2024 AQA IAS-Level Biology (9610) Practice Paper with Answers

(a) The water potential of the 0.8 mol dm^{-3} sucrose solution is lower (more negative) than the water potential inside the potato cells. Water moves out of the potato cells by osmosis down a water potential gradient across the selectively permeable cell membrane. This loss of water decreases the turgor pressure of the cells, causing them to flaccidly shrink, which results in a overall decrease in mass and length of the tissue.

(b) Plot a graph with sucrose concentration on the x-axis and percentage change in mass on the y-axis. Draw a line of best fit through the plotted points. Find the concentration where the line crosses the x-axis (where percentage change in mass is zero). At this point, the water potential of the sucrose solution is equal to the water potential of the potato tissue. Use a textbook or reference table to find the water potential value that corresponds to this specific sucrose concentration.

(c) Decrease in length = 40.0 mm - 37.2 mm = 2.8 mm.
Percentage decrease = (2.8 / 40.0) * 100 = 7.0%.

(d) Mass takes into account changes throughout the entire volume of the potato cylinder (all three dimensions), whereas length only measures change along a single dimension. Additionally, potato cylinders may not shrink uniformly along their length, and water loss can occur from all surfaces, making mass a more comprehensive and accurate measurement of water movement.

(a) 1 mark for identifying the water potential gradient (sucrose solution is lower/more negative than potato). 1 mark for stating water moves out by osmosis. 1 mark for referring to selectively permeable membrane or loss of turgidity.

(b) 1 mark for plotting percentage change in mass against concentration and drawing a line of best fit. 1 mark for identifying the x-intercept (where change is zero). 1 mark for stating that this concentration has a water potential equal to the potato and using a reference table to convert concentration to water potential.

(c) 1 mark for calculating correct difference of 2.8 mm. 1 mark for correct calculation of 7.0%.

(d) 1 mark for stating that mass changes reflect changes in all three dimensions/volume. 1 mark for noting that water loss occurs from all surface areas. 0.7 marks for mentioning that length changes may not be uniform or are subject to greater measurement error with a standard ruler.

(a) A phylogenetic hierarchy is a system of grouping organisms based on shared evolutionary history/common ancestry. It consists of nested groups (taxa) that do not overlap, where smaller groups are contained within larger, more inclusive groups.

(b) Cytochrome c is an essential protein found in all eukaryotic organisms, which means its gene has been conserved over evolutionary time. Mutations accumulate in the DNA sequence encoding this protein over time, leading to changes in the amino acid sequence. Species that share a more recent common ancestor will have had less time for mutations to occur, and will therefore show fewer differences in their cytochrome c amino acid sequence. A higher percentage similarity indicates a closer evolutionary relationship.

(c) Step 1: Calculate N (total number of organisms of all species):
N = 12 + 3 + 7 + 8 = 30.
Calculate N(N-1):
30 * 29 = 870.

Step 2: Calculate n(n-1) for each species:
Species A: 12 * 11 = 132
Species B: 3 * 2 = 6
Species C: 7 * 6 = 42
Species D: 8 * 7 = 56
Sum of n(n-1) = 132 + 6 + 42 + 56 = 236.

Step 3: Calculate d:
d = 870 / 236 = 3.6864... (3.69 to 3 significant figures).

(d) Monocultures provide only a single plant species over a large area, meaning there are very few niches, nesting sites, and limited food variety. Removing hedgerows destroys microhabitats and food plants that support diverse insect populations. Consequently, only a few insect species adapted to the single crop can survive and multiply, leading to low species richness and a low index of diversity.

(a) 1 mark for 'based on evolutionary origin/common ancestors'. 1 mark for 'groups within groups with no overlap'.

(b) 1 mark for stating that mutations change amino acid sequences over time. 1 mark for stating that fewer differences in amino acid sequence indicate a more recent common ancestor. 1 mark for explaining that similarity in sequence correlates with closer evolutionary relationship.

(c) 1 mark for correct calculation of N(N-1) = 870. 1 mark for correct sum of n(n-1) = 236. 1 mark for correct final answer of 3.69 (allow 3.7).

(d) 1 mark for stating that monocultures/hedgerow removal reduce plant species diversity. 1 mark for stating this reduces the variety of food sources/habitats/niches available. 0.7 marks for concluding this leads to fewer insect species being able to coexist (reducing species richness).

(a) Three structural differences are:
1. Eukaryotic nuclear DNA is linear, whereas prokaryotic DNA is circular.
2. Eukaryotic DNA is associated with histone proteins, whereas prokaryotic DNA is 'naked' (not associated with histones).
3. Eukaryotic DNA contains non-coding introns within genes, whereas prokaryotic DNA generally lacks introns.

(b) The enzyme is DNA polymerase. It catalyzes condensation reactions that form covalent phosphodiester bonds between the deoxyribose sugar of one nucleotide and the phosphate group of the adjacent nucleotide.

(c) In double-stranded DNA, the amount of cytosine (C) equals the amount of guanine (G).
Thus, C = 34%, so G = 34%.
Together, C + G = 68%.
The remaining bases must be Adenine (A) and Thymine (T):
100% - 68% = 32%.
Since A = T, the percentage of thymine is 32% / 2 = 16%.

(d) Three contrasts between DNA replication and transcription:
1. DNA replication uses DNA nucleotides (containing deoxyribose and thymine), whereas transcription uses RNA nucleotides (containing ribose and uracil).
2. DNA replication utilizes DNA polymerase to synthesize the new strand, while transcription utilizes RNA polymerase.
3. Replication copies both strands of the entire DNA molecule, whereas transcription only copies the template strand of a specific gene/section of DNA.

(a) 1 mark per correct comparison (e.g., linear vs circular, associated with histones vs naked, introns present vs absent) up to 3 marks.

(b) 1 mark for naming DNA polymerase. 1 mark for describing the formation of phosphodiester bonds via condensation reactions.

(c) 1 mark for showing that C = G = 34%, so remaining bases are 32%. 1 mark for the correct final percentage of thymine as 16%.

(d) 1 mark for each valid difference (up to 3 marks): e.g., nucleotides used (deoxyribose/T vs ribose/U), enzymes involved (DNA polymerase vs RNA polymerase), scale (whole genome/both strands vs single gene/one template strand). 0.7 marks for clarity and completeness of contrast.

(a) Each tRNA molecule has a specific amino acid binding site at one end and a specific anticodon loop at the other. tRNA transport specific amino acids to the ribosome. The tRNA anticodon binds to the complementary codon on the mRNA molecule via hydrogen bonding between complementary base pairs. This ensures that amino acids are aligned in the correct primary sequence as determined by the mRNA template.

(b) The codon on the mRNA is 5'-UUU-3'.
The complementary template DNA triplet is AAA (which templates the UUU transcription).
The complementary tRNA anticodon is AAA.

(c) 'Degenerate' means that more than one codon can code for the same amino acid (since there are 64 possible codons but only 20 amino acids). 'Non-overlapping' means that each nucleotide in the sequence is read only once as part of a specific triplet codon; the adjacent triplets do not share any base pairs.

(d) Since each codon consists of three nucleotides, the maximum number of codons is 900 / 3 = 300. This could translate to a maximum of 300 amino acids. However, the actual number might be fewer because the mature mRNA includes a stop codon that does not code for an amino acid, and post-translational modification may remove some amino acids (such as the start methionine) to form the final active protein.

(a) 1 mark for stating tRNA carries a specific amino acid to the ribosome. 1 mark for mentioning the anticodon pairing with the complementary codon on mRNA. 1 mark for stating this ensures the correct sequence of amino acids is joined.

(b) 1 mark for DNA triplet = AAA. 1 mark for tRNA anticodon = AAA.

(c) 1.5 marks for explaining degenerate (more than one triplet codes for one amino acid). 1.5 marks for explaining non-overlapping (each base is read once / triplets are read in sequence without sharing bases).

(d) 1 mark for calculating 300 amino acids. 1 mark for stating that a stop codon is present which does not code for an amino acid. 0.7 marks for mentioning post-translational modifications/removal of start methionine.

(a) Hemoglobin has a low affinity for oxygen when no oxygen is bound because its shape makes it difficult for the first oxygen molecule to bind. Once the first oxygen molecule binds, it changes the quaternary structure of the hemoglobin molecule, exposing the other oxygen-binding sites. This greatly increases the affinity for the second and third oxygen molecules, leading to a rapid increase in saturation (the steep part of the curve).

(b) The shrew's oxygen dissociation curve is shifted to the right compared to the human curve. Shrews have a high surface-area-to-volume ratio, meaning they lose heat rapidly and must maintain a very high metabolic rate to stay warm. A right-shifted curve means their hemoglobin has a lower affinity for oxygen at any given partial pressure of oxygen. This allows oxygen to be unloaded/released much more easily and rapidly to highly respiring tissues.

(c) The Bohr effect occurs when increased carbon dioxide concentration (partial pressure) in the blood lowers the pH. This causes hemoglobin's tertiary structure to change, reducing its affinity for oxygen. During vigorous exercise, active muscles produce large quantities of CO2. The resulting Bohr shift causes hemoglobin to release more oxygen to the exercising muscle tissue where it is urgently needed for aerobic respiration.

(d) Carbon monoxide binds irreversibly to the oxygen-binding iron sites on the heme groups of hemoglobin, forming carboxyhemoglobin. This directly reduces the maximum amount of oxygen that the blood can carry because those binding sites are no longer available for oxygen transport.

(a) 1 mark for stating that binding of the first oxygen molecule changes the shape/quaternary structure of hemoglobin. 1 mark for explaining that this conformational change makes it easier for subsequent oxygen molecules to bind.

(b) 1 mark for identifying that the shrew's curve is shifted to the right. 1 mark for linking the high surface-area-to-volume ratio to high heat loss/high metabolic rate. 1 mark for explaining that high metabolic rate requires rapid aerobic respiration. 1 mark for stating that a lower oxygen affinity allows easier unloading of oxygen to tissues.

(c) 1 mark for stating high CO2 levels lower pH/cause hemoglobin to change shape. 1 mark for stating this reduces hemoglobin's affinity for oxygen. 1 mark for linking this to increased oxygen delivery to highly active/respiring tissues during exercise.

(d) 1 mark for explaining that CO2/carbon monoxide binds to heme sites, reducing available oxygen binding locations. 0.7 marks for noting that this permanently decreases oxygen carrying capacity.

(a) The RER consists of a network of membrane-bound fluid-filled sacs (cisternae) covered with ribosomes, whereas the Golgi apparatus consists of a stack of flattened membrane-bound sacs without ribosomes. The role of the RER is the translation and folding of proteins synthesized by its ribosomes. The role of the Golgi apparatus is the chemical modification (e.g., adding carbohydrates to form glycoproteins), sorting, and packaging of these proteins into secretory vesicles for transport to the cell membrane.

(b) Step 1: Convert both measurements to the same unit (micrometers or millimeters).
Image length = 45 mm = 45,000 \mu m.
Actual length = 1.5 \mu m.
Step 2: Use the formula Magnification = Image size / Actual size.
Magnification = 45,000 / 1.5 = 30,000.
Therefore, the magnification is x30,000.

(c) The eukaryotic plant cell wall is composed of the polysaccharide cellulose, which forms microfibrils to provide high tensile strength. In contrast, the prokaryotic cell wall is composed of murein (a glycoprotein/peptidoglycan). Plant cell walls also contain middle lamellae with pectin, whereas prokaryotic cell walls do not.

(d) A TEM uses a beam of electrons instead of a beam of light. Electrons have a much shorter wavelength than light waves. A shorter wavelength allows for much higher resolution, meaning two close points can be distinguished as separate objects.

(a) 1 mark for structural contrast (ribosomes on RER vs no ribosomes on Golgi / cisternae network vs flattened stack). 1 mark for RER role (protein synthesis/folding). 1 mark for Golgi role (modifying/glycosylating proteins). 1 mark for Golgi role (packaging into vesicles for secretion).

(b) 1 mark for converting units correctly (45 mm to 45,000 \mu m, or 1.5 \mu m to 0.0015 mm). 1 mark for the correct final answer of 30,000 (or x30,000).

(c) 1 mark for plant cell wall composed of cellulose. 1 mark for prokaryotic cell wall composed of murein/peptidoglycan. 1 mark for mentioning structural differences (microfibrils in cellulose vs cross-linked peptidoglycan network).

(d) 1 mark for stating that electrons have a much shorter wavelength than light. 0.7 marks for explaining that shorter wavelength yields a higher resolving power.

(a) A competitive inhibitor has a molecular shape that is highly similar to that of the enzyme's substrate. It binds to the active site of the enzyme, blocking the substrate from entering and preventing the formation of enzyme-substrate complexes. The effect of competitive inhibition can be minimized by significantly increasing the concentration of the substrate. This increases the probability of a substrate molecule colliding with the active site rather than an inhibitor molecule.

(b) Rate of reaction = Change in product concentration / Change in time
Change in concentration = 0.8 mg dm^{-3} - 0.2 mg dm^{-3} = 0.6 mg dm^{-3}.
Change in time = 30 s - 10 s = 20 s.
Rate = 0.6 / 20 = 0.03 mg dm^{-3} s^{-1}.

(c) As temperature increases from 10 ^\circ C to 40 ^\circ C, the kinetic energy of both the enzyme and substrate molecules increases. This causes them to move faster, increasing the frequency of successful collisions between the substrate molecules and the enzyme's active site. Consequently, more enzyme-substrate complexes are formed per unit time, resulting in an increased rate of reaction.

(d) Denaturation involves the breaking of hydrogen bonds, ionic bonds, and disulfide bridges that stabilize the tertiary structure of the enzyme. This alters the specific 3D shape of the active site. As a result, the substrate is no longer complementary in shape to the active site, cannot bind to it, and enzyme-substrate complexes can no longer form.

(a) 1 mark for stating competitive inhibitor has a similar shape to substrate and binds to the active site. 1 mark for stating it prevents enzyme-substrate complexes from forming. 1 mark for stating that increasing substrate concentration overcomes the inhibition.

(b) 1 mark for showing correct changes: concentration difference of 0.6 and time difference of 20. 1 mark for correct calculation of 0.03 mg dm^{-3} s^{-1}.

(c) 1 mark for stating molecules gain kinetic energy. 1 mark for stating there is a higher frequency of successful collisions. 1 mark for stating this leads to more enzyme-substrate complexes forming per unit time.

(d) 1 mark for mentioning the breaking of hydrogen/ionic/disulfide bonds within the enzyme's tertiary structure. 1 mark for stating this changes the shape of the active site. 0.7 marks for explaining that the substrate is no longer complementary, so no enzyme-substrate complexes can form.

(a) Structure of HIV:
- Genetic material consisting of two single strands of RNA.
- Reverse transcriptase (and integrase) enzymes located in the core.
- An inner protein capsid surrounding the genetic material.
- An outer lipid envelope derived from the host cell's membrane.
- Glycoprotein attachment proteins (specifically gp120) embedded in the envelope.

(b) Replication cycle of HIV:
- The gp120 attachment proteins on the virus bind specifically to CD4 receptors on the membrane of a T-helper cell.
- The viral envelope fuses with the host cell membrane, releasing the capsid and viral RNA/enzymes into the cytoplasm.
- Reverse transcriptase converts the single-stranded viral RNA into a double-stranded viral DNA copy.
- Integrase inserts this viral DNA into the host cell's nuclear DNA/genome.
- Host cell enzymes transcribe the viral DNA into viral mRNA, which is translated by host ribosomes into viral proteins.
- New viral particles assemble and bud off from the host cell membrane, taking a portion of the lipid bilayer to form their outer envelope.

(c) Antibiotic vs. Antiretroviral action:
- Penicillin targets bacterial cell wall synthesis (peptidoglycan cross-links). Viruses like HIV do not have cell walls, cell membranes of their own, or metabolic machinery (like ribosomes), making penicillin completely ineffective.
- Antiretroviral drugs target specific components of viral replication. For example, reverse transcriptase inhibitors (like AZT) stop the conversion of viral RNA into DNA, or protease inhibitors prevent the cleavage of viral polyproteins into functional units, halting replication.

Part (a): Max 3 marks
- 1. Presence of two single strands of RNA (1 mark);
- 2. Contains reverse transcriptase (and integrase) inside a protein capsid (1 mark);
- 3. Outer lipid envelope containing glycoprotein / attachment proteins / gp120 (1 mark);

Part (b): Max 4 marks
- 1. gp120 / attachment glycoprotein binds specifically to CD4 receptors on T-helper cell membrane (1 mark);
- 2. Envelope fuses with host membrane and releases viral RNA/capsid into cell (1 mark);
- 3. Reverse transcriptase synthesizes viral DNA from RNA template (1 mark);
- 4. Viral DNA integrated into host genome / host transcribes and translates viral genes to produce new viral proteins and assemble new particles that bud off (1 mark);

Part (c): Max 3.7 marks
- 1. Antibiotics target cell wall synthesis / prokaryotic translation / metabolic pathways (1 mark);
- 2. Viruses lack cell walls / ribosomes / metabolic pathways (1 mark);
- 3. Antiretrovirals target viral-specific enzymes / reverse transcriptase / integrase / protease (1 mark);
- 4. This prevents replication / assembly of new virions, reducing the viral load (0.7 marks).

(a) Atheroma to Myocardial Infarction:
- Damage to the endothelium of a coronary artery (due to high blood pressure, toxins, etc.) triggers an inflammatory response.
- White blood cells (macrophages) and lipids (specifically LDL cholesterol) accumulate under the endothelium, forming an atheroma / fatty streak.
- Calcium deposits and fibrous tissue build up, forming a hard plaque that narrows the lumen of the coronary artery and restricts blood flow.
- The plaque may rupture, exposing collagen and triggering a clotting cascade that forms a thrombus (blood clot).
- If the thrombus completely blocks the coronary artery, blood flow to downstream cardiac muscle tissue (myocardium) is cut off.
- This deprives the cardiac muscle cells of oxygen and glucose, stopping aerobic respiration; cells are forced to respire anaerobically, run out of ATP, and die (myocardial infarction).

(b) Prospective Cohort Study Design:
- Recruit a large, representative sample of healthy individuals with no prior history of coronary heart disease.
- Quantify their baseline dietary saturated fat intake using food frequency questionnaires or food diaries.
- Monitor the participants over a long follow-up period (e.g., several years or decades).
- Record any new diagnoses of CHD or cardiovascular-related deaths within the cohort.
- Control or adjust for confounding variables (e.g., age, smoking status, physical activity level, genetics).
- Analyze the statistical relationship / relative risk between the levels of saturated fat intake and the incidence of CHD.

(c) Other Lifestyle Factors:
- Cigarette smoking: Contains nicotine, which stimulates adrenaline release, increasing heart rate and blood pressure, and making platelets stickier (increasing clotting risk). It also contains carbon monoxide, which binds irreversibly to hemoglobin, reducing oxygen carriage and forcing the heart to work harder.
- High blood pressure (hypertension) / lack of exercise: Increases the shear stress on arterial walls, directly damaging the endothelium and initiating the atherosclerotic process. Physical inactivity also increases resting blood pressure and LDL levels.

Part (a): Max 4 marks
- 1. Endothelial damage leads to lipid/cholesterol accumulation forming an atheroma (1 mark);
- 2. Plaque hardens and narrows the lumen, restricting coronary blood flow (1 mark);
- 3. Rupture of plaque initiates blood clotting cascade, forming a thrombus (1 mark);
- 4. Blockage of coronary artery deprives heart muscle (myocardium) of oxygen, preventing aerobic respiration and causing cell death/infarction (1 mark);

Part (b): Max 3 marks
- 1. Recruit large, healthy cohort free of CHD at the start (1 mark);
- 2. Assess baseline saturated fat intake and monitor over a long time period (1 mark);
- 3. Record CHD incidence and control for confounding variables / age / exercise / smoking (1 mark);

Part (c): Max 3.7 marks
- 1. Identify Factor 1 (e.g., smoking) and describe its biological effect (nicotine increases heart rate/constricts vessels OR carbon monoxide reduces oxygen capacity) (1.5 marks);
- 2. Identify Factor 2 (e.g., high stress / lack of exercise) and describe its biological effect (high blood pressure damages endothelium, accelerating plaque formation) (1.5 marks);
- 3. Clear link between endothelial damage/thrombosis and increased cardiovascular disease risk (0.7 marks).

(a) Pressure changes and valve actions:
- Atrial systole: The left atrium contracts, raising atrial pressure above ventricular pressure. This opens the bicuspid (atrioventricular) valve, forcing blood into the ventricle.
- Ventricular systole: The left ventricle contracts, rapidly increasing ventricular pressure. As soon as ventricular pressure exceeds atrial pressure, the bicuspid valve closes to prevent backflow of blood into the atrium.
- As ventricular contraction continues, ventricular pressure rises above the pressure in the aorta. This opens the semilunar valve, and blood is ejected into the aorta.
- Diastole: The ventricle relaxes, causing ventricular pressure to fall below aortic pressure. This pressure difference forces the semilunar valve to close, preventing backflow of blood into the ventricle.

(b) Artery vs. Vein structure-function relationships:
- Thick tunica media in arteries: Contains more smooth muscle and elastic fibers. Elastic fibers allow arteries to stretch and recoil to smooth out blood flow and maintain high blood pressure. Veins have a much thinner tunica media because they carry blood under low pressure.
- Lumen size: Arteries have a relatively narrow lumen to maintain high pressure, whereas veins have a wide lumen to minimize resistance to blood flow under low pressure.
- Valves: Veins contain pocket valves along their length to ensure blood flows unidirectionally back to the heart against gravity. Arteries do not have valves (except at the base of the aorta/pulmonary artery) because the high pressure of blood prevents backflow.

(c) Calculation:
- Formula: \(\text{Cardiac Output} = \text{Stroke Volume} \times \text{Heart Rate}\)
- Stroke volume in \(\text{dm}^3\): \(80\text{ cm}^3 / 1000 = 0.080\text{ dm}^3\)
- Calculation: \(0.080\text{ dm}^3 \times 72\text{ beats min}^{-1} = 5.76\text{ dm}^3\text{ min}^{-1}\)
- Working in \(\text{cm}^3\) first: \(80 \times 72 = 5760\text{ cm}^3\text{ min}^{-1}\), and dividing by 1000 gives \(5.76\text{ dm}^3\text{ min}^{-1}\).

Part (a): Max 4 marks
- 1. Atrial pressure exceeding ventricular pressure forces atrioventricular (AV) valve open (1 mark);
- 2. Ventricular contraction raises ventricular pressure above atrial pressure, closing the AV valve to prevent backflow (1 mark);
- 3. Ventricular pressure exceeding aortic pressure opens the semilunar valve (1 mark);
- 4. Ventricular relaxation causes ventricular pressure to fall below aortic pressure, closing the semilunar valve (1 mark);

Part (b): Max 3.7 marks
- 1. Artery has thicker elastic tissue / muscle layer to withstand and maintain high pressure (1 mark);
- 2. Vein has wider lumen to reduce resistance to blood flow (1 mark);
- 3. Veins contain pocket valves to prevent backflow under low pressure, which arteries lack (1 mark);
- 4. Correct contrast of structural layers (tunica media comparison) (0.7 marks);

Part (c): Max 3 marks
- 1. State correct formula: \(\text{Cardiac Output} = \text{Stroke Volume} \times \text{Heart Rate}\) (1 mark);
- 2. Correct conversion of units (e.g. dividing by 1000) (1 mark);
- 3. Correct final answer of \(5.76\) (1 mark).

(a) Phagocytosis and Antigen Presentation:
- Chemotaxis attracts the phagocyte (e.g., neutrophil or macrophage) to chemical signals released by the pathogen.
- Receptor proteins on the phagocyte cell-surface membrane bind to the foreign antigens on the pathogen's surface.
- The phagocyte extends pseudopodia to engulf the pathogen, trapping it in a membrane-bound vesicle called a phagosome.
- Lysosomes within the phagocyte cytoplasm fuse with the phagosome to form a phagolysosome, releasing hydrolytic enzymes (lysozymes) that digest the pathogen.
- The phagocyte processes the pathogen's antigens and displays them on its own cell-surface membrane using major histocompatibility complex (MHC) proteins, becoming an Antigen-Presenting Cell (APC).

(b) Humoral Immune Response:
- T-helper cells with complementary receptors bind to the antigen displayed on the APC.
- This binding activates the T-helper cells, stimulating them to secrete signaling chemicals called cytokines.
- Cytokines bind to complementary B-lymphocytes that have already bound the specific pathogen antigen (clonal selection).
- Activated B-lymphocytes undergo rapid mitotic division (clonal expansion) to produce two populations of cells:
- Plasma cells: synthesize and secrete large quantities of specific antibodies complementary to the pathogen's antigen.
- Memory B-cells: remain in circulation for long-term immunity, capable of rapid differentiation into plasma cells upon re-exposure.

(c) Active vs. Passive Immunity:
- Active immunity: The individual's own immune system is stimulated to produce antibodies and memory cells in response to foreign antigens. Example: Natural infection by a pathogen or artificial vaccination using weakened/dead pathogens. This provides long-term protection.
- Passive immunity: The individual receives pre-formed antibodies from an external source; no memory cells are produced. Example: Natural transfer of antibodies across the placenta or through colostrum/breast milk, or artificial injection of antivenoms/monoclonal antibodies. This provides immediate but short-term protection.

Part (a): Max 4 marks
- 1. Chemotaxis / binding of phagocyte receptors to foreign antigens (1 mark);
- 2. Engulfment of pathogen into a phagosome (1 mark);
- 3. Fusion of lysosome and digestion by hydrolytic enzymes / lysozymes (1 mark);
- 4. Insertion and display of foreign antigens on the phagocyte cell-surface membrane to form an APC (1 mark);

Part (b): Max 3.7 marks
- 1. T-helper cells bind to APC and secrete cytokines (1 mark);
- 2. Cytokines stimulate complementary B-cells to undergo clonal selection / mitosis (1 mark);
- 3. Plasma cells produce and secrete monoclonal antibodies (1 mark);
- 4. Memory B-cells produced for rapid secondary response (0.7 marks);

Part (c): Max 3 marks
- 1. Active immunity involves antibody production by self AND results in memory cell formation (1 mark);
- 2. Passive immunity involves introduction of external antibodies AND does not produce memory cells (1 mark);
- 3. Accurate examples for both types provided (e.g., vaccine for active, breast milk/injection for passive) (1 mark).

(a) Aphid feeding mechanism:
- Aphids target the phloem sieve tube elements, which transport organic solutes (mainly sucrose) through the plant.
- They utilize specialized piercing-sucking mouthparts called stylets to penetrate the plant stem tissues (epidermis, cortex) and enter a single sieve tube element.
- Active transport of sucrose into the phloem at source tissues lowers the water potential, causing water to enter by osmosis. This builds up high positive hydrostatic pressure inside the phloem sieve tubes.
- When the stylet punctures the phloem cell, this hydrostatic pressure is high enough to force the nutrient-rich phloem sap directly up the stylet and into the aphid's digestive system, requiring minimal active sucking by the insect.

(b) Non-persistent vs. Persistent virus transmission:
- Non-persistent viruses: These viral particles adhere to the cuticle lining the aphid's stylet/mouthparts during feeding. They can be immediately transmitted to a healthy plant when the aphid probes its tissue. The virus is quickly lost from the stylet within hours or after a few probes.
- Persistent (circulative) viruses: These must be ingested by the aphid during feeding. The virus passes through the gut wall into the hemolymph (circulatory fluid) and travels to the salivary glands. Transmission occurs when the aphid secretes infected saliva during subsequent feeding. This requires a latency/incubation period inside the vector, but the aphid remains infectious for days, weeks, or its entire lifespan.

(c) Biological and Chemical Control:
- Biological control: Release natural predators of aphids, such as ladybirds (*Coccinellidae*) or lacewings, or release parasitoid wasps (e.g., *Aphidius* species). These organisms feed on or parasitize the aphids, lowering their population density, which directly reduces the frequency of virus transmission between plants.
- Chemical control: Use systemic insecticides. These chemicals are absorbed by the plant and transported in the phloem. When aphids feed on the phloem sap, they ingest the toxic insecticide and die, preventing them from migrating and spreading the virus to healthy plants.

Part (a): Max 4 marks
- 1. Target tissue is phloem sieve tubes / elements (1 mark);
- 2. Piercing mouthparts called stylets are used to penetrate tissues (1 mark);
- 3. High hydrostatic pressure in phloem is generated by active loading of solutes/osmotic water entry (1 mark);
- 4. Hydrostatic pressure forces phloem sap into the aphid's alimentary canal (1 mark);

Part (b): Max 3.7 marks
- 1. Non-persistent: virus is carried on the surface of the stylet / mouthparts and transmitted rapidly during probing (1 mark);
- 2. Persistent: virus is ingested, passes through gut wall into hemolymph, and reaches salivary glands (1 mark);
- 3. Persistent requires an incubation / latency period before transmission can occur, but lasts much longer (1 mark);
- 4. Accurate contrast of the durability of infectivity (non-persistent lost quickly, persistent lasts for days/life) (0.7 marks);

Part (c): Max 3 marks
- 1. Biological: introduction of natural predators/parasitoids (e.g., ladybirds/wasps) to reduce vector populations (1 mark);
- 2. Chemical: application of systemic insecticides that kill aphids upon feeding (1 mark);
- 3. Explanation of how either control method reduces virus transmission rate (by reducing vector abundance / movement) (1 mark).

(a) Cohesion-Tension Theory:
- Water vapor evaporates from the wet cell walls of mesophyll cells into the air spaces, and diffuses out of the stomata (transpiration).
- This lowers the water potential of these leaf cells, pulling water from adjacent xylem vessels by osmosis.
- This pulling action creates tension (negative pressure) at the top of the xylem column.
- Water molecules are polar and form hydrogen bonds with one another (cohesion), creating a strong, continuous column of water extending from the roots up to the leaves.
- As water is pulled upwards, adhesion between water molecules and the hydrophilic cellulose in the xylem walls prevents the water column from dropping due to gravity, and prevents the vessel from collapsing.

(b) Phloem Loading and Mass Flow:
- Companion cells use ATP to actively pump hydrogen ions (\(\text{H}^+\)) out into the surrounding cell wall space.
- This establishes an electrochemical gradient. Hydrogen ions diffuse back into the companion cell through a co-transporter protein, bringing sucrose molecules with them against their concentration gradient.
- Sucrose then diffuses into the phloem sieve tube elements via plasmodesmata, lowering the water potential inside the sieve tube.
- Water moves from the nearby xylem vessels into the phloem by osmosis, creating high hydrostatic pressure at the source.
- At the sink, sucrose is unloaded and converted to starch or used in respiration, raising the water potential so water leaves the phloem. This creates low hydrostatic pressure.
- The resulting hydrostatic pressure gradient forces the phloem sap to flow from source to sink by mass flow.

(c) Potometer precautions:
- Cut the leafy shoot underwater: This prevents air from entering the xylem vessels, which would break the continuous column of water needed for cohesion-tension.
- Assemble the potometer completely underwater: This prevents any air bubbles from becoming trapped in the capillary tube or reservoir.
- Seal all joints with petroleum jelly (Vaseline): This ensures the apparatus is completely airtight and watertight, preventing leakage or air intake that would affect bubble movement.
- Dry the leaves after assembly: Leftover water on the leaves would artificially decrease the transpiration rate by lowering the concentration gradient of water vapor.

Part (a): Max 4 marks
- 1. Water evaporates / transpires from leaves via stomata, lowering water potential of mesophyll (1 mark);
- 2. Pulling force creates tension (negative pressure) in xylem (1 mark);
- 3. Cohesion due to hydrogen bonding between polar water molecules forms a continuous water column (1 mark);
- 4. Adhesion between water molecules and hydrophilic xylem walls supports the water column (1 mark);

Part (b): Max 3.7 marks
- 1. Active transport of \(\text{H}^+\) out of companion cells creates a gradient for sucrose co-transport (1 mark);
- 2. High sucrose concentration in sieve tube lowers water potential, causing water to enter from xylem by osmosis (1 mark);
- 3. This increases hydrostatic pressure at the source (1 mark);
- 4. Unloading of sucrose at the sink reduces hydrostatic pressure, establishing a pressure gradient for mass flow (0.7 marks);

Part (c): Max 3 marks
- 1. Cut shoot underwater to prevent air blocks in xylem vessels (1 mark);
- 2. Assemble potometer underwater / ensure no air bubbles are present except the measured indicator bubble (1 mark);
- 3. Seal joints with petroleum jelly / dry leaves to maintain normal humidity gradients (1 mark).

(a) Chromosome behavior in Prophase and Metaphase:
- Prophase: Chromatin condenses, shortening and thickening into visible chromosomes. Each chromosome consists of two genetically identical sister chromatids joined together at a region called the centromere. The nuclear envelope breaks down and the nucleolus disappears.
- Metaphase: Centrioles reach opposite poles of the cell and form a spindle apparatus made of microtubules. Spindle fibers attach to the centromere of each chromosome. The chromosomes are pulled by spindle fibers to line up individually along the equator (metaphase plate) of the cell.

(b) Mitosis vs. Binary Fission:
- Mitosis involves the division of linear DNA packaged into multiple chromosomes; binary fission involves the replication and division of a single circular DNA molecule and any accessory plasmids.
- Mitosis requires the formation of a spindle apparatus to pull sister chromatids apart; binary fission does not use a spindle apparatus (DNA molecules are attached to the plasma membrane as they replicate and separate).
- Mitosis involves the breakdown and reformation of a nuclear envelope; prokaryotes lack a nucleus, so no nuclear envelope breakdown occurs during binary fission.
- Cytokinesis in animal cells involves a cleavage furrow, whereas binary fission involves the inward growth of the plasma membrane and cell wall to form a septum, dividing the cell into two.

(c) Calculation:
- Determine the proportion of cells in metaphase:
\(\text{Proportion} = 32 / 400 = 0.08\) (or \(8\%\))
- Convert the total cell cycle duration from hours to minutes:
\(18\text{ hours} \times 60\text{ minutes/hour} = 1080\text{ minutes}\)
- Calculate the duration of metaphase:
\(\text{Duration} = 0.08 \times 1080\text{ minutes} = 86.4\text{ minutes}\).

Part (a): Max 4 marks
- 1. Prophase: Chromosomes condense / become visible (1 mark);
- 2. Chromosomes appear as sister chromatids held by a centromere (1 mark);
- 3. Nuclear envelope disintegrates (1 mark);
- 4. Metaphase: Chromosomes line up individually along the equator / metaphase plate of the cell (1 mark);
- 5. Spindle fibers attach to the centromeres (1 mark);

Part (b): Max 3.7 marks
- 1. Mitosis involves linear chromosomes/DNA; binary fission involves a single circular chromosome (1 mark);
- 2. Mitosis utilizes a spindle apparatus to separate chromatids; binary fission does not (1 mark);
- 3. Mitosis involves nuclear envelope breakdown; binary fission has no nucleus/envelope to break down (1 mark);
- 4. Binary fission involves plasmid replication, which is absent in mitosis (0.7 marks);

Part (c): Max 3 marks
- 1. Correct calculation of metaphase cell proportion (\(32 / 400 = 0.08\) or \(8\%\)) (1 mark);
- 2. Correct conversion of cell cycle time to minutes (\(18 \times 60 = 1080\text{ min}\)) (1 mark);
- 3. Correct final answer of \(86.4\) minutes (accept \(86\) or \(86.4\)) (1 mark).

Part (a): Total population size N = 350. Dark phenotype = 294. Light phenotype (homozygous recessive, dd) = 350 - 294 = 56. Frequency of homozygous recessive genotype \(q^2 = 56 / 350 = 0.16\). Therefore, \(q = \sqrt{0.16} = 0.40\). Since \(p + q = 1\), \(p = 1 - 0.40 = 0.60\). Part (b): Heterozygous frequency = \(2pq = 2 \times 0.60 \times 0.40 = 0.48\). Expected number of heterozygotes = \(0.48 \times 350 = 168\). Part (c): Conditions include: 1. Large population size to eliminate genetic drift. 2. Random mating within the population. 3. No mutations to introduce new alleles. 4. No natural selection favoring one phenotype over another. 5. No migration (no gene flow) into or out of the population.

Part (a) [3 marks]: 1 mark for calculating light mice count (56) and homozygous recessive frequency (0.16). 1 mark for correct calculation of recessive allele frequency (q = 0.40). 1 mark for correct calculation of dominant allele frequency (p = 0.60). Part (b) [3 marks]: 1 mark for stating or showing the heterozygote frequency calculation formula (2pq). 1 mark for obtaining the heterozygote frequency of 0.48. 1 mark for the final correct count of 168 mice. Part (c) [4.7 marks]: 1.5 marks each for any three valid conditions described in detail (total 4.5 marks) plus 0.2 marks for clear biological expression. Acceptable conditions: large population, random mating, no migration (isolation), no mutation, no selection.

Part (a): In ethanol fermentation, pyruvate is decarboxylated to ethanal, which is then reduced to ethanol by alcohol dehydrogenase. This reaction oxidizes reduced NAD (NADH) back to NAD. This free NAD is essential as a coenzyme and hydrogen acceptor in glycolysis (specifically for the oxidation of triose phosphate), allowing ATP generation to continue via substrate-level phosphorylation. Part (b): Glycolysis yields a net of 2 ATP directly. The Krebs cycle produces 2 ATP directly per molecule of glucose (1 ATP per turn of the cycle). Part (c): Oligomycin blocks the proton channel of ATP synthase. Protons (H+) can still be pumped into the intermembrane space by the electron transport chain, but they cannot flow back into the matrix. This leads to an accumulation of H+ in the intermembrane space, which decreases the pH. Without the flow of protons down their electrochemical gradient through ATP synthase, ADP cannot be phosphorylated to ATP, stopping ATP synthesis.

Part (a) [3 marks]: 1 mark for stating reduced NAD is oxidized back to NAD during the reduction of ethanal to ethanol. 1 mark for stating NAD is required as a hydrogen acceptor/coenzyme in glycolysis. 1 mark for stating this allows glycolysis to continue producing ATP via substrate-level phosphorylation. Part (b) [2.7 marks]: 1.35 marks for identifying net 2 ATP from glycolysis; 1.35 marks for identifying 2 ATP from the Krebs cycle. Part (c) [5 marks]: 1 mark for stating protons continue to be pumped into the intermembrane space by the electron transport chain. 1 mark for stating protons cannot pass back through ATP synthase. 1 mark for explaining this leads to an accumulation of protons, decreasing pH in the intermembrane space. 1 mark for stating the proton gradient increases or cannot be dissipated. 1 mark for explaining that without proton flow through ATP synthase, there is no energy for ADP phosphorylation, stopping ATP synthesis.

Part (a): Carbon dioxide (1C) combines with the five-carbon sugar ribulose bisphosphate (RuBP, 5C). This reaction is catalyzed by the enzyme rubisco (ribulose bisphosphate carboxylase). This produces an unstable six-carbon intermediate, which immediately splits into two molecules of the three-carbon compound glycerate 3-phosphate (GP). Part (b): Normally, in the dark, RuBP concentration decreases because there is no light-dependent reaction to produce ATP and reduced NADP needed to regenerate RuBP from GP, while RuBP continues to be converted to GP by reacting with CO2. However, if CO2 is reduced to zero, RuBP cannot be converted to GP (carbon fixation stops). Any remaining ATP and reduced NADP from the light period can still convert remaining GP to TP and then to RuBP, leading to an immediate increase in RuBP levels. Part (c): 5 out of every 6 molecules of TP (3C) are rearranged to form 3 molecules of RuBP (5C) using ATP. This regeneration is essential because without RuBP, carbon dioxide fixation cannot continue, causing the Calvin cycle and photosynthesis to stop.

Part (a) [4 marks]: 1 mark for CO2 combining with RuBP. 1 mark for enzyme Rubisco. 1 mark for formation of unstable 6C intermediate. 1 mark for split into two molecules of GP. Part (b) [3.7 marks]: 1.2 marks for stating that zero CO2 means no RuBP can react with CO2 to form GP. 1.2 marks for stating that any residual ATP/reduced NADP from the light reaction continues to convert remaining GP/TP into RuBP. 1.3 marks for concluding that the rate of RuBP synthesis exceeds its consumption, causing its concentration to rise. Part (c) [3 marks]: 1 mark for stating that 5/6 of TP molecules are recycled to regenerate RuBP. 1 mark for stating this process requires ATP. 1 mark for explaining that regeneration is critical to allow continuous fixation of carbon dioxide.

Part (a): Saprobionts decompose dead organic matter, urea, and feces. They secrete extracellular enzymes to break down proteins/nucleic acids into amino acids, which they absorb. They then release nitrogen as ammonium ions (NH4+) into the soil in a process called ammonification. Part (b): Nitrifying bacteria require aerobic (oxygen-rich, well-aerated) soil conditions. They carry out nitrification, converting ammonium ions to nitrites, and then nitrites to nitrates. Denitrifying bacteria require anaerobic (waterlogged, oxygen-depleted) soil conditions. They carry out denitrification, converting soil nitrates back into nitrogen gas (N2) which escapes into the atmosphere. Part (c): Legumes have a mutualistic relationship with nitrogen-fixing bacteria (Rhizobium) inside their root nodules. These bacteria fix atmospheric nitrogen gas into organic nitrogen compounds (amino acids/ammonium). When the legumes die, decay, or are ploughed back into the soil, saprobionts and nitrifying bacteria convert these organic nitrogen compounds into nitrates, significantly increasing soil fertility.

Part (a) [3 marks]: 1 mark for extracellular digestion/decomposition of organic matter (proteins/urea/DNA). 1 mark for absorption of digestion products. 1 mark for releasing ammonium ions (ammonification). Part (b) [4.7 marks]: 1.2 marks for nitrifying bacteria requiring aerobic/oxygen-rich soil and converting ammonium to nitrites/nitrates. 1.2 marks for denitrifying bacteria requiring anaerobic/waterlogged/oxygen-deficient soil and converting nitrates to nitrogen gas. 1.2 marks for contrasting the conditions explicitly (aerobic vs anaerobic). 1.1 marks for stating how farmers prevent denitrification (e.g. ploughing to aerate soil). Part (c) [3 marks]: 1 mark for identifying nitrogen-fixing bacteria in the root nodules of legumes. 1 mark for stating that these bacteria convert nitrogen gas to ammonium/amino acids. 1 mark for explaining that decay of legumes adds organic nitrogen which is converted to nitrates by nitrification, enriching the soil.

Part (a): Using the Lincoln Index formula: \(N = \frac{M \times C}{R}\), where M is the number in the first sample (80), C is the number in the second sample (75), and R is the number of marked individuals recaptured (15). \(N = \frac{80 \times 75}{15} = \frac{6000}{15} = 400\). Part (b): Assumptions include: 1. No significant migration (immigration or emigration) or births/deaths occurred during the 5 days. 2. The mark does not affect the survival rate or behavior of the beetles (e.g., does not make them more visible to predators). 3. The marked individuals have distributed themselves evenly throughout the population and have fully integrated before recapture. Part (c): Density-dependent factors are those whose effects on the size or growth of the population vary with the population density (e.g., disease transmission, competition for food). Density-independent factors affect population size regardless of population density (e.g., extreme weather events like a severe frost or forest fire).

Part (a) [3 marks]: 1 mark for correct formula or substituting correct values into the equation: (80 * 75) / 15. 1 mark for intermediate calculation of 6000. 1 mark for correct answer of 400. Part (b) [4.7 marks]: 1.5 marks each for any three valid assumptions described (total 4.5 marks) plus 0.2 marks for clear communication. Examples: no birth/death, no immigration/emigration, marking does not affect predation, marking is durable and does not rub off, marked beetles mix randomly with the population. Part (c) [3 marks]: 1 mark for defining density-dependent factors and providing an example (e.g., disease, competition). 1 mark for defining density-independent factors and providing an example (e.g., temperature, drought). 1 mark for contrasting their dependency on population density.

Part (a): \(NPP = GPP - R = 24000 - 14400 = 9600\text{ kJ m}^{-2}\text{ yr}^{-1}\). Part (b): Net productivity (N) of consumers is calculated as \(N = I - (F + R)\), where I is chemical energy in ingested food (1800), F is chemical energy lost in faeces and urine (540), and R is respiratory losses (900). \(N = 1800 - (540 + 900) = 1800 - 1440 = 360\text{ kJ m}^{-2}\text{ yr}^{-1}\). Part (c): Energy transfer between trophic levels is highly inefficient, typically only around 10% of energy is passed on. Energy is lost due to: 1. Parts of organisms not being consumed (e.g. roots, bones). 2. Indigestible material lost in faeces. 3. Heat lost to the environment during respiration. As a result, after four or five trophic levels, there is insufficient energy remaining to support another viable breeding population of consumers.

Part (a) [2 marks]: 1 mark for correct subtraction (24000 - 14400 = 9600). 1 mark for correct units (kJ m^-2 yr^-1 or kJ m^-2 y^-1). Part (b) [3.7 marks]: 1.2 marks for stating the correct equation N = I - (F + R). 1.2 marks for correct substitution: 1800 - (540 + 900). 1.3 marks for correct answer of 360 (or 360 kJ m^-2 yr^-1). Part (c) [5 marks]: 1 mark for stating that energy transfer is inefficient / only ~10% transfer. 1 mark for energy lost in faeces/urine (egested/excreted). 1 mark for energy lost as heat from respiration. 1 mark for some parts of organisms being uneaten. 1 mark for explaining that after 4-5 levels, there is too little energy left to sustain a viable population.

Part (a): 1. Geographic isolation prevents gene flow between the two populations. 2. The two populations experience different environmental conditions and therefore different selection pressures. 3. Random mutations occur independently in each population, generating new alleles. 4. Natural selection favors different alleles in each environment, altering allele frequencies. 5. Over time, genetic divergence occurs until the populations become reproductively isolated and can no longer interbreed to produce fertile offspring. Part (b): Allopatric speciation occurs when populations are geographically separated. Sympatric speciation occurs when populations remain in the same geographic area but become reproductively isolated. Mechanisms of sympatric speciation include: 1. Behavioral isolation (e.g., changes in courtship behaviors). 2. Temporal isolation (e.g., breeding at different times of day/year). 3. Ecological/niche isolation (exploiting different niches). Part (c): A gene pool is the complete set of all the alleles of all the genes present in a single population at any given time.

Part (a) [5 marks]: 1 mark for geographic isolation preventing gene flow. 1 mark for different environments having different selection pressures. 1 mark for mutation introducing new alleles. 1 mark for natural selection changing allele frequencies. 1 mark for genetic divergence leading to reproductive isolation (unable to breed to produce fertile offspring). Part (b) [3.7 marks]: 1.2 marks for contrasting geographic separation (allopatric) vs same geographic area (sympatric). 1.2 marks for describing one specific sympatric mechanism (e.g., behavioral, temporal, or ecological isolation). 1.3 marks for explaining how this mechanism prevents interbreeding/gene flow. Part (c) [2 marks]: 1 mark for 'all alleles'. 1 mark for 'in a population / at a given time'.