2022 OCR A-Level Biology A - H420 Practice Paper with Answers

In the Calvin cycle, the conversion of glycerate 3-phosphate (GP) to triose phosphate (TP) is a reduction reaction. This reaction requires both hydrogen/electrons from reduced NADP and energy from ATP. The subsequent regeneration of the carbon dioxide acceptor, ribulose bisphosphate (RuBP), from TP does not require reduced NADP, but does require energy provided by the hydrolysis of ATP.

[1 mark] C - Reduced NADP is used to reduce GP to TP, while ATP provides energy for both the conversion of GP to TP and the regeneration of RuBP.

Active loading of sucrose begins with the active transport of protons (\(\text{H}^+\) ions) out of the companion cells into the apoplast (cell wall space) using ATP. This creates a high proton concentration in the cell wall space compared to the inside of the companion cell. Protons then diffuse back down their concentration gradient into the companion cells via co-transporter proteins. This movement of protons down their electrochemical gradient drives the co-transport of sucrose into the companion cells against its concentration gradient.

[1 mark] B - Sucrose is co-transported with protons into the companion cells down a proton concentration gradient.

Primary metabolites are compounds produced by the microorganism that are essential for its normal, active growth and reproduction. They are produced during the exponential/log phase of growth. Secondary metabolites are not essential for normal growth and are typically produced after the exponential phase, primarily during the stationary phase, often as a response to nutrient depletion or competition.

[1 mark] C - Primary metabolites are produced during the log phase of growth, whilst secondary metabolites are produced mainly during the stationary phase.

The Respiratory Quotient (RQ) is calculated as: \(\text{RQ} = \frac{\text{molecules of } \text{CO}_2 \text{ produced}}{\text{molecules of } \text{O}_2 \text{ consumed}}\).
Using the balanced equation: \(\text{RQ} = \frac{18}{26} \approx 0.692\), which rounds to 0.69.
Lipids (fatty acids) have an RQ value of approximately 0.7 because they contain a high proportion of hydrogen relative to oxygen, requiring more oxygen for complete oxidation compared to carbohydrates (which have an RQ of 1.0).

[1 mark] A - RQ = 0.69, lipid

The correct physiological sequence is:
1. An action potential depolarises the presynaptic membrane, opening voltage-gated calcium channels. Calcium ions diffuse into the presynaptic neurone (Event 1).
2. This influx of calcium ions causes synaptic vesicles containing acetylcholine to move towards and fuse with the presynaptic membrane (Event 4), releasing neurotransmitter into the cleft.
3. Acetylcholine diffuses across the synaptic cleft and binds to ligand-gated receptors on the postsynaptic membrane (Event 2).
4. This binding causes sodium channels to open, and sodium ions enter the postsynaptic neurone, causing depolarisation (Event 3).

[1 mark] A - 1 \(\rightarrow\) 4 \(\rightarrow\) 2 \(\rightarrow\) 3

A single PCR cycle involves three main steps:
1. Denaturation: heated to \(95^\circ\text{C}\) to break the hydrogen bonds between complementary bases, separating double-stranded DNA into single strands. Covalent phosphodiester bonds in the backbone remain intact.
2. Annealing: cooled to \(55^\circ\text{C}\) to allow primers to bind (anneal) to complementary sequences on the single strands.
3. Extension: heated to \(72^\circ\text{C}\), which is the optimum temperature for Taq polymerase to add free nucleotides to synthesise a new complementary DNA strand.

[1 mark] B - Denaturation occurs at \(95^\circ\text{C}\) to break hydrogen bonds; annealing occurs at \(55^\circ\text{C}\) to allow primers to bind; extension occurs at \(72^\circ\text{C}\) for Taq polymerase activity.

Adrenaline (the first messenger) binds to a specific, complementary transmembrane receptor on the hepatocyte cell surface. This binding activates an intracellular G-protein, which in turn activates the enzyme adenylyl cyclase. Adenylyl cyclase converts ATP into cyclic AMP (cAMP, the second messenger). cAMP then binds to and activates protein kinase enzymes, initiating an enzyme cascade that culminates in the breakdown of glycogen into glucose (glycogenolysis).

[1 mark] A - Adrenaline binds to cell-surface receptor \(\rightarrow\) G-protein activated \(\rightarrow\) adenylyl cyclase activated \(\rightarrow\) ATP converted to cAMP \(\rightarrow\) protein kinase activated \(\rightarrow\) glycogenolysis

Scenario 1: The baby receives ready-made antibodies of maternal origin via breast milk (natural source), without their own immune system producing them. This is natural passive immunity.
Scenario 2: The individual receives ready-made antibodies via an injection (medical/artificial source) to neutralise venom immediately. This is artificial passive immunity.
Scenario 3: The individual is injected with antigens/weakened pathogen (artificial source) to trigger their own immune response, making memory cells and antibodies. This is artificial active immunity.

[1 mark] A - Scenario 1 = Natural passive, Scenario 2 = Artificial passive, Scenario 3 = Artificial active

The correct answer is C. Immobilisation of enzymes in alginate beads introduces a physical barrier. Substrate (lactose) molecules must diffuse through the alginate matrix to reach the enzyme's active sites, which limits the rate of reaction compared to free enzymes in a well-mixed solution where substrate and enzyme collide freely. A is incorrect because alginate is an inert support matrix, not an inhibitor. B is incorrect because enzymes always lower the activation energy; immobilisation does not increase the activation energy of the reaction itself. D is incorrect because immobilisation in alginate typically stabilises the enzyme's tertiary structure rather than disrupting it.

1 mark for identifying the correct option (C). Reject all other options.

The correct answer is A. The balanced equation for the complete oxidation of stearic acid is: \(C_{18}H_{36}O_2 + 26 O_2 \rightarrow 18 CO_2 + 18 H_2O\). To balance the oxygen atoms: Products side has \((18 \times 2) + 18 = 54\) oxygen atoms. Reactants side has 2 (from stearic acid) + 52 (from \(26 O_2\)) = 54 oxygen atoms. \(\text{RQ} = \frac{\text{CO}_2 \text{ produced}}{\text{O}_2 \text{ consumed}} = \frac{18}{26} \approx 0.69\).

1 mark for calculating the correct RQ value of 0.69 (A). Reject all other options.

The correct answer is B. During active loading, companion cells use ATP-driven proton pumps to actively transport hydrogen ions (\(H^+\)) out of the cytoplasm into the cell wall (apoplast). This establishes an electrochemical gradient. The \(H^+\) ions then diffuse back into the companion cells down their gradient through co-transporter proteins, which simultaneously carry sucrose molecules against their concentration gradient into the companion cells. A is incorrect because hydrogen ions are pumped out of, not into, the companion cells. C is incorrect because sucrose transport is indirect (secondary active transport). D is incorrect because sucrose is loaded into companion cells first, and then moves into the sieve tube elements via plasmodesmata.

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

The correct answer is A. In resting muscle, tropomyosin blocks the myosin-binding sites on actin filaments. When \(Ca^{2+}\) binds to troponin, troponin undergoes a conformational change. This pulls tropomyosin away from the binding sites on the actin filament, allowing myosin heads to bind and form cross-bridges. B is incorrect because troponin is what binds calcium and shifts tropomyosin, and the binding sites are on actin, not myosin. C is incorrect because ATPase activity is intrinsic to the myosin head. D is incorrect because sarcoplasmic reticulum depolarisation releases calcium, which is the cause of calcium binding to troponin, not a direct result of it.

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

The correct answer is B. When lactose is present, it (or its isomer allolactose) acts as an inducer by binding to the repressor protein. This causes a conformational change in the repressor, making it unable to bind to the operator region. Consequently, RNA polymerase can bind to the promoter and transcribe the structural genes (lacZ, lacY, and lacA). A is incorrect because lactose does not bind to the promoter. C is incorrect because lactose binds to the repressor, not the operator. D is incorrect because the regulatory gene (lacI) is constitutively expressed; lactose affects the protein product, not the transcription of the regulatory gene itself.

1 mark for selecting B. Reject all other options.

The correct answer is B. Collagen is a fibrous structural protein made of three polypeptide chains wrapped around each other to form a triple helix. Haemoglobin is a globular transport protein made of four polypeptide chains (two alpha, two beta subunits), each containing a haem prosthetic group with an iron ion (\(Fe^{2+}\)). A is incorrect because collagen is fibrous and haemoglobin is globular. C is incorrect because collagen is insoluble and has hydrophobic residues on its surface, whereas haemoglobin is soluble and has hydrophilic residues on its outer surface. D is incorrect because collagen does not contain a prosthetic group.

1 mark for B. Reject all other options.

The correct answer is C. In cyclic photophosphorylation, electrons leaving PSI are recycled back into the electron transport chain to generate ATP via chemiosmosis; no water is photolysed because PSI retrieves its own electrons. In non-cyclic photophosphorylation, electrons from PSII are passed to PSI and ultimately reduce NADP, requiring photolysis of water at PSII to replace the lost electrons. A is incorrect because cyclic photophosphorylation involves only PSI, not PSII. B is incorrect because cyclic photophosphorylation produces only ATP, while non-cyclic produces both ATP and reduced NADP. D is incorrect because it describes the processes in reverse.

1 mark for C. Reject all other options.

1. **Advantage of immobilised enzyme:** The enzyme can be easily separated from the product, preventing contamination of the milk and allowing the enzyme to be recovered and reused. It also often exhibits increased stability under varying temperatures and pH levels.
2. **Calculation of bead volume:**
- Diameter \(d = 4.2\text{ mm}\), so radius \(r = 2.1\text{ mm}\).
- Substitute into the sphere volume formula: \(V = \frac{4}{3} \pi (2.1)^3\).
- \(V = \frac{4}{3} \pi (9.261) \approx 38.792\text{ mm}^3\).
- Rounding to 2 significant figures gives \(39\text{ mm}^3\).

Max 2.5 marks:
- **1 mark** for stating any valid advantage of using immobilised enzymes (e.g. easily recovered, no product contamination, reusable, or increased stability).
- **1 mark** for correct substitution into volume formula using radius \(r = 2.1\text{ mm}\) (i.e. \(V = \frac{4}{3} \times \pi \times 2.1^3\)).
- **0.5 mark** for correct final value rounded to 2 significant figures (\(39\) or \(39\text{ mm}^3\)).

1. **Calculation and interpretation of \(D\):**
- \(D = 1 - 0.24 = 0.76\).
- This relatively high value (close to 1.0) indicates high species diversity. This means the ecosystem is stable, resilient to environmental change, and has complex food webs with few dominant species.
2. **Limitation of species richness:**
- Species richness only counts the number of different species present. It does not provide information about species evenness (the relative abundance of each species), which is critical because an ecosystem dominated by a single species is less biodiverse than one where individuals are evenly distributed.

Max 2.5 marks:
- **0.5 mark** for correctly calculating \(D = 0.76\).
- **1 mark** for explaining that a high \(D\) value indicates high ecosystem stability/resilience or complex food webs (accept: not dominated by a single species).
- **1 mark** for stating that species richness does not take into account species evenness or the abundance/relative numbers of each species.

1. **Calculation of water uptake rate:**
- Diameter \(d = 0.80\text{ mm}\), so radius \(r = 0.40\text{ mm}\).
- Cross-sectional area of capillary tube: \(A = \pi r^2 = \pi \times (0.40)^2 \approx 0.50265\text{ mm}^2\).
- Volume of water taken up: \(V = A \times \text{distance} = 0.50265 \times 45 \approx 22.6195\text{ mm}^3\).
- Rate of water uptake per minute: \(\text{Rate} = \frac{22.6195}{15} \approx 1.50796\text{ mm}^3\text{ min}^{-1}\).
- Rounded to 2 decimal places: \(1.51\text{ mm}^3\text{ min}^{-1}\).
2. **Assumption:**
- It is assumed that the rate of water uptake is equal to the rate of transpiration (water loss), ignoring the small volume of water used in photosynthesis, respiration, or maintaining cell turgidity.

Max 2.5 marks:
- **1 mark** for calculating the total volume of water absorbed (\(22.62\text{ mm}^3\)) or showing the correct formula usage with radius \(r = 0.40\text{ mm}\).
- **0.5 mark** for the correct rate of \(1.51\text{ mm}^3\text{ min}^{-1}\) (allow ecf if diameter was used as radius, resulting in \(6.03\)).
- **1 mark** for stating the assumption that water uptake equals water loss via transpiration (or that no water is stored/used for photosynthesis/turgidity).

1. **In the absence of lactose:**
- The regulator gene (*lacI*) is constantly transcribed and translated to produce an active repressor protein.
- This repressor protein binds to the operator region of the operon.
- This physically blocks RNA polymerase from binding to the promoter region (or moving along the DNA), preventing transcription of the structural genes (*lacZ*, *lacY*, and *lacA*).
2. **When lactose is present:**
- Lactose acts as an inducer by binding to the repressor protein.
- This binding induces a conformational (tertiary structure) change in the repressor protein.
- The repressor protein can no longer bind to the operator region and detaches, allowing RNA polymerase to bind to the promoter and transcribe the structural genes.

Max 2.5 marks:
- **1 mark** for stating that the regulator gene produces a repressor protein that binds to the operator, blocking RNA polymerase/transcription in the absence of lactose.
- **1 mark** for explaining that lactose binds to the repressor protein, altering its tertiary structure/shape so it can no longer bind to the operator.
- **0.5 mark** for stating that this allows RNA polymerase to bind to the promoter and initiate transcription of the structural genes.

1. **Calculation of RQ:**
- \(\text{RQ} = \frac{\text{Volume of } CO_2 \text{ produced}}{\text{Volume of } O_2 \text{ consumed}}\)
- \(\text{RQ} = \frac{17.15}{24.5} = 0.70\).
2. **Identification of substrate:**
- An RQ value of approximately \(0.7\) is characteristic of lipids (fatty acids).
3. **Location of the link reaction:**
- The link reaction takes place in the mitochondrial matrix.

Max 2.5 marks:
- **1 mark** for correct calculation of RQ as \(0.70\) (or \(0.7\)).
- **0.5 mark** for identifying the substrate as a lipid / fatty acid (accept triglyceride).
- **1 mark** for identifying the location as the mitochondrial matrix.

1. **Structural differences:**
- Cellulose is composed of \(\beta\)-glucose monomers, whereas amylose is composed of \(\alpha\)-glucose monomers.
- In cellulose, alternate glucose monomers are rotated by \(180^\circ\) to allow glycosidic bonding, resulting in a straight, unbranched chain. In amylose, there is no rotation, resulting in a coiled, helical structure.
2. **Hydrogen bonding and function:**
- Many parallel cellulose chains run side by side, and hydrogen bonds form cross-links between their hydroxyl groups.
- This bundles the chains together into strong microfibrils and macrofibrils.
- This arrangement provides high tensile strength to the cell wall, allowing it to withstand high turgor pressure without bursting.

Max 2.5 marks:
- **1 mark** for stating two valid differences (0.5 mark each: e.g., \(\beta\)- vs \(\alpha\)-glucose; alternate monomer rotation vs no rotation; straight/unbranched vs coiled/helical).
- **1 mark** for explaining that hydrogen bonds cross-link adjacent cellulose chains to form microfibrils/macrofibrils.
- **0.5 mark** for linking this structure to high tensile strength / preventing cell lysis under turgor pressure.

1. **Clonal Selection:** This is the process where a specific B lymphocyte with a complementary cell surface receptor is identified because it binds to the specific foreign antigen.
2. **Clonal Expansion:** Once selected, this specific B lymphocyte is activated (often with the help of cytokines from helper T cells) and divides repeatedly by mitosis to produce a large clone of genetically identical cells (plasma cells and memory B cells).
3. **Antibody Binding Region:** The variable region (or antigen-binding site) contains a specific amino acid sequence and tertiary structure complementary to the shape of the antigen.

Max 2.5 marks:
- **1 mark** for explaining clonal selection (B cell with complementary receptor binds to a specific antigen).
- **1 mark** for explaining clonal expansion (the selected cell divides by mitosis to produce a clone of plasma/memory cells).
- **0.5 mark** for identifying the variable region (or antigen-binding site) of the antibody.

1. **Role of Calcium Ions:**
- Calcium ions bind to the protein troponin.
- This causes a conformational (shape) change in troponin.
- Troponin pulls the attached tropomyosin filament away from the myosin-binding sites on the actin filament, exposing them.
2. **Role of ATP in Myosin Movement:**
- ATP binds to the myosin head, causing it to detach from the actin filament after the power stroke.
- Hydrolysis of ATP to ADP and inorganic phosphate (\(\text{P}_i\)) by ATPase provides the energy required to reset ('recock') the myosin head back to its high-energy angle, allowing it to bind to a new site further along the actin filament.

Max 2.5 marks:
- **1 mark** for stating that calcium ions bind to troponin, changing its shape and pulling tropomyosin away to expose myosin-binding sites on actin.
- **1 mark** for explaining that ATP hydrolysis provides energy to reset/recock the myosin head.
- **0.5 mark** for stating that the binding of a new ATP molecule causes the myosin head to detach from the actin filament.

1. Triglycerides have a high ratio of carbon-hydrogen (C-H) bonds to carbon atoms compared to carbohydrates, meaning they release significantly more energy per gram when oxidized during respiration.
2. During esterification, a covalent ester bond forms between glycerol and a fatty acid. This is a condensation reaction where a hydrogen atom from the hydroxyl group on glycerol and a hydroxyl group from the carboxyl group of the fatty acid are removed to release a molecule of water (\(H_2O\)).

Part 1 (Max 1 mark):
- Triglycerides have a higher ratio/proportion of C-H bonds than carbohydrates / contain more hydrogen and less oxygen per carbon. (1)

Part 2 (Max 1.5 marks):
- Reaction between the hydroxyl / \(-OH\) group of glycerol and carboxyl / \(-COOH\) group of a fatty acid. (1)
- Release/removal/elimination of a water / \(H_2O\) molecule. (0.5)
[Do not accept 'water is added']

1. Batch fermentation is a closed system where all nutrients are introduced at the start of the culture and products/wastes are only harvested at the end of the run. Continuous fermentation is an open system where fresh nutrient medium is constantly pumped in, and spent medium containing wastes and products is continuously siphoned off at the same rate to maintain a constant volume.
2. Primary metabolites are substances synthesized by microorganisms during their normal active growth (log phase). Because cells are dividing rapidly and metabolizing at their peak rate during the log phase, the synthesis of these essential growth-related substances is at its highest.

Part 1 (Max 1.5 marks):
- Batch: nutrients added at the start AND waste/products removed at the end of the run only. (0.5)
- Continuous: nutrients added continuously AND waste/products removed continuously. (1)

Part 2 (Max 1 mark):
- Cells are dividing at their maximum rate / metabolic activity is highest during active growth, when primary metabolites (which are essential for growth) are naturally produced. (1)

1. Gravity causes auxin (specifically IAA) to accumulate on the lower side of the horizontally placed root. Because roots are highly sensitive to auxin, this high concentration inhibits cell elongation on the lower side. The cells on the upper side, having a lower concentration of auxin, continue to elongate at a normal/faster rate, causing the root to curve downwards towards gravity.
2. In contrast, auxin has the opposite effect on shoot tissues, where high concentrations of auxin promote cell elongation rather than inhibiting it. Thus, the lower side of a horizontal shoot elongates faster, causing the shoot to bend upwards (negative gravitropism).

Part 1 (Max 1.5 marks):
- Auxin accumulates / is transported to the lower side of the root (due to gravity). (0.5)
- High auxin concentration inhibits cell elongation in roots, so cells on the upper side elongate faster (causing downward bending). (1)

Part 2 (Max 1 mark):
- In shoots, high auxin concentration stimulates/promotes cell elongation, so the lower side elongates faster (causing upward bending). (1)

1. Simpson's Index of Diversity ranges from 0 (no diversity) to 1 (infinite diversity). A high value of 0.74 at Site A indicates high species richness and species evenness; the community is stable, dominated by many species, and resilient to environmental changes. A low value of 0.21 at Site B indicates low diversity, dominated by only one or a few species, making it highly vulnerable to abiotic or biotic changes.
2. To ensure representative sampling, bias must be eliminated, which is achieved by randomizing the sample sites (e.g., using a random number generator to select coordinates on a grid). Alternatively, ensuring a large sample size/repeats makes the sample representative and reduces the impact of anomalous data.

Part 1 (Max 1.5 marks):
- Site A has higher species diversity / evenness / richness AND Site B is dominated by few species. (1)
- Site A is more stable / less vulnerable to environmental changes OR Site B is unstable / highly vulnerable to change. (0.5)

Part 2 (Max 1 mark):
- Use random sampling / random number generator for coordinates (to prevent bias). (1)
- OR collect a large number of samples / carry out many repeats (to minimize effect of anomalies). (1)
[Do not accept 'do it carefully']

1. Hydrostatic pressure is generated by the contraction of the heart ventricles and forces fluid and small dissolved solutes out of the capillary walls. Oncotic pressure is created by osmotic movement due to plasma proteins remaining in the blood, which draws fluid back in. At the arteriole end, the hydrostatic pressure (approx. 4.6 kPa) exceeds the oncotic pressure (approx. -3.3 kPa), creating a net filtration pressure of +1.3 kPa that drives fluid out into the surrounding intercellular space.
2. Because capillary walls have tiny pores, large substances cannot pass through. Red blood cells (erythrocytes), platelets, and large plasma proteins are too large to pass through the capillary wall and remain in the plasma, so they are absent or extremely scarce in tissue fluid.

Part 1 (Max 1.5 marks):
- Hydrostatic pressure forces/pushes fluid out of the capillary AND oncotic pressure draws/pulls fluid back into the capillary. (0.5)
- At the arteriole end, hydrostatic pressure is greater than oncotic pressure, creating a net outward flow of fluid. (1)

Part 2 (Max 1 mark):
- Any two of: Red blood cells / erythrocytes, platelets, large plasma proteins / albumin / globulins. (1)
[Award 0.5 for one correct answer. Do not accept 'white blood cells' as they can squeeze through capillary walls via diapedesis]

1. In yeast, anaerobic respiration occurs via ethanol fermentation, yielding ethanol and carbon dioxide (\(CO_2\)). In mammalian muscle cells, lactate fermentation converts pyruvate directly into lactate.
2. Lactate (lactic acid) is an acidic metabolite. As it accumulates in muscle tissue, it dissociates and releases hydrogen ions (\(H^+\)), which lowers the pH of the intracellular environment. This change in pH disrupts hydrogen and ionic bonds within the tertiary structure of cellular proteins and enzymes (such as those involved in glycolysis and muscle contraction). This reduces enzyme activity/denatures them, leading to muscle fatigue and cramping, meaning the process cannot be sustained.

Part 1 (Max 1 mark):
- Yeast: ethanol AND carbon dioxide / \(CO_2\) (0.5)
- Mammalian: lactate / lactic acid (0.5)

Part 2 (Max 1.5 marks):
- Lactate is acidic / releases \(H^+\) ions, which lowers pH of tissue/cells. (0.5)
- Low pH denatures / changes shape of active site of enzymes OR reduces the activity of enzymes (e.g. glycolytic enzymes). (1)
[Do not accept 'enzymes die']

1. The enzyme RuBisCO (ribulose bisphosphate carboxylase-oxygenase) catalyzes the fixation of carbon dioxide by attaching it to the 5-carbon acceptor molecule RuBP (ribulose bisphosphate), forming an unstable 6-carbon compound that immediately splits into two molecules of GP (glycerate 3-phosphate).
2. The Calvin cycle depends directly on products generated by the light-dependent reactions of photosynthesis: ATP (which provides energy for reduction and regeneration phases) and reduced NADP / NADPH (which provides reducing power / hydrogen to convert GP to TP). If a plant is kept in darkness, the light-dependent stage ceases, ATP and reduced NADP concentrations drop to zero, and the Calvin cycle stops because GP cannot be converted to TP, nor can RuBP be regenerated from TP.

Part 1 (Max 1 mark):
- Enzyme: RuBisCO (0.5)
- Acceptor molecule: RuBP / ribulose bisphosphate (0.5)

Part 2 (Max 1.5 marks):
- Calvin cycle requires ATP and reduced NADP / NADPH. (0.5)
- These are produced in the light-dependent stage (which requires light). (0.5)
- Without them, GP cannot be reduced to TP / RuBP cannot be regenerated (0.5)

1. When a pathogen is first encountered, the primary immune response is initiated, characterized by a slow clonal selection and expansion phase (long lag period) and low antibody concentration. The secondary response is initiated upon reinfection; it is much faster (short lag phase), produces a far higher peak concentration of antibodies, and these antibodies remain in the blood for a longer duration.
2. Memory cells (specifically memory B and memory T lymphocytes) are formed during the primary response and remain dormant in the blood and lymphatic system for many years. When they encounter the same antigen again, they immediately recognize it without the need for lengthy clonal selection. They undergo rapid clonal expansion, differentiating into plasma cells (which secrete large volumes of specific antibodies) and active T helper/cytotoxic cells, destroying the pathogen before symptoms develop.

Part 1 (Max 1 mark):
- Any two differences: Secondary response has a shorter lag phase/is faster; secondary response produces a higher concentration of antibodies; secondary response antibody levels stay elevated for longer. (1)
[Accept a comparison table format, 0.5 for each correct comparison]

Part 2 (Max 1.5 marks):
- Memory cells persist/remain in the blood/lymphatic system for a long time. (0.5)
- Upon re-exposure, they rapidly divide / undergo clonal expansion and differentiate into plasma cells (to produce antibodies quickly). (1)
[Do not accept 'memory cells produce antibodies directly' as plasma cells do]

Glycogen is the primary storage polymer of glucose in animals. Its highly branched structure (containing both alpha-1,4 and alpha-1,6 glycosidic bonds) ensures there are numerous terminal glucose residues available for rapid enzymatic cleavage during glycogenolysis. Being insoluble, it prevents osmotic issues such as water entering the cell via osmosis. Its compact nature ensures high-density energy storage within the cytoplasm.

1 mark: Highly branched structure with 1,6-glycosidic bonds provides many terminal ends for rapid hydrolysis to release glucose. 1 mark: Insoluble in water, so it has no osmotic effect / does not alter the water potential of the cytoplasm. 0.5 marks: Compact shape allows high density of energy storage per unit volume.

Abscisic acid causes stomata to close to conserve water, but this prevents carbon dioxide diffusion into the mesophyll. In the Calvin cycle (light-independent stage), carbon dioxide is fixed to ribulose bisphosphate (RuBP) to form glycerate 3-phosphate (GP). Without carbon dioxide, RuBP cannot be carboxylated, leading to a build-up of RuBP and a decline in GP and triose phosphate (TP) production.

1 mark: Stomatal closure stops or severely limits the diffusion of carbon dioxide into the leaf. 1 mark: Prevents carboxylation of ribulose bisphosphate (RuBP) / combination of RuBP with carbon dioxide. 0.5 marks: Leads to a decrease in glycerate 3-phosphate (GP) and triose phosphate (TP) levels / a temporary increase in RuBP levels.

In the absence of oxygen, oxidative phosphorylation cannot occur because oxygen is not available to act as the final electron acceptor, meaning the electron transport chain and Link reaction/Krebs cycle shut down. To maintain ATP production, the cell relies on glycolysis. Converting pyruvate to lactate by lactate dehydrogenase oxidises reduced NAD back to NAD. This oxidized NAD is a necessary coenzyme for the triose phosphate dehydrogenase step in glycolysis, permitting continuous substrate-level phosphorylation of ADP to ATP.

1 mark: Regeneration / reoxidation of NAD / NAD+ from reduced NAD (NADH). 1 mark: NAD is required as a coenzyme for glycolysis (specifically the oxidation of triose phosphate) to continue. 0.5 marks: Allows the continued net yield of 2 ATP molecules per glucose molecule by substrate-level phosphorylation under anaerobic conditions.

Inside red blood cells, carbonic anhydrase catalyses: \(\text{CO}_2 + \text{H}_2\text{O} \rightleftharpoons \text{H}_2\text{CO}_3\). Carbonic acid then dissociates: \(\text{H}_2\text{CO}_3 \rightleftharpoons \text{H}^+ + \text{HCO}_3^-\). The accumulation of \(\text{H}^+\) ions lowers intracellular pH, forcing haemoglobin to bind \(\text{H}^+\) to form haemoglobinic acid (acting as a buffer). This binding induces a conformational change that decreases its affinity for oxygen, shifting the oxygen dissociation curve to the right (the Bohr effect) and facilitating more oxygen release to active respiring tissues.

1 mark: Carbonic anhydrase catalyses the conversion of carbon dioxide and water to carbonic acid (which dissociates into H+ and HCO3-). 1 mark: Hydrogen ions bind to haemoglobin to form haemoglobinic acid, decreasing haemoglobin's affinity for oxygen. 0.5 marks: This triggers the Bohr effect, causing more oxygen to be unloaded at actively respiring tissues.

T helper cells and T killer cells are both T lymphocytes but have distinct effector functions. T helper cells express CD4 receptors and act as coordinators of the immune response by releasing cytokines (interleukins) to activate macrophages, B cells, and cytotoxic T cells. In contrast, T killer cells express CD8 receptors and directly target and kill infected, abnormal, or foreign host cells by binding to foreign antigens presented on MHC Class I molecules and releasing cytotoxic proteins (like perforin and granzymes) that induce cell death.

1 mark: T helper cells secrete cytokines / interleukins to activate / stimulate clonal expansion of B cells, T killer cells, and macrophages. 1 mark: T killer cells target and destroy infected or abnormal host cells. 0.5 marks: T killer cells kill by secreting perforin / granzymes / cytolytic substances that cause lysis or apoptosis of the infected cell.

In a closed system (batch culture), growth resources are finite. During the exponential phase, bacteria divide at their maximum rate. As the population size climbs, nutrients such as carbon sources (e.g., glucose) and nitrogen sources become depleted, which limits macromolecular synthesis needed for binary fission. Concurrently, the metabolic activities of the bacteria result in the accumulation of toxic waste products (e.g., ethanol, lactic acid, or other organic compounds) which alters the local pH or directly poisons enzymes. These factors slow down division (deceleration phase) until the birth rate equals the death rate (stationary phase).

1 mark: Depletion / limitation of essential nutrients / carbon sources / nitrogen sources / oxygen. 1 mark: Accumulation of toxic metabolic waste products (such as ethanol, lactic acid, or carbon dioxide). 0.5 marks: Alteration of pH beyond the optimum for bacterial enzymes OR increased competition for physical space.

To set up the photosynthesiometer: 1. Cut a fresh piece of Elodea canadensis stem at an angle and place it cut-end upwards in a boiling tube containing sodium hydrogencarbonate solution. 2. Place the boiling tube in a water bath to maintain a constant temperature. 3. Position an LED light source at a specific starting distance from the tube. 4. Allow the plant to equilibrate for 5 to 10 minutes. 5. Use the syringe to draw the oxygen bubble produced into the capillary tube and measure its length against the scale. 6. Repeat at different distances, using \(1/d^2\) to calculate relative light intensity. 7. Repeat each measurement to identify anomalies and calculate a mean. Biochemical explanation of decreased light intensity: 1. A decrease in light intensity reduces the light-dependent reaction, leading to less photophosphorylation and photolysis. 2. This results in decreased production of ATP and reduced NADP. 3. In the light-independent stage (Calvin cycle), ATP and reduced NADP are needed to convert GP to triose phosphate (TP). 4. Therefore, GP cannot be reduced to TP, causing GP levels to rise or accumulate. 5. RuBP is regenerated from TP using ATP. Since less TP and ATP are available, RuBP regeneration decreases, causing RuBP levels to fall. 6. RuBP also continues to fix carbon dioxide to form GP, further depleting RuBP.

Level 3 (5-6 marks): Detailed and accurate description of the experimental setup and procedure (including control of temperature and carbon dioxide concentration), AND a clear, correct explanation of the biochemical changes in GP and RuBP when light intensity decreases, linking this to the light-dependent and light-independent reactions. Level 2 (3-4 marks): Clear description of the experimental setup/procedure (with some mention of controls) OR a clear explanation of the biochemical changes in GP and RuBP (with minor omissions or errors). Level 1 (1-2 marks): Basic description of the experimental setup OR basic mention of GP/RuBP changes without detailed biochemistry. Indicative scientific content: [Experimental Setup] - Cut pondweed stem placed in sodium hydrogencarbonate (carbon dioxide source). - Constant temperature water bath used. - Use LED light (minimal heat) at measured distances. - Equilibration period allowed. - Gas bubble length measured in capillary tube. - Repeated trials for reliability. - Rate calculated as length/volume per unit time. [Biochemical Explanation] - Low light reduces ATP and reduced NADP production from light-dependent stage. - GP cannot be reduced to TP due to lack of ATP and reduced NADP, so GP accumulates. - RuBP regeneration decreases due to lack of TP and ATP, so RuBP levels fall.

Methods of enzyme immobilization: 1. Adsorption: Enzymes are attached by weak forces (hydrophobic interactions, ionic or hydrogen bonds) to a solid support such as clay, silica, or alginate. 2. Covalent bonding: Enzymes are covalently linked to an insoluble support (e.g. cellulose, collagen) using a cross-linking agent like glutaraldehyde. 3. Entrapment: Enzymes are trapped within a gel lattice or polymer network (e.g. silica gel or alginate beads) through which substrates and products can diffuse. 4. Membrane separation / Encapsulation: Enzymes are enclosed by a semi-permeable membrane. Evaluation: Advantages: 1. The product is not contaminated with the enzyme, reducing purification and downstream processing costs. 2. Enzymes can be easily recovered and reused, reducing operating costs. 3. Immobilization increases the thermal and pH stability of enzymes, as the matrix protects their tertiary structure. 4. Allows continuous flow processing, which is more efficient for large-scale production. Disadvantages: 1. High initial set-up costs for immobilization equipment and materials. 2. Immobilization can reduce reaction rates because active sites may be blocked or distorted. 3. Diffusion rates of substrates and products through the matrix/membrane can limit the overall reaction rate.

Level 3 (5-6 marks): Detailed description of at least three immobilization methods (covering both physical entrapment/adsorption and chemical bonding), AND a balanced evaluation covering multiple advantages and disadvantages of using immobilized enzymes in industrial processes. Level 2 (3-4 marks): Description of at least two immobilization methods AND some advantages/disadvantages, or a highly detailed evaluation with weaker description of methods. Level 1 (1-2 marks): Basic description of immobilization methods OR simple advantages/disadvantages of immobilized enzymes. Indicative scientific content: [Methods] - Adsorption (surface attachment via weak bonds). - Covalent bonding (strong chemical bonds to support/cross-linking). - Entrapment (trapped in gel/matrix). - Encapsulation (membrane barrier). [Advantages] - No downstream purification/contamination. - Reuse/recovery of enzymes. - Increased thermal/pH stability. - Continuous flow processes possible. [Disadvantages] - High setup costs. - Lower catalytic activity/blocked active sites. - Diffusion limitations.

Entrapment in a silica gel matrix holds the enzyme in place physically without chemically modifying it, which preserves the shape of the active site. However, the substrate molecules must diffuse through the physical matrix to reach the active site, which can limit the rate of diffusion and therefore the overall rate of reaction. Covalent bonding can cause conformational changes; adsorption is weak and prone to leakage; membrane separation does not require chemical modification.

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

The Respiratory Quotient (RQ) is calculated using the formula: \(\text{RQ} = \frac{\text{Volume of } \text{CO}_2 \text{ produced}}{\text{Volume of } \text{O}_2 \text{ consumed}}\).

Substituting the values: \(\text{RQ} = \frac{17.2}{24.5} \approx 0.70\).

An RQ value of approximately 0.7 is characteristic of the aerobic respiration of lipids (triglycerides). Carbohydrates have an RQ of 1.0 and proteins have an RQ of approximately 0.8 to 0.9.

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

During muscle contraction, the thin actin filaments slide over the thick myosin filaments. The I-band (which contains only actin filaments) and the H-zone (which contains only myosin filaments) both decrease in width as overlap increases. The A-band (which spans the entire length of the thick myosin filament) remains at a constant width. Therefore, only statements 2 and 3 are correct.

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

Active loading of sucrose is achieved by companion cells actively pumping hydrogen ions (protons) out of their cytoplasm into the surrounding cell wall spaces (apoplast) using ATP-dependent proton pumps. This creates a steep proton gradient. Hydrogen ions then diffuse back down their concentration gradient into the companion cells via co-transporter proteins, which simultaneously carry sucrose molecules into the cell against the sucrose concentration gradient.

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

Both glycogen and amylopectin are branched storage polymers of \(\alpha\)-glucose linked by both \(\alpha\)-1,4-glycosidic bonds (in the linear chains) and \(\alpha\)-1,6-glycosidic bonds (at the branch points). Glycogen is much more highly branched than amylopectin, which makes it more compact and allows rapid hydrolysis to release glucose in animal cells. Amylopectin is found in plants, not animals.

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

Neurone W is myelinated and has the largest diameter, giving it the fastest speed of conduction. Myelination enables saltatory conduction, where action potentials jump from one Node of Ranvier to the next, which is significantly faster than continuous propagation. A larger axon diameter reduces the internal resistance of the cytoplasm to the flow of local currents (ions), further increasing the propagation velocity of the impulse.

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

Adrenaline is a hydrophilic, amino-acid-derived hormone and cannot diffuse through the lipid bilayer of the cell surface membrane. It acts as a first messenger by binding to specific complementary G-protein coupled receptors on the cell surface of target cells (hepatocytes). This binding activates a G-protein, which subsequently activates the enzyme adenylyl cyclase. Adenylyl cyclase catalyses the conversion of ATP into cyclic AMP (cAMP), which acts as the second messenger to trigger an enzyme cascade.

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

During the light-independent stage (Calvin cycle) of photosynthesis, ribulose bisphosphate (RuBP) fixes carbon dioxide (\(\text{CO}_2\)) under the catalytic action of the enzyme RuBisCO to form glycerate 3-phosphate (GP). GP is subsequently reduced to triose phosphate (TP). This reduction reaction requires energy from ATP and reducing power from reduced NADP (NADPH), both of which are products of the light-dependent stage.

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

In batch culture, the system is closed and all nutrients are added at the start of the fermentation process. Microorganisms go through all stages of the population growth curve (lag, log, stationary, and death phases). Secondary metabolites, which are not essential for normal growth (such as penicillin), are produced during the stationary phase when nutrients become limiting and are harvested at the end of the run.

1 mark for the correct option (C).
- Reject A: Continuous culture, not batch, maintains organisms in the exponential phase.
- Reject B: This describes continuous culture.
- Reject D: Fermenters are sealed to maintain aseptic conditions and prevent contamination; pH and temperature are regulated automatically via internal probes.

Rubisco catalyses the fixation of carbon dioxide by combining it with RuBP to form GP. If Rubisco is inhibited, RuBP is no longer consumed, so its concentration increases (as it continues to be regenerated from existing triose phosphate using ATP). At the same time, no new GP is formed, but existing GP continues to be converted into triose phosphate (TP), leading to a decrease in GP concentration.

1 mark for the correct option (B).
- Reject A: RuBP accumulates because it is not being combined with carbon dioxide.
- Reject C: RuBP increases and GP decreases, not the other way around.
- Reject D: GP must decrease because its production stops while its conversion to TP continues.

During muscle contraction, according to the sliding filament model, the actin (thin) filaments slide past the myosin (thick) filaments. This causes the sarcomere to shorten. The H zone (the region containing only myosin) and the I band (the region containing only actin) both decrease in width. The A band (representing the total length of the myosin filaments) remains constant, and the individual myofilaments do not change length.

1 mark for the correct option (B).
- Reject A: The A band remains constant in width.
- Reject C: Actin and myosin filaments themselves do not shorten; they slide past each other.
- Reject D: The distance between adjacent Z lines (sarcomere length) decreases during contraction.

The Bohr effect describes the reduction in haemoglobin's affinity for oxygen in the presence of carbon dioxide. Carbon dioxide dissolves in blood water to form carbonic acid, which dissociates into hydrogen ions and hydrogencarbonate ions. The increased concentration of hydrogen ions (lower pH) causes haemoglobin to release oxygen more readily, making more oxygen available to actively respiring tissues that have high metabolic rates.

1 mark for the correct option (B).
- Reject A: High partial pressures of carbon dioxide decrease, rather than increase, haemoglobin's affinity for oxygen.
- Reject C: Low pH reduces the affinity of haemoglobin for oxygen.
- Reject D: Carbon monoxide, not carbon dioxide, binds irreversibly to haem groups causing carbon monoxide poisoning; carbon dioxide binds to amine groups.

To calculate Simpson's Index of Diversity (\(D = 1 - \sum(\frac{n}{N})^2\)):
1. Find the total number of organisms (\(N\)): \(15 + 25 + 40 + 20 = 100\).
2. Calculate \(\frac{n}{N}\) for each species:
- Species A: \(15/100 = 0.15\)
- Species B: \(25/100 = 0.25\)
- Species C: \(40/100 = 0.40\)
- Species D: \(20/100 = 0.20\)
3. Square each fraction: \(0.15^2 = 0.0225\), \(0.25^2 = 0.0625\), \(0.40^2 = 0.1600\), \(0.20^2 = 0.0400\).
4. Sum the squared values: \(0.0225 + 0.0625 + 0.1600 + 0.0400 = 0.285\).
5. Subtract from 1: \(1 - 0.285 = 0.715\).

1 mark for the correct option (B).
- Reject A: This is \(\sum(\frac{n}{N})^2\) before subtracting from 1.
- Reject C and D: Incorrect mathematical processing.

Adrenaline acts as a first messenger and cannot pass through the plasma membrane. It binds to a specific, complementary transmembrane receptor. This binding activates a G-protein, which in turn activates the enzyme adenylyl cyclase. Adenylyl cyclase converts ATP to cyclic AMP (cAMP, the second messenger). cAMP then binds to and activates protein kinase A, triggering an intracellular phosphorylation cascade that breaks down glycogen.

1 mark for the correct option (A).
- Reject B: Adrenaline does not enter the cell or bind to nuclear receptors (which is characteristic of steroid hormones).
- Reject C: G-protein activates adenylyl cyclase, not protein kinase A directly.
- Reject D: Adenylyl cyclase is activated (not inactivated), increasing cAMP concentration.

Density-independent factors affect the population size regardless of how many individuals are present in the population. Abiotic factors, such as extreme weather events (e.g., severe frost, floods, or volcanic eruptions), exert their effects equally on high-density and low-density populations. In contrast, competition, disease transmission, and predation are density-dependent because their impacts increase as the population density increases.

1 mark for the correct option (C).
- Reject A: Competition is density-dependent as resources become scarcer at higher densities.
- Reject B: Disease transmission rate is density-dependent because closer contact at higher densities facilitates spread.
- Reject D: Predation rates are density-dependent as predators focus more on abundant prey species.

Immobilised enzymes are physically or chemically bound to an inert matrix, preventing them from mixing freely with the reactant/product solution. This ensures the output product is free of enzyme contamination, eliminating the need for expensive downstream purification. Additionally, the bound state stabilizes the enzyme's tertiary structure, rendering it more resistant to thermal or pH-induced denaturation, and allowing the enzymes to be recovered and reused repeatedly in batch or continuous processes.

1 mark for: Product is not contaminated with enzyme / no downstream purification needed.
1 mark for: Enzymes can be easily recovered and reused (reducing running costs).
1 mark for: Greater tolerance to temperature/pH changes / less easily denatured.

Organisms possess common proteins like cytochrome c because they share evolutionary origins. Comparing the sequence of amino acids in cytochrome c or the sequence of nucleotides in DNA allows scientists to estimate when species diverged. A higher percentage similarity indicates a more recent common ancestor, while more differences suggest they diverged longer ago.

1 mark for: All organisms share certain molecules (such as cytochrome c) due to common ancestry.
1 mark for: Closer evolutionary relationship / more recent common ancestor correlates with fewer differences in sequence (DNA or amino acids).
1 mark for: Mutations accumulate over time, so greater differences indicate longer time since divergence.

A belt transect is used to study systematic changes in communities along an environmental gradient. A tape measure is laid down along the gradient (e.g. from shoreline inland). Quadrat sampling is performed at fixed intervals (e.g., every 5 meters) along the line. Inside each quadrat, the species present, their percentage cover, or species frequency are recorded.

1 mark for: Lay down a tape measure/line along the environmental gradient (from shore inland).
1 mark for: Place quadrats at regular / systematic intervals (e.g., every 5 metres) along the line.
1 mark for: Record species presence / abundance / percentage cover in each quadrat.

Homeobox genes are regulatory genes containing a highly conserved 180-base-pair sequence called a homeobox. They code for homeodomain proteins, which act as transcription factors to turn on or off other genes that dictate the anatomical structure (body plan) of the organism. In contrast, structural genes code for proteins that perform physical, biochemical, or structural functions in the cell (e.g., enzymes like amylase, or structural proteins like keratin) rather than controlling the expression of other genes.

1 mark for: Homeobox genes are regulatory genes that control the body plan / developmental patterns of an organism.
1 mark for: Homeobox genes code for transcription factors / homeodomain proteins that turn other genes on/off.
1 mark for: Structural genes code for proteins (e.g., enzymes, structural proteins) required for cell structure or metabolism (not gene regulation).

Lipids (specifically fatty acids) have many more carbon-hydrogen bonds than carbohydrates. When broken down, they generate a significantly greater number of hydrogen atoms. These hydrogens are transferred to NAD and FAD to form reduced NAD and reduced FAD. Consequently, many more electrons enter the electron transport chain, creating a larger proton gradient across the inner mitochondrial membrane, which drives the synthesis of a larger amount of ATP per gram.

1 mark for: Lipids contain a higher ratio of hydrogen atoms (or C-H bonds) per gram than carbohydrates.
1 mark for: More reduced NAD / reduced FAD are produced (per gram of lipid).
1 mark for: More protons flow through ATP synthase / greater proton gradient is set up, producing more ATP.

In PCR, primers are required because DNA polymerase cannot initiate the synthesis of a DNA strand from scratch; it can only add nucleotides to an existing chain. Primers are designed to be complementary to the 3' ends of the target DNA sequence, defining the region to be amplified. Taq polymerase is used because it is isolated from thermophilic bacteria (Thermus aquaticus), allowing it to remain active and functional at high denaturation temperatures (e.g., 95 °C) without denaturing, and it synthesises new DNA strands at its optimum temperature of around 72 °C.

1 mark for: Primers bind to complementary target sequences at the 3' end to provide a starting point for DNA replication.
1 mark for: Taq polymerase synthesises new DNA strands by joining free complementary DNA nucleotides.
1 mark for: Taq polymerase is thermostable / does not denature at high temperatures (e.g. 95 degrees C).

The primary immune response occurs upon first exposure to an antigen. It has a lag phase (several days) because clonal selection and expansion of specific B and T lymphocytes must occur. It produces fewer antibodies, and antibody concentration declines quickly. The secondary response occurs on re-exposure to the same antigen. Because memory cells are already circulating, they quickly detect the antigen, clonal expansion is rapid, and a much larger concentration of antibodies is produced much faster, often preventing symptoms.

1 mark for: Primary response has a lag phase / is slower and produces lower antibody concentration because clonal selection/expansion must occur.
1 mark for: Secondary response is much faster and produces a higher antibody concentration.
1 mark for: Secondary response depends on memory cells already present in the blood, which rapidly clonal expand/differentiate into plasma cells.

When an action potential depolarises the sarcolemma, calcium ions are released from the sarcoplasmic reticulum. These calcium ions bind to troponin, causing a conformational change that moves tropomyosin away from the myosin-binding sites on the actin filament. This allows myosin heads to bind to actin, forming cross-bridges. ATP then binds to the myosin head to cause its detachment from actin. Hydrolysis of ATP by ATPase provides the energy to reset ("cock") the myosin head, allowing the cycle to repeat and pull the actin filaments further, shortening the sarcomere.

1 mark for: Calcium ions bind to troponin, changing its shape, which pulls tropomyosin off the myosin-binding sites on actin.
1 mark for: ATP binding causes myosin heads to detach from actin (breaking the cross-bridge).
1 mark for: ATP hydrolysis provides the energy to reset/cock the myosin head (allowing the power stroke to repeat).

Immobilised enzymes are physically confined or localised in a defined region of space, such as being trapped in alginate beads or bound to an inert support. This offers several industrial advantages: 1. Product purity: The enzyme does not mix with the milk, so no downstream purification is required to remove lactase. 2. Reuse: The lactase can be easily recovered and used for multiple batches, making the process highly cost-effective. 3. Stability: Binding the enzyme to a matrix makes it more robust and resistant to denaturation by temperature or pH fluctuations.

Award 1 mark for each correct point up to a maximum of 3 marks:
- Enzyme can be easily recovered and recycled / reused [1]
- Product (lactose-free milk) is not contaminated with the enzyme / no downstream processing needed to remove enzyme [1]
- Enzyme is more stable / less susceptible to denaturation by temperature or pH changes [1]
- Process can be run continuously [1]

A high Simpson's Index of Diversity (\(D\)) close to 1 (such as 0.85 in Community A) indicates high species richness and high species evenness. This means the community has many different species with relatively equal abundance, and is dominated by no single species. Such a community is highly stable and resilient to environmental changes because if one species is affected, others can fill its ecological niche. In contrast, a low value close to 0 (such as 0.23 in Community B) indicates low diversity, dominance by one or a few species, and high vulnerability to environmental changes.

Award 1 mark for each correct point up to a maximum of 3 marks:
- Community A has higher species richness AND species evenness (or Community B has lower richness and evenness / is dominated by one or few species) [1]
- Community A is more stable / less susceptible to environmental change OR Community B is highly vulnerable/susceptible to environmental change [1]
- In Community A, an environmental change is less likely to affect the whole ecosystem because other species can adapt / fill the niche (or vice versa for B) [1]

Cytochrome c is an essential protein in the electron transport chain of aerobic respiration and is present in almost all eukaryotic organisms. Because it is highly conserved, differences in its amino acid sequence accumulate slowly through mutations over evolutionary time. By comparing the primary structure (sequence of amino acids) of cytochrome c between two species, scientists can count the number of differences. Fewer differences indicate that the two species shared a common ancestor more recently, meaning they are more closely related (have a closer phylogenetic relationship).

Award 1 mark for each correct point up to a maximum of 3 marks:
- Cytochrome c is a highly conserved protein / present in all eukaryotes [1]
- Compare the sequence of amino acids / primary structure of cytochrome c between the two species [1]
- Fewer differences in amino acid sequence indicate a more recent common ancestor / closer phylogenetic relationship (or vice versa) [1]

Industrial fermenters must be sterilised using superheated steam before a batch process begins. If foreign microorganisms enter, they compete with the cultured organism for essential nutrients (like glucose, amino acids) and oxygen, which decreases the growth rate and overall yield of the desired product. Furthermore, they may secrete waste products, toxins, or enzymes that alter the pH, degrade the desired product, or contaminate the final product, making downstream processing difficult or expensive.

Award 1 mark for each correct point up to a maximum of 3 marks:
- Reason for sterility: To prevent competition for nutrients / oxygen OR to ensure only the desired organism grows [1]
- Consequence 1: Contaminants compete for resources, decreasing growth rate / yield of product [1]
- Consequence 2: Contaminants may produce toxic metabolites / unwanted chemicals that taint the product or make it unsafe [1]
- Consequence 3: Contaminants may secrete enzymes that destroy/digest the desired product [1]

To study a gradient or change across an environmental transition (an ecotone), random sampling is not appropriate. Instead, systematic sampling using a line or belt transect must be used. A line is established using a tape measure (the transect line) from the woodland edge into the open field. Frame quadrats are placed at regular, systematic intervals along this line (e.g., every 2 or 5 metres). In each quadrat, the student records the species present and estimates their abundance (using percentage cover or ACFOR scale) to map changes along the gradient.

Award 1 mark for each correct point up to a maximum of 3 marks:
- Identify the use of a (systematic) belt transect / line transect [1]
- Lay down a tape measure/transect line along the environmental gradient (from woodland into field) [1]
- Place quadrats at regular / fixed / systematic intervals (e.g. every 2m) along the tape and record species abundance / percentage cover in each quadrat [1]

Carl Woese proposed the three-domain system in 1990 based on molecular phylogeny, specifically looking at ribosomal RNA (rRNA) genes. Evidence included differences in ribosomal RNA (rRNA) nucleotide sequencing, which showed that prokaryotes were split into two distinct, unrelated groups (Archaea and Bacteria), differences in cell membrane lipid structure (ether linkages in Archaea vs ester linkages in Bacteria/Eukarya), and differences in the structure of RNA polymerase. The three domains proposed are Archaea, Bacteria, and Eukarya.

Award 1 mark for each correct point up to a maximum of 3 marks:
- Evidence (max 1 mark): Differences in ribosomal RNA (rRNA) / nucleotide sequence of rRNA OR differences in membrane lipid structure (ether vs ester bonds) OR differences in RNA polymerase [1]
- Identification of Archaea and Bacteria domains [1]
- Identification of Eukarya / Eukaryota domain [1]

In entrapment, enzymes are physically trapped within a polymer network (such as calcium alginate beads) or silica gel, meaning the enzyme itself is not bound. Covalent bonding involves chemically linking enzyme molecules directly to an inorganic support (like clay or glass beads) or cross-linking them to each other. A key disadvantage of entrapment is that the rate of reaction can be limited by the rate of diffusion of substrate into the matrix and product out of the matrix. Additionally, smaller enzyme molecules may leak out of the matrix over time.

Award 1 mark for each correct point up to a maximum of 3 marks:
- Entrapment: Enzymes are physically trapped/confined within a gel matrix / polymer network without chemical bonding [1]
- Covalent bonding: Enzymes are chemically bonded / linked directly to an insoluble support / solid carrier [1]
- Disadvantage: Rate of reaction is limited by the rate of diffusion of substrate into the matrix / enzymes can leak out of the matrix [1]

Botanic gardens conserve plants outside of their natural habitat, which is defined as ex situ conservation. Advantages include: 1. Protected environment: Plants are safeguarded from natural threats such as pests, pathogens, herbivores, extreme weather, and human activity. 2. Propagation: Fast replication of rare species via cloning, tissue culture, or seed banks to allow reintroduction programs.

Award 1 mark for each correct point up to a maximum of 3 marks:
- Identify as ex situ conservation [1]
- Advantage 1 (max 2 marks): Protected from pests / pathogens / herbivores / competition / climate extremes [1]
- Advantage 2: Can be easily propagated / bred (via tissue culture / seed collection) to increase numbers for reintroduction [1]
- Advantage 3: Can be used for research / education of the public [1]

Meadow A, with a high Simpson's Index of Diversity (0.78), has high species richness and evenness. This high biodiversity means the ecosystem is stable and resilient to environmental changes like drought, because if one species dies out, others can fill its ecological role. Meadow B, with a low index (0.21), is dominated by one or a few species and has low biodiversity. It is unstable, and a drought that affects the dominant species could cause the ecosystem to collapse.

[1 mark] Meadow A has higher biodiversity (species richness and evenness) than Meadow B. [1 mark] Meadow A is more stable and resilient to drought because alternative species can maintain ecological functions. [1 mark] Meadow B is unstable because it is dominated by few species, making it highly vulnerable to collapse if a dominant species is affected.

Cytochrome c is a highly conserved respiratory protein found in all eukaryotic organisms. By comparing the primary structure (sequence of amino acids) of cytochrome c between the three mammal species, evolutionary relationships can be determined. Species with fewer differences in their amino acid sequences share a more recent common ancestor, whereas those with more differences diverged longer ago in evolutionary history.

[1 mark] State that cytochrome c is a conserved protein present in all three species. [1 mark] Identify that fewer differences in the amino acid sequence indicate a more recent common ancestor (closer evolutionary relationship). [1 mark] Identify that more differences indicate a more distant common ancestor (earlier evolutionary divergence).

Penicillin is a secondary metabolite, which means it is not required for normal growth and is only produced during the stationary phase of the fungal growth cycle, often in response to nutrient limitation. A continuous culture is designed to keep microorganisms in the log/exponential growth phase, where they primarily produce primary metabolites. In contrast, a batch culture allows the culture to progress naturally through all growth phases, reaching the stationary phase where penicillin production is turned on and can be harvested at the end of the run.

[1 mark] Explain that penicillin is a secondary metabolite only produced during the stationary phase (when nutrients become limiting). [1 mark] State that continuous culture maintains the population in the exponential/log growth phase, preventing secondary metabolite production. [1 mark] State that batch culture allows the population to reach the stationary phase to maximize penicillin yield before harvesting.

Using immobilised lactase in a continuous flow reactor offers several advantages: 1) The enzyme remains fixed in the reactor and can be reused multiple times, significantly reducing enzyme costs. 2) The end product (lactose-free milk) does not contain the enzyme, eliminating the need for expensive downstream processing to separate the enzyme. 3) Immobilisation often stabilises the enzyme's tertiary structure, making it more resistant to changes in temperature and pH, allowing the process to run at higher temperatures without denaturing the enzyme.

[1 mark] Enzyme can be easily recovered and reused, lowering overall costs. [1 mark] The product (milk) is free from enzyme contamination, saving on downstream purification costs. [1 mark] Immobilisation increases the enzyme's thermal/pH stability, protecting it from denaturation.

In eDNA metabarcoding, universal PCR primers are designed to bind to highly conserved, identical regions of DNA flanking a standard taxonomic 'barcode' gene (such as mitochondrial cytochrome c oxidase subunit I, or COI) present in all target fish species. PCR then amplifies the highly variable region within this gene across all species present in the sample. Once sequenced, these variable PCR products are run through a bioinformatics database (such as GenBank) containing known reference sequences; high-percentage sequence alignment matches allow precise identification of each fish species present.

[1 mark] Primers bind to conserved flanking regions of a specific barcode gene (e.g., COI) to initiate PCR. [1 mark] PCR amplifies the unique, variable sequence within this barcode gene from the diverse DNA in the sample. [1 mark] Amplified sequences are compared with reference sequences in DNA databases to determine matches and identify species.

The temperature coefficient (\(Q_{10}\)) represents the factor by which the rate of a reaction increases for every \(10^\circ\text{C}\) rise in temperature. Since the rate doubles, \(Q_{10} = 2.0\). Above \(50^\circ\text{C}\), which is higher than the optimum of \(45^\circ\text{C}\), the thermal energy is high enough to disrupt the weak hydrogen and ionic bonds stabilizing the enzyme's tertiary structure. This causes the enzyme to denature, changing the shape of its active site so that substrates can no longer bind, causing a dramatic drop in reaction rate and therefore a drop in the \(Q_{10}\) value.

[1 mark] State that \(Q_{10} = 2.0\) (or 2). [1 mark] Explain that above \(50^\circ\text{C}\), thermal energy breaks hydrogen and ionic bonds holding the tertiary structure. [1 mark] State that the active site changes shape/denatures, preventing substrate binding and decreasing the rate of reaction.

The high salinity of the marine environment acts as a strong selection pressure. Mussels with the Lap94 allele are better able to regulate their internal osmotic pressure and survive the physiological stress of high salinity compared to those without it. Consequently, individuals with the Lap94 allele have a selective advantage, meaning they are more likely to survive to reproductive age and pass the Lap94 allele to their offspring. Over many generations, this directional selection increases the frequency of the Lap94 allele in the marine population, while in low-salinity estuaries, different alleles are favored or selected for.

[1 mark] Identify high salinity as the environmental selection pressure. [1 mark] State that individuals with the Lap94 allele are better adapted to regulate osmotic pressure, survival rate is higher, and they reproduce. [1 mark] State that the Lap94 allele is passed on, increasing its frequency in the marine population over generations.

Conserving plants as seeds in a seed bank offers significant benefits over maintaining live adult specimens: 1) Space efficiency: Seeds are extremely small, allowing thousands of individuals to be stored in a very compact area, which maximizes the genetic diversity that can be preserved. 2) Cost and maintenance: Seeds are kept in a dormant state (frozen and dehydrated) and do not require constant watering, soil maintenance, or space, reducing long-term labor and resource costs. 3) Protection: Unlike outdoor botanical gardens, stored seeds are completely shielded from environmental hazards such as pests, pathogens, weather extremes, and natural disasters.

[1 mark] Space efficiency: allows greater genetic diversity/more individual specimens to be stored in a small space. [1 mark] Low cost/maintenance: dormant seeds require minimal care, water, and labor once frozen. [1 mark] Security/protection: insulated from environmental threats such as pests, diseases, and extreme weather events.

1. **Advantages of Continuous Flow over Batch Culture:**
- In batch culture, the fermentation must be stopped, the vessel emptied, cleaned, and sterilized before starting a new run. Continuous flow operates indefinitely, meaning there is no down-time, leading to higher overall productivity.
- Lactose-free milk contains glucose and galactose. If these products accumulate (as they would in batch culture), they can competitively or non-competitively inhibit the lactase enzyme. Continuous removal of product prevents this feedback inhibition.
- It is easier to maintain a constant, optimal temperature and pH (steady-state conditions) in a continuous system.

2. **Limitation of Entrapment:**
- The enzyme is trapped inside a physical matrix (calcium alginate gel). The substrate (lactose) is a relatively large disaccharide and must diffuse through the pores of the matrix to reach the active sites of the immobilised enzymes. This diffusion barrier slows down the rate of product formation compared to free enzymes or enzymes bound directly to a surface.

Award up to 3 marks in total.

**Suitability of continuous flow (max 2 marks):**
* **Mark 1:** Continuous harvesting / no down-time / no need to stop to empty and clean the bioreactor (unlike batch).
* **Mark 2:** Product is constantly removed, which prevents product/feedback inhibition of the lactase enzyme (by glucose/galactose).
* *Allow:* Easier to automate / maintain constant physical conditions (pH/temperature).

**Limitation of entrapment (max 1 mark):**
* **Mark 3:** Diffusion of substrate (lactose) into the beads (or product out of the beads) is slow / is a rate-limiting step.
* *Allow:* Some enzyme active sites may be inaccessible / buried deep inside the alginate matrix.
* *Reject:* The enzyme denatures due to entrapment.

Traditional vs. Molecular Classification:
1. Traditional classification is based on morphology, anatomy, and shared physical characteristics. This approach is highly subjective and can be misleading due to convergent evolution (analogous structures), where organisms in similar niches (like caves) evolve similar adaptations (e.g., loss of eyes, elongated limbs) despite having different ancestors.
2. Molecular phylogeny uses genetic material (DNA/RNA) and proteins (such as cytochrome c) to identify evolutionary relationships. It is highly quantitative and objective, measuring the accumulation of mutations over time.

Why Molecular Evidence is More Reliable:
1. It looks directly at the genetic code rather than phenotypic expression, which can be influenced by environmental factors.
2. Cytochrome c is an essential protein found in almost all eukaryotic organisms; comparing its amino acid sequence allows scientists to trace evolutionary divergence over long periods.
3. DNA sequencing allows for the comparison of non-coding regions, which accumulate mutations at a steady rate, acting as a molecular clock to estimate the time of divergence.

Level 3 (5–6 marks):
- Detailed and balanced comparison of traditional (morphological) vs. molecular methods.
- Clear explanation of the limitations of morphology (specifically referencing convergent evolution and analogous structures in cave environments).
- Explains why molecular evidence (DNA/cytochrome c) is more objective and quantitative.
- Applies understanding directly to the cave beetle scenario.

Level 2 (3–4 marks):
- Describes both traditional and molecular methods.
- Mentions convergent evolution or DNA/protein comparison, but the explanation of why one is more reliable than the other is less detailed or not fully linked to the cave context.

Level 1 (1–2 marks):
- Simple statements identifying that traditional is physical and molecular is genetic.
- Minimal explanation of evolutionary relationships or reliability.

Methods of Immobilisation:
1. Adsorption: Enzymes are bound to a supporting carrier (such as clay, silica, or glass beads) via weak hydrophobic interactions or ionic bonds. Highly accessible active sites, but enzymes can easily detach (leach).
2. Covalent Bonding: Enzymes are chemically bound to a support (like cellulose) or cross-linked using agents like glutaraldehyde. Very secure, but can distort the active site shape.
3. Entrapment: Enzymes are trapped inside a gel matrix (e.g., alginate beads) or silica gel. They do not bind chemically, but substrate must diffuse in and product must diffuse out.
4. Membrane Separation (Microencapsulation): Enzymes are physically separated from the reaction mixture by a semi-permeable membrane.

Comparison of Advantages and Disadvantages:
- Advantages: Easily separated from products (pure product, lower downstream processing costs); reusable (highly cost-effective for expensive enzymes); more stable at extreme temperatures and pH (reduces denaturation).
- Disadvantages: High initial capital and technical setup costs; diffusion rates of substrates to the active sites can be lower (reducing overall rate of reaction); the immobilisation process itself may denature or alter the active site of the enzyme.

Level 3 (5–6 marks):
- Accurately describes at least two distinct methods of enzyme immobilisation (e.g., adsorption, entrapment, covalent bonding).
- Compares immobilised enzymes to free enzymes by balancing both biological factors (stability, reaction rates, diffusion) and economic factors (reusability, downstream processing costs, setup costs).
- Structure is highly logical and uses correct scientific terminology throughout.

Level 2 (3–4 marks):
- Describes at least one method of immobilisation.
- Discusses some advantages and disadvantages, but the comparison lacks balance between biological and economic factors, or lacks depth.

Level 1 (1–2 marks):
- Identifies what immobilised enzymes are.
- Mentions simple advantages (e.g., reusable) or disadvantages (e.g., expensive) without detailed mechanism or method description.

1. Lacking mitochondria means that erythrocytes cannot perform aerobic respiration. 2. Consequently, they do not consume any of the oxygen they are transporting bound to oxyhaemoglobin, maximizing the efficiency of oxygen delivery to tissues. 3. To meet their metabolic energy demands (such as maintaining active transport across their cell surface membrane via sodium-potassium pumps), they generate ATP solely via anaerobic respiration or glycolysis in the cytoplasm.

Mark 1: Lacking mitochondria means the cell does not consume the oxygen it is transporting (for aerobic respiration) / maximizes the oxygen payload delivered to respiring tissues. Mark 2: Erythrocytes synthesize ATP via glycolysis / anaerobic respiration / lactate fermentation (in the cytoplasm). Mark 3: ATP is required to maintain active transport / ion pumps (e.g., \(Na^+/K^+\) ATPase) to maintain cell shape / membrane potential.

A neuromuscular junction links a motor neurone to a muscle fiber (sarcolemma), whereas a cholinergic synapse links two neurones (pre-synaptic to post-synaptic neurone). In a healthy individual, transmission at a neuromuscular junction is always excitatory, whereas synaptic transmission between neurones can be excitatory or inhibitory. Structurally, the post-synaptic membrane of a neuromuscular junction has folded structures (junctional folds) to increase surface area for acetylcholine receptors, whereas the post-synaptic neurone membrane is relatively smooth.

MP1: Neuromuscular junction connects a motor neurone to a muscle fiber (sarcolemma) whereas a cholinergic synapse connects two neurones. MP2: Neuromuscular junction is always excitatory / causes depolarization leading to contraction, whereas a cholinergic synapse can be excitatory or inhibitory. MP3: The post-synaptic membrane (sarcolemma) is highly folded / has junctional folds to increase ACh receptor density, whereas the post-synaptic membrane of a neurone is not folded.

An increase in temperature above the optimum increases the kinetic energy of the thylakoid membrane phospholipids, causing the membrane to become highly fluid and leaky to protons (\(H^+\)). Consequently, protons diffuse directly across the phospholipid bilayer down their electrochemical gradient, bypassing ATP synthase. This collapses or reduces the proton motive force, meaning fewer protons flow through ATP synthase, resulting in a significantly reduced rate of photophosphorylation (ATP synthesis).

MP1: High temperature increases kinetic energy of phospholipids, increasing membrane permeability / making the thylakoid membrane 'leaky'. MP2: Protons (\(H^+\)) diffuse directly across the phospholipid bilayer, collapsing / reducing the proton gradient / proton motive force. MP3: Less / no diffusion of protons through ATP synthase, resulting in reduced / no ATP synthesis (photophosphorylation).

During the denaturation step of PCR, the mixture is heated to approximately \(95^\circ\text{C}\) to break hydrogen bonds between complementary strands of DNA. Standard DNA polymerases would denature at this temperature because the thermal energy breaks their weak intramolecular bonds, altering the active site. Taq polymerase is thermostable and remains functional over multiple cycles. Its tertiary structure is adapted through a high density of stabilizing bonds (hydrogen bonds, ionic bonds/salt bridges, and disulfide bridges) that resist thermal disruption.

MP1: PCR requires a high temperature (\(~95^\circ\text{C}\)) to denature / separate double-stranded DNA; standard DNA polymerases would denature at this temperature. MP2: Taq polymerase does not denature at high temperatures, so it remains functional across multiple thermal cycles / does not need to be replaced. MP3: Its tertiary structure is stabilized by an increased number of hydrogen bonds, disulfide bonds, or ionic bonds / salt bridges (which require more kinetic energy to break).

Ultrafiltration is achieved through three layers: the capillary endothelium, the basement membrane, and the podocytes. The endothelium contains small pores (fenestrations) that allow all constituents of blood plasma except blood cells and platelets to pass through. The basement membrane acts as a second, finer molecular sieve composed of collagen and glycoproteins, preventing proteins with a molecular mass greater than 69,000 Da from passing. The podocytes are specialized epithelial cells with finger-like projections (pedicels) that wrap around the capillaries, leaving narrow filtration slits that allow the filtered small molecules to pass without resistance into the Bowman's capsule space.

MP1 (Endothelium): Endothelial cells of the glomerular capillaries have pores / fenestrations that allow water and solutes to pass but prevent the passage of blood cells / platelets. MP2 (Basement membrane): The basement membrane (mesh of collagen/glycoproteins) acts as the main / second filter that prevents the passage of large plasma proteins (molecular mass > 69,000 Da) / negatively charged proteins. MP3 (Podocytes): Podocytes have finger-like projections / pedicels that wrap around capillaries, creating filtration slits that allow the filtered substances to pass freely into the Bowman's capsule.

Because estrogen is lipid-soluble, it easily diffuses through the hydrophobic core of the cell membrane's phospholipid bilayer. Inside the cell, it binds to a specific intracellular receptor. The resulting estrogen-receptor complex enters the nucleus, where it binds to DNA and acts directly as a transcription factor to regulate gene expression. In contrast, adrenaline is water-soluble and cannot cross the membrane. It binds to a specific complementary receptor on the cell-surface membrane, activating a G-protein. This G-protein activates adenylyl cyclase, producing the second messenger cyclic AMP (cAMP), which triggers a cascade of intracellular enzyme activations.

MP1 (Membrane crossing & receptors): Estrogen diffuses directly through the phospholipid bilayer to bind to intracellular receptors, whereas adrenaline cannot cross the membrane and must bind to cell-surface receptors. MP2 (Estrogen action): The estrogen-receptor complex enters the nucleus to act directly as a transcription factor / regulate gene transcription. MP3 (Adrenaline action): Adrenaline binding activates a G-protein, triggering a second messenger (e.g., cAMP) / an enzyme cascade.

Codominance is expressed when both alleles in a heterozygote are active and expressed in the phenotype. In a heterozygote with genotype \(I^AI^B\), transcription and translation of both alleles occur. The \(I^A\) allele codes for a functional glycosyltransferase enzyme that attaches N-acetylgalactosamine to the precursor H-antigen, producing Antigen A. The \(I^B\) allele codes for a different functional transferase enzyme that attaches galactose, producing Antigen B. Since both enzymes are fully synthesized and active, both Antigen A and Antigen B are present on the glycoproteins and glycolipids of the erythrocyte cell surface membrane.

MP1: Both alleles are transcribed and translated, so both functional transferase enzymes are produced in the same cell. MP2: The \(I^A\) allele produces an enzyme that synthesizes Antigen A, while the \(I^B\) allele produces an enzyme that synthesizes Antigen B. MP3: Both Antigen A and Antigen B glycoproteins / glycolipids are present simultaneously on the erythrocyte cell surface membrane.

When a plant detects a pathogen, callose (a \(\beta\)-1,3-glucan polysaccharide) is rapidly synthesized and deposited in cell walls, plasmodesmata between cells, and in the sieve plates/sieve pores of the phloem. This deposition blocks the connections between cells, physically sealing off the infected area and preventing the systemic spread of the pathogen (such as viruses or bacteria) through the plant's vascular system. However, blocking the sieve pores also prevents the mass flow of phloem sap, thereby stopping the translocation of organic solutes (such as sucrose) through the affected phloem sieve tubes.

MP1: Callose is deposited in the cell walls / plasmodesmata (between adjacent cells) AND in the sieve plates / sieve pores of the phloem. MP2: The deposition seals off infected tissues, preventing the systemic spread of the pathogen through the vascular tissue / plasmodesmata. MP3: Blocking the sieve pores prevents the mass flow of phloem sap / translocation of organic solutes (e.g., sucrose) through the phloem.

During aerobic respiration in the mitochondrial matrix, carbon dioxide is released in the link reaction and Krebs cycle. It diffuses out of the muscle cells, into the blood plasma, and then into erythrocytes. Inside erythrocytes, the enzyme carbonic anhydrase catalyses the reaction between carbon dioxide and water to form carbonic acid. Carbonic acid is unstable and dissociates into hydrogencarbonate ions \(\text{HCO}_3^-\)\ and hydrogen ions \(\text{H}^+\)\. The hydrogencarbonate ions diffuse out of the erythrocyte into the plasma down a concentration gradient, while chloride ions diffuse in to maintain electrical neutrality (the chloride shift). The hydrogen ions are buffered by binding to oxyhaemoglobin, triggering the release of oxygen and forming haemoglobinic acid, thus maintaining the pH of the blood.

1. CO2 reacts with H2O to form carbonic acid catalysed by carbonic anhydrase AND dissociates to \(\text{H}^+\)\ and \(\text{HCO}_3^-\)\ [1 mark]; 2. \(\text{HCO}_3^-\)\ ions diffuse out of erythrocytes into plasma in exchange for chloride ions / chloride shift [1 mark]; 3. \(\text{H}^+\)\ ions bind to haemoglobin to form haemoglobinic acid, which acts as a buffer to prevent a decrease in pH [1 mark]

Both plant and animal action potentials are transient depolarization events across the cell membrane that follow an all-or-nothing principle once a threshold is reached. However, mammalian action potentials in myelinated neurones are much faster (up to 120 m/s) due to saltatory conduction between nodes of Ranvier and are driven by the influx of sodium ions \(\text{Na}^+\)\. In contrast, plant action potentials are much slower (often less than 1 cm/s), do not involve myelin, and are typically driven by the efflux of chloride ions \(\text{Cl}^-\)\ and influx of calcium ions \(\text{Ca}^{2+}\)\

Award up to 2 marks for similarities: - Both involve a temporary change in membrane potential / depolarization / depolarizing phase followed by repolarization [1 mark]; - Both are all-or-nothing responses / propagate along the membrane once threshold is reached [1 mark]. Award 1 mark for difference: - Mammalian action potentials are myelinated / show saltatory conduction / are much faster OR mammalian action potentials rely on \(\text{Na}^+\)\ influx while plant action potentials rely on \(\text{Cl}^-\)\ efflux or \(\text{Ca}^{2+}\)\ influx [1 mark]

HIV infects helper T cells by binding its surface glycoprotein gp120 to both the CD4 receptor and the CCR5 co-receptor on the host cell membrane. Knocking out CCR5 prevents this binding, thereby blocking viral entry and replication, preserving the helper T cell population and preventing the immunodeficiency that leads to AIDS. Mature helper T cells have a limited lifespan and do not self-renew. If only mature T cells are modified, they will eventually die, and the patient's remaining unmodified stem cells will produce vulnerable T cells. Targeting hematopoietic stem cells in the bone marrow ensures that all descendant helper T cells produced throughout the patient's life will lack CCR5, providing a permanent cure.

1. Loss of CCR5 prevents HIV envelope glycoprotein (gp120) binding / prevents viral entry into host helper T cells [1 mark]; 2. Mature helper T cells are short-lived / will eventually die and be replaced [1 mark]; 3. Modifying hematopoietic stem cells ensures all newly differentiated helper T cells carry the knockout / provides permanent resistance [1 mark]

RuBisCO has an active site that can bind both carbon dioxide and oxygen. When oxygen binds instead of carbon dioxide, it acts as a competitive inhibitor. This reduces the rate of carboxylation of ribulose bisphosphate (RuBP). Instead of forming two molecules of glycerate 3-phosphate (GP), photorespiration produces one molecule of GP and one molecule of phosphoglycolate, which must be recycled in an energy-expensive pathway. This significantly reduces the net synthesis of GP and subsequent triose phosphate (TP), thereby lowering the overall rate of glucose production in photosynthesis.

1. Oxygen acts as a competitive inhibitor by binding to the active site of RuBisCO [1 mark]; 2. This prevents carbon dioxide from binding / reduces the rate of carboxylation of RuBP [1 mark]; 3. Less GP / glycerate 3-phosphate (and subsequently less TP / glucose) is produced [1 mark]

Sucrose loading is an active process that occurs at the companion cells. First, proton pumps (active transport proteins) in the plasma membrane of the companion cell use energy from ATP hydrolysis to actively pump hydrogen ions (protons, \(\text{H}^+\)\) out of the cell into the surrounding apoplast (cell wall space). This creates a high concentration of protons outside the companion cell compared to the inside. Next, these protons diffuse back down their electrochemical gradient into the companion cell through co-transporter proteins. The co-transporter protein couples the movement of protons with the movement of sucrose, bringing sucrose into the companion cell against its concentration gradient.

1. Proton pumps actively transport \(\text{H}^+\)\ ions (protons) out of the companion cell into the cell wall/apoplast using ATP [1 mark]; 2. This establishes a high concentration gradient of \(\text{H}^+\)\ ions outside the companion cell [1 mark]; 3. \(\text{H}^+\)\ ions diffuse back into the companion cell down their concentration gradient via a co-transporter protein, co-transporting sucrose against its concentration gradient [1 mark]

Because thyroxine is lipid-soluble, it readily diffuses through the phospholipid bilayer of both the plasma and nuclear membranes. Inside the nucleus, it binds to a specific thyroid hormone receptor. The resulting thyroxine-receptor complex acts as a transcription factor. It binds to specific hormone response elements (promoters) on the DNA. This binding stimulates (or represses) the transcription of target genes into messenger RNA (mRNA). The mRNA is then transported out of the nucleus and translated by ribosomes into specific proteins, such as respiratory enzymes, which increases the metabolic rate of the cell.

1. Thyroxine-receptor complex acts as a transcription factor [1 mark]; 2. It binds to specific DNA sequences / promoter regions to stimulate transcription of target genes into mRNA [1 mark]; 3. This leads to increased translation / synthesis of specific proteins (e.g. metabolic/respiratory enzymes) which alter cell activity [1 mark]

Waterlogged soils suffer from a lack of oxygen because water fills the air spaces, creating anaerobic conditions. Nitrifying bacteria (such as Nitrosomonas and Nitrobacter), which convert ammonium ions to nitrites and then nitrates, require oxygen for aerobic respiration and are therefore inhibited. Conversely, anaerobic conditions stimulate denitrifying bacteria (such as Pseudomonas), which use nitrates as an alternative electron acceptor in anaerobic respiration. These bacteria convert nitrates back into nitrogen gas (\(\text{N}_2\)\), which escapes into the atmosphere. This shift in the bacterial population dramatically decreases the concentration of nitrates in the soil, leaving crops with insufficient nitrogen for amino acid and protein synthesis.

1. Waterlogging creates anaerobic / low-oxygen conditions in the soil [1 mark]; 2. This inhibits aerobic nitrifying bacteria (reducing nitrate production) AND stimulates anaerobic denitrifying bacteria [1 mark]; 3. Denitrifying bacteria convert nitrates into nitrogen gas (\(\text{N}_2\)\), reducing available soil nitrates for plant uptake [1 mark]

One key advantage of DNA barcoding is that it does not rely on the presence of specific morphological features, which may be absent in damaged specimens, juvenile/larval stages, or in closely related 'cryptic' species that look identical. Once the DNA barcode sequences are obtained, phylogenetic classification compares the sequence of nucleotides. Organisms with highly similar sequences share a more recent common ancestor. By calculating the percentage divergence between sequences, biologists can construct phylogenetic trees that show evolutionary distances and clarify the taxonomic relationships among the rainforest species.

1. Advantage: Allows identification of species from incomplete/damaged specimens, larval stages, or cryptic species (which lack distinct morphological differences) [1 mark]; 2. Comparison of nucleotide / base sequences of the barcode gene allows quantification of genetic differences [1 mark]; 3. Fewer differences in DNA sequence indicate a more recent common ancestor / closer evolutionary relationship (allowing construction of a phylogenetic tree) [1 mark]

During anaerobic respiration, lactic acid dissociates into lactate and hydrogen ions (\(H^+\)), which lowers the pH of the blood plasma in the microenvironment of the active muscle tissue. This increase in acidity alters the tertiary structure of haemoglobin, reducing its affinity for oxygen. Consequently, the oxygen dissociation curve shifts to the right, a phenomenon known as the Bohr effect/shift. The physiological benefit is that haemoglobin more readily dissociates from and releases oxygen at any given partial pressure of oxygen (\(pO_2\)), delivering crucial oxygen to the highly active, respiring muscle tissues to sustain aerobic pathways and minimise further oxygen debt.

Mark 1: Hydrogen ions (\(H^+\)) / lactate lower the blood pH in active tissues.
Mark 2: Lower pH shifts the oxygen dissociation curve to the right / causes the Bohr effect / reduces the affinity of haemoglobin for oxygen.
Mark 3: (Benefit) Oxygen is more easily / rapidly unloaded / released to the respiring muscle tissues (helping to maintain aerobic respiration).

Cyanide inhibits cytochrome c oxidase, stopping the electron transport chain and oxidative phosphorylation in the mitochondria of companion cells, which prevents ATP synthesis. Without ATP, proton pumps in the companion cell membrane cannot actively transport hydrogen ions (\(H^+\)) out into the apoplast. This abolishes the proton gradient needed to co-transport sucrose into the companion cells and sieve tube elements. As a result, the sucrose concentration in the phloem does not rise, preventing the osmotic entry of water and the generation of high hydrostatic pressure at the source, thus halting mass flow.

Mark 1: Inhibition of oxidative phosphorylation / ETC reduces ATP synthesis in companion cells.
Mark 2: Active transport of hydrogen ions (\(H^+\)) out of companion cells stops, which prevents the co-transport of sucrose into the phloem.
Mark 3: Solute concentration does not increase, so water does not enter by osmosis, preventing the generation of a hydrostatic pressure gradient / mass flow.

Immobilising lactase within calcium alginate beads physically confines the enzyme molecules, restricting their movement and vibrational energy. This structural support helps to stabilise the enzyme's tertiary structure, preventing the hydrogen and ionic bonds from breaking easily at elevated temperatures, thus preserving the active site shape and resisting denaturation. An economic advantage of this setup is that the enzyme remains trapped in the beads and does not contaminate the final milk product, making it easy to recover and reuse multiple times, which significantly lowers production costs.

Mark 1: Alginate matrix / beads support the enzyme structure, restricting conformational movement / vibration at higher temperatures.
Mark 2: This prevents hydrogen / ionic bonds from breaking, thus protecting the tertiary structure / active site shape from denaturing.
Mark 3: (Economic advantage) The enzyme can be recovered and reused OR there is less downstream processing / purification required because the product is not contaminated with enzyme.

ADH circulates in the blood and binds to specific, complementary receptors located on the cell surface membrane of the collecting duct cells (facing the interstitial fluid). This binding activates an intracellular secondary messenger cascade (involving cAMP), which signals vesicles containing aquaporins (water channel proteins) to migrate towards the luminal / apical membrane (facing the filtrate). These vesicles fuse with the luminal membrane via exocytosis, inserting the aquaporins and greatly increasing the membrane's water permeability. Water then moves down its water potential gradient by osmosis out of the tubule and into the cells.

Mark 1: ADH binds to specific / complementary receptors on the basolateral / outer cell surface membrane of the collecting duct cells.
Mark 2: Vesicles containing aquaporins (water channels) are stimulated to move to and fuse with the luminal / apical membrane via exocytosis.
Mark 3: This increases the water permeability of the luminal membrane, allowing water to move down its water potential gradient into the cell / blood by osmosis.

To identify a deletion mutation, researchers would extract DNA and use PCR to amplify the region of the specific Hox gene. Primers are designed to bind specifically to the regions flanking the gene of interest. Once amplified, the PCR products of the patient and a normal (wild-type) control are loaded into an agarose gel. An electric current is applied, and the negatively charged DNA fragments migrate towards the positive anode. Because smaller DNA fragments move faster and further through the gel matrix than larger fragments, a deletion mutation will be detected as a band that has travelled further compared to the normal control band.

Mark 1: Use PCR to amplify the specific Hox gene region using sequence-specific primers flanking the target sequence (from patient and control).
Mark 2: Separate the amplified PCR fragments using gel electrophoresis, where fragments move towards the positive electrode / anode based on size / mass.
Mark 3: A deletion mutation is identified if the patient's DNA fragment is shorter / has migrated further / faster through the gel compared to the normal control fragment.

During times of water stress, the plant hormone abscisic acid (ABA) is produced to minimise transpirational water loss by closing stomata:

1. **Mechanism of Stomatal Closure:**
- ABA binds to specific cell-surface receptors on the plasma membrane of the guard cells.
- This binding activates a signalling cascade that opens calcium ion (
\(\text{Ca}^{2+}\)
) channels, allowing calcium ions to enter the cytoplasm from the vacuole and extracellular space.
- The rise in cytosolic
\(\text{Ca}^{2+}\)
acts as a second messenger, which inhibits proton pumps and activates anion channels.
- This leads to the rapid efflux of potassium ions (
\(\text{K}^+\)
) and negative anions (such as chloride and malate) out of the guard cells into the surrounding extracellular space.
- The loss of these solutes increases the water potential (
\(\psi\)
) inside the guard cells (making it less negative) relative to the surrounding epidermal cells.
- Consequently, water leaves the guard cells down a water potential gradient via osmosis.
- The guard cells lose turgidity, become flaccid, and the stomatal pore closes.

2. **Impact on the Calvin Cycle:**
- Stomatal closure severely restricts the diffusion of carbon dioxide (
\(\text{CO}_2\)
) through the stomata into the sub-stomatal air space and the chloroplast stroma.
- As a result, the concentration of
\(\text{CO}_2\)
in the stroma drops significantly.
- RuBP (ribulose bisphosphate) cannot be carboxylated at normal rates due to the lack of
\(\text{CO}_2\)
, even though the enzyme RuBisCO is present.
- This leads to a decrease in the synthesis of glycerate 3-phosphate (GP).
- Because GP levels fall, less GP is reduced to triose phosphate (TP), despite the availability of ATP and reduced NADP from the light-dependent stage.
- Initially, RuBP levels may rise or remain stable because it continues to be regenerated from existing TP but is not being consumed by carboxylation. However, as the overall cycle slows down due to the depletion of TP, the regeneration of RuBP eventually declines as well.

**Level 3 (5–6 marks)**
- Detailed and accurate description of both the cellular mechanism of ABA-induced stomatal closure (including receptor binding, ion fluxes, and water potential changes) AND a clear explanation of its impact on the concentrations of Calvin cycle intermediates (GP, TP, and RuBP).
- There is a well-developed, logical line of reasoning which is clear and biologically correct.

**Level 2 (3–4 marks)**
- Explains the mechanism of stomatal closure (e.g. movement of
\(\text{K}^+\)
ions and osmosis) and makes a clear link to how reduced
\(\text{CO}_2\)
levels affect the Calvin cycle, but may omit some specific steps (such as calcium acting as a second messenger or the exact fate of all three intermediates: GP, TP, and RuBP).
- The information is presented with some structure and logical sequence.

**Level 1 (1–2 marks)**
- Mentions that ABA causes stomatal closure by water leaving guard cells, and states that photosynthesis/Calvin cycle is reduced due to a lack of carbon dioxide. No detailed biochemical mechanism of ion transport or specific cycle intermediates is provided.
- The explanation may be unstructured or lack scientific terminology.

**0 marks**
- No response worthy of credit.

**Indicative Scientific Content:**
- **ABA mechanism:**
- ABA binds to receptors on guard cell membranes.
- Calcium ions (
\(\text{Ca}^{2+}\)
) enter cytoplasm / act as second messengers.
- Potassium ions (
\(\text{K}^+\)
) leave the guard cells.
- Water potential of guard cells increases / becomes less negative.
- Water leaves by osmosis.
- Guard cells become flaccid and pore closes.
- **Calvin Cycle impact:**
- Stomatal closure stops/reduces
\(\text{CO}_2\)
entry.
- Less
\(\text{CO}_2\)
to combine with RuBP (catalyzed by RuBisCO).
- Concentration of GP decreases.
- Concentration of TP decreases.
- RuBP concentration initially accumulates (as it is not being used to fix
\(\text{CO}_2\)
) but ultimately falls when TP is exhausted.