binding to the promoter. Mark 3: Prevents transcription of the gene into mRNA / synthesis of mRNA.

1. During the refractory period, sodium ion channels are closed/inactive and cannot be opened. 2. This prevents depolarisation/action potentials from being generated in the region of the membrane that has just been depolarised. 3. Therefore, the action potential can only propagate forward into the region with resting (receptive) sodium channels.

Mark 1: Voltage-gated sodium channels are inactivated/closed during the refractory period. Mark 2: Prevents depolarization/action potential initiation in the membrane section behind the impulse. Mark 3: Restricts propagation to one direction (forward) along the axon.

1. Active transport requires carrier proteins (Na+/K+-ATPase) to undergo conformational/shape changes. 2. A fluid membrane allows the lipid bilayer to yield to these shape changes, facilitating transport. 3. If the membrane is too rigid/viscous, the protein is restricted, preventing efficient transport of sodium and potassium ions.

Mark 1: Active transport depends on carrier proteins changing shape/conformation. Mark 2: High membrane fluidity allows the lipid bilayer to be flexible, supporting these conformational changes. Mark 3: Rigid membranes restrict protein movement/conformational change, reducing the rate of active transport.

1. Primers are short single strands of DNA that have a specific base sequence complementary to the ends of the target DNA sequence. 2. They bind/anneal to the target DNA to provide a starting point for DNA polymerase. 3. Incorrect primers would not anneal to the target gene, meaning DNA polymerase cannot bind, and the specific gene would not be amplified.

Mark 1: Primers must have a complementary base sequence to the start/end of the target DNA. Mark 2: Primers anneal to the target DNA to provide a double-stranded region/starting point for DNA polymerase. Mark 3: Specificity of primers ensures only the target gene is amplified, avoiding amplification of non-target DNA.

1. Geographic isolation prevents gene flow/interbreeding between the sub-populations. 2. Different sub-populations experience different environmental selection pressures (or genetic drift). 3. Beneficial mutations are selected for, causing changes in allele frequencies over time. 4. The populations become genetically distinct, leading to reproductive isolation (they can no longer interbreed to produce fertile offspring).

Mark 1: Physical barrier prevents gene flow/interbreeding between populations. Mark 2: Different environmental conditions exert different selection pressures, changing allele frequencies (accept genetic drift/mutation). Mark 3: Development of reproductive isolation so they can no longer interbreed to produce fertile offspring.

1. The formula is: Cardiac Output = Stroke Volume x Heart Rate (or Stroke Volume = Cardiac Output / Heart Rate). 2. Calculation: Stroke Volume = 4.5 / 15. 3. Stroke Volume = 0.3 cm^3 (accept 0.3 ml).

Mark 1: Correct formula: Cardiac Output = Stroke Volume x Heart Rate (or rearranged). Mark 2: Correct calculation: 4.5 / 15 = 0.3. Mark 3: Correct units: cm^3 (or cm^3 beat^-1, or ml).

1. A reduction in oxidative phosphorylation means less ATP is produced aerobically. 2. To maintain the ATP required for muscle contraction, the rate of glycolysis (anaerobic respiration) must increase. 3. This results in a higher rate of pyruvate production. 4. Pyruvate is then reduced to lactate by accepting hydrogen from reduced NAD (NADH), which regenerates oxidized NAD (NAD+) to allow glycolysis to continue.

1. Less ATP produced by oxidative phosphorylation / aerobic respiration (1); 2. (Therefore) rate of glycolysis / anaerobic respiration increases (to meet ATP demand) (1); 3. Pyruvate is reduced / converted to lactate to regenerate oxidized NAD / pyruvate accepts hydrogen from NADH (1).

1. DNA hypomethylation (less methylation) of the promoter region prevents chromatin condensation or allows transcription factors and RNA polymerase to bind more easily to the promoter. 2. This increases the rate of transcription of the PGC-1alpha gene into mRNA. 3. The increased abundance of PGC-1alpha mRNA leads to higher rates of translation at ribosomes, resulting in an increased synthesis of the PGC-1alpha protein.

1. Less methylation of DNA allows transcription factors / RNA polymerase to bind (to promoter) (1); 2. (Increased) transcription of (PGC-1alpha) gene to produce mRNA (1); 3. (More) mRNA is translated (at ribosomes) into protein (1). [Do not accept: transcription of protein]

1. Myelin sheath, produced by Schwann cells, acts as an electrical insulator preventing ion movement across the axon membrane. 2. Depolarisation and action potentials can only occur at the gaps in the myelin sheath, known as the nodes of Ranvier, where sodium voltage-gated channels are concentrated. 3. This causes the action potential to jump from one node of Ranvier to the next, a process called saltatory conduction, which significantly increases conduction velocity compared to unmyelinated neurones.

1. Myelin / Schwann cells act as an electrical insulator / prevent movement of ions (1); 2. Depolarisation / action potentials only occur at the nodes of Ranvier (where sodium channels are located) (1); 3. (Impulse / action potential) jumps from node to node / saltatory conduction (which increases speed) (1). [Reject: saltatory conduction makes the impulse travel slower]