2025 AQA IAL Biology (9610) 模擬試題連答案詳解

(a) A triglyceride is formed by condensation reactions between one glycerol molecule and three fatty acid molecules. Each reaction forms an ester bond with the release of a water molecule. (b) A phospholipid contains one glycerol molecule bonded to two fatty acid chains and one phosphate group, unlike a triglyceride which has three fatty acids. The phosphate head is polar and hydrophilic, while the fatty acid tails are non-polar and hydrophobic. This causes phospholipids to arrange into a bilayer in cell membranes, with heads facing outwards towards water and tails facing inwards, creating a hydrophobic core that regulates the movement of water-soluble substances. (c) First, add ethanol to the sample in a clean test tube and shake thoroughly to dissolve any lipids. Then, pour the liquid mixture into a test tube containing water and mix. A positive result is indicated by the formation of a milky white/cloudy emulsion. (d) Triglycerides are highly insoluble in water because they are non-polar, hydrophobic molecules that cannot form hydrogen bonds with water. They are soluble in ethanol because ethanol is an organic solvent with non-polar characteristics that can dissolve hydrophobic substances.

(a) 1 mark for condensation reaction (or loss of 3 water molecules), 1 mark for 1 glycerol and 3 fatty acids, 1 mark for formation of 3 ester bonds.
(b) 1 mark for identifying phosphate group replaces one fatty acid, 1 mark for polar/hydrophilic head and non-polar/hydrophobic tails, 1 mark for forming a bilayer (heads facing out, tails in), 1 mark for creating a barrier to water-soluble/polar substances.
(c) 1 mark for adding ethanol and shaking/mixing, 1 mark for adding the mixture to water, 1.5 marks for describing the positive observation as a white/milky/cloudy emulsion.
(d) 1 mark for stating insoluble in water due to being non-polar/hydrophobic, 1 mark for stating soluble in ethanol because it is an organic/less polar solvent.

(a) Water evaporates from the mesophyll cell walls into the air spaces and diffuses out through stomata (transpiration). This lowers water potential in mesophyll cells, causing water to move from xylem by osmosis. Water molecules are cohesive due to hydrogen bonding, forming a continuous column of water up the xylem. Adhesion between water molecules and the xylem wall supports the column. This creates tension (negative pressure) that pulls the water column upwards. (b) Diameter = \(0.8\text{ mm}\), so radius \(r = 0.4\text{ mm}\). Cross-sectional area of capillary tube \(A = \pi r^2 = 3.14 \times (0.4)^2 = 0.5024\text{ mm}^2\). Volume of water taken up \(V = A \times \text{distance} = 0.5024 \times 45 = 22.608\text{ mm}^3\). Rate of water uptake = \(22.608 \div 15 = 1.5072\text{ mm}^3\text{ min}^{-1}\), which rounds to \(1.51\text{ mm}^3\text{ min}^{-1}\). (c) The potometer measures water uptake, but not all water absorbed is lost by transpiration. Some water is used in photosynthesis to produce glucose, some is used to maintain cell turgor pressure (support), and some is produced as a byproduct of aerobic respiration. (d) Higher temperature increases the kinetic energy of water molecules, increasing the rate of evaporation and diffusion. High wind speed blows away humid air near stomata, keeping the air dry and maintaining a steep water potential gradient between the leaf interior and atmosphere.

(a) 1 mark for transpiration of water from leaves, 1 mark for hydrogen bonding creating cohesion between water molecules, 1 mark for continuous water column in xylem, 1 mark for adhesion of water to xylem walls, 1 mark for tension/negative pressure pulling the column up.
(b) 1 mark for calculating radius (0.4 mm), 1 mark for calculating cross-sectional area (0.5024 mm^2), 1 mark for calculating total volume of water (22.608 mm^3), 0.5 marks for final rate calculation (1.51 mm^3 min^-1; accept 1.5).
(c) 1 mark for stating water is used in photosynthesis / maintaining cell turgidity, 1 mark for stating water is produced during respiration.
(d) 1 mark for stating higher temperature increases kinetic energy of water molecules (increasing rate of diffusion), 1 mark for stating wind speed removes water vapor, maintaining a steep water potential gradient.

(a) Sodium ions (\(\text{Na}^+\)) are actively transported out of the epithelial cells into the blood by the sodium-potassium pump. This actively lowers the concentration of sodium ions inside the cell, establishing a concentration gradient between the lumen and the cell. Sodium ions in the lumen bind to a co-transporter protein and diffuse down their concentration gradient into the epithelial cell, bringing glucose molecules into the cell alongside them against the glucose concentration gradient. Glucose then leaves the epithelial cell into the blood down its concentration gradient via facilitated diffusion. (b) Simple diffusion involves the passive net movement of small, non-polar molecules directly through the phospholipid bilayer. Facilitated diffusion is also passive but requires specific carrier or channel proteins to transport larger, polar, or charged molecules that cannot pass through the hydrophobic core of the membrane. (c) Change in mass = \(2.16\text{ g} - 2.40\text{ g} = -0.24\text{ g}\). Percentage change = \((-0.24 \div 2.40) \times 100 = -10.0\%\). Since the tissue lost mass, water must have moved out of the tissue into the solution by osmosis. This means the water potential of the solution was lower (more negative) than the water potential of the tissue.

(a) 1 mark for sodium ions actively transported out of cell into blood (by sodium-potassium pump), 1 mark for creating a low concentration of sodium ions in the cell / establishing a concentration gradient, 1 mark for sodium ions diffusing from the lumen into the cell through a co-transporter protein, 1 mark for glucose being carried in with sodium ions against its concentration gradient, 1 mark for glucose moving into blood by facilitated diffusion.
(b) 1.5 marks for stating simple diffusion occurs directly through bilayer whereas facilitated diffusion requires specific channel/carrier proteins, 1 mark for simple diffusion of non-polar/lipid-soluble molecules vs facilitated of polar/charged/large molecules, 1 mark for simple diffusion rate is linear with concentration whereas facilitated diffusion plateaus when proteins are saturated.
(c) 1 mark for change in mass calculation (-0.24 g), 1 mark for percentage change calculation (-10% or 10% loss), 1 mark for deducing that water potential of the solution was lower (more negative) than the tissue, 1 mark for stating water moved out of tissue by osmosis down water potential gradient.

(a) Similarities: both eukaryotic and prokaryotic DNA are constructed from nucleotide monomers containing deoxyribose sugar, a phosphate group, and nitrogenous bases (A, T, C, G) joined by phosphodiester bonds. Differences: eukaryotic nuclear DNA is linear, associated with histone proteins, contains non-coding regions called introns, and is longer; prokaryotic DNA is circular, 'naked' (not associated with histones), lacks introns, and is shorter. (b) In double-stranded DNA, cytosine (C) pairs with guanine (G). Therefore, if \(C = 28\%\), then \(G = 28\%\). Combined, \(C + G = 28\% + 28\% = 56\%\). The remaining bases must be adenine (A) and thymine (T), so \(A + T = 100\% - 56\% = 44\%\). Since A pairs with T, \(A = T\), so \(A = 44\% \div 2 = 22\%\). (c) The genetic code is 'degenerate' because there are \(64\) possible triplet codons but only \(20\) amino acids, meaning some amino acids are coded for by more than one triplet codon. It is 'non-overlapping' because each base is part of only one triplet and is read once in sequence, meaning adjacent triplets do not share any bases.

(a) 1 mark for similarity (both have double helix / nucleotides / phosphodiester bonds), 1 mark for eukaryotic being linear vs prokaryotic circular, 1 mark for eukaryotic associated with histones vs prokaryotic naked, 1 mark for eukaryotic containing non-coding introns vs prokaryotic lacking them, 1 mark for eukaryotic being longer/larger genome.
(b) 1 mark for recognizing C = G (therefore G = 28%), 1 mark for calculating total G + C = 56%, 1 mark for calculating total A + T = 44%, 0.5 marks for correct answer of 22% adenine.
(c) 2 marks for degenerate explanation (some amino acids are coded for by more than one triplet codon, reducing the effect of mutations), 2 marks for non-overlapping explanation (each base is only part of one triplet codon and is read in sequence without sharing bases with adjacent triplets).

(a) First calculate total number of organisms \(N = 12 + 8 + 15 + 5 = 40\). Thus, \(N(N-1) = 40 \times 39 = 1560\). Next, calculate \(n(n-1)\) for each species:
Species A: \(12 \times 11 = 132\)
Species B: \(8 \times 7 = 56\)
Species C: \(15 \times 14 = 210\)
Species D: \(5 \times 4 = 20\)
\∑ n(n-1) = 132 + 56 + 210 + 20 = 418\).
Index of Diversity \(D = 1560 \div 418 = 3.732\), which rounds to \(3.73\). (b) A monoculture wheat field has only one plant species. This provides a very low variety of ecological niches and habitats, and fewer food sources. Consequently, only a few insect species adapted to wheat can survive (low species richness), and these dominant species will be present in very high numbers (low evenness), resulting in a low index of diversity. (c) Hedgerows introduce a wider variety of plant species into the agricultural area. They provide physical habitats, shelter, and nesting sites for birds, insects, and mammals. They also provide varied food resources such as nectar, berries, and seeds, and act as biological corridors that connect isolated habitats, allowing species to migrate and disperse safely.

(a) 1 mark for total N = 40 and N(N-1) = 1560, 1.5 marks for calculating individual n(n-1) values and summing them to 418, 1 mark for the division step (1560 / 418), 1 mark for correct final answer of 3.73 (accept 3.7).
(b) 1 mark for monoculture having only one plant species, 1 mark for fewer niches/habitats, 1 mark for fewer food sources/types of food, 1 mark for leading to fewer species of insects (low species richness) or dominance of one/two species.
(c) 1 mark for providing shelter/nesting sites, 1 mark for providing more diverse food sources (such as nectar/seeds), 1 mark for acting as corridors allowing dispersal of organisms, 1 mark for introducing different plant species which support more insect/animal species.

(a) Fish gills have many gill filaments covered in numerous secondary lamellae, which vastly increases the overall surface area for gas exchange. The lamellae have a very thin epithelium (often only one cell thick), which provides a short diffusion distance between the water and the blood. In addition, a extensive network of blood capillaries ensures that a steep concentration gradient is maintained. (b) The counter-current system means that water and blood flow over the gill lamellae in opposite directions. This ensures that water with a higher concentration of oxygen always meets blood with a lower concentration of oxygen along the entire length of the lamella. This maintains a concentration gradient across the whole exchange surface, enabling maximum diffusion of oxygen into the blood (extracting up to 80% of oxygen from water). (c) The oxygen dissociation curve is sigmoid (S-shaped) due to cooperative binding: the binding of the first oxygen molecule alters the hemoglobin's quaternary shape, making it easier for subsequent oxygen molecules to bind. During exercise, respiring muscles release carbon dioxide, which dissolves to form carbonic acid, lowering the blood pH. This acidic environment alters the shape of hemoglobin, shifting the curve to the right (the Bohr effect). This decreases its affinity for oxygen, causing hemoglobin to unload oxygen more readily to the hard-working, respiring muscle tissues.

(a) 1 mark for many gill filaments and secondary lamellae to increase surface area, 1 mark for thin epithelium/walls of lamellae for short diffusion distance, 1 mark for rich capillary network to maintain concentration gradient, 1 mark for continuous ventilation/flow of water to maintain concentration gradient.
(b) 1 mark for water and blood flowing in opposite directions, 1 mark for maintaining a concentration gradient of oxygen between water and blood, 1.5 marks for this gradient being maintained along the entire length of the gill lamella, 1 mark for ensuring maximum oxygen extraction (approx 80% vs 50% in concurrent).
(c) 1 mark for describing the shape of the oxygen dissociation curve as sigmoid/S-shaped, 1 mark for identifying that increased CO2 during exercise lowers blood pH, 1 mark for stating this shifts the curve to the right, 1 mark for explaining that hemoglobin has a lower affinity for oxygen and unloads/associates oxygen more readily to respiring tissues.

Part (a): 1. Water evaporates from the cell walls of mesophyll cells into the sub-stomatal air spaces, and then diffuses out of the stomata. 2. This lowers the water potential of the mesophyll cells, causing water to move from neighboring cells and eventually from the xylem by osmosis. 3. This creates tension (pulling force) on the water column in the xylem. 4. Water molecules are cohesive due to hydrogen bonding, forming a continuous column of water from the roots to the leaves. 5. Water molecules also adhere to the xylem walls, preventing the column from breaking. Part (b): 1. Radius of capillary tube = \(0.8\text{ mm} / 2 = 0.4\text{ mm}\). 2. Cross-sectional area = \(\pi r^2 = 3.14 \times (0.4)^2 = 0.5024\text{ mm}^2\). 3. Volume of water moved = \(\text{area} \times \text{distance} = 0.5024 \times 45 = 22.608\text{ mm}^3\). 4. Rate of uptake = \(\text{volume} / \text{time} = 22.608 / 15 = 1.5072\text{ mm}^3\text{ min}^{-1}\). This rounds to 1.51 (or 1.5). Part (c): 1. Some water is used in photosynthesis. 2. Some water is used to maintain cell turgidity (support). 3. Some water is produced during respiration. Part (d): 1. High salt concentration lowers the water potential of the soil solution. 2. This reduces the water potential gradient between the root hair cells and the soil. 3. Less water enters the root hair cells by osmosis, reducing the overall volume of water moving up the plant and lowering transpiration.

Part (a) [4 marks]: 1. Transpiration/evaporation from leaves creates tension/pulling force; 2. Water molecules are cohesive / form hydrogen bonds between each other; 3. Forms a continuous, unbroken water column in xylem; 4. Adhesion of water molecules to xylem walls (supports the column). Part (b) [3 marks]: 1. Correct calculation of radius (0.4 mm) and area (0.5024 mm^2) [1 mark]; 2. Correct calculation of volume (22.608 mm^3) [1 mark]; 3. Correct calculation of rate = 1.51 (or 1.5) mm^3 min^{-1} [1 mark]. Part (c) [2 marks]: Any two from: 1. Water used in photosynthesis; 2. Water used for turgidity/support; 3. Water produced during respiration. Part (d) [2 marks]: 1. Lowers water potential of soil (reducing the water potential gradient) [1 mark]; 2. Less water absorbed by osmosis [1 mark].

Part (a): 1. Carbon dioxide (\(^{14}\text{CO}_2\)) is fixed during photosynthesis (specifically during the light-independent reaction / Calvin cycle). 2. It is incorporated into glucose or triose phosphate, which is subsequently converted into sucrose for translocation. Part (b): 1. Sucrose is actively loaded into sieve tube elements at the source (by co-transport with hydrogen ions). 2. This lowers the water potential inside the sieve tube elements. 3. Water enters the sieve tube elements from the xylem and surrounding cells by osmosis. 4. This increases the volume of liquid inside the sieve tube, generating high hydrostatic pressure. Part (c): 1. Lower temperature reduces the kinetic energy of enzymes and substrates involved in respiration. 2. This reduces the rate of respiration, leading to less ATP production. 3. Active transport of sucrose (loading) requires ATP, so translocation rate decreases. Part (d): 1. Companion cells contain many mitochondria to produce ATP for active transport/loading of sucrose. 2. They synthesize proteins and carry out metabolic functions needed to support the sieve tube elements, which lack organelles.

Part (a) [2 marks]: 1. \(^{14}\text{CO}_2\) is used in photosynthesis/light-independent reaction [1 mark]; 2. To produce glucose/triose phosphate which is converted to sucrose [1 mark]. Part (b) [4 marks]: 1. Active transport of sucrose into sieve tube elements [1 mark]; 2. Lowers water potential inside sieve tube [1 mark]; 3. Water enters by osmosis (from xylem/companion cells) [1 mark]; 4. Increases volume/hydrostatic pressure [1 mark]. Part (c) [3 marks]: 1. Lower temperature decreases kinetic energy of respiratory enzymes/substrates [1 mark]; 2. Less respiration occurs, meaning less ATP is produced [1 mark]; 3. Less ATP available for active transport/loading of sucrose [1 mark]. Part (d) [2 marks]: 1. Provide ATP/energy (from mitochondria) for active loading [1 mark]; 2. Maintain/support sieve tubes / synthesize proteins [1 mark].

Part (a): 1. Increased (hyper)methylation occurs in the promoter region of a tumor suppressor gene. 2. This prevents transcription factors from binding (or attracts proteins that condense chromatin). 3. The gene is not transcribed (silenced), so the tumor suppressor protein is not produced. 4. Without this protein, cell division (mitosis) is no longer regulated/inhibited, leading to uncontrolled cell division and tumor formation. Part (b): 1. Proto-oncogenes stimulate cell division / mitosis when activated by growth factors. 2. Tumor suppressor genes slow down/inhibit cell division, repair DNA errors, or trigger apoptosis. 3. Proto-oncogenes promote progression through the cell cycle, while tumor suppressor genes act as brakes/checkpoints. Part (c): 1. A base substitution changing UCG to UAG is a nonsense mutation. 2. This introduces a premature stop codon during translation. 3. The polypeptide chain is terminated early, resulting in a shorter/truncated protein. 4. The tertiary structure is significantly altered, meaning the spindle protein cannot function correctly, which disrupts mitosis/chromosome separation.

Part (a) [4 marks]: 1. Methylation occurs at promoter region of tumor suppressor gene [1 mark]; 2. Prevents binding of transcription factors / RNA polymerase [1 mark]; 3. Gene expression/transcription is silenced / protein not produced [1 mark]; 4. Uncontrolled cell division/mitosis occurs [1 mark]. Part (b) [3 marks]: 1. Proto-oncogenes stimulate cell division / accelerate cell cycle [1 mark]; 2. Tumor suppressor genes inhibit cell division / repair DNA / cause apoptosis [1 mark]; 3. Contrasting mechanism (e.g., proto-oncogene is active when division is needed, tumor suppressor is active to prevent division) [1 mark]. Part (c) [4 marks]: 1. Nonsense mutation / premature stop codon introduced [1 mark]; 2. Translation stops early [1 mark]; 3. Truncated/shorter polypeptide chain produced [1 mark]; 4. Loss of tertiary structure / non-functional spindle protein (disrupts spindle assembly/mitosis) [1 mark].

Part (a): 1. The semi-lunar valves open when the pressure in the left ventricle exceeds the pressure in the aorta (or right ventricle pressure exceeds pulmonary artery pressure). 2. This occurs during ventricular systole. 3. It forces blood out of the ventricles into the arteries. Part (b): 1. Duration of one cardiac cycle = 60 / 72 = 0.83 s (accept 0.83 or 0.833). 2. Cardiac output = Heart rate \\times Stroke volume = 72 \\times 75 = 5400 \(\text{cm}^3\text{ min}^{-1}\). 3. Convert to \(\text{dm}^3\text{ min}^{-1}\): 5400 / 1000 = 5.4 \(\text{dm}^3\text{ min}^{-1}\). Part (c): 1. High blood pressure increases the hydrostatic pressure at the arterial end of the capillary. 2. This increases the pressure gradient forcing fluid/water out of the capillary into the surrounding tissue spaces. 3. More tissue fluid is formed than can be reabsorbed at the venule end (or drained by lymph vessels), leading to accumulation (edema). Part (d): 1. The wall of the left ventricle has a much thicker muscular wall (myocardium) than the right ventricle. 2. This is significant because the left ventricle must pump blood under high pressure around the entire body (systemic circulation), whereas the right ventricle only pumps blood under lower pressure to the lungs (pulmonary circulation).

Part (a) [3 marks]: 1. Pressure in ventricle is greater than pressure in aorta/pulmonary artery [1 mark]; 2. Ventricular systole [1 mark]; 3. Forces blood into arteries [1 mark]. Part (b) [3 marks]: 1. Correct cycle duration: 0.83 s (or 0.833 s) [1 mark]; 2. Correct multiplication: 72 \\times 75 = 5400 [1 mark]; 3. Correct conversion to 5.4 dm^3 min^{-1} (with units) [1 mark]. Part (c) [3 marks]: 1. High hydrostatic pressure at arterial end of capillaries [1 mark]; 2. Larger outward pressure gradient / more fluid forced out [1 mark]; 3. Less fluid reabsorbed / exceeds lymphatic drainage capacity [1 mark]. Part (d) [2 marks]: 1. Left ventricle has thicker muscular wall [1 mark]; 2. Left ventricle needs to contract with greater force / pump blood under higher pressure to the whole body [1 mark].

Part (a): 1. Once inside the helper T cell, the viral RNA is released. 2. The viral enzyme reverse transcriptase copies the viral single-stranded RNA into double-stranded DNA. 3. This viral DNA is transported into the nucleus and integrated into the host cell's DNA using the enzyme integrase. 4. The host cell's RNA polymerase transcribes the integrated viral DNA to produce viral mRNA and new viral RNA genomes. 5. Host ribosomes translate the viral mRNA into viral proteins (polyproteins), which are cleaved by protease. 6. New viral particles assemble near the cell membrane and bud off, taking a portion of the host cell membrane as their envelope. Part (b): 1. HIV infects and replicates inside helper T cells, eventually killing or lysing them. 2. This severely reduces the number of functional helper T cells. 3. Without helper T cells, B cells cannot be activated to produce antibodies, and cytotoxic T cells cannot be activated. This weakens both humoral and cellular immunity, leaving the body unable to fight pathogens. Part (c): 1. HIV antigens are bound to the bottom of a well in a microtiter plate. 2. The patient's blood sample (serum) is added. If HIV antibodies are present, they bind to the HIV antigens to form antigen-antibody complexes. The plate is then washed to remove unbound antibodies. 3. A second antibody with an enzyme attached (enzyme-linked antibody) is added, which binds to the patient's HIV antibody. The well is washed again. 4. A substrate is added; the enzyme acts on the substrate to produce a color change, indicating a positive result.

Part (a) [5 marks]: 1. Reverse transcriptase makes DNA copy of viral RNA [1 mark]; 2. Integrase inserts viral DNA into host genome/DNA [1 mark]; 3. Host RNA polymerase transcribes viral DNA to viral mRNA [1 mark]; 4. Ribosomes translate viral mRNA to make viral proteins [1 mark]; 5. Viral particles assemble and bud off (taking host membrane envelope) [1 mark]. Part (b) [3 marks]: 1. Destroys/reduces helper T cell count [1 mark]; 2. B cells are not stimulated/activated (to secrete antibodies) [1 mark]; 3. Cytotoxic T cells / phagocytes are not stimulated, leaving immune system unable to fight pathogens [1 mark]. Part (c) [3 marks]: 1. HIV antigen bound to well, add patient serum (wash to remove unbound antibodies) [1 mark]; 2. Add enzyme-linked antibody (wash again to remove unbound) [1 mark]; 3. Add substrate and look for color change [1 mark].

Part (a): 1. A macrophage engulfs the pathogen via phagocytosis into a phagosome. 2. Lysosomes fuse with the phagosome, and hydrolytic enzymes digest the pathogen. 3. The macrophage separates the pathogen's antigens and places them on its outer cell surface membrane, presenting them alongside MHC (major histocompatibility complex) proteins to become an antigen-presenting cell. Part (b): 1. Antigen-presenting cells (like macrophages) bind to specific helper T cells, activating them. 2. B cells bind directly to complementary antigens on pathogens and process them to present on their surface. 3. Activated helper T cells bind to the B cells presenting the antigen and release cytokines (interleukins). 4. These cytokines stimulate the B cells to divide by mitosis (clonal selection) and differentiate into plasma cells that secrete specific antibodies (and memory B cells). Part (c): 1. A monoclonal antibody is an antibody produced from a single clone of B cells (or hybridoma cells). 2. They are all genetically identical and have an identical tertiary structure. 3. They are highly specific because their variable regions are complementary to only one specific antigen (or epitope). 4. They can be conjugated to dyes or enzymes to locate or quantify specific proteins in a sample with high sensitivity and no cross-reactivity.

Part (a) [3 marks]: 1. Pathogen engulfed/ingested into phagosome/vesicle [1 mark]; 2. Lysosomes fuse and release enzymes to hydrolyze/digest pathogen [1 mark]; 3. Antigens from pathogen are presented on the macrophage cell surface membrane [1 mark]. Part (b) [4 marks]: 1. Helper T cells activated by antigen-presenting cells [1 mark]; 2. B cells present antigen on their surface [1 mark]; 3. Activated helper T cells bind to B cell / release cytokines [1 mark]; 4. Stimulates clonal selection/division of B cells into plasma cells that secrete antibodies [1 mark]. Part (c) [4 marks]: 1. Antibodies produced from a single clone of cells (or hybridoma) [1 mark]; 2. Genetically identical with identical variable region [1 mark]; 3. Bind specifically to one single antigen/epitope [1 mark]; 4. High sensitivity/specificity allows detection of trace amounts of target proteins [1 mark].

Part (a): 1. Bile salts emulsify large lipid droplets into tiny droplets (micelles). 2. This significantly increases the surface area of lipids available for lipase action. 3. Pancreatic lipase hydrolyzes the ester bonds in triglycerides. 4. This produces fatty acids and monoglycerides. Part (b): 1. Micelles are temporary structures made of fatty acids, monoglycerides, and bile salts. 2. They make these hydrophobic lipid breakdown products soluble in water. 3. They transport the fatty acids and monoglycerides to the brush border/membrane of the epithelial cells. 4. At the membrane, they break down, releasing the fatty acids and monoglycerides, which can diffuse directly across the phospholipid bilayer. Part (c): 1. Sodium ions (Na+) are actively transported out of the epithelial cells into the blood by the sodium-potassium pump. 2. This creates and maintains a low concentration of Na+ inside the epithelial cell compared to the lumen of the ileum. 3. Na+ ions diffuse down their concentration gradient from the lumen into the epithelial cell through a co-transporter protein. 4. As Na+ enters, it carries glucose molecules against their concentration gradient into the epithelial cell via the same carrier protein.

Part (a) [4 marks]: 1. Bile salts emulsify lipids into smaller droplets [1 mark]; 2. Increases surface area for lipase action [1 mark]; 3. Lipase hydrolyzes ester bonds [1 mark]; 4. To produce fatty acids and monoglycerides [1 mark]. Part (b) [3 marks]: 1. Micelles make fatty acids/monoglycerides soluble in water [1 mark]; 2. Transport fatty acids/monoglycerides to the cell membrane of epithelial cells [1 mark]; 3. Fatty acids/monoglycerides are released and diffuse directly across bilayer [1 mark]. Part (c) [4 marks]: 1. Na+ actively pumped out of epithelial cell into blood (by Na+/K+ pump) [1 mark]; 2. Establishes/maintains concentration gradient of Na+ (lower in cell than lumen) [1 mark]; 3. Na+ enters cell by facilitated diffusion through co-transporter protein [1 mark]; 4. This pulls/carries glucose into cell against its concentration gradient [1 mark].

Part (a):
RuBP acts as the carbon dioxide acceptor in the light-independent reaction. It reacts with \(\text{CO}_2\) in a carboxylation reaction catalyzed by the enzyme Rubisco to produce an unstable 6-carbon intermediate, which immediately splits into two molecules of glycerate 3-phosphate (GP).

Part (b):
Carbon dioxide is a reactant in the light-independent stage. Low \(\text{CO}_2\) concentration means less RuBP is carboxylated to form GP, resulting in less GP to be reduced to triose phosphate (TP) using ATP and reduced NADP from the light-dependent stage. Consequently, regeneration of RuBP is reduced, and the production of organic substances (like glucose) slows down, making \(\text{CO}_2\) the limiting factor despite abundant light.

Part (c):
1. Relationship: 2 molecules of TP (3C) are required to produce 1 molecule of glucose (6C).
2. Calculate moles of glucose produced: \(4.2\text{ mmol dm}^{-2}\text{h}^{-1} / 2 = 2.1\text{ mmol dm}^{-2}\text{h}^{-1}\).
3. Convert to mass (mg): \(2.1\text{ mmol} \times 180\text{ mg mmol}^{-1} = 378\text{ mg dm}^{-2}\text{h}^{-1}\).

Part (a) (Max 3 marks):
- RuBP binds/combines with CO2 (carbon fixation) (1 mark)
- Catalyzed by Rubisco (1 mark)
- To produce 2 molecules of GP / glycerate 3-phosphate (1 mark)

Part (b) (Max 3.7 marks):
- Less carbon dioxide means less GP formed (1 mark)
- Less GP means less TP formed (1 mark)
- Light-dependent products (ATP and reduced NADP) cannot be used efficiently / build up (1 mark)
- Thus, rate of synthesis of organic substances is limited by CO2 availability, not light (0.7 marks)

Part (c) (Max 4 marks):
- Correct division of TP rate by 2 to find glucose rate: \(2.1\text{ mmol dm}^{-2}\text{h}^{-1}\) (1 mark)
- Correct multiplication of glucose rate by molar mass (180): \(2.1 \times 180\) (1 mark)
- Correct final numerical answer of 378 (1 mark)
- Correct unit: \(\text{mg dm}^{-2}\text{h}^{-1}\) (1 mark)
(Accept alternative calculation method: TP mass = \(4.2 \times 90 = 378\text{ mg}\), since all TP is converted, mass of glucose produced is equal to mass of TP used, leading to \(378\text{ mg dm}^{-2}\text{h}^{-1}\) - award full marks for correct final answer regardless of method)

Part (a):
During aerobic respiration, organisms absorb oxygen and release an equal volume of carbon dioxide. KOH absorbs the carbon dioxide produced. This ensures that any change in the volume of gas (and thus pressure) inside the chamber is solely due to the uptake of oxygen, allowing the liquid in the capillary tube to move.

Part (b):
1. Calculate volume of oxygen consumed: \(\text{Volume} = \text{cross-sectional area} \times \text{distance} = 0.8\text{ mm}^2 \times 15\text{ mm} = 12\text{ mm}^3\).
2. Calculate rate of consumption per minute: \(\text{Rate} = 12\text{ mm}^3 / 20\text{ minutes} = 0.6\text{ mm}^3\text{min}^{-1}\).

Part (c):
In yeast (ethanol pathway):
- Pyruvate is decarboxylated to form ethanal, releasing CO2.
- Ethanal is then reduced to ethanol by reduced NAD.
- This reaction is irreversible.
In muscle cells (lactate pathway):
- Pyruvate is directly reduced to lactate by reduced NAD.
- No carbon dioxide is produced.
- This reaction is reversible (lactate can be converted back to pyruvate).

Part (a) (Max 3 marks):
- Absorbs carbon dioxide produced by respiration (1 mark)
- Prevents CO2 from replacing oxygen volume / keeps pressure changes due only to O2 (1 mark)
- Allows the movement of the liquid to represent only oxygen uptake (1 mark)

Part (b) (Max 3.7 marks):
- Correct formula/calculation for volume: \(0.8 \times 15 = 12\text{ mm}^3\) (1 mark)
- Correct division by time: \(12 / 20\) (1 mark)
- Correct final answer: 0.6 (1 mark)
- Correct units: \(\text{mm}^3\text{min}^{-1}\) (0.7 marks)

Part (c) (Max 4 marks):
- Decarboxylation / CO2 release occurs in yeast but not in muscle cells (1 mark)
- Ethanal is an intermediate in yeast but not in muscle cells (1 mark)
- Lactate is produced in muscle cells, whereas ethanol is produced in yeast (1 mark)
- Yeast pathway is irreversible, whereas muscle pathway is reversible / lactate converted back when O2 is available (1 mark)

Part (a):
This interaction is called epistasis (specifically, recessive epistasis). Gene A is epistatic to Gene B because the homozygous recessive genotype \(aa\) suppresses the expression of the hypostatic gene B (which determines the deep blue color). If the precursor cannot be converted to the intermediate because of \(aa\), the plant is white regardless of the alleles at the B locus.

Part (b):
The expected phenotypic ratio is 9 blue : 3 purple : 4 white.
- Blue (\(A\_B\_\)): \(9/16\)
- Purple (\(A\_bb\)): \(3/16\)
- White (\(aa\_\_\)): \(4/16\)

Part (c):
1. Calculate expected values (E) for total \(N = 160\):
- Expected Blue = \(160 \times 9/16 = 90\)
- Expected Purple = \(160 \times 3/16 = 30\)
- Expected White = \(160 \times 4/16 = 40\)
2. Calculate \((O - E)^2 / E\) for each class:
- Blue: \((92 - 90)^2 / 90 = 4 / 90 \approx 0.044\)
- Purple: \((28 - 30)^2 / 30 = 4 / 30 \approx 0.133\)
- White: \((40 - 40)^2 / 40 = 0\)
3. Sum the values to find \(\chi^2\):
- \(\chi^2 = 0.044 + 0.133 + 0 = 0.178\) (or \(0.18\)).
4. Conclusion:
Since the calculated \(\chi^2\) value of \(0.18\) is less than the critical value of \(5.99\), we accept the null hypothesis. There is no significant difference between the observed and expected phenotypic ratios, so the results support the epistatic inheritance model.

Part (a) (Max 3.7 marks):
- Epistasis / epistatic interaction (1 mark)
- Recessive epistasis identified (1 mark)
- Explanation: homozygous recessive allele 'a' prevents production of the precursor/intermediate, masking the expression of gene B (1.7 marks)

Part (b) (Max 2 marks):
- Correct phenotypic ratio: 9 blue : 3 purple : 4 white (2 marks, 1 mark if ratio is correct but labels are missing/incorrect)

Part (c) (Max 5 marks):
- Correct expected numbers calculated (90, 30, 40) (1 mark)
- Correct application of \((O-E)^2/E\) for all classes (1 mark)
- Correct calculated \(\chi^2\) value of 0.18 (or 0.178) (1 mark)
- Comparison with critical value: \(0.18 < 5.99\) (1 mark)
- Correct conclusion: No significant difference / accept null hypothesis / observed results fit expected ratio (1 mark)

Part (a):
Any three from:
1. No mutations occur.
2. No selection occurs (all genotypes have equal reproductive success).
3. The population is large.
4. Mating within the population is random.
5. The population is isolated / no gene flow (no migration into or out of the population).

Part (b):
1. Identify \(q^2\): The frequency of affected individuals (recessive genotype) is \(q^2 = 1 / 2500 = 0.0004\).
2. Calculate \(q\): \(q = \sqrt{0.0004} = 0.02\).
3. Calculate \(p\): \(p = 1 - q = 1 - 0.02 = 0.98\).
4. Calculate heterozygous frequency (\(2pq\)): \(2 \times 0.98 \times 0.02 = 0.0392\).
5. Convert to percentage: \(0.0392 \times 100 = 3.92\%\).

Part (c):
Without medical treatment, individuals homozygous for the cystic fibrosis allele (recessive phenotype) would have a severely reduced lifespan and would be less likely to reproduce. This means the recessive allele is selected against, and fewer copies of the allele are passed on to the next generation. However, because the allele can hide in healthy heterozygous carriers, the allele is never completely eliminated from the gene pool, but its frequency will decrease and stabilize at a very low level.

Part (a) (Max 3 marks):
- Any three correct conditions: No mutations, no selection, large population, random mating, no migration/gene flow (1 mark per point up to 3)

Part (b) (Max 3.7 marks):
- Correct identification of \(q^2 = 0.0004\) or \(q = 0.02\) (1 mark)
- Correct calculation of \(p = 0.98\) (1 mark)
- Correct formula usage for carriers: \(2pq = 2(0.98)(0.02)\) (1 mark)
- Correct final percentage: \(3.92\%\) (0.7 marks)

Part (c) (Max 4 marks):
- Homozygous recessive individuals die before reproducing / have low reproductive success (1 mark)
- Recessive allele not passed on by affected individuals / selective pressure against allele (1 mark)
- Heterozygotes survive and reproduce / carriers are unaffected (1 mark)
- Frequency of allele decreases but remains in the population because it is masked in heterozygotes (1 mark)

Part (a):
Geographical isolation physically separates the original population into two, preventing gene flow between them. Each population experiences different environmental conditions, such as climate, food availability, or predators. Consequently, different selective pressures act on each population. Random mutations occur independently in each population, and those that offer a selective advantage are selected for, increasing in frequency. Over time, the allele frequencies of the two populations diverge significantly, leading to reproductive isolation where they can no longer interbreed.

Part (b):
Scientists could attempt to breed individuals from both populations together. If they cannot produce offspring, or if the offspring produced are sterile (not fertile), then they are confirmed to be separate species.

Part (c):
In small populations, the gene pool is much smaller. Consequently, random fluctuations in allele frequencies (e.g., due to chance events like the death of a few individuals) can have a massive impact on the overall percentage of that allele in the population. In contrast, in a large population, random events are buffered by the large number of individuals, making allele frequencies much more stable and less prone to change purely by chance.

Part (a) (Max 5 marks):
- Geographic barrier prevents gene flow/interbreeding (1 mark)
- Different environmental conditions / selective pressures in each area (1 mark)
- Mutation occurs in one/both populations (1 mark)
- Beneficial alleles selected for / survive and reproduce (natural selection) (1 mark)
- Leads to changes in allele frequencies over time (1 mark)
- Results in reproductive isolation (1 mark)

Part (b) (Max 2 marks):
- Breed individuals from both populations together (1 mark)
- Observe if they produce fertile offspring (if not, they are separate species) (1 mark)

Part (c) (Max 3.7 marks):
- Genetic drift is the random change in allele frequencies due to chance (1 mark)
- In a small population, a chance event can cause a larger proportional change in allele frequency (1.7 marks)
- Alleles can easily be lost or fixed by chance in a small gene pool (1 mark)

Part (a):
Gross primary productivity (GPP) is the total chemical energy store in plant biomass in a given area or volume, in a given time, resulting from photosynthesis. Net primary productivity (NPP) is the chemical energy store remaining in plant biomass after taking into account energy losses due to respiration (R). It is calculated as \(NPP = GPP - R\).

Part (b):
1. Calculate GPP:
\(GPP = 1.5 \times 10^6 \times 0.012 = 18,000\text{ kJ m}^{-2}\text{year}^{-1}\).
2. Calculate energy lost in respiration (R):
\(R = 18,000 \times 0.55 = 9,900\text{ kJ m}^{-2}\text{year}^{-1}\).
3. Calculate NPP:
\(NPP = GPP - R = 18,000 - 9,900 = 8,100\text{ kJ m}^{-2}\text{year}^{-1}\).
(Alternatively: \(NPP = GPP \times (1 - 0.55) = 18,000 \times 0.45 = 8,100\text{ kJ m}^{-2}\text{year}^{-1}\))

Part (c):
Confining animals indoors increases energy transfer efficiency by reducing respiratory losses in the following ways:
1. Warm temperatures indoors reduce the energy the animals must spend maintaining their body temperature.
2. Restricted movement reduces muscle contraction, thereby reducing the rate of respiration.
3. This means more energy is directed towards growth and biomass production rather than lost as heat. Consequently, a higher proportion of the consumed energy is stored as biomass (meat/milk) that humans can eat.

Part (a) (Max 3 marks):
- GPP is total chemical energy produced / stored in plant biomass by photosynthesis (1 mark)
- NPP is the chemical energy store remaining after respiratory losses (1 mark)
- State the equation: \(NPP = GPP - R\) (1 mark)

Part (b) (Max 3.7 marks):
- Correct calculation of GPP: \(18,000\text{ kJ m}^{-2}\text{year}^{-1}\) (1 mark)
- Correct subtraction or factor of \(45\%\) (1 mark)
- Correct final answer: 8,100 (1 mark)
- Correct units: \(\text{kJ m}^{-2}\text{year}^{-1}\) (0.7 marks)

Part (c) (Max 4 marks):
- Movement restricted so less respiration/muscle contraction (1 mark)
- Warm environment so less heat loss / less energy used to maintain body temperature (1 mark)
- Less respiratory loss means more energy converted into biomass/growth (1 mark)
- Higher percentage of energy transferred to humans (1 mark)

Part (a):
Saprobionts decompose organic waste, such as dead organisms, faeces, and urea. They secrete extracellular enzymes to break down nitrogen-containing compounds (like proteins, DNA, and urea) and release ammonia/ammonium ions into the soil, in a process called ammonification.

Part (b):
Waterlogged soils lack oxygen, creating anaerobic conditions. Under these anaerobic conditions, anaerobic denitrifying bacteria flourish. They convert nitrates back into gaseous nitrogen (\(\text{N}_2\)) which escapes into the atmosphere, thereby reducing the concentration of nitrates in the soil.

Part (c):
1. Nitrate runoff causes rapid growth of algae, leading to an algal bloom on the water surface.
2. The algal bloom blocks sunlight from reaching aquatic plants deeper in the water, preventing photosynthesis so they die.
3. Saprobiontic bacteria decompose the dead plants and multiply rapidly.
4. Due to high respiration by these decomposing bacteria, the oxygen concentration in the water drops drastically, causing fish to die due to lack of oxygen for aerobic respiration.

Part (a) (Max 3 marks):
- Decompose dead matter/waste (1 mark)
- Break down nitrogenous compounds (proteins/DNA/urea) (1 mark)
- Release ammonium ions / ammonification (1 mark)

Part (b) (Max 3.7 marks):
- Waterlogged soil is anaerobic / lacks oxygen (1 mark)
- Denitrifying bacteria active (1 mark)
- Convert nitrates to nitrogen gas (1.7 marks)

Part (c) (Max 4 marks):
- Algae bloom blocks light, causing deeper plants to die (1 mark)
- Saprobionts / decomposers break down dead plants (1 mark)
- Decomposers respire aerobically, depleting oxygen levels (1 mark)
- Low oxygen causes death of fish / aerobic organisms (1 mark)

(a) Active transport of sodium ions (\(Na^+\)) out of the axon and potassium ions (\(K^+\)) into the axon by the sodium-potassium pump, using ATP (1 mark). Three \(Na^+\) are pumped out for every two \(K^+\) pumped in (1 mark). The membrane is more permeable to potassium ions than to sodium ions because there are more open non-gated potassium channels (1 mark). Potassium ions diffuse out of the axon down their concentration gradient, leaving a net negative charge inside relative to the outside (1 mark).

(b) Depolarization cannot occur (1 mark) because sodium ions cannot enter the axon through the voltage-gated channels (1 mark). Thus, the threshold potential is not reached, preventing the initiation of an action potential (1 mark).

(c) The inhibitory neurotransmitter binds to specific receptors on the post-synaptic membrane (1 mark). This causes ligand-gated chloride ion (\(Cl^-\)) channels to open, allowing \(Cl^-\) to diffuse into the post-synaptic neurone (1 mark). It also causes potassium ion (\(K^+\)) channels to open, allowing \(K^+\) to diffuse out (1 mark). This makes the inside of the post-synaptic neurone more negative, causing hyperpolarization, meaning a larger sodium influx is required to reach the threshold potential (1 mark).

(a)
1. Active transport of \(Na^+\) out and \(K^+\) in via the sodium-potassium pump using ATP;
2. 3 \(Na^+\) out for every 2 \(K^+\) in;
3. Membrane is more permeable to \(K^+\) than \(Na^+\) (due to open leak channels);
4. \(K^+\) diffuses out down its concentration gradient, leaving a net negative charge inside.

(b)
1. No depolarization occurs;
2. Because \(Na^+\) ions cannot enter the axon (via voltage-gated channels);
3. Threshold potential is not reached / no action potential generated.

(c)
1. Neurotransmitter (GABA) binds to specific post-synaptic receptors;
2. Causes chloride channels to open and \(Cl^-\) to enter;
3. Causes potassium channels to open and \(K^+\) to leave;
4. Results in hyperpolarization / membrane potential becomes more negative than resting potential, so threshold is harder to reach.

(a) Pressure deforms the lamellae of the Pacinian corpuscle (1 mark). This deforms/stretches the sensory neurone membrane (1 mark), which stretches and opens the stretch-mediated sodium ion channels (1 mark). Sodium ions (\(Na^+\)) diffuse rapidly into the neurone down their electrochemical gradient, causing depolarization which establishes a generator potential (1 mark).

(b) High sensitivity: Many rod cells share/connect to a single bipolar neurone (retinal convergence) (1 mark). This allows spatial summation, where multiple small generator potentials combine to exceed the threshold needed to trigger an action potential in the bipolar cell (1 mark). Low acuity: Because multiple rod cells send impulses down the same single bipolar neurone, the brain cannot distinguish which individual rod cell was stimulated, meaning two close objects are perceived as one (1 mark). Rhodopsin is also broken down easily in low light (1 mark).

(c) High acuity: Each cone cell is connected to its own individual bipolar cell (1:1 ratio) (1 mark). Therefore, if two adjacent cone cells are stimulated, the brain receives two separate impulses and can resolve them as two separate points of light (1 mark). Low sensitivity: There is no retinal convergence/spatial summation, so a high light intensity is required to break down iodopsin and generate a generator potential large enough to trigger an action potential in the bipolar cell (1 mark).

(a)
1. Pressure deforms/squeezes the lamellae;
2. Stretches the membrane of the sensory neurone;
3. Opens stretch-mediated sodium channels;
4. \(Na^+\) enters the neurone causing depolarization / generator potential.

(b)
1. High sensitivity: Retinal convergence / many rods connect to one bipolar neurone;
2. Spatial summation allows threshold to be reached in low light;
3. Low acuity: Brain cannot identify which specific rod cell was stimulated;
4. Two points of light close together cannot be resolved as separate.

(c)
1. High acuity: Each cone connects to a single bipolar cell (no convergence);
2. Brain receives separate impulses from adjacent cones to resolve two points;
3. Low sensitivity: No summation, so high light intensity needed to trigger action potential.

(a) Calcium ions (\(Ca^{2+}\)) are released from the sarcoplasmic reticulum and bind to troponin (1 mark). This causes a conformational change that moves tropomyosin, exposing the myosin-binding sites on the actin filament (1 mark). This allows myosin heads to bind to actin, forming actomyosin cross-bridges (1 mark). Calcium ions also activate the ATPase enzyme on myosin (1 mark). ATP hydrolysis provides the energy for the myosin head to bend/rotate (the power stroke), pulling the actin filament along, and the binding of a new ATP molecule is required for the myosin head to detach from actin (1 mark).

(b) The I-band (actin only) and the H-zone (myosin only) both decrease in width/shorten (1 mark). The distance between the Z-lines decreases (1 mark). The A-band remains constant because it represents the total length of the myosin (thick) filaments, which do not shorten; instead, actin filaments slide over them (1 mark).

(c) Muscle contraction would be inhibited/stopped because there would be no energy released from ATP hydrolysis to power the movement of the myosin heads (power stroke) (1 mark). Muscle relaxation would also be prevented (1 mark) because ATP binding and hydrolysis is needed to break the actomyosin cross-bridges and pump \(Ca^{2+}\) back into the sarcoplasmic reticulum, leaving the muscle in a state of continuous spasm/rigor (1 mark).

(a)
1. \(Ca^{2+}\) binds to troponin, moving tropomyosin;
2. Exposes myosin-binding sites on actin;
3. Allows formation of actomyosin cross-bridges;
4. \(Ca^{2+}\) activates ATPase;
5. ATP hydrolysis provides energy for myosin head power stroke AND ATP binding causes detachment of myosin head from actin.

(b)
1. I-band and H-zone shorten;
2. Z-lines move closer together;
3. A-band remains constant because myosin filaments do not change length/only actin slides.

(c)
1. Contraction stops because no energy is released for the power stroke;
2. Relaxation is prevented / muscle remains rigid;
3. Because actomyosin cross-bridges cannot be broken / \(Ca^{2+}\) cannot be pumped back into sarcoplasmic reticulum.

(a) Unilateral light causes the plant growth regulator IAA to move/migrate from the light side to the shaded side of the shoot tip (1 mark). This results in a higher concentration of IAA on the shaded side (1 mark). In shoots, IAA stimulates cell elongation (1 mark). Therefore, cells on the shaded side elongate faster/more than those on the light side, causing the shoot to bend towards the light source (1 mark).

(b) Gravity causes IAA to accumulate on the lower side of both the root and the shoot (1 mark). In shoots, the high concentration of IAA on the lower side stimulates cell elongation, causing the shoot to bend upwards (negative gravitropism) (1 mark). In roots, a high concentration of IAA inhibits cell elongation (1 mark). Therefore, cells on the upper side of the root elongate faster than those on the lower side, causing the root to bend downwards (positive gravitropism) (1 mark).

(c) IAA is transported over short distances by diffusion (1 mark) and by active transport (1 mark). This active transport is facilitated by specific carrier proteins called PIN proteins / efflux carriers in the cell membranes (1 mark).

(a)
1. IAA migrates from the illuminated side to the shaded side of the shoot tip;
2. High concentration of IAA on the shaded side;
3. In shoots, IAA stimulates cell elongation;
4. Cells on the shaded side elongate more, causing bending towards light.

(b)
1. Gravity causes IAA accumulation on the lower side of both organs;
2. In shoots, high IAA stimulates elongation, causing upward bending / negative gravitropism;
3. In roots, high IAA concentration inhibits cell elongation;
4. Upper cells elongate faster, causing downward bending / positive gravitropism.

(c)
1. Diffusion;
2. Active transport;
3. Via specific carrier/efflux (PIN) proteins.

(a) Glucagon binds to specific receptors on the cell surface membrane of target cells (hepatocytes) (1 mark). The binding causes a conformational change that activates the transmembrane enzyme adenylate cyclase inside the cell (1 mark). Activated adenylate cyclase converts ATP into cyclic AMP (cAMP), which acts as the second messenger (1 mark). cAMP binds to and activates protein kinase A (an enzyme) (1 mark). Active protein kinase A initiates a cascade of reactions that activates other enzymes responsible for converting glycogen to glucose (glycogenolysis) (1 mark).

(b) Insulin binds to specific receptors on the cell membrane of target cells (1 mark). This triggers a signaling cascade that causes vesicles containing glucose transporter proteins (GLUT4) to fuse with the cell surface membrane, increasing the number of glucose channel proteins (1 mark). This increases the permeability of the cell membrane, so more glucose enters the cells by facilitated diffusion (1 mark). Insulin also activates enzymes that convert glucose to glycogen (glycogenesis) and increases the rate of respiration in cells, lowering blood glucose concentration (1 mark).

(c) The competitive inhibitor binds to the active site of adenylate cyclase, preventing it from converting ATP to cAMP (1 mark). Without cAMP, protein kinase A cannot be activated, so the enzyme cascade that breaks down glycogen to glucose is blocked, reducing glycogenolysis (1 mark).

(a)
1. Glucagon binds to specific receptors on hepatocytes;
2. Activates adenylate cyclase;
3. Converts ATP to cyclic AMP (cAMP) / second messenger;
4. cAMP binds to and activates protein kinase A;
5. Active protein kinase A activates glycogen phosphorylase / enzymes that catalyze glycogenolysis.

(b)
1. Insulin binds to specific glycoprotein receptors;
2. Vesicles containing GLUT4 transporters fuse with the membrane;
3. Increases facilitated diffusion/entry of glucose into cells;
4. Activates enzymes for glycogenesis / converts glucose to glycogen.

(c)
1. Inhibitor prevents conversion of ATP to cAMP;
2. Lack of second messenger means protein kinase A remains inactive, so glycogenolysis cascade is blocked.

(a) Cells that naturally express the desired gene at a high rate are selected, and their mRNA is extracted (1 mark). Reverse transcriptase is added along with free DNA nucleotides (1 mark). The enzyme uses the mRNA as a template to synthesize a single-stranded complementary DNA (cDNA) molecule. DNA polymerase is then used to synthesize the second complementary DNA strand to form double-stranded DNA (1 mark).

(b) PCR is used to amplify/produce many copies of the DNA fragment (1 mark).
- Step 1: Reactants are heated to 95 °C to break the hydrogen bonds between bases, separating the double-stranded DNA into single strands (1 mark).
- Step 2: The mixture is cooled to around 55 °C to allow primers to bind/anneal to their complementary sequences on the single strands of DNA (1 mark).
- Step 3: The temperature is raised to 72 °C because this is the optimum temperature for the thermostable DNA polymerase (Taq polymerase) (1 mark). Taq polymerase aligns free DNA nucleotides along the templates to synthesize complementary strands (1 mark).

(c) Not all host cells will successfully take up the plasmid/recombinant DNA (the process is inefficient) (1 mark). Marker genes are placed in the same plasmid vector as the target gene (1 mark). They allow scientists to easily identify and select only the cells that have taken up the recombinant plasmid (e.g., by observing fluorescence under UV light for GFP, or survival on selective media containing antibiotics) (1 mark).

(a)
1. Extract mRNA from cells expressing the target gene;
2. Use reverse transcriptase to synthesize single-stranded cDNA from mRNA template;
3. Use DNA polymerase to synthesize the complementary strand to make it double-stranded.

(b)
1. PCR amplifies DNA;
2. 95 °C: Denaturation / separation of DNA strands by breaking hydrogen bonds;
3. 55 °C: Annealing of primers to complementary sequences;
4. 72 °C: Extension / DNA synthesis;
5. Optimum temperature for thermostable Taq polymerase.

(c)
1. Transformation/gene uptake efficiency is low / not all cells take up the gene;
2. Marker gene is co-transformed with target gene;
3. Allows selection/identification of successfully transformed cells.

(a) Oestrogen is lipid-soluble, so it diffuses across the phospholipid bilayer of the cell surface membrane (1 mark). Inside the cytoplasm, oestrogen binds to a specific receptor on an inactive transcription factor (1 mark). This binding causes a conformational change in the transcription factor, which releases an inhibitor and exposes its DNA-binding site (1 mark). The active transcription factor enters the nucleus through a nuclear pore, binds to the promoter region of the target gene, and stimulates the recruitment of RNA polymerase to begin transcription (1 mark).

(b) Double-stranded siRNA is cleaved into smaller fragments, and one of the single strands associates with an enzyme complex called RISC (1 mark). The single-stranded siRNA guides the RISC complex to a target mRNA molecule by complementary base pairing (A with U, C with G) (1 mark). Once bound, the enzyme in the complex cuts/cleaves the mRNA molecule into fragments (1 mark). This cleaved mRNA can no longer be translated by ribosomes, so the expression of the gene is blocked/silenced (1 mark).

(c) Increased methylation of DNA involves adding methyl groups to cytosine bases in DNA. This attracts proteins that condense chromatin, preventing transcription factors from binding, thus inhibiting transcription (1 mark). In contrast, increased acetylation of histones involves adding acetyl groups to histones, which reduces their positive charge (1 mark). This decreases the attraction between histones and negatively charged DNA, loosening the chromatin (euchromatin) and allowing transcription factors to bind, thus stimulating transcription (1 mark).

(a)
1. Oestrogen diffuses through the cell membrane;
2. Binds to an intracellular receptor, causing a conformational change;
3. Exposes the DNA-binding site of the transcription factor / activates it;
4. Transcription factor enters nucleus and binds to promoter of target gene, initiating transcription (via RNA polymerase recruitment).

(b)
1. Double-stranded RNA is cut into siRNA;
2. Single-strand siRNA binds to an enzyme complex;
3. siRNA guides enzyme to target mRNA via complementary base pairing;
4. Enzyme cuts mRNA, preventing translation.

(c)
1. Increased DNA methylation inhibits transcription by condensing chromatin;
2. Increased histone acetylation stimulates transcription by reducing positive charge on histones;
3. This decreases DNA-histone binding, making chromatin looser / more accessible to transcription factors.

a) Water is lost from the leaves by transpiration, creating tension. Cohesion between water molecules allows them to be pulled up as a continuous column. b) High wind speed removes the boundary layer of water vapor, maintaining a steep water potential gradient. Low humidity increases the water potential gradient between inside and outside the leaf. c) Area conversion: \(250\text{ cm}^2 = 2.5\text{ dm}^2\). Rate: \(5.0\text{ g} / 2\text{ hours} = 2.5\text{ g hour}^{-1}\). Rate per area: \(2.5\text{ g hour}^{-1} / 2.5\text{ dm}^2 = 1.0\text{ g dm}^{-2}\text{ hour}^{-1}\).

a) 1. Water evaporates from the mesophyll cell walls and diffuses out of stomata; 2. This lowers water potential of mesophyll cells; 3. Water is pulled up the xylem by tension; 4. Cohesion between water molecules due to hydrogen bonding; 5. Forms a continuous, unbroken water column; 6. Adhesion of water molecules to xylem walls supports the column. (Max 6 marks)

b) 1. High wind speed blows away the saturated air/boundary layer around stomata; 2. Maintaining a steep water potential gradient; 3. Low humidity means low water potential in surrounding air; 4. Increasing the water potential gradient, leading to faster diffusion. (Max 4 marks)

c) 1. Converts area correctly: \(250\text{ cm}^2 = 2.5\text{ dm}^2\) (1 mark); 2. Divides water mass by hours and area (0.5 marks); 3. Correct final answer of \(1.0\text{ g dm}^{-2}\text{ hour}^{-1}\) (1 mark). (Total 2.5 marks)

a) Active loading of sucrose at source decreases water potential, causing water to enter and raise hydrostatic pressure. unloading at sink lowers pressure. b) Ringing shows sucrose accumulation above phloem removal. Tracers show the path of carbon-14. c) Radius \(r = 6\ \mu\text{m} = 0.006\text{ mm}\). Area \(A = 3.14 \times (0.006)^2 = 0.00011304\text{ mm}^2\). Velocity per hour: \(0.5\text{ mm s}^{-1} \times 3600\text{ s} = 1800\text{ mm hour}^{-1}\). Volume: \(0.00011304 \times 1800 = 0.203472\text{ mm}^3\text{ hour}^{-1}\), which is \(2.03 \times 10^{-1}\text{ mm}^3\text{ hour}^{-1}\).

a) 1. Sucrose is actively loaded into sieve tubes by companion cells at source; 2. This lowers water potential in sieve tubes; 3. Water enters sieve tubes from xylem by osmosis; 4. This increases hydrostatic pressure at the source; 5. At the sink, sucrose is unloaded and used/stored; 6. This increases water potential at sink, so water leaves by osmosis; 7. This lowers hydrostatic pressure at the sink, establishing a hydrostatic pressure gradient from source to sink. (Max 7 marks)

b) Ringing: 1. Removing a ring of bark (phloem) leads to swelling above the ring containing sugar; 2. Proves phloem transports organic solutes. Tracers: 3. Plants exposed to carbon-14 produce radioactive sugars; 4. Autoradiography tracks movement, showing radioactive sugars move down phloem. (Max 4 marks)

c) 1. Calculates cross-sectional area: \(1.13 \times 10^{-4}\text{ mm}^2\) (0.5 marks); 2. Calculates hourly velocity: \(1800\text{ mm hour}^{-1}\) (0.5 marks); 3. Correct final answer of \(2.03 \times 10^{-1}\text{ mm}^3\text{ hour}^{-1}\) (0.5 marks). (Total 1.5 marks)

a) Starch/glycogen are polymers of alpha-glucose; cellulose is beta-glucose. Starch/glycogen have 1,4 and 1,6 glycosidic bonds leading to branching, making them compact and easily hydrolyzed; cellulose has only 1,4 glycosidic bonds, forming straight unbranched chains. Chains run parallel, held by hydrogen bonds to form microfibrils which provide tensile strength to cell walls. b) Non-competitive inhibitors bind to the allosteric site of amylase, altering its tertiary structure and changing the active site shape so it is no longer complementary to starch, preventing enzyme-substrate complexes from forming.

a) 1. Starch and glycogen made of alpha-glucose, cellulose made of beta-glucose; 2. Starch (amylopectin) and glycogen have branched structures due to 1,6-glycosidic bonds; 3. Branching allows rapid hydrolysis to release glucose for respiration; 4. Both are insoluble, meaning they do not affect water potential/osmotic balance of cells; 5. Cellulose forms long, straight, unbranched chains of beta-glucose (alternate monomers rotated 180 degrees); 6. Multiple parallel chains are held together by hydrogen bonds to form strong microfibrils; 7. This provides high tensile strength to plant cell walls; 8. Starch and glycogen are compact storage molecules. (Max 8 marks)

b) 1. Inhibitor binds to the enzyme at an allosteric site / site other than the active site; 2. This alters the tertiary structure / 3D shape of the enzyme; 3. This changes the shape of the active site; 4. Active site is no longer complementary to the substrate (starch); 5. Fewer enzyme-substrate complexes form. (Max 4.5 marks)

a) Genes are located on the same chromosome (linked). The alleles R and D are on one chromosome, r and d on the homologous chromosome in the F1 parent (RD/rd). Test cross is RD/rd x rd/rd. Parental genotypes RD/rd and rd/rd are produced in higher numbers (Red cut, Yellow potato). Recombinants Rd/rd (Red potato) and rD/rd (Yellow cut) are produced in lower numbers due to crossing over. b) Recombinants = 68 + 72 = 140. Total offspring = 1000. Recombination frequency = (140 / 1000) * 100 = 14.0%. c) Crossing over occurs in Prophase I of meiosis. Homologous chromosomes pair (bivalent formation), non-sister chromatids cross over at chiasmata, exchange alleles, creating new allele combinations.

a) 1. Genes are linked / located on the same chromosome; 2. This prevents independent assortment during meiosis; 3. F1 genotype is heterozygous (RD/rd); 4. Most gametes produced by F1 are parental types (RD or rd); 5. Test cross partner (rrdd) only produces rd gametes; 6. Small numbers of recombinants (Red/potato and Yellow/cut) are produced due to crossing over; 7. Correct genetic diagram showing linked alleles in F1 parent (e.g., RD/rd x rd/rd) and offspring genotypes. (Max 7 marks)

b) 1. Identifies recombinant phenotypes (Red, potato leaf and Yellow, cut leaf) and adds their counts: 68 + 72 = 140 (1 mark); 2. Divides recombinants by total offspring (140 / 1000) and multiplies by 100 (0.5 marks); 3. Correct answer: 14% or 14.0% (1 mark). (Total 2.5 marks)

c) 1. Homologous chromosomes pair up to form bivalents during prophase I; 2. Non-sister chromatids break and rejoin at chiasmata; 3. This exchanges alleles between maternal and paternal chromosomes, creating recombinant chromatids. (Max 3 marks)

a) Glycolysis involves: phosphorylation of glucose to glucose phosphate (using 2 ATP), splitting into triose phosphate, oxidation to pyruvate with reduction of NAD and net production of 2 ATP. In anaerobic conditions, oxygen is not present as the terminal electron acceptor in the electron transport chain, so Link reaction, Krebs cycle, and oxidative phosphorylation cannot occur. Glycolysis must continue, meaning pyruvate is reduced to lactate to regenerate NAD. b) Aerobic respiration yields much more ATP (32 ATP) per glucose than anaerobic respiration (2 ATP). To maintain the same level of ATP required for cellular processes, the yeast must consume glucose at a rate 16 times higher. Since anaerobic respiration in yeast (ethanol fermentation) produces CO2 as a byproduct, this massive increase in glucose consumption rate leads to an increase in CO2 production rate. c) Aerobic ATP = 15 * 32 = 480. Anaerobic ATP = 15 * 2 = 30. Difference = 480 - 30 = 450 ATP.

a) 1. Glucose is phosphorylated using phosphate from ATP to form glucose phosphate; 2. Glucose phosphate is split into triose phosphate; 3. Triose phosphate is oxidized to pyruvate, forming NADH and 4 ATP (net 2 ATP); 4. Oxygen is absent, so it cannot act as final electron acceptor; 5. Thus, Link, Krebs, and Oxidative Phosphorylation stop; 6. Pyruvate is reduced to lactate using reduced NAD, regenerating NAD to allow glycolysis to continue. (Max 6 marks)

b) 1. Aerobic respiration is much more efficient / produces ~32 ATP per glucose; 2. Anaerobic respiration is inefficient / produces only 2 ATP per glucose; 3. Yeast must respire glucose at a much higher rate to maintain metabolic activity / generate equivalent ATP; 4. Since ethanol fermentation produces CO2, the increased rate of glycolysis/fermentation results in higher CO2 production. (Max 4 marks)

c) 1. Calculates aerobic yield: 15 x 32 = 480 ATP (1 mark); 2. Calculates anaerobic yield: 15 x 2 = 30 ATP (0.5 marks); 3. Calculates the difference: 480 - 30 = 450 ATP (1 mark). (Total 2.5 marks)

a) Proto-oncogenes mutate into oncogenes, causing hyperactive/continuous signal transduction which stimulates rapid cell division. Tumor suppressor genes, which normally produce proteins that slow cell cycle or promote apoptosis, are inactivated by mutations. If both alleles of a TSG are mutated, cell division occurs unchecked, leading to tumor formation. b) Epigenetics refers to heritable changes in gene expression without altering DNA base sequence. Hypermethylation of DNA at CpG islands in promoter regions of tumor suppressor genes prevents transcription factor binding, silencing the gene. Deacetylation of histones increases positive charges on histones, causing them to bind more tightly to DNA (heterochromatin structure), preventing RNA polymerase access and silencing tumor suppressor genes, leading to uncontrolled cell division.

a) 1. Proto-oncogenes mutate into oncogenes; 2. Oncogenes are permanently activated / overexpressed, leading to continuous stimulation of cell division / mitosis; 3. Tumor suppressor genes (TSG) normally produce proteins that slow down cell division, repair DNA, or initiate apoptosis; 4. Mutated TSGs are inactivated / do not produce a functional protein; 5. This removes the 'brake' on cell division, allowing uncontrolled mitosis; 6. Mention that proto-oncogene mutations are dominant (only one allele mutated needed) while tumor suppressor mutations are recessive (both alleles must be mutated) (1.5 marks). (Max 6.5 marks)

b) 1. Epigenetics involves changes in gene expression without altering DNA sequence; 2. Increased methylation of DNA (at CpG islands/promoter regions) of tumor suppressor genes blocks transcription factors/RNA polymerase; 3. Decreased acetylation of histones causes chromatin to condense / histones to bind more tightly to DNA; 4. This prevents transcription of tumor suppressor genes, leading to uncontrolled division. (Max 4 marks)

c) Quality of Written Communication (2 marks):
- 2 marks: Logical, structured, clear argument with correct use of key terms (oncogene, tumor suppressor, methylation, acetylation, transcription, cell division).
- 1 mark: Some structure and correct terms, but minor gaps or errors in expression.
- 0 marks: Unstructured, major errors in biological terminology.