2024 Edexcel A-Level Biology B (9BI0) 模擬試題連答案詳解

Inhibitor X is a competitive inhibitor because it increases the substrate concentration required to reach half-maximal velocity (\(K_m\)) but does not affect the maximum rate of reaction (\(V_{max}\)), as high substrate concentrations can displace the inhibitor from the active site. Inhibitor Y is a non-competitive inhibitor because it reduces the overall rate of reaction (lowering \(V_{max}\)) by binding to an allosteric site, but does not alter the affinity of the unbound enzymes for the substrate (leaving \(K_m\) unchanged).

Award 1 mark for selecting option A. Reject all other options.

Antibiotic P is penicillin, which acts by inhibiting transpeptidase enzymes, preventing the cross-linking of peptidoglycan chains in the bacterial cell wall, which leads to cell lysis under osmotic pressure. Antibiotic Q is tetracycline, which inhibits translation by binding to the 30S subunit of bacterial ribosomes, preventing the binding of aminoacyl-tRNA.

Award 1 mark for selecting option A. Reject all other options.

First, calculate the total length of the mature mRNA after splicing: \(4 \times 180\text{ nucleotides} = 720\text{ nucleotides}\). Splicing removes all introns completely. Since three nucleotides code for one amino acid (a codon), the total number of codons in the mature mRNA is \(720 / 3 = 240\text{ codons}\). Since the translation is terminated by a stop codon, which does not code for any amino acid, the maximum length of the polypeptide is \(240 - 1 = 239\text{ amino acids}\).

Award 1 mark for selecting option A. Reject all other options.

One sucrose molecule yields two hexose monosaccharides (1 glucose and 1 fructose). Each hexose undergoes glycolysis to yield 2 molecules of pyruvate (4 pyruvate molecules total per sucrose). Each pyruvate undergoes the link reaction to form 1 acetyl-CoA and 1 \(\text{CO}_2\) (4 acetyl-CoA and 4 \(\text{CO}_2\) total from the link reaction). Each of the 4 acetyl-CoA molecules enters the Krebs cycle, where 2 molecules of \(\text{CO}_2\) are released per turn. Therefore, the Krebs cycle produces \(4 \times 2 = 8\) molecules of \(\text{CO}_2\) in total.

Award 1 mark for selecting option C. Reject all other options.

1. The specific solute molecule or ion binds to the complementary binding site on the carrier protein.
2. ATP binds to the carrier protein and is hydrolysed to ADP and inorganic phosphate (Pi), causing phosphorylation of the carrier protein.
3. This provides energy that triggers a conformational (shape) change in the carrier protein, moving the solute across the membrane and releasing it on the opposite side against its concentration gradient.

MP1: Solute (molecule or ion) binds to a specific / complementary binding site on the carrier protein (1)
MP2: Hydrolysis of ATP (to ADP and Pi) / phosphorylation of the carrier protein (1)
MP3: Triggers a conformational / shape change in the carrier protein, moving and releasing the solute across the membrane against its concentration gradient (1)

1. The sodium-potassium pump uses ATP to actively transport sodium ions (\(Na^+\)) out of the epithelial cell and into the blood.
2. This maintains a concentration gradient where the concentration of sodium ions is lower inside the cell than in the lumen of the ileum.
3. Sodium ions diffuse down their electrochemical gradient into the cell through a co-transport protein, bringing glucose molecules with them into the cell against their concentration gradient.

MP1: Sodium ions (\(Na^+\)) are actively transported out of the epithelial cell (into the blood) by the sodium-potassium pump using ATP (1)
MP2: This establishes / maintains a concentration gradient of sodium ions (lower inside the cell than in the lumen of the ileum) (1)
MP3: Sodium ions diffuse down their concentration gradient into the cell through a co-transport protein, carrying glucose with them against its concentration gradient (1)

1. Water is split by light energy (photolysis) in the thylakoid space to produce protons (\(H^+\)), electrons (\(e^-\)), and oxygen (\(O_2\)).
2. The electrons from water replace the excited electrons lost from the primary pigment reaction centre of photosystem II (chlorophyll a).
3. Protons accumulate inside the thylakoid lumen to establish a proton gradient used by ATP synthase, and are also used to reduce NADP to NADPH.

MP1: Photolysis of water splits water molecules into protons (\(H^+\)), electrons (\(e^-\)), and oxygen (\(O_2\)) using light energy (1)
MP2: Electrons replace those lost by excited chlorophyll / photosystem II (1)
MP3: Protons contribute to the proton gradient across the thylakoid membrane / are used to reduce NADP (to NADPH) (1)

1. Decreased light intensity reduces the rate of the light-dependent reaction, leading to a deficiency of ATP and reduced NADP (NADPH).
2. The regeneration of ribulose bisphosphate (RuBP) from glyceraldehyde 3-phosphate (GALP) requires ATP, so RuBP cannot be regenerated.
3. However, any existing RuBP continues to combine with carbon dioxide to form glycerate 3-phosphate (GP) in the carbon fixation step, leading to its depletion.

MP1: Less ATP and reduced NADP (NADPH) are produced from the light-dependent reactions (1)
MP2: Regeneration of RuBP from GALP / GP is prevented because this reaction requires ATP (1)
MP3: RuBP continues to be converted into GP as it combines with carbon dioxide (until the existing pool of RuBP is exhausted) (1)

1. Pyruvate is decarboxylated, which removes a carbon atom in the form of carbon dioxide (\(CO_2\)).
2. Pyruvate is oxidised (dehydrogenated), and the lost hydrogen atoms are transferred to NAD to produce reduced NAD (NADH).
3. The resulting two-carbon acetate group combines with Coenzyme A (CoA) to form acetyl coenzyme A (acetyl CoA).

MP1: Pyruvate is decarboxylated to produce carbon dioxide (\(CO_2\)) (1)
MP2: Pyruvate is oxidised / dehydrogenated, and NAD is reduced to form reduced NAD (NADH) (1)
MP3: The resulting acetate (2C) group combines with Coenzyme A to form acetyl Coenzyme A (acetyl CoA) (1)

1. Oxygen acts as the final (terminal) electron acceptor at the end of the electron transport chain.
2. It combines with protons (\(H^+\)) from the mitochondrial matrix and electrons from the electron transport chain to produce water: \(\frac{1}{2}\text{O}_2 + 2\text{H}^+ + 2e^- \rightarrow \text{H}_2\text{O}\).
3. Removing electrons from the final carrier molecule allows electron flow to continue down the chain, which maintains the active transport of protons into the intermembrane space, preserving the proton gradient needed for ATP synthesis via ATP synthase.

MP1: Oxygen acts as the terminal / final electron acceptor (1)
MP2: Oxygen combines with protons and electrons to form water / equation: \(\frac{1}{2}\text{O}_2 + 2\text{H}^+ + 2e^- \rightarrow \text{H}_2\text{O}\) (1)
MP3: This maintains the flow of electrons along the electron transport chain / allows the proton gradient to be maintained to drive ATP synthesis (1)

1. The values show that Woodland A has a higher biodiversity (higher species richness and evenness) than Woodland B.
2. A lower Simpson's Index (Woodland B) indicates that the community is dominated by one or a few species.
3. Consequently, the ecosystem is less stable and more vulnerable to environmental changes (such as climate change or disease), because a threat to the dominant species could severely impact the entire food web.

MP1: Woodland A has higher species diversity / biodiversity / evenness than Woodland B (or vice versa) (1)
MP2: A low index (in B) indicates that the ecosystem / community is dominated by one or a few species (1)
MP3: This makes the ecosystem less stable / more vulnerable to environmental change or disease because there are fewer alternative species to maintain community function (1)

1. Random genetic mutations within a population create new alleles, resulting in phenotypic variation.
2. An environmental selection pressure acts on the population, giving individuals with advantageous alleles a selective advantage, making them more likely to survive and reproduce.
3. These surviving individuals pass on their advantageous alleles to their offspring, which increases the frequency of these alleles in the gene pool over generations.

MP1: Mutation creates genetic variation / new alleles (1)
MP2: Selection pressures lead to differential survival and reproductive success (individuals with advantageous alleles / phenotypes are more likely to survive and reproduce) (1)
MP3: Advantageous alleles are passed on to offspring, increasing the frequency of these alleles in the gene pool over time / generations (1)

First, sodium ions (\(Na^+\)) are actively transported out of the epithelial cells into the blood by the sodium-potassium pump, maintaining a low concentration of sodium ions inside the cell. This creates a concentration gradient between the lumen of the ileum and the cell. Sodium ions then diffuse down their concentration gradient into the epithelial cell via a co-transporter protein in the apical membrane. As sodium ions bind to this co-transporter, they carry glucose molecules into the cell with them, against the glucose concentration gradient.

1. Active transport of sodium ions out of the epithelial cell (into the blood/capillary) via the sodium-potassium pump, which lowers the intracellular sodium concentration [1 mark].
2. Sodium ions bind to a co-transporter protein and diffuse down their concentration/electrochemical gradient from the lumen into the epithelial cell [1 mark].
3. Glucose is co-transported (carried simultaneously) into the cell against its concentration gradient [1 mark].

Both cell types are placed in a hypotonic environment (distilled water has a higher water potential than the cytoplasm), so water enters both cells by osmosis down a water potential gradient. Since red blood cells have no cell wall, the uptake of water causes them to swell and burst (undergo haemolysis) due to the lack of structural support. In contrast, plant palisade cells possess a rigid cell wall made of cellulose. This cell wall exerts an opposing turgor pressure, preventing further water entry and stopping the cell from bursting, leaving the plant cells turgid.

1. Water enters both cell types by osmosis down a water potential gradient (from high water potential to low water potential) [1 mark].
2. Red blood cells swell and burst / lyse because they lack a cell wall to withstand the internal hydrostatic pressure [1 mark].
3. Plant cells become turgid but do not burst because the rigid cellulose cell wall resists expansion / withstands the turgor pressure [1 mark].

If carbon dioxide is reduced, less carbon dioxide is available to combine with ribulose bisphosphate (RuBP) via the enzyme RuBisCO. Consequently, RuBP is not being converted to glycerate 3-phosphate (GP), leading to an accumulation and thus an increase in RuBP concentration. At the same time, the concentration of GP decreases because its production from RuBP and CO2 ceases, while the remaining GP continues to be reduced to glyceraldehyde 3-phosphate (GALP/TP) using ATP and reduced NADP from the light-dependent reactions.

1. RuBP concentration increases AND GP concentration decreases [1 mark].
2. RuBP increases because less of it is used up in the fixation of carbon dioxide / carboxylation reaction [1 mark].
3. GP decreases because less is formed due to lack of carbon dioxide, but it continues to be converted to GALP/triose phosphate (using ATP and reduced NADP) [1 mark].

Water plays a vital role in non-cyclic photophosphorylation through the process of photolysis. Light energy absorbed by photosystem II (PSII) causes the splitting of water molecules into protons (\(H^+\)), electrons (\(e^-\)), and oxygen (\(O_2\)). The electrons produced immediately replace the excited electrons lost from the primary pigment reaction centre of PSII. The protons accumulate within the thylakoid lumen, generating a proton gradient across the thylakoid membrane used by ATP synthase to make ATP. Oxygen is released as a metabolic by-product.

1. Water undergoes photolysis (is split by light energy) to produce protons (\(H^+\)), electrons (\(e^-\)), and oxygen (\(O_2\)) [1 mark].
2. Electrons from water replace the excited electrons lost from the chlorophyll a / reaction centre of photosystem II (PSII) [1 mark].
3. Protons accumulate inside the thylakoid lumen to help establish the electrochemical / proton gradient required for ATP synthesis (via chemiosmosis / ATP synthase) [1 mark].

Normally, protons are pumped into the intermembrane space and can only return to the mitochondrial matrix through ATP synthase, driving ATP production. DNP allows protons (\(H^+\)) to leak directly across the inner membrane back into the matrix. This destroys or drastically reduces the proton concentration gradient (proton motive force). Consequently, little to no ATP is synthesised. The potential energy stored within the electrochemical gradient is not captured to form chemical bonds in ATP and is instead lost to the surroundings as thermal energy, causing a rapid increase in body temperature.

1. Protons (\(H^+\)) diffuse/leak directly back into the mitochondrial matrix, bypassing ATP synthase [1 mark].
2. This dissipates/reduces the proton gradient / proton motive force, resulting in a severe decrease in ATP synthesis [1 mark].
3. The energy from electron transport/proton gradient is not conserved as ATP and is instead released as heat energy, increasing body temperature [1 mark].

During the oxidation reactions in the Krebs cycle, hydrogen atoms (consisting of protons and electrons) are removed from intermediates by dehydrogenase enzymes. The coenzymes NAD and FAD act as hydrogen acceptors, becoming reduced to NADH and \(\text{FADH}_2\). These reduced coenzymes then migrate to the inner mitochondrial membrane (cristae), where they transfer their electrons and protons to the electron transport chain, generating the proton gradient required for ATP synthesis in oxidative phosphorylation.

1. Act as hydrogen / electron acceptors and are reduced (to NADH and \(\text{FADH}_2\)) during dehydrogenation/oxidation reactions in the Krebs cycle [1 mark].
2. Transport protons (\(H^+\)) and high-energy electrons to the inner mitochondrial membrane / electron transport chain [1 mark].
3. Re-oxidation of these coenzymes provides the energy / proton gradient required to drive the synthesis of ATP during oxidative phosphorylation [1 mark].

First, calculate the total number of individuals, \(N\):
\(N = 10 + 15 + 5 + 20 = 50\).

Then calculate \(N(N-1)\):
\(N(N-1) = 50 \times 49 = 2450\).

Next, calculate \(n(n-1)\) for each species:
- Species W: \(10 \times 9 = 90\)
- Species X: \(15 \times 14 = 210\)
- Species Y: \(5 \times 4 = 20\)
- Species Z: \(20 \times 19 = 380\)

Sum of all \(n(n-1)\):
\(\sum n(n-1) = 90 + 210 + 20 + 380 = 700\).

Now, calculate the index:
\(D = 1 - \frac{700}{2450} = 1 - 0.2857... = 0.7142...\)

Rounded to 2 decimal places, \(D = 0.71\).

1. Correct calculation of \(N = 50\) and \(N(N-1) = 2450\) [1 mark].
2. Correct calculation of \(\sum n(n-1) = 700\) (by summing 90, 210, 20, 380) [1 mark].
3. Correct calculation of \(D = 0.71\) [1 mark].
Note: Accept only 0.71. Do not accept 0.7 or 0.714 (must be rounded to exactly 2 decimal places).

A random mutation in the grass population results in a new allele that confers tolerance to high copper concentrations in the soil. In the vicinity of the copper mine, the soil contains toxic levels of copper, which acts as a powerful environmental selection pressure. Grasses without the tolerance allele cannot survive and die before reproducing. Those individuals possessing the copper-tolerance allele are able to survive, grow, and reproduce. They pass the advantageous copper-tolerance allele to their offspring, causing the frequency of this allele to increase in the gene pool over successive generations.

1. Recognition of pre-existing genetic variation / mutation in the population that creates a copper-tolerance allele [1 mark].
2. High soil copper concentration acts as a selection pressure, causing differential survival (only plants with the tolerance allele survive and reproduce) [1 mark].
3. The survivors pass on the advantageous allele to their offspring, leading to an increase in the allele frequency in the gene pool / population over generations [1 mark].

Facilitated diffusion of glucose requires a carrier protein because glucose is a polar, hydrophilic molecule and cannot pass directly through the hydrophobic fatty acid tails of the phospholipid bilayer. The carrier protein has a specific three-dimensional tertiary structure with a binding site complementary to the shape of a glucose molecule. When glucose binds to this site, it causes a conformational (shape) change in the carrier protein. This shape change translocates the glucose molecule across the membrane, releasing it on the other side. This is a passive process that does not require ATP, occurring down a concentration gradient.

1. Reference to glucose being polar / hydrophilic / large so it cannot pass through the hydrophobic phospholipid bilayer by simple diffusion [1 mark]. 2. Carrier protein has a specific or complementary binding site that only binds to glucose [1 mark]. 3. Binding of glucose triggers a conformational change / change in shape of the protein to transport glucose across the membrane [1 mark].

Photolysis uses light energy to split water molecules into protons, electrons, and oxygen. First, the electrons produced replace the electrons lost from the primary pigment (chlorophyll a) in photosystem II when it is photoactivated by light, allowing the electron transport chain to continue. Second, the protons (hydrogen ions) accumulate in the thylakoid lumen to create a proton/electrochemical gradient across the thylakoid membrane. This proton motive force drives ATP synthesis via ATP synthase (chemiosmosis). Finally, these protons and electrons are used to reduce the coenzyme NADP to reduced NADP (NADPH), which is required in the light-independent stage (Calvin cycle) to reduce GP to GALP.

1. Light-driven splitting of water provides electrons to replace those lost from photosystem II / chlorophyll [1 mark]. 2. Provides protons (H+ ions) to establish an electrochemical gradient across the thylakoid membrane to drive ATP synthesis / chemiosmosis [1 mark]. 3. Provides protons and electrons to reduce NADP to reduced NADP (NADPH) for the light-independent stage [1 mark].

In normal oxidative phosphorylation, electrons moving through the electron transport chain release energy used to pump protons into the intermembrane space. This creates an electrochemical proton gradient. Protons must flow down this gradient through ATP synthase to drive ATP synthesis. DNP makes the inner mitochondrial membrane permeable to protons, allowing them to leak directly back into the mitochondrial matrix without passing through ATP synthase. This collapses the proton gradient, drastically reducing the rate of ATP synthesis. Because the potential energy of the electrochemical gradient is not harnessed to make ATP, it is instead dissipated and released into the cell as thermal energy (heat).

1. Protons leak across the inner mitochondrial membrane back into the matrix, bypassing ATP synthase [1 mark]. 2. The proton gradient / proton motive force is collapsed or reduced, resulting in decreased ATP synthesis [1 mark]. 3. Energy from the electron transport chain / proton gradient is dissipated and released as heat [1 mark].

A Simpson's Index of Diversity (\(D\)) ranges from 0 to 1, where values closer to 1 represent high species diversity. Therefore, Meadow B has a significantly higher species diversity (high species richness and species evenness) than Meadow A, which has low diversity. Because of this higher diversity, Meadow B is much more ecologically stable. In Meadow A, the ecosystem is likely dominated by only one or a few species, making it fragile. If an environmental change or disturbance occurs (such as a disease, pest, or drought), Meadow B is more likely to survive because some species will have adaptations to tolerate the change, whereas Meadow A is vulnerable to total collapse if its dominant species is affected.

1. Meadow B has a higher species diversity / species richness and evenness than Meadow A (or vice versa) [1 mark]. 2. Meadow B is more ecologically stable / less hostile than Meadow A (or vice versa, Meadow A is dominated by few species / unstable) [1 mark]. 3. Meadow B is more resilient to environmental change because if one species is affected, others can survive, whereas Meadow A is highly vulnerable to collapse [1 mark].

DNP is lipid-soluble and binds to protons in the intermembrane space, transporting them directly across the inner mitochondrial membrane into the matrix. This bypasses ATP synthase, which dissipates the proton motive force (the electrochemical proton gradient). Because fewer protons flow down their gradient through ATP synthase, the phosphorylation of ADP to ATP is severely decreased or stopped. However, the electron transport chain continues to operate and may even speed up as it attempts to re-establish the proton gradient. This continuous electron flow increases the oxidation of reduced NAD and FAD, requiring a higher rate of oxygen consumption as oxygen acts as the terminal electron acceptor. Since the energy stored in the proton gradient is not captured as chemical energy in ATP, it is instead released as heat, raising the cell's temperature.

Award 1 mark for each of the following points, up to a maximum of 6 marks: 1. DNP increases the permeability of the inner mitochondrial membrane to protons / allows protons to bypass ATP synthase. 2. This dissipates or reduces the proton gradient / electrochemical gradient / proton motive force. 3. Decreased flow of protons through ATP synthase (or stalked particles) results in reduced ATP synthesis. 4. The electron transport chain continues to function / rate of electron transport increases to try and restore the gradient. 5. Increased oxidation of reduced NAD / reduced FAD leads to an increased rate of oxygen consumption (as oxygen is the terminal electron acceptor). 6. Energy from the proton gradient that is not used to synthesize ATP is dissipated/released as heat. [Maximum 6 marks. Answer must show a logical, coherent line of reasoning to access higher marks, e.g. linking the disruption of the proton gradient directly to both ATP decrease and heat increase.]

Sub-lethal antibiotic concentrations act as a selective pressure, but instead of killing the entire population, they permit the survival of bacteria with minor mutations that confer partial resistance (e.g., increased efflux pump activity). These surviving bacteria replicate by binary fission, passing the resistance genes vertically to their offspring. Furthermore, sub-lethal stress can trigger the bacterial SOS response, which increases the mutation rate, potentially generating more effective resistance mutations. Horizontal gene transfer is also accelerated under stress; bacteria can share resistance genes carried on plasmids or transposons with neighboring cells through conjugation. This allows resistance to spread rapidly through the population and even between different bacterial species, increasing the overall frequency of resistance genes in the gene pool.

Award 1 mark for each of the following points, up to a maximum of 6 marks: 1. Sub-lethal concentrations act as a selection pressure (but do not kill all bacteria). 2. Bacteria with pre-existing mutations / new mutations conferring partial resistance survive and reproduce (vertical transmission). 3. Sub-lethal antibiotic exposure can increase mutation rates / trigger the SOS response, leading to more genetic variation. 4. Resistance genes (often located on plasmids or transposons) can be transferred horizontally via conjugation. 5. Conjugation allows rapid spread of resistance to other bacteria in the population (including different species). 6. This increases the frequency of the resistance alleles in the bacterial gene pool. [Maximum 6 marks. Answer must show a logical, coherent line of reasoning to access higher marks, clearly distinguishing between the roles of natural selection (survival of the fittest mutations) and horizontal gene transfer (sharing of genetic material).]

First, convert the temperature of \(20^\circ\text{C}\) to Kelvin:
\(T = 20 + 273.15 = 293.15\text{ K}\)

Next, convert the concentration from \(\text{mol dm}^{-3}\) to \(\text{mol m}^{-3}\):
\(C = 0.35\text{ mol dm}^{-3} \times 1000 = 350\text{ mol m}^{-3}\)

Use the formula to calculate \(\psi_s\):
\(\psi_s = - (1 \times 350 \times 8.31 \times 293.15) = -852,625.025\text{ Pa}\)

Convert Pascals to Megapascals by dividing by \(10^6\):
\(\psi_s = -0.8526\text{ MPa}\), which rounds to \(-0.85\text{ MPa}\).

Award 1 mark for correct calculation leading to option B.
- Correct conversion of units (temperature to K, concentration to mol m^-3).
- Correct arithmetic to get solute potential in MPa.

To produce one molecule of glucose (\(\text{C}_6\text{H}_{12}\text{O}_6\)), 6 molecules of \(\text{CO}_2\) must be fixed, requiring 6 turns of the Calvin cycle.

1. For 6 molecules of \(\text{CO}_2\) fixed, 12 molecules of GP are produced.
2. Reduction of 12 GP to 12 TP requires 12 molecules of ATP and 12 molecules of reduced NADP.
3. Out of the 12 TP produced, 2 TP leave the cycle to form 1 molecule of glucose, while 10 TP are recycled to regenerate 6 molecules of RuBP.
4. Regeneration of 6 RuBP from 10 TP requires an additional 6 molecules of ATP.

Total ATP required = \(12\text{ (reduction)} + 6\text{ (regeneration)} = 18\text{ ATP}\).
Total reduced NADP required = \(12\text{ reduced NADP}\).

Award 1 mark for correct selection of option B.
- Correctly accounts for 6 turns of the cycle to synthesize one 6C glucose.
- Correctly identifies total ATP and NADPH requirements for reduction and regeneration.

Let's evaluate each option:
- **a** is incorrect because oxygen acts as the final electron acceptor in oxidative phosphorylation (forming water), not in the Krebs cycle.
- **b** is incorrect because the ATP produced directly in the Krebs cycle is formed via substrate-level phosphorylation, not oxidative phosphorylation.
- **c** is incorrect because FADH\(_2\) is oxidized at Complex II, not Complex I, and yields less ATP per molecule than NADH.
- **d** is correct because decarboxylation (loss of \(\text{CO}_2\)) occurs in both the link reaction (pyruvate to acetyl CoA) and the Krebs cycle (isocitrate to \(\alpha\)-ketoglutarate, and \(\alpha\)-ketoglutarate to succinyl CoA), but does not occur during oxidative phosphorylation.

Award 1 mark for correct selection of option D.

First, calculate \(D\) for Habitat A:
- \(N = 55\), so \(N(N-1) = 55 \times 54 = 2970\).
- \(\sum n(n-1) = 45(44) + 5(4) + 3(2) + 2(1) = 1980 + 20 + 6 + 2 = 2008\).
- \(D_A = 1 - \frac{2008}{2970} = 1 - 0.676 = 0.324 \approx 0.32\).

Second, calculate \(D\) for Habitat B:
- \(N = 60\), so \(N(N-1) = 60 \times 59 = 3540\).
- \(\sum n(n-1) = 15(14) + 16(15) + 14(13) + 15(14) = 210 + 240 + 182 + 210 = 842\).
- \(D_B = 1 - \frac{842}{3540} = 1 - 0.238 = 0.762 \approx 0.76\).

Since a higher value of \(D\) indicates a higher species diversity (taking into account both species richness and species evenness), Habitat B has a higher biodiversity than Habitat A.

Award 1 mark for correct selection of option A.
- Correctly calculates Simpson's Index for Habitat A as 0.32.
- Correctly calculates Simpson's Index for Habitat B as 0.76.
- Identifies Habitat B as having higher biodiversity.

At the arteriole end of a capillary, the hydrostatic pressure is high (due to ventricular contraction) and is greater than the oncotic pressure (which draws fluid into the capillary by osmosis due to plasma proteins). This results in a net filtration pressure, forcing fluid out of the capillary into the surrounding tissue spaces. At the venule end, the hydrostatic pressure has dropped due to friction along the narrow capillary and the loss of fluid, making it less than the oncotic pressure. This results in net fluid movement back into the capillary.

Award 1 mark for correct selection of option A.

Active transport of sodium ions out of the epithelial cells into the blood lowers the concentration of sodium ions inside the cytoplasm of the epithelial cells. This creates a steep concentration gradient for sodium ions between the tubule lumen and the cell. Sodium ions then diffuse down their concentration gradient into the epithelial cells via a co-transport protein, carrying glucose molecules with them against their concentration gradient.

1. Active transport of sodium ions out of the cell lowers intracellular sodium ion concentration / creates a concentration gradient. (1)
2. Sodium ions enter the cell down their concentration gradient via a co-transport protein. (1)
3. This facilitates/provides energy for the movement of glucose into the cell against its concentration gradient. (1)

Under low carbon dioxide levels, the light-independent reaction (Calvin cycle) slows down, reducing the demand for NADPH. However, the plant still requires ATP for other metabolic functions and cellular maintenance. If non-cyclic photophosphorylation continued at a high rate, the pool of NADP+ would become fully reduced, leading to a bottleneck in electron transport. This would cause the formation of reactive oxygen species (ROS) that damage the thylakoid membrane. Cyclic photophosphorylation allows ATP production to continue without accumulating excess NADPH, protecting the photosynthetic apparatus from photo-oxidative damage.

1. Low CO2 levels reduce the rate of the Calvin cycle, meaning less NADPH is oxidized back to NADP+. (1)
2. Cyclic photophosphorylation produces ATP only (and no NADPH), allowing the cell to meet its ATP requirements. (1)
3. This prevents the over-reduction of electron carriers, avoiding the production of damaging reactive oxygen species (ROS) / preventing photoinhibition. (1)

Uncoupling proteins provide an alternative pathway for protons to flow down their electrochemical gradient from the intermembrane space back into the mitochondrial matrix, bypassing ATP synthase. Consequently, the proton motive force is dissipated without driving the synthesis of ATP, resulting in a dramatic decrease in ATP production. Instead, the potential energy stored in the proton gradient is released directly as thermal energy (heat), which is crucial for thermoregulation in newborns and hibernating mammals.

1. UCPs act as channels that allow protons to diffuse down their electrochemical gradient into the matrix, bypassing ATP synthase. (1)
2. This reduces the proton motive force, resulting in less ATP being synthesised. (1)
3. The potential energy of the proton gradient is dissipated/released as heat energy. (1)

A high Simpson's Index of Diversity indicates that the ecosystem has both high species richness and high species evenness. In such a diverse ecosystem, the food webs are highly complex with many interconnected trophic links. If one species experiences a population decline due to disease, predators, or environmental changes, consumers can easily switch to alternative prey species. This buffers the community against collapse. In contrast, an ecosystem with low diversity relies on a few dominant species, making it fragile and susceptible to drastic changes if any key species is lost.

1. High index value indicates high species richness and evenness / high species diversity. (1)
2. This results in complex food webs with multiple alternative food sources / ecological niches filled. (1)
3. Therefore, the ecosystem is more resilient to environmental changes / less likely to collapse if one species is affected. (1)

The scientist can distinguish between the two mechanisms by adding a respiratory inhibitor such as cyanide or azide to block ATP synthesis. Since active transport requires ATP, its rate of uptake will decrease to near zero, whereas facilitated diffusion, being a passive process, will continue unaffected. Alternatively, the scientist can monitor the internal concentration of the substance over time; active transport will accumulate the substance against its concentration gradient until the internal concentration exceeds the external concentration, whereas facilitated diffusion will stop once net dynamic equilibrium is reached.

1. Measure uptake in the presence of a metabolic/respiratory inhibitor (e.g., cyanide) or under anaerobic conditions. (1)
2. Active transport will be inhibited/stopped because it requires ATP, whereas facilitated diffusion will be unaffected because it is passive. (1)
3. Monitor accumulation of the substance; active transport can move the substance against its concentration gradient (internal concentration > external), whereas facilitated diffusion only moves it down the gradient until equilibrium. (1)

Different photosynthetic pigments have different physical properties, namely their solubility in the mobile phase (the running solvent) and their affinity for/adsorption to the stationary phase (the chromatography paper). A pigment that is highly soluble in the solvent and has low affinity for the paper will travel further and faster up the paper. Conversely, a pigment that is less soluble in the solvent and has a higher affinity for the cellulose fibers of the paper will adsorb more strongly and travel a shorter distance. This results in distinct Rf values for each pigment.

1. Pigments have different solubilities in the solvent/mobile phase. (1)
2. Pigments have different affinities for/adsorption to the paper/stationary phase. (1)
3. Pigments with higher solubility and lower affinity travel further up the paper, resulting in different Rf values. (1)

An RQ value of 0.7 indicates that the germinating seed is primarily using lipids (fats or oils) as its respiratory substrate. Lipids contain a higher ratio of hydrogen to carbon atoms compared to carbohydrates, meaning they are highly reduced molecules. Consequently, more oxygen is required to completely oxidize them relative to the amount of carbon dioxide produced during respiration. To metabolise lipids, they are first hydrolysed into glycerol and fatty acids. Glycerol is converted into a glycolysis intermediate, while fatty acids undergo beta-oxidation in the matrix of the mitochondria to form acetyl CoA, which directly enters the Krebs cycle.

1. An RQ of 0.7 indicates that lipids/fats/fatty acids are the primary respiratory substrate. (1)
2. Lipids have a higher hydrogen-to-carbon ratio / are more reduced, requiring more oxygen for complete oxidation relative to CO2 produced. (1)
3. Lipids are hydrolysed to glycerol (enters glycolysis) and fatty acids (undergo beta-oxidation to form acetyl CoA for the Krebs cycle). (1)

Geographic isolation physically separates a rodent population into two or more subpopulations, preventing any gene flow between them (reproductive isolation). Over many generations, the separated subpopulations will experience different environmental conditions, such as different climates, food sources, or predators. These distinct environments exert different selection pressures, leading to the natural selection of different advantageous alleles in each population. Random mutations and genetic drift will also occur independently in each group. Over time, these genetic changes accumulate until the populations become genetically distinct, preventing them from interbreeding to produce fertile offspring even if reunited.

1. Geographic isolation prevents gene flow between the sub-populations (reproductive isolation). (1)
2. Different environmental conditions exert different selection pressures, leading to natural selection of different advantageous alleles / independent mutations/genetic drift. (1)
3. Genetic divergence occurs over time until the populations can no longer interbreed to produce fertile offspring (speciation). (1)

Sodium-potassium pump proteins actively transport sodium ions (\(\text{Na}^+\)) out of the epithelial cells and into the blood. This maintains a lower concentration of sodium ions inside the cell cytoplasm than in the intestinal lumen. As a result, sodium ions diffuse down their electrochemical gradient into the epithelial cell through a sodium-glucose co-transporter (symport) protein. This movement provides the energy to transport glucose into the cell against its concentration gradient.

1. MP1: Sodium ions are actively transported out of the epithelial cell (into the blood/interstitial fluid) by the sodium-potassium pump / active transport requiring ATP. [1 mark]
2. MP2: This lowers the concentration of sodium ions inside the cell, establishing a concentration gradient between the lumen and the cell. [1 mark]
3. MP3: Sodium ions diffuse into the cell (from the lumen) down their concentration gradient via a co-transporter protein, carrying glucose into the cell against its concentration gradient. [1 mark]

Under low light intensity, the light-dependent reactions of photosynthesis slow down, resulting in a reduced production of ATP and reduced NADP (\(\text{NADPH}\)). These products are required to convert glycerate 3-phosphate (GP) to triose phosphate (TP), and ATP is further required to regenerate RuBP from TP. Without ATP and reduced NADP, RuBP cannot be regenerated, while any existing RuBP continues to fix carbon dioxide to form GP, leading to a decline in RuBP levels.

1. MP1: Less light results in less ATP and reduced NADP produced during the light-dependent stage. [1 mark]
2. MP2: (Therefore) less glycerate 3-phosphate (GP) is converted to triose phosphate (TP) / less TP is available to regenerate RuBP (which also requires ATP). [1 mark]
3. MP3: RuBP continues to combine with carbon dioxide to form GP (catalyzed by RuBisCO), resulting in a net decrease in RuBP concentration. [1 mark]

The accumulation of protons in the intermembrane space normally creates an electrochemical gradient that drives ATP synthesis as protons flow through ATP synthase. DNP allows these protons to bypass ATP synthase and leak directly back into the mitochondrial matrix. Since the gradient is dissipated without driving the mechanical rotation of ATP synthase, the potential energy is released as heat instead of being captured in the bonds of ATP, raising the body temperature.

1. MP1: Protons (\(\text{H}^+\)) leak/diffuse back into the mitochondrial matrix bypassing ATP synthase / uncoupling the electron transport chain from phosphorylation. [1 mark]
2. MP2: Less/no ATP is produced by chemiosmosis / ATP synthase. [1 mark]
3. MP3: The potential energy / electrochemical gradient energy is released / dissipated as heat energy (increasing body temperature). [1 mark]

Species richness simply counts the total number of species present in a habitat. However, it does not distinguish between a woodland dominated by one species where all other species are rare, and a woodland where all species are equally abundant. Simpson’s Index of Diversity takes into account both the number of species (richness) and the relative abundance of each species (evenness), providing a more realistic assessment of community stability and biodiversity.

1. MP1: Species richness only measures the number of different species present, but ignores their relative abundance / species evenness. [1 mark]
2. MP2: Simpson's Index takes into account both species richness and species evenness (the relative abundance of individuals within each species). [1 mark]
3. MP3: A high index indicates a more stable ecosystem (less dominated by a single species) / a low index indicates a highly vulnerable habitat dominated by one or two species. [1 mark]

Once in the mitochondrial matrix, pyruvate (a 3-carbon compound) undergoes decarboxylation, releasing a molecule of carbon dioxide. It is then oxidized by losing hydrogen atoms (dehydrogenation), which are accepted by \(\text{NAD}^+\) to form reduced NAD (\(\text{NADH} + \text{H}^+\)). This produces a 2-carbon acetate group, which combines with coenzyme A (CoA) to form acetyl coenzyme A.

1. MP1: Pyruvate is decarboxylated / carbon dioxide (\(\text{CO}_2\)) is removed. [1 mark]
2. MP2: Pyruvate is oxidized / hydrogen is removed, which reduces \(\text{NAD}^+\) to reduced \(\text{NAD}\) (\(\text{NADH}\)). [1 mark]
3. MP3: The resulting acetate group combines with coenzyme A to form acetyl coenzyme A. [1 mark]

Aquaporins facilitate the movement of water using a specialized 3D channel structure. The channel is lined with hydrophilic amino acid residues, allowing polar water molecules to form temporary hydrogen bonds and pass through in a single-file line. Selective transport is maintained by a very narrow pore constriction that physically excludes larger molecules (size exclusion) and the presence of positively charged amino acids (like arginine) that electrostatically repel protons and other cations.

1. MP1: The channel has a hydrophilic lining / polar amino acids that interact with water molecules (allowing them to pass in single file). [1 mark]
2. MP2: The narrow diameter / constriction of the pore physically blocks larger molecules (size exclusion). [1 mark]
3. MP3: Positively charged amino acids (e.g. arginine) within the pore repel cations / protons (\(\text{H}^+ / \text{H}_3\text{O}^+\)), preventing their passage. [1 mark]

During thin-layer chromatography, pigments are dissolved in a mobile phase (solvent) and pass over a stationary phase (TLC plate). Pigments that are highly soluble in the solvent and have a low affinity/adsorption to the stationary phase will migrate the furthest up the plate. To identify these pigments, the student must calculate the Rf value of each spot (\(\text{distance moved by pigment} \div \text{distance moved by solvent front}\)) and compare these values to established literature standards obtained under identical conditions.

1. MP1: Pigments are separated based on their solubility in the mobile phase/solvent AND their adsorption/affinity to the stationary phase. [1 mark]
2. MP2: More soluble pigments (or those with lower affinity/adsorption to the stationary phase) travel further up the plate. [1 mark]
3. MP3: Calculate the Rf value (\(\text{distance moved by pigment} \div \text{distance moved by solvent front}\)) and compare it with known reference Rf values (under identical conditions). [1 mark]

A low heterozygosity index indicates that individuals in the population are highly homozygous, reflecting very low genetic diversity within the gene pool. If a new pathogen or disease enters the habitat, the lack of genetic variation means it is highly unlikely that any individuals possess an allele that confers resistance to this specific pathogen. Consequently, almost all individuals are highly susceptible, which could result in high mortality rates, reproductive failure, and the possible extinction of the entire population.

1. MP1: A low heterozygosity index indicates a small gene pool / low genetic variation / high proportion of homozygous loci. [1 mark]
2. MP2: There is a lower probability that any individuals possess an allele that confers resistance to the new pathogen/disease. [1 mark]
3. MP3: If the disease strikes, most/all individuals may be susceptible, leading to widespread mortality / a lack of individuals surviving to reproduce and adapt. [1 mark]

First, calculate the Simpson's Index of Diversity using the provided numbers: \(d = 1 - \frac{3120}{120 \times 119}\). Since \(120 \times 119 = 14280\), we have \(d = 1 - \frac{3120}{14280} = 1 - 0.2185 = 0.7815\). To two decimal places, this is 0.78. A value close to 1 indicates high species richness and evenness (high biodiversity). This means the ecosystem is highly stable and resilient to environmental changes, as there are alternative food sources and complex food webs.

MP1: Correct calculation of Simpson's Index of Diversity as 0.78 (accept 0.78 or 0.782) [1 mark]. MP2: Explains that a value close to 1 indicates high biodiversity / species richness and evenness [1 mark]. MP3: Explains that higher biodiversity increases ecosystem stability / resilience to environmental change (due to alternative food sources or complex food webs) [1 mark].

RuBisCO is responsible for catalyzing the carboxylation reaction, where carbon dioxide is added to the 5-carbon compound ribulose bisphosphate (RuBP). This reaction forms an unstable 6-carbon intermediate compound. Because it is highly unstable, it immediately splits into two 3-carbon molecules of glycerate 3-phosphate (GP).

MP1: RuBisCO catalyzes the reaction of carbon dioxide with ribulose bisphosphate (RuBP) / 5C compound [1 mark]. MP2: This forms an unstable 6-carbon intermediate [1 mark]. MP3: The 6C compound splits into two molecules of glycerate 3-phosphate (GP) / 3C compound [1 mark].

During glycolysis, the link reaction, and the Krebs cycle, NAD and FAD act as hydrogen carriers and are reduced (forming reduced NAD and reduced FAD). They transport these hydrogen atoms (protons and high-energy electrons) to the electron transport chain located on the inner mitochondrial membrane (cristae). The oxidation of these coenzymes releases electrons that pass along the transport chain, driving the pumping of protons and creating a proton gradient that powers ATP synthase to synthesize ATP.

MP1: NAD and FAD are reduced (accept hydrogen atoms / protons and electrons) during glycolysis, the link reaction, or the Krebs cycle [1 mark]. MP2: Reduced NAD / reduced FAD transport electrons and protons to the electron transport chain (on the inner mitochondrial membrane) [1 mark]. MP3: The transfer of electrons / proton movement through ATP synthase (chemiosmosis) drives the synthesis of ATP from ADP and inorganic phosphate [1 mark].

The relationship between GPP, NPP, and respiration is given by the formula: \(\text{NPP} = \text{GPP} - R\). Substituting the values: \(\text{NPP} = 24\,800 - 14\,200 = 10\,600\text{ kJ m}^{-2}\text{ yr}^{-1}\). NPP is always less than GPP because plants must use a portion of the organic compounds produced during photosynthesis for their own respiration to release energy for metabolic activities, such as active transport and cell division, releasing this energy as heat.

MP1: Calculates NPP correctly as \(10\,600\text{ kJ m}^{-2}\text{ yr}^{-1}\) (accept 10600) [1 mark]. MP2: Explains that GPP is the total chemical energy fixed by photosynthesis, but some of this energy is lost/used up during autotrophic respiration (R) [1 mark]. MP3: Explains that NPP is the remaining chemical energy stored as plant biomass, which is available to consumers/decomposers [1 mark].

When an action potential arrives, it depolarises the presynaptic membrane. This depolarisation causes voltage-gated calcium channels to open, allowing calcium ions to diffuse into the presynaptic knob. The influx of calcium ions triggers synaptic vesicles to fuse with the presynaptic membrane, releasing the neurotransmitter acetylcholine into the synaptic cleft via exocytosis. Acetylcholine diffuses across the cleft and binds to specific receptor proteins on ligand-gated sodium channels of the postsynaptic membrane, opening these channels, allowing sodium ions to enter and depolarise the postsynaptic membrane to generate a new action potential.

MP1: Depolarisation of the presynaptic membrane opens voltage-gated calcium channels, causing calcium ions (\(\text{Ca}^{2+}\)) to enter the presynaptic neurone [1 mark]. MP2: Calcium ions trigger the fusion of synaptic vesicles with the presynaptic membrane, releasing acetylcholine into the synaptic cleft by exocytosis [1 mark]. MP3: Acetylcholine diffuses across the synaptic cleft and binds to receptors on the postsynaptic membrane, opening sodium (\(\text{Na}^+\)) channels, causing depolarisation / action potential [1 mark].

A random mutation creates genetic variation in the grass population, resulting in some individuals having an allele for copper tolerance. On soils contaminated with heavy metals, copper toxicity acts as a selective pressure. Non-tolerant plants die, while copper-tolerant plants have a selective advantage, surviving to maturity and successfully reproducing. They pass on the advantageous copper-tolerance allele to their offspring, causing the frequency of this allele to increase in the population over generations.

MP1: Genetic variation exists due to random mutation, producing a copper-tolerance allele [1 mark]. MP2: Contaminated soil acts as a selection pressure, giving copper-tolerant plants a selective advantage so they survive and reproduce (while non-tolerant plants die) [1 mark]. MP3: The surviving plants pass on the advantageous allele to offspring, leading to an increase in the frequency of the copper-tolerance allele in the gene pool / population over time [1 mark].

1. When the concentration of \(\text{CO}_2\) is suddenly reduced:
- The rate of carbon fixation decreases because there is less \(\text{CO}_2\) to react with RuBP.
- Therefore, the concentration of RuBP increases because it continues to be regenerated from triose phosphate (TP) but is not being used up to fix \(\text{CO}_2\).
- The concentration of GP decreases rapidly because less GP is produced from carbon fixation, while the existing GP continues to be converted into TP (and then RuBP).

2. When the light source is switched off:
- The light-dependent reactions stop, which prevents the production of ATP and reduced NADP (NADPH).
- The conversion of GP to TP requires both ATP and reduced NADP, so this reaction stops. This causes GP to accumulate, so its concentration increases.
- The regeneration of RuBP from TP requires ATP. Without ATP, RuBP cannot be regenerated, so its concentration falls.

3. Evidence for a cyclic process:
- The reciprocal relationship (when one goes up, the other goes down under different limiting factors) shows they are metabolically linked.
- The fact that RuBP is regenerated from the products of GP conversion shows that the starting material of carbon fixation is continuously reformed, completing a cyclic pathway (the Calvin cycle).

Mark scheme / Guidance:

Level 1 (1–2 marks):
- Describes the basic change in either RuBP or GP in response to one of the changes (e.g., GP decreases when \(\text{CO}_2\) is reduced, or GP increases when light is turned off).
- Explanation of the role of carbon dioxide or light is limited or contains inaccuracies.

Level 2 (3–4 marks):
- Explains the biochemical changes in both RuBP and GP for one of the environmental shifts (either \(\text{CO}_2\) reduction or light removal) with clear links to carbon fixation or ATP/reduced NADP availability.
- There is some attempt to explain the other shift, or a basic link is made to the cyclic nature of the pathway.

Level 3 (5–6 marks):
- Provides a comprehensive, detailed biochemical explanation for the changes in both RuBP and GP under both conditions (\(\text{CO}_2\) reduction and light removal).
- Clearly explains how these reciprocal changes provide evidence for a cyclic process (specifically highlighting that RuBP is regenerated from GP/TP intermediates, meaning it acts as both a reactant and a product in a self-sustaining cycle).

Key scientific points to look for:
- Reduction in \(\text{CO}_2\) results in RuBP accumulation (no fixation) and GP decrease (used up to make TP/RuBP).
- Turning off light results in GP accumulation (no ATP/reduced NADP to reduce it to TP) and RuBP decrease (no ATP to regenerate it from TP).
- Reciprocal changes demonstrate they are intermediates in the same pathway.
- Regeneration of RuBP from products of GP/TP proves the pathway is a cycle.

Oligomycin A:
- ATP Synthesis: The rate of ATP synthesis decreases to near zero. This is because oligomycin A directly blocks the proton channel of ATP synthase, preventing protons from flowing down their electrochemical gradient into the matrix to power the phosphorylation of ADP to ATP.
- Oxygen Consumption: The rate of oxygen consumption decreases significantly. When proton flow through ATP synthase is blocked, protons accumulate in the intermembrane space, creating an extremely steep proton gradient. The work required to pump more protons against this highly concentrated gradient becomes too high for the electron transport chain (ETC). As a result, electron flow along the ETC slows down or stops, reducing the rate at which oxygen acts as the terminal electron acceptor and is reduced to water.

DNP (Dinitrophenol):
- ATP Synthesis: The rate of ATP synthesis decreases to near zero. Because DNP carries protons directly across the inner mitochondrial membrane into the matrix, it dissipates (destroys) the proton gradient (proton motive force). Without this gradient, there is no driving force to power ATP synthase.
- Oxygen Consumption: The rate of oxygen consumption increases significantly. Since the proton gradient is constantly dissipated by DNP, the ETC can pump protons into the intermembrane space without facing any resistance (there is no steep gradient opposing proton pumping). The cell attempts to restore the gradient by running the ETC at its maximum possible rate. Therefore, electron transport is highly active, and oxygen is consumed at a very high rate, though the energy is lost entirely as heat.

Mark scheme / Guidance:

Level 1 (1–2 marks):
- Identifies that both oligomycin A and DNP result in a decrease in ATP synthesis.
- Simple statement about the change in oxygen consumption for either drug, with limited explanation.

Level 2 (3–4 marks):
- Explains the effect of oligomycin A on both ATP synthesis and oxygen consumption, linking the lack of proton flow through ATP synthase to the build-up of the proton gradient and subsequent stall of the electron transport chain.
- OR explains the effect of DNP on both ATP synthesis and oxygen consumption, linking the dissipation of the proton gradient to the uncoupled, rapid operation of the electron transport chain.

Level 3 (5–6 marks):
- Provides a comprehensive, detailed explanation for both oligomycin A and DNP.
- For oligomycin A: Explains how blocking ATP synthase causes proton accumulation in the intermembrane space, increasing the gradient, which inhibits the ETC and thus decreases oxygen consumption.
- For DNP: Explains how permeabilising the membrane to protons dissipates the electrochemical gradient, which stops ATP synthesis but removes resistance to proton pumping, leading to a maximal rate of electron transport and increased oxygen consumption.

To find the rate of diffusion:
1. Submerge the pink agar cubes (which contain NaOH and phenolphthalein) into dilute hydrochloric acid for a fixed, measured duration (e.g., 5 minutes or 300 seconds).
2. Remove the cubes, blot them dry with a paper towel, and carefully slice them in half using a scalpel to expose the cross-section.
3. Measure the width, length, and height of the remaining pink core using a digital caliper or ruler.
4. Calculate the volume of this pink core: \(V_{\text{core}} = \text{length} \times \text{width} \times \text{height}\).
5. Calculate the volume of the clear region (where acid has diffused and neutralized the NaOH): \(V_{\text{diffused}} = V_{\text{total}} - V_{\text{core}}\).
6. Convert the volume from \(\text{cm}^3\) to \(\text{mm}^3\) by multiplying by 1000.
7. Calculate the rate of diffusion: \(\text{Rate} = \frac{V_{\text{diffused}}}{\text{time in seconds}}\).

MP1: Submerge agar cubes in acid for a standardized length of time (e.g. 5 minutes/300 seconds) and blot dry.
MP2: Slice cubes in half using a scalpel to expose the internal colored core.
MP3: Measure the dimensions of the remaining pink core using a ruler/calipers and calculate its volume.
MP4: Subtract the core volume from the total volume to find the volume of the diffused region, convert to \(\text{mm}^3\), and divide by the time in seconds.

To determine the isotonic point, the student must plot a graph showing the percentage change in mass on the y-axis against the concentration of sucrose on the x-axis. A line of best fit is drawn through the data points. The point where this line intersects the x-axis (y = 0) represents the concentration where there is no net movement of water into or out of the cells by osmosis, which is the isotonic concentration.

Using percentage change in mass is essential because it is impossible to cut all potato cylinders to exactly the same initial mass. If absolute change in mass were used, cylinders with a larger initial mass would show larger absolute mass changes, making comparisons unfair. Percentage change standardizes the data, controlling for this variable and allowing a valid comparison.

MP1: Plot a calibration curve / line graph with percentage change in mass on the y-axis and sucrose concentration on the x-axis.
MP2: Draw a line of best fit and locate the concentration where the line intersects the x-axis (where percentage change in mass is zero).
MP3: Explain that initial masses of the potato cylinders were not identical / varied.
MP4: State that calculating percentage change standardizes the data, allowing a valid comparison between the samples.

There are several physiological and physical reasons why the volume of gas collected does not perfectly match the rate of photosynthesis:
1. Respiration: Plant cells respire continuously. Some of the oxygen produced during photosynthesis is immediately consumed by the plant's mitochondria during aerobic respiration, meaning less gas is released.
2. Dissolution: Oxygen is soluble in water. Some of the oxygen gas produced remains dissolved in the surrounding water rather than forming gas bubbles that can be collected.
3. Gas composition: Some of the gas in the bubble may be nitrogen or other dissolved gases that have diffused out of solution into the bubble, overestimating the oxygen produced.

To minimize the respiration limitation: Keep the plant in the dark for a set time, measure the volume of oxygen consumed (respiration rate), and add this value to the volume of gas collected in the light to get the true (gross) rate of photosynthesis.

MP1: (Reason 1) Some of the oxygen produced is immediately used by the plant for aerobic respiration.
MP2: (Reason 2) Some of the oxygen remains dissolved in the water/solution and does not form bubbles, or other dissolved gases (e.g., nitrogen) diffuse into the bubble.
MP3: (Minimization) Measure oxygen consumption in the dark to determine the respiration rate.
MP4: (Minimization continued) Add the respiration rate to the measured rate in the light to calculate the gross rate of photosynthesis. (Accept: Pre-saturate the water with oxygen to prevent further dissolution of the produced gas).

The \(R_{\text{f}}\) value is calculated by dividing the distance traveled by the center of the pigment spot by the distance traveled by the solvent front, both measured from the origin (pencil start line):
\(R_{\text{f}} = \frac{\text{distance moved by pigment}}{\text{distance moved by solvent front}}\)

Essential practical precautions include:
1. Drawing the start line in pencil: Ink from a pen contains pigments that would dissolve in the chromatography solvent and separate alongside the plant pigments, obscuring the results.
2. Applying a concentrated but small spot: The pigment extract should be applied using a capillary tube, letting it dry, and repeating. Keeping the spot small prevents the pigment bands from smearing or overlapping during development.
3. Keeping the solvent level below the start line: If the solvent is deeper than the start line, the pigments will dissolve directly into the solvent reservoir at the bottom of the beaker rather than traveling up the plate.

MP1: State the formula: \(R_{\text{f}} = \frac{\text{distance moved by pigment}}{\text{distance moved by solvent front}}\), both measured from the start line.
MP2: Draw the start line in pencil because pen ink is soluble in chromatography solvent and would separate, interfering with the chromatogram.
MP3: Apply the pigment extract as a small, concentrated spot to prevent the spots from spreading sideways/overlapping/smearing.
MP4: Ensure the solvent level in the development chamber is below the pencil line so the pigments do not wash off/dissolve directly into the solvent reservoir.

During aerobic respiration, the germinating peas consume oxygen gas and release carbon dioxide gas.
1. Role of KOH: The potassium hydroxide solution absorbs the carbon dioxide gas as soon as it is produced. Consequently, any reduction in gas volume and pressure inside the sealed tube is entirely due to the uptake of oxygen gas. This pressure drop causes the colored liquid in the capillary tube to move towards the chamber containing the peas.
2. Calculation of rate:
- Measure the distance (\(d\)) moved by the liquid in a given time (\(t\) in minutes).
- Measure the internal diameter of the capillary tube to find the radius (\(r\)).
- Calculate the volume of oxygen consumed using the cylinder formula: \(V = \pi r^2 d\) (converting units to \(\text{cm}^3\)).
- Record the mass of the peas (\(m\) in grams).
- Calculate the rate of respiration: \(\text{Rate} = \frac{V}{t \times m}\) in \(\text{cm}^3\text{ min}^{-1}\text{ g}^{-1}\).

MP1: State that potassium hydroxide (KOH) absorbs carbon dioxide (\(\text{CO}_2\)) produced by respiration.
MP2: Explain that this ensures any change in volume/pressure of gas (and thus liquid movement) is solely due to oxygen consumption.
MP3: Measure the distance (\(d\)) moved by the colored liquid over a recorded time interval.
MP4: Calculate volume using \(\pi r^2 d\) (where \(r\) is capillary radius) and divide by time (minutes) and mass of the germinating peas (grams).

The liquid paraffin layer acts as a physical barrier. Because paraffin is less dense than water and immiscible with it, it sits on top of the yeast-glucose mixture, preventing oxygen from the air from dissolving into the solution. This deprives the yeast of oxygen, forcing it to switch to anaerobic respiration.

To confirm anaerobic conditions, the student can add a redox indicator such as methylene blue to the yeast-glucose mixture before layering with paraffin. Methylene blue is blue when oxidized (in the presence of oxygen) and turns colorless when reduced (once oxygen is fully depleted by initial aerobic respiration of yeast). When the solution turns completely colorless, anaerobic conditions are confirmed.

MP1: State that the liquid paraffin acts as a barrier that prevents oxygen from the air from dissolving into the yeast-glucose mixture.
MP2: Explain that this forces the yeast to respire anaerobically.
MP3: Add methylene blue (or Janus Green B / resazurin) indicator to the yeast suspension.
MP4: Confirm that the solution has turned from blue to colorless, indicating that oxygen has been fully depleted and anaerobic conditions are established.

To collect unbiased data:
1. Lay out two tape measures at right angles to create a grid system over the meadow.
2. Use a random number generator to select coordinates within the grid to place the quadrats, avoiding subjective bias.
3. Count the number of individuals of each plant species (or estimate percentage cover) within each quadrat, repeating this for a large number of quadrats (at least 20) to ensure a representative sample.

To calculate Simpson's Index of Diversity (\(D\)):
- Sum the total number of individuals of all species to find \(N\).
- For each individual species, find the total number of organisms \(n\) and calculate the ratio \(\frac{n}{N}\).
- Square this ratio for each species, sum the squared ratios, and subtract this sum from 1: \(D = 1 - \sum \left(\frac{n}{N}\right)^2\) (or use \(D = 1 - \frac{\sum n(n-1)}{N(N-1)}\)).

MP1: Lay out two perpendicular tape measures to form a coordinate grid over the sample area.
MP2: Use a random number generator to determine coordinates for quadrat placement to prevent bias.
MP3: Record the count / abundance of each plant species in a large, representative number of quadrats (e.g. 15-20).
MP4: Calculate Simpson's Index of Diversity using the formula \(D = 1 - \sum \left(\frac{n}{N}\right)^2\) or \(D = 1 - \frac{\sum n(n-1)}{N(N-1)}\), defining \(n\) as the number of individuals of a species and \(N\) as the total number of individuals of all species.

1. Lay out a tape measure perpendicular to the footpath, extending into the meadow to serve as a transect line.
2. Place quadrats systematically at regular, fixed intervals along the tape measure (e.g., every 1 or 2 meters).
3. In each quadrat, record the percentage cover or density of the target plant species and record the distance of the quadrat from the footpath.
4. Repeat this process by laying down multiple parallel transects starting at different points along the footpath to obtain reliable, averaged data.
5. To analyze the relationship, use Spearman's rank correlation coefficient (test) because the data consists of two continuous variables (distance and abundance) and the student is testing for a significant correlation/association between them.

MP1: Lay out a tape measure perpendicular to the footpath to act as a transect line.
MP2: Place quadrats at regular, systematic intervals along the tape (e.g., every 1 or 2 meters).
MP3: Record the percentage cover / density of the plant species in each quadrat alongside the exact distance from the footpath, repeating along multiple parallel transects.
MP4: Name Spearman's rank correlation coefficient as the appropriate statistical test to analyze the association between distance and abundance.

To find the isotonic concentration, the student must first plot a graph of the percentage change in mass of the potato cylinders on the y-axis against the concentration of sucrose on the x-axis. By drawing a line of best fit, they can identify the point where the line crosses the x-axis (where the percentage change in mass is \(0\%\)). At this point, there is no net movement of water, indicating the solution is isotonic to the tissue.

To ensure validity:
1. The temperature must be controlled (e.g., using a temperature-controlled water bath) because temperature affects the kinetic energy of water molecules and membrane permeability, which would otherwise alter the rate of osmosis.
2. The potato cylinders must be blotted dry with a paper towel before weighing at the end of the experiment to remove excess surface liquid, which would otherwise artificially increase the measured final mass.

1. Plot a graph of percentage change in mass against sucrose concentration (1)
2. Identify the concentration where the line of best fit crosses the x-axis / percentage change in mass is \(0\%\) (1)
3. Control temperature using a water bath because temperature affects rate of diffusion / kinetic energy of molecules / membrane permeability (1)
4. Blot cylinders dry before weighing to remove excess surface solution that would affect final mass (1)

Beetroot cell vacuoles contain a red pigment called betalain. A blue filter is selected because blue is the complementary color to red, meaning the red pigment strongly absorbs light of this wavelength (around 470 nm), maximizing the sensitivity of the absorbance readings.

To calibrate the colorimeter, the student must use a cuvette containing distilled water (a blank). They place this in the colorimeter and adjust the instrument to read zero absorbance (or 100% transmission) so that any subsequent absorbance readings are solely due to the released betalain pigment.

1. Beetroot cells contain a red pigment called betalain (1)
2. A blue filter is used as it is the complementary color / wavelength of light most strongly absorbed by the red pigment (1)
3. Calibrate using a cuvette of distilled water / a blank (1)
4. Set the colorimeter reading to zero absorbance / 100% transmission (1)

To calculate the \(R_f\) value, the student must measure the distance from the point of origin (where the pigment extract was applied) to the center of the resolved pigment spot. They then measure the distance from the origin to the solvent front (the furthest point reached by the solvent). The \(R_f\) value is calculated by dividing the distance moved by the pigment spot by the distance moved by the solvent front.

TLC is preferred over paper chromatography because the stationary phase in TLC (typically silica gel) allows for sharper, more distinct separation of pigments with less lateral spreading (diffusion) of the spots. Additionally, the separation occurs much more rapidly.

1. Measure the distance from the origin to the center of the pigment spot (1)
2. Measure the distance from the origin to the solvent front (1)
3. Divide the distance moved by the pigment by the distance moved by the solvent front (1)
4. TLC provides better/sharper separation / less diffusion of spots / is a faster process (1)

During the light-dependent stage of photosynthesis, water is split (photolysis), releasing electrons. In intact chloroplasts, these electrons are normally accepted by NADP. In this investigation, DCPIP acts as an artificial electron acceptor, intercepting these electrons. When DCPIP accepts electrons (is reduced), it changes color from blue to colourless.

To ensure validity, variables that must be kept constant include:
1. Temperature, which must be controlled (e.g., using a cold-water shield or water bath) to keep the kinetic energy and enzyme activity (such as the oxygen-evolving complex) constant.
2. The concentration and volume of the chloroplast suspension, ensuring that the number of active reaction centers is consistent across trials.

1. DCPIP acts as an artificial electron acceptor / replaces NADP (1)
2. It is reduced by electrons from the light-dependent reaction / photolysis of water, causing a color change from blue to colourless (1)
3. Control temperature (using a water bath/glass shield) because temperature affects enzyme activity / rate of reaction (1)
4. Control chloroplast concentration/volume OR pH of the buffer solution (1)

Potassium hydroxide (KOH) absorbs the carbon dioxide (\(\text{CO}_2\)) produced by the respiring germinating seeds. This ensures that any change in the volume of gas inside the chamber is solely due to the uptake of oxygen (\(\text{O}_2\)) by the seeds.

To calculate the rate of oxygen consumption:
1. Measure the distance (\(d\)) moved by the colored liquid drop in the capillary tube over a set period of time (\(t\)).
2. Calculate the volume of oxygen consumed using the formula for the volume of a cylinder: \(\pi r^2 d\), where \(r\) is the internal radius of the capillary tube.
3. Divide this volume by the mass of the germinating seeds and by the time (\(t\)) to express the rate in the required units.

1. Potassium hydroxide absorbs the carbon dioxide released during respiration (1)
2. Measure the distance moved by the colored liquid bubble in the capillary tube in a set time (1)
3. Calculate the volume of oxygen using \(\pi r^2 d\) (where \(r\) is the capillary tube radius and \(d\) is the distance) (1)
4. Divide the calculated volume by the mass of the seeds and the time in minutes (1)

The layer of liquid paraffin is necessary because it acts as a physical barrier that prevents oxygen from diffusing from the air into the yeast suspension. This ensures that the environment remains completely anoxic, forcing the yeast to respire anaerobically rather than aerobically.

To determine the rate of carbon dioxide production using a gas syringe, the student should connect the delivery tube of the reaction vessel to the gas syringe. They must record the volume of gas collected at regular, timed intervals (e.g., every minute) for a specified duration. By plotting a graph of volume against time, they can calculate the gradient of the initial linear portion of the curve to determine the rate of carbon dioxide production.

1. Liquid paraffin acts as a barrier to prevent oxygen from dissolving/diffusing into the yeast mixture (1)
2. This ensures anaerobic conditions are maintained / forces yeast to respire anaerobically (1)
3. Record the volume of gas in the syringe at regular, timed intervals (1)
4. Calculate the rate by finding the gradient of the graph of volume vs time / dividing volume by time (1)

To collect representative data, the student should use a grid system over each woodland area and use a random number generator to select coordinates for placing the pitfall traps, ensuring unbiased sampling. The trapping effort must be standardised by using identical traps and leaving them open for the same length of time (e.g., 24 hours).

Ethical measures must be taken: traps should be checked frequently to minimize stress and mortality of captured organisms, and they should be covered with a raised lid or stone to shield them from rain (preventing drowning) and from being visible to predators. Finally, the student must identify and count the number of individuals of each species (\(n\)) and the total number of individuals across all species (\(N\)) to calculate the index.

1. Use random sampling (grid and random number generator) to position traps to avoid bias (1)
2. Standardise trapping effort by using identical traps and keeping them open for the same duration (1)
3. Cover traps with a raised lid/stone to prevent drowning/predation AND check them frequently to ensure ethical treatment (1)
4. Identify and count the abundance of each species (\(n\)) and the total number of individuals of all species (\(N\)) (1)

To systematically study the distribution, the student should lay out a tape measure perpendicular to the pathway edge, extending into the woodland to serve as a line transect. Quadrats should be placed at regular, predetermined intervals along the tape measure (e.g., every 1 meter or 2 meters). In each quadrat, the percentage cover or frequency of each plant species is recorded.

To measure abiotic factors that could influence this distribution: they should use a light meter held at a standardized height and angle to measure light intensity at each quadrat location. They should also use a soil moisture probe or soil pH meter inserted to a consistent depth to measure soil abiotic properties.

1. Lay a tape measure along the environmental gradient / from the pathway into the woodland (1)
2. Place quadrats at regular intervals along the tape measure and record percentage cover/frequency of plant species (1)
3. Use a light meter (held at standard height/angle) to measure light intensity at each quadrat (1)
4. Use a soil probe to measure soil moisture or pH at a standardised depth at each quadrat (1)

1. Calculate the percentage change in mass for each potato cylinder using the formula: \(\text{Percentage change} = \frac{\text{Final mass} - \text{Initial mass}}{\text{Initial mass}} \times 100\). This accounts for differences in the initial masses of the potato cylinders.
2. Plot a graph of the mean percentage change in mass on the y-axis against the concentration of sucrose solution on the x-axis.
3. Draw a line of best fit (or curve) and identify the sucrose concentration at which the line intersects the x-axis (where the percentage change in mass is \(0.0\%\)). At this point, there is no net movement of water by osmosis because the water potential of the solution is equal to the water potential of the potato cells.
4. Look up this specific sucrose concentration on a standard calibration chart or table that relates sucrose concentration to water potential at the temperature of the experiment to determine the water potential of the tissue.

Mark 1: Calculate percentage change in mass for each cylinder to standardise the starting mass / allow comparison. (Accept formula: \(\frac{\text{final} - \text{initial}}{\text{initial}} \times 100\))
Mark 2: Plot a graph of (percentage) change in mass against concentration of sucrose and draw a line/curve of best fit.
Mark 3: Identify the concentration where the line crosses the x-axis / where change in mass is zero / where there is no net water loss or gain.
Mark 4: Use a reference table/calibration curve/known values to find the solute/water potential corresponding to that specific sucrose concentration (at the same temperature).
[Reject: 'Plot mass change instead of percentage mass change' for Mark 1]

1. To standardise the beetroot preparation: Use a cork borer of a specific size to ensure all cylinders have the same diameter. Use a ruler and scalpel to cut each cylinder to the exact same length (e.g., \(20\text{ mm}\)), ensuring equal surface area and volume.
2. Wash the cut cylinders thoroughly under running distilled water and pat them dry with a paper towel. This step is crucial to remove any betalain pigment already leaked from cells damaged by the cutting process, ensuring any measured absorbance is solely due to the temperature treatment.
3. To calibrate the colorimeter: Fill a clean cuvette with distilled water and insert it into the colorimeter to calibrate/zero the instrument (setting absorbance to \(0.00\) or transmission to \(100\%\)).
4. Select a suitable green filter (approximately \(520\text{ nm}\) to \(550\text{ nm}\)) because beetroot pigment (betalain) absorbs green light most strongly, maximizing the sensitivity of the readings.

Mark 1 (Preparation): Use a cork borer and ruler/scalpel to cut cylinders to the same diameter and length to ensure equal surface area/volume.
Mark 2 (Preparation): Wash cylinders in distilled water/running water and pat dry to remove pigment released from cells damaged during cutting.
Mark 3 (Calibration): Zero the colorimeter using a blank cuvette containing only distilled water.
Mark 4 (Calibration): Use a green filter / filter in range 500-550 nm (as betalain absorbs green light / this wavelength range).

1. During the light-dependent stage of photosynthesis, light absorption by photosystem II leads to the photolysis of water, releasing electrons. These electrons are transferred along an electron transport chain.
2. DCPIP acts as an artificial electron acceptor, substituting for the natural coenzyme NADP. When DCPIP accepts these high-energy electrons (and hydrogen ions), it becomes reduced. The oxidised state of DCPIP is blue, whereas its reduced state is colourless; thus, the rate of decolourisation reflects the rate of the light-dependent reactions.
3. Control: A tube containing the chloroplast suspension and DCPIP but completely wrapped in aluminium foil / kept in darkness.
4. This control is essential to prove that the reduction (and decolourisation) of DCPIP is dependent on light driving the photosystems, rather than being caused by spontaneous chemical reduction or mitochondrial respiratory activity.

Mark 1: DCPIP acts as an electron acceptor (substituting for NADP) and is reduced by electrons.
Mark 2: Electrons are generated from photolysis of water / excited from chlorophyll (by light) which changes the colour of DCPIP from blue to colourless.
Mark 3: Control tube: Chloroplast suspension + DCPIP kept in the dark / wrapped in foil (or a tube with DCPIP and light but NO chloroplasts / boiled chloroplasts).
Mark 4: Explanation of control: It shows that light is required for the reaction / shows that living chloroplasts are required to reduce the DCPIP.

1. To prepare the chromatogram, draw a straight origin line in pencil (not ink, as ink will dissolve and separate in the solvent) about \(1.5\text{ cm}\) from the bottom of the TLC plate. Spot the pigment extract onto this line using a fine capillary tube, allowing the spot to dry completely before applying more to build up a small, highly concentrated, and compact spot.
2. Place the plate into a chromatography jar containing the solvent, ensuring that the level of the solvent is below the pencil origin line (so the pigment spots do not dissolve into the solvent pool). Seal the jar with a lid to saturate the atmosphere with solvent vapour, which prevents evaporation from the plate and ensures even movement.
3. Allow the solvent to travel up the plate until it is near the top. Remove the plate and immediately mark the solvent front with a pencil before the solvent evaporates.
4. To calculate the \(R_{\text{f}}\) value of a pigment spot: measure the distance from the pencil origin line to the centre of the pigment spot, and divide this by the distance from the origin line to the solvent front. Use the formula: \(R_{\text{f}} = \frac{\text{distance moved by pigment}}{\text{distance moved by solvent front}}\).

Mark 1: Draw origin line in pencil and build up a small, highly concentrated spot of pigment extract using a capillary tube (drying between applications).
Mark 2: Ensure the solvent level in the chromatography tank is below the pencil origin line and seal the tank with a lid.
Mark 3: Mark the solvent front in pencil immediately after removing the plate before it dries.
Mark 4: State the formula: \(R_{\text{f}} = \frac{\text{distance moved by pigment}}{\text{distance moved by solvent front}}\) (accept measurements from the origin line).

1. During aerobic respiration, woodlice consume oxygen (\(\text{O}_2\)) and release carbon dioxide (\(\text{CO}_2\)) in equal volume if the respiratory quotient is \(1.0\). The potassium hydroxide (\(\text{KOH}\)) solution absorbs all the \(\text{CO}_2\) produced by the respiring woodlice.
2. As a result, the consumption of oxygen by the organisms creates a net decrease in the volume of gas inside the chamber, which reduces the internal pressure. This pressure drop draws the coloured liquid along the capillary tube toward the respiration chamber, meaning the movement of the liquid directly represents the volume of oxygen consumed.
3. To calculate the volume of oxygen consumed: Measure the distance (\(d\)) moved by the coloured liquid in a known time interval. Use the formula for the volume of a cylinder, \(V = \pi r^2 d\), where \(r\) is the internal radius of the capillary tube.
4. To find the rate per gram per minute: Weigh the woodlice before the experiment to determine their mass (\(m\)). Divide the calculated volume of oxygen by the mass of the woodlice and by the duration of the trial in minutes (\(t\)): \(\text{Rate} = \frac{\pi r^2 d}{m \times t}\).

Mark 1: State that potassium hydroxide absorbs carbon dioxide produced during respiration.
Mark 2: Explain that this creates a decrease in volume/pressure inside the tube because oxygen is taken up (so movement of liquid is due only to oxygen consumption).
Mark 3: Calculate volume of oxygen consumed using the formula for a cylinder: \(\pi r^2 d\) where \(r\) is capillary radius and \(d\) is distance moved by liquid.
Mark 4: Divide volume by the mass of woodlice and the time taken (units e.g., \(\text{cm}^3\text{ g}^{-1}\text{ min}^{-1}\)).

1. Two key variables to control are:
- The concentration and volume of the yeast suspension and glucose solution. (This ensures that substrate availability and the number of respiring cells are constant).
- The pH of the mixture, which should be controlled using a buffer solution to prevent changes in pH from affecting enzyme-controlled respiratory pathways.
- The volume and concentration of the methylene blue indicator added.
2. To determine the endpoint objectively:
- Rather than relying on human eyesight to judge when the blue colour has completely disappeared, the student should use a colorimeter.
- By taking regular readings of light absorbance (using a red filter, approximately \(600\text{–}660\text{ nm}\), where methylene blue absorbs strongly) or transmission, they can accurately determine when the reaction is complete. The endpoint is reached when the absorbance value decreases to a stable, minimum plateau (representing the fully reduced, colourless state).
- Alternatively, the student can compare the tubes against a pre-prepared reference standard tube containing a fully decolourised yeast-glucose mixture (without methylene blue or with methylene blue fully reduced by heating beforehand).

Mark 1: Control variable (any one): concentration/volume of yeast, concentration/volume of glucose, pH (using a buffer), or volume/concentration of methylene blue.
Mark 2: Control variable (second one): any other valid variable from the list above.
Mark 3: Objective endpoint (colorimeter): Use a colorimeter to measure absorbance/transmission of light over time.
Mark 4: Analysis of endpoint: The endpoint is reached when the absorbance falls to a constant minimum value / transmission reaches a constant maximum / or use a colour reference standard tube for visual comparison to eliminate subjective bias.

1. To collect unbiased data, the student must use random sampling. Lay out two long tape measures at right angles along the boundary of the meadow to establish a coordinate grid. Use a random number generator to select pairs of coordinates where sampling will occur.
2. Place a quadrat of known size (e.g., \(0.5\text{ m} \times 0.5\text{ m}\)) at each selected coordinate. Identify and count the individual abundance of each plant species present within the quadrat. This gives \(n\) (the number of individuals of a particular species) and allows calculation of \(N\) (the total number of individuals of all species found).
3. Repeat this sampling process numerous times (e.g., at least 10–15 times per meadow) to obtain a representative sample and calculate a reliable mean.
4. Simpson's Index of Diversity (\(D\)) ranges from \(0\) to \(1\). A high value of \(D\) (close to \(1\)) indicates high species richness and high species evenness (equitability). In such a community, the food web is complex, with many interconnections. If one species is affected by a disease, predator, or climate change, there are alternative species to maintain the structure of the ecosystem, making it highly stable and resilient.

Mark 1: Set up a grid using tape measures at right angles and use a random number generator to obtain coordinates to avoid bias.
Mark 2: Place quadrats at coordinates to identify species and record the number of individuals of each species (\(n\)) and total individuals (\(N\)).
Mark 3: Repeat sampling many times / at least 10 times to ensure the sample is representative of the area.
Mark 4: High \(D\) value indicates high richness and evenness, meaning the ecosystem has complex food webs / is resilient to environmental changes (as organisms are not dependent on a single species).

1. To calculate the estimated population size using the Lincoln Index formula:
\[N = \frac{n_1 \times n_2}{m_2}\]
where:
- \(n_1 = 80\) (number of individuals caught and marked in the first sample)
- \(n_2 = 60\) (total number of individuals caught in the second sample)
- \(m_2 = 15\) (number of marked individuals recaptured in the second sample)
\[N = \frac{80 \times 60}{15} = \frac{4800}{15} = 320\text{ snails}\]
2. Assumptions of the method (any two of the following):
- **No births, deaths, immigration, or emigration** occurred within the snail population during the week between sampling (i.e., the population is closed).
- **The marking method is non-toxic and non-conspicuous**, meaning the waterproof paint does not harm the snail or make it more visible to predators, which would otherwise reduce its chances of survival.
- **Marked individuals mix thoroughly and randomly** with the rest of the unmarked population once released, giving them an equal chance of being recaptured.
- **The marks do not fall off** or rub away during the investigation period.

Mark 1 (Calculation): Correct working or final answer of 320 snails. (Formula: \(\frac{80 \times 60}{15}\))
Mark 2 (Assumption 1): No births, deaths, or migration (immigration/emigration) / population is closed.
Mark 3 (Assumption 2): The mark does not affect survival / predation / behaviour of the snails.
Mark 4 (Assumption 3): Marked snails mix randomly and fully back into the population / marks do not rub off/disappear.
[Accept any two assumptions for Marks 2, 3, 4]

To set up a reliable control:
1. Replace the germinating seeds with an equal volume of non-respiring material, such as boiled (dead) seeds of the same type or glass/plastic beads.
2. Keep all other variables identical: use the same volume and concentration of potassium hydroxide (\(\text{KOH}\)) solution, and expose the control respirometer to the same temperature (\(20^\circ\text{C}\)) and pressure.

To use the control data:
3. The control tube acts as a thermobarometer to monitor fluctuations in gas volume caused by changes in ambient temperature and atmospheric pressure.
4. Subtract the displacement of the liquid droplet in the control tube from the displacement in the experimental tube (algebraically) to obtain the true volume change caused solely by oxygen consumption.

1. Set up a tube containing an equal volume of non-respiring or inert material (such as glass beads, plastic beads, or boiled, sterilised seeds) [1 mark]
2. Keep all other conditions identical, including using the same volume/concentration of \(\text{KOH}\) and incubating at the same temperature [1 mark]
3. Identify that the control acts as a thermobarometer to measure volume/pressure changes caused by ambient temperature or atmospheric pressure fluctuations [1 mark]
4. Subtract the volume change (or distance moved by the fluid) in the control tube from the volume change in the experimental tube to calculate the actual oxygen consumption [1 mark]

[Accept: add/subtract control movement from experimental movement depending on the direction of change to correct for environmental factors]

To determine the water potential of the potato tissue:
1. The student must plot the percentage change in mass on the vertical y-axis and the sucrose concentration on the horizontal x-axis, then draw a line or curve of best fit through the data points.
2. The student then identifies the sucrose concentration where there is no net movement of water into or out of the cells, which corresponds to the point where the line of best fit intersects the x-axis (percentage change in mass is exactly \(0\%\)).
3. Finally, the student converts this specific sucrose concentration into a water potential value by referring to a pre-established calibration curve or lookup table that lists the water potential of known sucrose concentrations.

Award 1 mark for each of the following points, up to a maximum of 3 marks:
- **MP1 (Graph Construction):** Plot percentage change in mass on the y-axis and sucrose concentration on the x-axis, and draw a line/curve of best fit. [1 mark]
- **MP2 (Determining Intercept):** Identify the sucrose concentration where the line of best fit crosses the x-axis / where the percentage change in mass is \(0\%\). [1 mark]
- **MP3 (Use of Calibration):** Use a calibration curve/reference table (of sucrose concentration against water potential) to find the water potential equivalent to that sucrose concentration. [1 mark]

To find the initial rate of reaction from a progress curve of absorbance against time:
1. Draw a straight tangent line to the curve starting at time zero (\(t = 0\)), and calculate the gradient of this tangent (\(\Delta \text{absorbance} \div \Delta \text{time}\)).
2. The initial rate is the most valid measure because at the start of the reaction, none of the reactants (such as oxidized DCPIP or electron donors) have become limiting factors, and the maximum rate of reaction is achieved.
3. Over the full 10-minute period, the rate of absorbance change decreases significantly as the oxidized DCPIP is depleted (fully reduced) and chloroplasts may deteriorate, making an average rate unrepresentative of the true photosynthetic capacity of the extract.

Award 1 mark for each of the following points, up to a maximum of 3 marks:
- **MP1 (Calculation method):** Draw a tangent to the curve at time zero (\(t = 0\)) and calculate its gradient (change in absorbance divided by change in time). [1 mark]
- **MP2 (Validity of initial rate):** State that the initial rate represents the maximum rate when reactants/substrates (such as DCPIP or water) are not limiting. [1 mark]
- **MP3 (Limitation of average rate):** Explain that the rate decreases over the 10-minute period as reactants are used up / DCPIP is reduced / chloroplasts deteriorate, making the average rate an underestimation. [1 mark]

To calculate the volume of oxygen consumed per unit time:
1. First, find the gradient of the linear section of the graph of distance against time (\(\text{gradient} = \Delta y \div \Delta x\)). This gradient represents the rate of movement of the liquid bubble in units of \(\text{mm min}^{-1}\).
2. Use the formula for the volume of a cylinder: \(V = \pi r^2 h\), where \(h\) is the height (or distance) moved.
3. Substitute the gradient for \(h\) in the equation. Therefore, multiplying the gradient (\(\text{mm min}^{-1}\)) by \(\pi r^2\) (where \(r\) is the internal radius of the capillary tube in mm) gives the rate of oxygen consumption in \(\text{mm}^3\text{ min}^{-1}\).

Award 1 mark for each of the following points, up to a maximum of 3 marks:
- **MP1 (Gradient Calculation):** Calculate the gradient of the line of best fit / linear region of the graph to obtain the rate of movement of the bubble in \(\text{mm min}^{-1}\). [1 mark]
- **MP2 (Volume Formula):** Reference the volume of a cylinder formula: \(V = \pi r^2 h\). [1 mark]
- **MP3 (Synthesis):** State that the rate of oxygen consumption is calculated by multiplying the gradient by \(\pi r^2\) (to obtain \(\text{mm}^3\text{ min}^{-1}\)). [1 mark]

Biological Explanation: Ethanol is an organic solvent that dissolves the phospholipids in the plasma membrane and tonoplast, increasing membrane fluidity and disruption. It also disrupts hydrogen and ionic bonds within membrane-bound proteins, causing them to denature and create channels for leakage. Consequently, the vacuolar pigment betalain leaks out of the cells by diffusion. Higher concentrations of ethanol cause more widespread disruption, resulting in more pigment release and thus a higher mean absorbance. Methodological Evaluation and Improvements: 1. Cylinder size was only 'approximate'. This should be improved by measuring precisely with a digital caliper or ruler, and weighing each cylinder to ensure a standardized mass and surface area to volume ratio. 2. Cylinders were rinsed only 'once'. This fails to remove all betalain from cells damaged during cutting. The cylinders should be washed thoroughly under running water for at least 30 minutes, or soaked in distilled water overnight. 3. Temperature was not controlled ('room temperature'). Temperature affects kinetic energy and membrane permeability. A water bath should be used to maintain a constant temperature (e.g., 25 degrees C). 4. Replicates were not explicitly built into the protocol steps. Multiple replicates (at least 5) per concentration should be processed to calculate standard deviations and check reliability. Statistical Analysis: To evaluate the correlation between ethanol concentration and absorbance, calculate the Spearman's rank correlation coefficient and compare it against the critical value at \(p = 0.05\). Alternatively, perform a Student's t-test to compare the mean absorbance between individual concentrations to see if the differences are statistically significant.

Marking scheme structured around three main assessment areas: biological science explanation, experimental evaluation/improvements, and statistical analysis. Level 1 (1-3 marks): Student identifies that ethanol disrupts the membrane or lipids, causing pigment release. Identifies at least one basic improvement (e.g., repeating the experiment or controlling temperature). Level 2 (4-6 marks): Student explains that ethanol dissolves phospholipids and denatures membrane proteins, causing leakage. Evaluates at least two specific limitations and matches them to logical improvements (e.g., using a water bath for temperature, precise cylinder measurements, or thorough rinsing). Mentions a relevant statistical test. Level 3 (7-9 marks): Student provides a detailed, synoptic biological explanation of both lipid bilayer dissolution and protein denaturation disrupting tonoplast and plasma membranes. Thoroughly evaluates the protocol, offering precise, realistic improvements for washing, cylinder dimensions, temperature, and replication. Recommends and explains a specific statistical test (such as Spearman's rank or Student's t-test) to analyze the significance of the findings.

Biological Explanation: Ethanol is an organic solvent that dissolves the phospholipids in the plasma membrane and tonoplast, increasing membrane fluidity and disruption. It also disrupts hydrogen and ionic bonds within membrane-bound proteins, causing them to denature and create channels for leakage. Consequently, the vacuolar pigment betalain leaks out of the cells by diffusion. Higher concentrations of ethanol cause more widespread disruption, resulting in more pigment release and thus a higher mean absorbance. Methodological Evaluation and Improvements: 1. Cylinder size was only 'approximate'. This should be improved by measuring precisely with a digital caliper or ruler, and weighing each cylinder to ensure a standardized mass and surface area to volume ratio. 2. Cylinders were rinsed only 'once'. This fails to remove all betalain from cells damaged during cutting. The cylinders should be washed thoroughly under running water for at least 30 minutes, or soaked in distilled water overnight. 3. Temperature was not controlled ('room temperature'). Temperature affects kinetic energy and membrane permeability. A water bath should be used to maintain a constant temperature (e.g., 25 degrees C). 4. Replicates were not explicitly built into the protocol steps. Multiple replicates (at least 5) per concentration should be processed to calculate standard deviations and check reliability. Statistical Analysis: To evaluate the correlation between ethanol concentration and absorbance, calculate the Spearman's rank correlation coefficient and compare it against the critical value at \(p = 0.05\). Alternatively, perform a Student's t-test to compare the mean absorbance between individual concentrations to see if the differences are statistically significant.

Marking scheme structured around three main assessment areas: biological science explanation, experimental evaluation/improvements, and statistical analysis. Level 1 (1-3 marks): Student identifies that ethanol disrupts the membrane or lipids, causing pigment release. Identifies at least one basic improvement (e.g., repeating the experiment or controlling temperature). Level 2 (4-6 marks): Student explains that ethanol dissolves phospholipids and denatures membrane proteins, causing leakage. Evaluates at least two specific limitations and matches them to logical improvements (e.g., using a water bath for temperature, precise cylinder measurements, or thorough rinsing). Mentions a relevant statistical test. Level 3 (7-9 marks): Student provides a detailed, synoptic biological explanation of both lipid bilayer dissolution and protein denaturation disrupting tonoplast and plasma membranes. Thoroughly evaluates the protocol, offering precise, realistic improvements for washing, cylinder dimensions, temperature, and replication. Recommends and explains a specific statistical test (such as Spearman's rank or Student's t-test) to analyze the significance of the findings.