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(a) Total lipid mass = \( 2.40 \times 10^{-4} \text{ g} = 240 \ \mu\text{g} \). Percentage of cholesterol = \( 100\% - (58.5\% + 24.2\%) = 17.3\% \). Mass of cholesterol = \( 17.3\% \times 240 \ \mu\text{g} = 41.52 \ \mu\text{g} \). Rounded to three significant figures, this is \( 41.5 \ \mu\text{g} \). (b) Both molecules contain glycerol and ester bonds. However, a phospholipid contains only two fatty acids and has a phosphate group, while a triglyceride contains three fatty acids and no phosphate. Phospholipids are amphipathic (polar/hydrophilic head and non-polar/hydrophobic tails), whereas triglycerides are entirely hydrophobic. (c) Homogenize the tissue, add ethanol and shake to dissolve lipids, then add water and shake again. A white, milky emulsion indicates the presence of lipids.

(a) 3.7 marks total: 1 mark for finding percentage of cholesterol (17.3%); 1 mark for converting mass to micrograms (240 \(\mu\text{g}\)); 1 mark for correct mass calculation (41.52); 0.7 marks for rounding to 3 significant figures with units (41.5 \(\mu\text{g}\)). (b) 4 marks total: 1 mark for similarity (both have glycerol/ester bonds/fatty acids); 1 mark for contrast 1 (phospholipids have 2 fatty acids, triglycerides have 3); 1 mark for contrast 2 (phospholipids have a phosphate group, triglycerides do not); 1 mark for contrast 3 (phospholipids have hydrophilic and hydrophobic regions, triglycerides are fully hydrophobic). (c) 3 marks total: 1 mark for adding ethanol and shaking; 1 mark for adding water; 1 mark for observing a white/milky emulsion.

(a) Convert the image size to the same unit as actual size: \( 4.8 \text{ cm} = 48 \text{ mm} = 48,000 \ \mu\text{m} \). Magnification = Image size / Actual size = \( 48,000 / 1.5 = 32,000 \). Written with the multiplication prefix, this is \( \times 32,000 \). (b) First, the tissue is homogenised to break open cell membranes and release organelles. The homogenate is filtered to remove unbroken cells and large debris. The mixture must be kept ice-cold (to stop enzyme activity), isotonic (to prevent osmotic damage/bursting of organelles), and buffered (to prevent pH-driven denaturation of proteins). The filtrate is centrifuged at a low speed to pellet the heaviest organelles (nuclei). The supernatant is poured off and centrifuged at a higher speed to pellet the mitochondria. (c) Mitochondria possess cristae (folded inner membrane) while chloroplasts have thylakoids/grana; chloroplasts contain chlorophyll/pigments while mitochondria do not; chloroplasts may contain starch grains while mitochondria do not.

(a) 3.7 marks total: 1 mark for converting image size to micrometers (48,000 \(\mu\text{m}\)); 1 mark for using Magnification = Image / Actual; 1 mark for calculating 32,000; 0.7 marks for writing the answer with \(\times\) prefix and no units (\(\times 32,000\)). (b) 4 marks total: 1 mark for homogenisation and filtration; 1 mark for stating and explaining the required conditions (ice-cold, isotonic, buffered); 1 mark for first centrifugation at low speed to remove nuclei; 1 mark for subsequent centrifugation of supernatant at higher speed to isolate mitochondria. (c) 3 marks total: 1 mark for each valid structural difference (up to 3).

(a) At low concentrations (0 to 4 \( \text{mmol dm}^{-3} \)), the rate of uptake increases linearly, showing it is proportional to concentration. However, at higher concentrations, the rate plateaus because all specific carrier or channel proteins are fully occupied (saturated). In simple diffusion, the rate would continue to rise linearly without a plateau. (b) Active transport involves sodium ions binding to specific sites on a carrier protein. ATP is hydrolyzed to ADP and inorganic phosphate, releasing energy that causes the carrier protein to change shape. This translocates the sodium ions across the membrane against their concentration gradient. (c) Sodium-potassium pumps actively transport sodium ions out of epithelial cells into the blood, creating a low concentration of sodium inside the cell. Sodium ions then diffuse down their concentration gradient from the lumen into the cell through a co-transport protein, bringing glucose molecules with them against their concentration gradient. Glucose then exits into the blood down its concentration gradient via facilitated diffusion.

(a) 3.7 marks total: 1 mark for noting the proportional rate increase at low concentrations; 1 mark for identifying that the rate plateaus/levels off at higher concentrations; 1 mark for explaining that this is due to carrier/channel protein saturation; 0.7 marks for contrasting with simple diffusion (which does not plateau). (b) 4 marks total: 1 mark for sodium binding to a specific site on carrier protein; 1 mark for ATP binding and hydrolysis; 1 mark for shape/conformational change of protein; 1 mark for transport against the concentration gradient. (c) 3 marks total: 1 mark for active transport of sodium out of cells; 1 mark for facilitated co-transport of sodium and glucose into cells; 1 mark for facilitated diffusion of glucose into blood.

(a) Because the DNA is double-stranded, Chargaff's rules apply. The percentage of cytosine (C) equals the percentage of guanine (G), so G = 34.2%. Thus, C + G = 34.2% + 34.2% = 68.4%. The remaining bases must be adenine (A) and thymine (T), so A + T = 100% - 68.4% = 31.6%. Since A = T, the percentage of adenine = \( 31.6\% / 2 = 15.8\% \). (b) Eukaryotic DNA is linear, associated with histone proteins, enclosed inside a nucleus, and contains introns (non-coding sequences). Prokaryotic DNA is circular, 'naked' (not associated with histones), free within the cytoplasm (not in a nucleus), and lacks introns. (c) A gene is a section of DNA containing the nucleotide sequence that codes for the amino acid sequence of a specific polypeptide or a functional RNA. A locus is the fixed position that a particular gene occupies on a DNA molecule or chromosome.

(a) 3.7 marks total: 1 mark for stating C = G, so G = 34.2%; 1 mark for calculating C + G = 68.4%; 1 mark for calculating A + T = 31.6%; 0.7 marks for dividing by 2 to get 15.8% with the percentage sign. (b) 4 marks total: 1 mark for linear vs circular; 1 mark for associated with histones vs naked; 1 mark for enclosed in nucleus vs free in cytoplasm; 1 mark for containing introns vs no introns. (c) 3 marks total: 2 marks for gene definition (1 mark for section of DNA, 1 mark for coding for polypeptide/functional RNA); 1 mark for locus definition (specific position of gene on chromosome).

(a) Each codon consists of 3 nucleotides, so \( 1200 / 3 = 400 \) codons. One of these must be a stop codon, which does not code for any amino acid, giving a maximum of \( 400 - 1 = 399 \) amino acids. The actual number of amino acids may be lower because mature mRNA contains untranslated regions (UTRs) at both ends that do not code for proteins, and post-translational modification may remove the starting amino acid (methionine) or cleave the chain. (b) Transcription begins when DNA helicase unwinds the double helix by breaking hydrogen bonds, exposing the template strand. Free RNA nucleotides align along the template strand by complementary base pairing. RNA polymerase then joins these RNA nucleotides together by forming phosphodiester bonds, synthesizing the pre-mRNA transcript until it reaches a stop signal. (c) mRNA acts as a template containing codons that carry the genetic instructions from DNA to the ribosome. tRNA molecules are adapters, carrying specific amino acids to the ribosome and pairing their anticodons with the complementary codons on the mRNA to assemble the amino acids in the correct sequence.

(a) 3.7 marks total: 1 mark for dividing nucleotides by 3 (400 codons); 1 mark for subtracting stop codon (399 amino acids); 1 mark for explaining UTRs/non-coding regions on mature mRNA; 0.7 marks for mentioning post-translational cleavage or start codon removal. (b) 4 marks total: 1 mark for helicase breaking hydrogen bonds/unwinding; 1 mark for complementary pairing of RNA nucleotides with template; 1 mark for RNA polymerase forming phosphodiester bonds; 1 mark for process stopping at a terminator sequence. (c) 3 marks total: 1 mark for mRNA carrying the genetic code/codons; 1 mark for tRNA carrying specific amino acids; 1 mark for tRNA anticodon pairing with mRNA codon.

(a) The absolute difference in oxygen saturation is \( 78\% - 62\% = 16\% \). The percentage increase compared to maternal hemoglobin is \( (16 / 62) \times 100\% = 25.80645\% \). Rounded to three significant figures, this is \( 25.8\% \). (b) Fetal hemoglobin has a higher affinity for oxygen, meaning its dissociation curve is shifted to the left of the maternal curve. At the placenta, where the partial pressure of oxygen is relatively low, maternal hemoglobin dissociates and releases oxygen. The fetal hemoglobin can readily bind to this released oxygen at the same partial pressure, ensuring the fetus receives an adequate oxygen supply for respiration. (c) Respiring tissues release carbon dioxide, which dissolves to form carbonic acid, lowering the pH of the blood. The hydrogen ions interact with hemoglobin, altering its tertiary structure and reducing its affinity for oxygen. This causes the dissociation curve to shift to the right, facilitating the unloading of oxygen to the tissues that need it most.

(a) 3.7 marks total: 1 mark for calculating absolute difference (16); 1 mark for setting up percentage change relative to maternal (16/62); 1 mark for calculation of 25.806...; 0.7 marks for rounding to 3 significant figures with correct % unit (25.8%). (b) 4 marks total: 1 mark for stating fetal hemoglobin has higher oxygen affinity / curve shifted left; 1 mark for stating maternal hemoglobin unloads oxygen at placenta; 1 mark for stating fetal hemoglobin loads oxygen at lower partial pressures; 1 mark for linking to aerobic respiration/survival of fetus. (c) 3 marks total: 1 mark for high CO2 reducing blood pH; 1 mark for change in hemoglobin shape reducing affinity; 1 mark for curve shifting to the right / easier unloading.

(a) The change in \( K_{\mathrm{m}} = 6.5 - 2.5 = 4.0 \text{ mmol dm}^{-3} \). The percentage change is \( (4.0 / 2.5) \times 100\% = 160\% \). This represents a 160% increase in the Michaelis constant. (b) A competitive inhibitor has a molecular shape that is highly similar to the substrate, allowing it to bind to the active site of the enzyme. This blocks substrate molecules from entering, preventing the formation of enzyme-substrate complexes. If substrate concentration is increased, the ratio of substrate to inhibitor molecules increases, making it much more likely for a substrate molecule to collide with and bind to an active site, thereby overcoming the inhibition and achieving the maximum reaction rate (\( V_{\max} \)). (c) Competitive inhibitors bind specifically to the active site, while non-competitive inhibitors bind to an allosteric site. Non-competitive binding alters the tertiary structure of the enzyme, changing the shape of the active site so that substrates can no longer bind. While competitive inhibition can be overcome by increasing substrate concentration, non-competitive inhibition cannot, and the maximum rate of reaction (\( V_{\max} \)) is reduced.

(a) 3.7 marks total: 1 mark for calculating the absolute change (4.0); 1 mark for dividing change by original value (4.0 / 2.5); 1 mark for calculating 160; 0.7 marks for correctly identifying it as an 'increase' or '+' with the % unit (160% increase). (b) 4 marks total: 1 mark for competitive inhibitor having complementary shape to active site; 1 mark for blocking substrate / preventing enzyme-substrate complex; 1 mark for stating that increasing substrate outnumbers inhibitor; 1 mark for stating that Vmax can still be achieved. (c) 3 marks total: 1 mark for active site vs allosteric site binding; 1 mark for active site shape change (non-competitive) vs no shape change (competitive); 1 mark for effect of substrate concentration (overcomes competitive, does not overcome non-competitive).

(a) The macrophage recognizes foreign antigens on the bacterial cell wall. The macrophage cell membrane invaginates / extends pseudopodia to engulf the bacterium into a vesicle called a phagosome. Lysosomes fuse with the phagosome to form a phagolysosome. Hydrolytic enzymes (such as lysozymes) are released into the vesicle to digest and destroy the bacterium. (b) An antibody is a Y-shaped protein with variable regions that possess a highly specific tertiary structure. These variable regions are complementary to specific antigens on the pathogen, allowing the antibody to bind to them. The constant region of the antibody binds to specific receptors on the cell surface membrane of phagocytes. This links the pathogen to the phagocyte (opsonisation), making the pathogen easier to detect and engulf. Additionally, because antibodies have multiple binding sites, they can cause agglutination (clumping) of pathogens, allowing multiple pathogens to be engulfed at once. (c) Difference in rate = \(4.8 \times 10^5 - 1.2 \times 10^4 = 480,000 - 12,000 = 468,000\) bacteria per hour. Percentage increase = \(\frac{468,000}{12,000} \times 100 = 3900\%\).

(a) 1 mark for recognition of non-self antigens. 1 mark for engulfment to form a phagosome. 1 mark for fusion of lysosomes with the phagosome. 1 mark for action of lysozymes/hydrolytic enzymes in digesting the pathogen. (b) 1 mark for variable regions having a specific tertiary structure complementary to the antigen. 1 mark for constant region binding to receptors on phagocytes. 1 mark for concept of opsonisation (marking/targeting the pathogen). 1 mark for agglutination / clumping due to multiple antigen-binding sites. (c) 1 mark for calculating the absolute difference (\(468,000\)). 1 mark for dividing by the original value (\(12,000\)). 1 mark for correct final answer of 3900%.

(a) Sucrose is actively transported from companion cells into sieve tube elements at the source. This lowers the water potential inside the sieve tube elements, causing water to enter from the xylem by osmosis. This entry of water increases the hydrostatic pressure at the source. At the sink (e.g. roots/growing shoots), sucrose is unloaded and used in respiration or stored as starch. This increases the water potential inside the sieve tubes, causing water to leave by osmosis, which lowers the hydrostatic pressure. This creates a hydrostatic pressure gradient from source to sink, driving the mass flow of organic substances along the phloem. (b) The researcher would expose a leaf to radioactive carbon dioxide (\(^{14}\text{CO}_2\)), which is incorporated into organic sugars (like sucrose) during photosynthesis. At timed intervals, sections of the plant stem are cut. The stem sections are pressed against photographic film (autoradiography). The position of the radioactive sugars is detected as dark spots on the developed film. By measuring the distance the radioactive sugars traveled along the stem and dividing by the time elapsed, the rate of translocation (e.g. in \(\text{cm hr}^{-1}\)) can be calculated. (c) DNP is an inhibitor of aerobic respiration/ATP synthesis in mitochondria. The loading of sucrose into companion cells requires ATP for active transport (via the proton pump). Without ATP, active loading stops, which prevents the development of the hydrostatic pressure gradient needed for mass flow.

(a) 1 mark for active transport of sucrose into sieve tubes at the source. 1 mark for osmosis of water from xylem lowering the water potential. 1 mark for increased hydrostatic pressure at the source. 1 mark for unloading of sucrose at the sink causing water to leave and lowering hydrostatic pressure. 1 mark for hydrostatic pressure gradient driving mass flow. (b) 1 mark for providing leaf with \(^{14}\text{CO}_2\) to form labeled sugars. 1 mark for cutting sections of stem at known/timed intervals. 1 mark for using photographic film/x-ray film to detect radioactivity (autoradiography). 1 mark for dividing the distance traveled by the time taken to calculate the rate. (c) 1 mark for DNP stopping ATP production/respiration. 1 mark for linking lack of ATP to the failure of active loading of sucrose.

(a) Sodium ions (\(\text{Na}^+\)) are actively transported out of the epithelial cells into the blood by the sodium-potassium pump, using ATP. This maintains a low concentration of sodium ions inside the cytoplasm of the epithelial cell. Consequently, a concentration gradient of sodium ions is established between the lumen of the ileum (high concentration) and the cell cytoplasm (low concentration). Sodium ions diffuse down their concentration gradient into the epithelial cells through a co-transporter protein in the microvilli membrane. As they do so, they couple with glucose molecules, carrying them into the cell against their concentration gradient. (b) Active transport of sodium ions out of the cell maintains the low concentration of sodium ions inside the cell. Without this active transport, sodium ions would accumulate in the cytoplasm, eliminating the sodium ion concentration gradient between the lumen and the cell. If there is no gradient, sodium ions cannot diffuse into the cell, and co-transport of glucose would cease. (c) The rate of glucose absorption would significantly decrease. This is because the concentration of sodium ions in the lumen (mucosal fluid) is now much lower, which reduces or reverses the sodium ion concentration gradient between the lumen and the epithelial cell. As a result, fewer sodium ions bind to and move through the co-transporter protein, meaning less glucose is transported into the cell.

(a) 1 mark for active transport of sodium ions out of the cell into the blood. 1 mark for use of ATP/sodium-potassium pump. 1 mark for establishing/maintaining a sodium ion concentration gradient. 1 mark for sodium ions diffusing through a co-transporter protein. 1 mark for glucose being carried into the cell against its concentration gradient alongside sodium. (b) 1 mark for maintaining low intracellular sodium concentration. 1 mark for maintaining the sodium concentration gradient. 1 mark for stating that without the pump, intracellular sodium would rise and glucose co-transport would stop. (c) 1 mark for predicting a decrease in glucose absorption. 1 mark for explaining that the sodium concentration gradient is reduced/removed. 1 mark for stating fewer sodium ions enter via the co-transporter, bringing less glucose with them.

(a) During ventricular systole, the left ventricle contracts, causing the pressure within it to rise. Once the ventricular pressure exceeds the pressure inside the aorta, the semi-lunar valves are forced open, allowing blood to flow into the aorta. During ventricular diastole, the ventricle relaxes, and ventricular pressure falls below aortic pressure. The high pressure in the aorta relative to the ventricle forces blood backward, which fills the valve pockets and forces the semi-lunar valves to close, preventing backflow. (b) Formula: \(\text{Cardiac Output} = \text{Heart Rate} \times \text{Stroke Volume}\). Stroke volume = \(75\text{ cm}^3 = 0.075\text{ dm}^3\). \(\text{Cardiac Output} = 72 \times 0.075 = 5.4\text{ dm}^3\text{ min}^{-1}\). (c) The left ventricle must pump blood through the systemic circulation to the entire body, which is a much longer distance/has higher resistance. In contrast, the right ventricle only pumps blood to the lungs (pulmonary circulation), which is a short distance/has low resistance. Therefore, the left ventricle must generate a much higher hydrostatic pressure than the right ventricle. The thicker muscular wall contains more muscle fibers to produce a more powerful contraction to generate this high pressure, while preventing damage to the delicate capillaries in the lungs which receive lower pressure blood from the right ventricle.

(a) 1 mark for semi-lunar valves opening when ventricular pressure exceeds aortic pressure. 1 mark for semi-lunar valves closing when ventricular pressure falls below aortic pressure. 1 mark for explaining that closure is driven by blood flowing backwards and filling the pockets of the valves. (b) 1 mark for converting stroke volume to \(\text{dm}^3\) (\(0.075\text{ dm}^3\)). 1 mark for correct formula/working (\(72 \times 0.075\)). 1 mark for correct final answer of 5.4 with units of \(\text{dm}^3\text{ min}^{-1}\). (c) 1 mark for left ventricle pumping blood to the whole body / systemic circulation while right ventricle pumps to lungs / pulmonary circulation. 1 mark for systemic circulation having higher resistance / longer distance. 1 mark for left ventricle needing to generate higher pressure. 1 mark for thicker wall having more muscle to contract with more force. 1 mark for mentioning that lower pressure from the right ventricle prevents damage to capillaries in lungs.

(a) HIV consists of a core containing two single strands of RNA (genetic material) and key enzymes, including reverse transcriptase and integrase. This core is surrounded by a protein coat called a capsid. Around the capsid is a lipid envelope, derived from the host cell membrane during budding. Projecting from the lipid envelope are attachment glycoproteins (such as gp120) which allow the virus to bind to host cells. (b) Once inside the helper T cell, the viral RNA is released. The enzyme reverse transcriptase uses the viral RNA as a template to synthesize a complementary single strand of DNA, and then double-stranded viral DNA. This viral DNA is transported into the nucleus where the enzyme integrase inserts it into the host cell's DNA. The host cell's RNA polymerase transcribes the integrated viral DNA into viral mRNA. This viral mRNA is translated by host cell ribosomes to synthesize viral proteins. These proteins and viral RNA assemble at the cell membrane to form new virus particles, which bud off from the helper T cell. (c) HIV antigen is fixed to the bottom of a well in a microtiter plate. The patient's blood serum is added. If HIV antibodies are present in the serum, they bind to the HIV antigen. The well is washed to remove any unbound antibodies. A secondary antibody, which is conjugated to an enzyme, is added; this antibody binds specifically to the patient's primary antibody. The well is washed again to remove any unbound secondary antibody. Finally, a colorless substrate for the enzyme is added. The enzyme acts on the substrate to produce a colored product, indicating a positive result. The intensity of the color can indicate the concentration of HIV antibodies.

(a) 1 mark for RNA core (two single strands) and reverse transcriptase. 1 mark for protein capsid. 1 mark for lipid envelope. 1 mark for attachment glycoproteins/gp120 on the outside of the envelope. (b) 1 mark for reverse transcriptase making viral DNA from viral RNA. 1 mark for integrase inserting viral DNA into host cell DNA. 1 mark for host cell transcribing viral DNA to mRNA and translating it to viral proteins. 1 mark for assembly and budding of new virus particles from host membrane. (c) 1 mark for fixing HIV antigen to the well and adding serum to allow primary antibody binding, followed by washing. 1 mark for adding an enzyme-linked secondary antibody that binds to the primary antibody, followed by washing. 1 mark for adding a substrate that changes color in the presence of the enzyme.

(a) Proto-oncogenes normally stimulate cell division / mitosis when growth factors bind to receptors on the cell membrane, promoting cell growth and survival. Tumor suppressor genes normally inhibit cell division, repair DNA damage, and trigger programmed cell death (apoptosis) when DNA damage cannot be repaired. (b) Hypermethylation involves the addition of excess methyl groups to cytosine bases within the promoter region of a gene. This high level of methylation prevents the binding of transcription factors and RNA polymerase to the promoter. Consequently, transcription of the tumor suppressor gene is silenced / blocked, so the corresponding protective protein is not translated. Without the tumor suppressor protein to regulate the cell cycle, cell division becomes uncontrolled, leading to the rapid growth of a tumor. (c) Gene mutations alter the DNA base sequence permanently; repairing them would require complex gene-editing techniques. In contrast, epigenetic modifications do not alter the base sequence, only how the DNA is read. Because they are reversible, drugs can be designed to inhibit methyltransferase enzymes (demethylating agents). This can remove the excess methyl groups from the promoter regions of silenced tumor suppressor genes, restoring their normal expression so they can synthesize proteins that stop uncontrolled cell division.

(a) 1 mark for proto-oncogenes stimulating cell division/mitosis. 1 mark for tumor suppressor genes inhibiting cell division. 1 mark for proto-oncogenes reacting to growth factors. 1 mark for tumor suppressor genes being involved in DNA repair / inducing apoptosis. (b) 1 mark for methylation occurring at the promoter region of the gene. 1 mark for preventing transcription factors/RNA polymerase from binding. 1 mark for silencing transcription/translation of the tumor suppressor protein. 1 mark for loss of cell cycle regulation leading to uncontrolled cell division/tumor. (c) 1 mark for stating mutations are permanent alterations in base sequence, making them hard to fix. 1 mark for stating epigenetics does not alter base sequence. 1 mark for explaining that drugs can demethylate/reactivate silenced tumor suppressor genes to stop tumor growth.

(a) Cut a small section (approx. 1-2 mm) from the very tip of a growing garlic root. Place the root tip in dilute hydrochloric acid and heat gently in a water bath to macerate/soften the tissue by breaking down the middle lamella. Rinse the root tip thoroughly with water on a watch glass. Place the root tip on a clean microscope slide and add a few drops of a stain (such as acetic orcein or toluidine blue) to stain the chromosomes. Gently lower a coverslip over the specimen and press down firmly (squash) with a thumb or pencil eraser without rolling the coverslip, to spread the cells into a single, thin layer. (b) First, calculate the total number of cells undergoing mitosis: \(35 + 12 + 8 + 15 = 70\) cells. The total number of cells observed is 250. Mitotic Index = \(\frac{\text{Number of cells in mitosis}}{\text{Total number of cells}} = \frac{70}{250} = 0.28\) (or \(28\%\)). (c) Eukaryotic mitosis involves the breakdown of a nuclear envelope, whereas prokaryotes do not have a nucleus. Mitosis involves the alignment and separation of sister chromatids using spindle fibers, whereas binary fission does not use spindle fibers. Mitosis replicates and separates multiple linear chromosomes, whereas binary fission involves the replication of a single circular DNA molecule and plasmids.

(a) 1 mark for cutting the active growing root tip (1-2 mm). 1 mark for acid treatment to macerate/soften tissue. 1 mark for rinsing and applying a stain (e.g. acetic orcein/toluidine blue). 1 mark for squashing the specimen to obtain a single thin layer of cells. 1 mark for emphasizing not rolling/moving the coverslip to avoid damage. (b) 1 mark for summing cells in mitosis (70). 1 mark for dividing by total cells (\(70 / 250\)). 1 mark for correct final answer of 0.28 or 28%. (c) 1 mark for nucleus breakdown in mitosis vs no nucleus in binary fission. 1 mark for use of spindle fibers in mitosis vs no spindle fibers in binary fission. 1 mark for separation of linear chromosomes vs replication of single circular chromosome/plasmids in binary fission.

(a) The expected phenotypic ratio is 12 white : 3 yellow : 1 green. (b) The cross is WwYy (white) x wwyy (green). The gametes produced by WwYy are WY, Wy, wY, and wy, each with a probability of 0.25. The gamete produced by wwyy is wy. The offspring genotypes will be: 1/4 WwYy (white), 1/4 Wwyy (white), 1/4 wwYy (yellow), and 1/4 wwyy (green). Therefore, the probability of obtaining white-fruited offspring (WwYy and Wwyy) is 2/4 = 0.5 (or 50%). (c) 1. Formulate a null hypothesis stating that there is no significant difference between the observed and expected phenotypic ratio (2 white : 1 yellow : 1 green). 2. Use the chi-squared formula: \chi^2 = \sum \frac{(O - E)^2}{E} where O is the observed number and E is the expected number for each phenotype. 3. Compare the calculated \chi^2 value to the critical value at the p = 0.05 significance level with 2 degrees of freedom (number of phenotypic classes minus 1). (d) Epistasis occurs when the allele of one gene masks the expression of another gene. This means multiple different genotypes (e.g., W_Y_ and W_yy) produce the same phenotype (white). This reduces the number of distinct phenotypic classes in the population, thereby reducing phenotypic variation.

(a) 1 mark for correct phenotypes linked to ratios (white, yellow, green); 1 mark for correct ratio: 12 white : 3 yellow : 1 green. (b) 1 mark for identifying the genotypes of the offspring (WwYy, Wwyy, wwYy, wwyy); 1 mark for identifying that both WwYy and Wwyy are white; 1 mark for correct probability of 0.5 / 50% / 1 in 2. (c) 1 mark for stating the null hypothesis (no significant difference between observed and expected 2:1:1 ratio); 1 mark for the correct formula of chi-squared; 1 mark for comparing to a critical table value at p = 0.05 with df = 2. (d) 1 mark for explaining that epistasis involves one gene masking another; 1 mark for noting that different genotypes result in the same phenotype; 1 mark for concluding that this limits the number of phenotypic classes.

(a) Succinate acts as a respiratory substrate, donating electrons to the electron transport chain (ETC). This flow of electrons leads to active proton pumping into the intermembrane space, creating a proton gradient. When ADP and P_i are added, ATP synthase is able to channel protons back into the matrix to synthesize ATP, which dissipates the proton gradient. This allows the ETC to continue flowing rapidly, thus increasing the consumption of oxygen as the terminal electron acceptor. (b) (i) The rate of ATP production will stop or decrease to zero because protons can no longer flow through the blocked ATP synthase channel. (ii) The rate of oxygen consumption will decrease because without proton flow through ATP synthase, the proton gradient across the inner membrane becomes too steep, which inhibits further proton pumping and slows down the ETC. (c) ATP synthesis will decrease or stop because protons leak across the inner membrane instead of passing through ATP synthase, meaning no proton motive force is harnessed. Oxygen consumption will increase significantly because the proton gradient is constantly dissipated, removing any resistance to proton pumping and allowing the ETC to run at its maximum rate.

(a) 1 mark for identifying that ADP and P_i are substrates for ATP synthase; 1 mark for stating that proton flow through ATP synthase dissipates the proton gradient; 1 mark for linking this to an increased rate of the electron transport chain and higher oxygen usage as the final acceptor. (b) 1 mark for predicting ATP production stops; 1 mark for explaining that protons cannot pass through ATP synthase; 1 mark for predicting oxygen consumption decreases; 1 mark for explaining that accumulation of protons in the intermembrane space restricts electron flow. (c) 1 mark for predicting ATP synthesis stops/decreases; 1 mark for explaining that protons leak across the membrane bypass ATP synthase; 1 mark for predicting oxygen consumption increases; 1 mark for explaining that the lack of a proton gradient allows maximum, unchecked electron transport.

(a) Accessory pigments absorb wavelengths of light that chlorophyll a cannot absorb, transferring this light energy to the primary reaction center (chlorophyll a) to increase photosynthetic efficiency. They also protect chlorophyll from damage by excess light. (b) Without VDE, the mutant plant cannot dissipate excess light energy as heat. This excess energy causes photoinhibition/damage to Photosystem II (PSII). Damage to PSII proteins reduces the photolysis of water and slows down electron transport. Consequently, the production of ATP and reduced NADP (NADPH) in the light-dependent reaction is severely reduced. This limits the light-independent reaction (Calvin cycle) as there is less ATP and NADPH available to reduce glycerate 3-phosphate (GP) to triose phosphate (TP), thus lowering the rate of carbon dioxide fixation. (c) Temperatures above 40 degrees C lead to the denaturation of key enzymes in the Calvin cycle, such as Rubisco, as their active sites change shape and can no longer bind substrates. Additionally, high temperatures damage the thylakoid membranes, making them more permeable to protons and reducing the proton gradient, which limits ATP production. Stomata may also close to prevent transpiration, reducing the diffusion of carbon dioxide into the leaf.

(a) 1 mark for absorbing different wavelengths of light and transferring energy to chlorophyll a; 1 mark for photoprotection / protecting chlorophyll from damage. (b) 1 mark for stating that excess energy damages PSII / causes photoinhibition; 1 mark for stating this reduces the photolysis of water; 1 mark for stating this slows electron transport; 1 mark for identifying that less ATP and NADPH are produced; 1 mark for linking this to reduced conversion of GP to TP / carbon dioxide fixation in the Calvin cycle. (c) 1 mark for denaturation of enzymes (such as Rubisco); 1 mark for loss of active site shape preventing substrate binding; 1 mark for thylakoid membrane damage reducing the proton gradient / ATP synthesis; 1 mark for stomatal closure limiting carbon dioxide entry.

(a) The curve is oscillatory, showing repeating peaks and troughs where the predator population peaks lag behind the prey population peaks. When prey numbers are high, predators have more food, so their survival and reproduction increase, causing the predator population to rise. As predator numbers rise, they consume more prey, causing the prey population to crash. With fewer prey, predators starve, and their population falls, allowing the prey population to recover and restart the cycle. (b) Without predators, the rodent population is still limited by the ecosystem's carrying capacity. Rodents will face intense intraspecific competition for limited resources such as food, water, and nesting sites/shelters. Additionally, high population density increases the transmission of diseases or attracts other opportunistic predators, raising the mortality rate. (c) Two species occupying the same niche compete for identical limited resources (e.g., food, space). One of the species will have a slight competitive advantage (due to better adaptations). This species will obtain resources more efficiently, leading to higher survival and reproduction rates. Over time, the less adapted species will experience a decline in its population as its death rate exceeds its birth rate, eventually leading to its local extinction or exclusion from that niche.

(a) 1 mark for describing the cycling/oscillating nature of the curve; 1 mark for stating that predator peaks lag behind prey peaks; 1 mark for explaining that high prey numbers lead to predator increases; 1 mark for explaining that high predator numbers drive prey decreases, which then causes predator decreases. (b) 1 mark for reference to carrying capacity; 1 mark for identifying intraspecific competition for resources (e.g., food/nesting sites); 1 mark for suggesting density-dependent factors like disease spread or other predators. (c) 1 mark for stating that resources are limited and both species compete for them; 1 mark for explaining that one species is better adapted / has a competitive advantage; 1 mark for stating the advantaged species has higher survival/reproduction; 1 mark for concluding that the disadvantaged species population declines to local extinction/exclusion.

(a) Geographic isolation by physical barriers (dry land between springs) prevents gene flow/interbreeding between the isolated populations. Each spring has different biotic and abiotic conditions, representing distinct selection pressures (e.g., temperature, salinity, predators). Random mutations occur independently in each population. Natural selection acts on these mutations, favoring different alleles in different springs, which changes allele frequencies over time. Genetic drift may also occur, especially in small spring populations. Over many generations, the populations become genetically distinct. (b) Scientists can bring individuals from both springs together in a laboratory setting and attempt to mate them. If they cannot successfully interbreed to produce fertile offspring, they are classified as separate species. Alternatively, they can sequence and compare their DNA to measure the degree of genetic divergence. (c) Allopatric speciation occurs when populations are geographically isolated by a physical barrier. Sympatric speciation occurs within the same geographic area without any physical isolation, usually driven by behavioral, ecological, or genetic barriers (e.g., polyploidy or different mating seasons).

(a) 1 mark for stating that geographic isolation prevents gene flow; 1 mark for mentioning different selection pressures / environments; 1 mark for random mutations occurring independently; 1 mark for natural selection altering allele frequencies; 1 mark for genetic divergence over time preventing future interbreeding. (b) 1 mark for attempting to breed individuals together; 1 mark for checking if they can produce fertile offspring; 1 mark for DNA analysis/sequencing to compare genomes. (c) 1 mark for defining allopatric speciation as requiring a physical/geographic barrier; 1 mark for defining sympatric speciation as occurring in the same geographic area; 1 mark for attributing sympatric speciation to reproductive, behavioral, or genetic barriers.

(a) Any three of: 1. Large population size. 2. Random mating. 3. No mutations. 4. No selection (all genotypes have equal survival/reproductive success). 5. No migration (no gene flow). (b) (i) Let q^2 be the frequency of the homozygous recessive genotype (cystic fibrosis). q^2 = 1 / 2500 = 0.0004. Therefore, the frequency of the recessive allele (q) = \sqrt{0.0004} = 0.02. (ii) The frequency of the dominant allele (p) = 1 - q = 1 - 0.02 = 0.98. The frequency of heterozygous carriers is represented by 2pq. 2pq = 2 \times 0.98 \times 0.02 = 0.0392. Expressed as a percentage, this is 3.92%. (c) Endangered animal populations in small reserves are very small, making them highly susceptible to genetic drift, which causes random, rapid changes in allele frequencies. Mating is often non-random due to inbreeding depression or managed breeding programs. There may be strong selective pressures due to restricted environments. Additionally, physical barriers prevent natural migration (gene flow) into or out of the reserve, violating the assumption of no migration.

(a) 1 mark for each of any three valid conditions: large population, random mating, no mutation, no selection, no migration. (b) 1 mark for calculating q^2 = 0.0004; 1 mark for calculating q = 0.02; 1 mark for calculating p = 0.98; 1 mark for calculating carrier frequency as 3.92% (or 0.0392). (c) 1 mark for stating small populations are highly prone to genetic drift / random changes in allele frequencies; 1 mark for noting mating is not random (inbreeding occurs); 1 mark for mentioning strong natural selection or active breeding selection; 1 mark for stating that barriers prevent gene flow / migration.

(a) Waterlogged soils have low oxygen concentrations, creating anaerobic conditions. Under these anaerobic conditions, denitrifying bacteria thrive and actively convert nitrates (NO_3^-) in the soil into gaseous nitrogen (N_2), which escapes into the atmosphere, reducing soil nitrate levels. (b) Saprobionts decompose dead organic matter (such as dead plants, animals, and waste products like urea/proteins) by secreting extracellular digestive enzymes. They absorb the products of this digestion. During this process, they release nitrogenous waste in the form of ammonium ions (NH_4^+) into the soil. This process is called ammonification. (c) The nitrogen-fixing bacteria (e.g., Rhizobium in root nodules) reduce atmospheric nitrogen gas into ammonia or amino acids, which they provide to the host plant. In return, the leguminous plant provides the bacteria with organic carbohydrates (e.g., sucrose/glucose produced via photosynthesis) as an energy source, along with a protected, low-oxygen microenvironment suitable for the nitrogenase enzyme to function.

(a) 1 mark for stating waterlogging creates anaerobic conditions / lacks oxygen; 1 mark for identifying that this favors denitrifying bacteria; 1 mark for stating that nitrates are converted into nitrogen gas. (b) 1 mark for stating saprobionts secrete extracellular enzymes; 1 mark for stating they decompose proteins/DNA/urea from dead organic matter/waste; 1 mark for stating they absorb the products of digestion; 1 mark for stating they release ammonium ions (NH_4^+) into the soil (ammonification). (c) 1 mark for stating bacteria convert nitrogen gas into ammonia / nitrogen-containing compounds; 1 mark for stating the plant receives these compounds for growth/protein synthesis; 1 mark for stating the plant provides carbohydrates / photosynthetic products to the bacteria; 1 mark for stating the plant provides a protected / low-oxygen habitat.

Part (a):
- To trigger an action potential, the depolarisation of the postsynaptic membrane must reach a specific threshold potential (usually around \(-55\text{ mV}\)).
- A single stimulus of neurone X only produces a depolarisation of \(+4\text{ mV}\), which is sub-threshold, so no action potential is generated.

Part (b):
- A single stimulus of X releases a small quantity of neurotransmitter that is insufficient to reach the threshold.
- When two stimuli are applied in rapid succession, more neurotransmitter is released into the synaptic cleft before the previous amount has been broken down or reabsorbed.
- This leads to the summation (adding together) of the postsynaptic potentials, producing a larger depolarisation of \(+12\text{ mV}\), which is more likely to trigger an action potential.

Part (c):
- Calcium ions normally enter the synaptic knob to trigger vesicle fusion and are then rapidly pumped back out by active transport to terminate the signal.
- If active transport of calcium ions is inhibited, calcium ion concentration inside the synaptic knob remains high.
- This causes continuous fusion of synaptic vesicles and prolonged, uncontrolled release of neurotransmitter, leading to continuous depolarisation of the postsynaptic membrane.

Part (d):
- The arrival of the action potential depolarises the presynaptic membrane, causing voltage-gated calcium ion channels to open, allowing calcium ions to diffuse into the synaptic knob.
- This influx of calcium ions causes synaptic vesicles containing acetylcholine to fuse with the presynaptic membrane, releasing the neurotransmitter into the synaptic cleft by exocytosis.
- Acetylcholine diffuses across the cleft and binds to specific receptor proteins on sodium ion channels in the postsynaptic membrane, causing them to open and allowing sodium ions to enter, depolarising the postsynaptic membrane.

Part (a) [2 marks]:
- 1 mark: Threshold potential must be reached to trigger an action potential.
- 1 mark: Depolarisation from a single stimulus of X (\(+4\text{ mV}\)) is insufficient/sub-threshold.

Part (b) [3 marks]:
- 1 mark: Idea that neurotransmitter is released in rapid succession before previous neurotransmitter is removed.
- 1 mark: Postsynaptic potentials add together / summate.
- 1 mark: Leading to a larger depolarisation (\(+12\text{ mV}\)) which exceeds threshold.

Part (c) [2.7 marks]:
- 1 mark: Calcium ion concentration remains high inside the presynaptic knob / cannot be pumped out.
- 1 mark: Continuous / prolonged fusion of vesicles and release of neurotransmitter.
- 0.7 mark: Results in continuous depolarisation / excitation of the postsynaptic membrane.

Part (d) [3 marks]:
- 1 mark: Calcium channels open and calcium ions enter the synaptic knob.
- 1 mark: Vesicles containing acetylcholine fuse with the membrane and release acetylcholine by exocytosis.
- 1 mark: Acetylcholine diffuses across the cleft, binds to receptors on sodium channels, opening them and allowing sodium ions to enter, causing depolarisation.

Part (a):
- Glucagon binds to specific, complementary receptors on the cell surface membrane of liver cells (hepatocytes).
- This binding activates the transmembrane enzyme adenylate cyclase.
- Adenylate cyclase catalyses the conversion of ATP into cyclic AMP (cAMP) inside the cytoplasm.
- cAMP acts as a second messenger, binding to and activating protein kinase A, which initiates an enzyme cascade that converts inactive glycogen phosphorylase into its active form to catalyse glycogenolysis.

Part (b):
- As blood glucose concentration increases, plasma insulin concentration also increases. This is because high blood glucose is detected by beta (\(\beta\)) cells in the Islets of Langerhans, which secrete insulin.
- Insulin acts on target cells (liver and muscle) to increase glucose uptake (via GLUT4 translocation) and promote glycogenesis, which lowers blood glucose.
- As blood glucose levels fall back towards the normal set point, insulin secretion decreases via negative feedback.

Part (c):
- A mutation in the insulin receptor gene alters the sequence of amino acids (primary structure) of the receptor protein.
- This changes the tertiary structure (and shape) of the receptor, meaning its binding site is no longer complementary to insulin.
- Insulin cannot bind to the receptor, so the intracellular signalling cascade is not triggered, GLUT4 glucose transporter proteins are not translocated to the cell membrane, and glucose uptake from the blood remains low (causing hyperglycemia).

Part (a) [4 marks]:
- 1 mark: Glucagon binds to specific receptors on target cell/hepatocyte membrane.
- 1 mark: Activates adenylate cyclase.
- 1 mark: Converts ATP to cAMP (second messenger).
- 1 mark: cAMP activates protein kinase (A), initiating a cascade that activates glycogen phosphorylase / glycogen breakdown.

Part (b) [3.7 marks]:
- 1 mark: Positive correlation described: as blood glucose concentration increases, insulin concentration increases.
- 1 mark: Beta (\(\beta\)) cells in the pancreas detect high glucose and secrete insulin.
- 1 mark: Insulin causes uptake of glucose, which lowers blood glucose levels.
- 0.7 mark: Reference to negative feedback loop returning blood glucose to normal and reducing insulin secretion.

Part (c) [3 marks]:
- 1 mark: Mutation alters primary structure / amino acid sequence, changing the tertiary structure of the receptor.
- 1 mark: Receptor shape is altered, so insulin cannot bind.
- 1 mark: Intracellular cascade is blocked, preventing translocation of GLUT4 transporters, so glucose uptake does not occur / insulin resistance occurs.

Part (a):
- Group D (agar block) will show phototropic curvature towards the unilateral light because IAA is a water-soluble chemical messenger produced in the tip that can diffuse through the agar block down into the growing region.
- Group E (mica sheet) will not show curvature (remains vertical) because the mica sheet is impermeable to water-soluble substances, preventing the diffusion of IAA from the tip down to the growing region below.

Part (b):
- Unilateral light causes the lateral transport of IAA (indoleacetic acid) from the illuminated side of the shoot tip to the shaded side.
- This creates a higher concentration of IAA on the shaded side of the shoot.
- IAA stimulates cell elongation in shoots (by activating proton pumps to pump \(H^{+}\) ions into the cell wall, loosening the structure).
- The cells on the shaded side elongate faster than those on the illuminated side, causing the shoot to bend towards the light source.

Part (c):
- In roots, a high concentration of IAA inhibits cell elongation.
- During gravitropism, gravity causes IAA to accumulate on the lower side of the root, which inhibits the elongation of cells on the lower side while cells on the upper side continue to elongate normally.
- This causes the root to bend downwards (positive gravitropism), which is beneficial because it anchors the plant in the soil and allows the root system to reach water and mineral ions deeper in the ground.

Part (a) [3.7 marks]:
- 1 mark: Group D shows curvature / bends towards light because IAA can diffuse through agar.
- 1 mark: Group E does not show curvature / remains straight because mica is impermeable.
- 1.7 marks: Relates this directly to the water-soluble nature of IAA / diffusion of hormone down into the elongation zone.

Part (b) [4 marks]:
- 1 mark: Unilateral light causes lateral transport of IAA to the shaded side.
- 1 mark: Leads to a higher concentration of IAA on the shaded side.
- 1 mark: IAA stimulates cell elongation (in shoots) / increases cell wall extensibility.
- 1 mark: Uneven elongation (shaded side elongating more than light side) causes bending towards the light.

Part (c) [3 marks]:
- 1 mark: High concentration of IAA inhibits cell elongation in roots.
- 1 mark: Gravity causes IAA to accumulate on the lower side of the root, causing the root to bend downwards.
- 1 mark: Advantage: Anchors plant / ensures root grows into soil to absorb water / minerals.

Part (a):
- The H-zone (which contains only myosin) narrows or completely disappears because actin filaments are pulled further towards the centre of the sarcomere, increasing the overlap between actin and myosin.
- The I-band (which contains only actin) narrows/shortens because the actin filaments slide over the myosin filaments, pulling the Z-lines closer together.
- The A-band (the width of the myosin filaments) remains constant in width because the actual length of the myosin filaments does not change during muscle contraction.

Part (b):
- Calcium ions (\(Ca^{2+}\)) are released from the sarcoplasmic reticulum and bind to troponin, causing a conformational change that pulls tropomyosin away from the myosin-binding sites on the actin filament.
- This allows myosin heads to bind to actin, forming actinomyosin cross-bridges.
- ATP is hydrolysed by ATPase (which is activated by calcium ions) to provide the energy required for the myosin head to bend/rotate (the power stroke), pulling the actin filament along the myosin.
- A new molecule of ATP binds to the myosin head, causing it to detach from the actin filament, and is hydrolysed to reset the myosin head to its high-energy position.

Part (c):
- Phosphocreatine acts as an immediate reserve of inorganic phosphate groups (\(P_i\)).
- It can rapidly donate a phosphate group to ADP to regenerate ATP: \(\text{Phosphocreatine} + \text{ADP} \rightarrow \text{Creatine} + \text{ATP}\).
- This reaction is anaerobic and does not require oxygen, allowing the muscle to maintain maximum contraction (such as during sprinting) for several seconds before aerobic or anaerobic pathways take over.

Part (a) [3.7 marks]:
- 1 mark: H-zone narrows / disappears because actin slides further into the center.
- 1 mark: I-band narrows because the Z-lines are pulled closer together / actin and myosin overlap more.
- 1 mark: A-band remains constant because myosin filaments do not shorten / change length.
- 0.7 mark: Clear explanation that filaments slide over each other rather than contracting themselves.

Part (b) [4 marks]:
- 1 mark: Calcium ions bind to troponin, causing tropomyosin to move / expose binding sites on actin.
- 1 mark: Allows myosin heads to bind to actin / form actinomyosin cross-bridges.
- 1 mark: ATP hydrolysis provides energy for the power stroke / myosin head bending.
- 1 mark: ATP binding to myosin head is required for detachment of the head from actin.

Part (c) [3 marks]:
- 1 mark: Phosphocreatine acts as a source of phosphate / phosphoryl groups.
- 1 mark: Rapidly phosphorylates ADP to ATP (anaerobically).
- 1 mark: Allows high-intensity muscle contraction to continue before respiration rate increases.

Part (a):
- ADH is produced in the hypothalamus (specifically by neurosecretory cells).
- It is transported down axons to the posterior pituitary gland, from where it is secreted into the blood.

Part (b):
- A decrease in blood water potential is detected by osmoreceptors in the hypothalamus. Water leaves these cells by osmosis, causing them to shrink.
- This triggers nerve impulses to the posterior pituitary gland, stimulating the release of ADH into the bloodstream.
- ADH travels in the blood to the kidneys, where it binds to specific receptors on the cell surface membrane of the cells lining the distal convoluted tubule (DCT) and the collecting duct.
- This binding activates an enzyme cascade (via cAMP) that causes vesicles containing aquaporins (water channel proteins) to move to and fuse with the luminal (apical) membrane of the cells.
- This significantly increases the permeability of the collecting duct and DCT walls to water.
- As the filtrate flows down the collecting duct through the renal medulla (which has a very low water potential due to the countercurrent multiplier of the loop of Henle), water moves out of the filtrate by osmosis into the interstitial space and is reabsorbed into blood capillaries.
- This results in a small volume of highly concentrated (hypertonic) urine being excreted.

Part (c):
- Symptoms: Excretion of a very large volume of highly dilute urine (polyuria) and extreme/constant thirst (polydipsia).
- Explanation: In the absence of ADH, aquaporins are not inserted into the luminal membranes of the collecting duct cells, so the collecting duct walls remain impermeable to water. Water cannot be reabsorbed from the filtrate by osmosis, leading to massive fluid loss, dehydration, and a strong stimulus to drink water to restore blood water potential.

Part (a) [2 marks]:
- 1 mark: Produced in the hypothalamus.
- 1 mark: Secreted from the posterior pituitary gland.

Part (b) [5.7 marks]:
- 1 mark: Osmoreceptors shrink due to water loss by osmosis, stimulating ADH release from posterior pituitary.
- 1 mark: ADH binds to specific receptors on collecting duct / DCT cell membranes.
- 1 mark: Causes vesicles containing aquaporins to fuse with the luminal membrane.
- 1 mark: Increases permeability of the collecting duct wall to water.
- 1 mark: Water moves out of filtrate by osmosis down a water potential gradient into the medulla.
- 0.7 mark: Results in low volume, highly concentrated urine.

Part (c) [3 marks]:
- 1 mark: Large volume of dilute urine produced (polyuria) / extreme thirst (polydipsia).
- 1 mark: Lack of ADH means aquaporins are not inserted into collecting duct membranes.
- 1 mark: Collecting duct remains impermeable, so water is not reabsorbed, leading to severe dehydration.

Part (a):
- An increase in blood carbon dioxide concentration causes a decrease in blood pH (due to the formation of carbonic acid, which dissociates to release hydrogen ions).
- This drop in pH is detected by chemoreceptors located in the walls of the carotid arteries (carotid bodies) and the aorta (aortic bodies).
- These chemoreceptors increase the frequency of nerve impulses sent along sensory neurones to the cardiovascular centre (specifically the acceleratory centre) in the medulla oblongata.
- The medulla oblongata responds by sending a higher frequency of nerve impulses along sympathetic nerve fibres to the sinoatrial node (SAN).
- This sympathetic stimulation causes the SAN to increase the frequency of electrical impulses (waves of excitation) it generates, resulting in an increased heart rate.

Part (b):
- High blood pressure (hypertension) is detected by baroreceptors (pressure receptors) located in the walls of the carotid sinuses and the aorta.
- The baroreceptors send a higher frequency of nerve impulses along sensory neurones to the inhibitory centre in the medulla oblongata.
- The medulla oblongata sends more nerve impulses along parasympathetic (vagus) nerve fibres (and fewer along sympathetic fibres) to the SAN.
- Parasympathetic stimulation reduces the frequency of electrical impulses generated by the SAN, which slows down the heart rate, reducing cardiac output and lowering blood pressure back to normal.

Part (c):
- Dual innervation (antagonistic control) allows for rapid and precise homeostatic adjustments to heart rate.
- The sympathetic nervous system can rapidly increase heart rate and cardiac output to meet oxygen demands during exercise or a 'fight-or-flight' response, while the parasympathetic nervous system can quickly lower it back to resting levels, conserving metabolic energy and preventing potential damage to blood vessels from prolonged high blood pressure.

Part (a) [5 marks]:
- 1 mark: Increased \(CO_2\) lowers blood pH.
- 1 mark: Detected by chemoreceptors in carotid / aortic bodies.
- 1 mark: Impulses sent to the (acceleratory centre of) medulla oblongata.
- 1 mark: Medulla oblongata sends a higher frequency of impulses via sympathetic nerve fibres.
- 1 mark: To the SAN, which increases the frequency of electrical waves / heart rate.

Part (b) [3.7 marks]:
- 1 mark: High blood pressure detected by baroreceptors in the aorta / carotid sinuses.
- 1 mark: Higher frequency of impulses sent to the (inhibitory centre of) medulla oblongata.
- 1 mark: Medulla sends more impulses via parasympathetic / vagus nerve fibres to the SAN.
- 0.7 mark: SAN decreases electrical wave frequency, lowering heart rate / blood pressure.

Part (c) [2 marks]:
- 1 mark: Provides rapid / precise homeostatic control via antagonistic systems.
- 1 mark: Sympathetic meets immediate metabolic demands (exercise/stress), parasympathetic conserves energy at rest / protects blood vessels.

Part (a):
- Oestrogen is a lipid-soluble steroid hormone that diffuses directly through the phospholipid bilayer of the cell surface membrane into the cytoplasm of target cells.
- In the cytoplasm, oestrogen binds to a specific, complementary receptor on an inactive transcription factor.
- This binding induces a conformational (shape) change in the transcription factor, activating its DNA-binding site.
- The activated transcription factor-oestrogen complex enters the nucleus via a nuclear pore and binds to a specific promoter region of DNA upstream of the target gene, facilitating the binding of RNA polymerase and initiating transcription.

Part (b):
- Double-stranded RNA is cut into short, double-stranded fragments of siRNA by the enzyme Dicer.
- One of the strands of the double-stranded siRNA is incorporated into an enzyme complex called RISC (RNA-induced silencing complex), while the other strand is degraded.
- The single-stranded siRNA acts as a guide, leading the RISC complex to a target mRNA molecule by complementary base pairing.
- Once bound, the enzyme within the RISC complex cleaves (cuts) the mRNA molecule into smaller fragments.
- The cleaved mRNA is no longer complete and is degraded by cellular enzymes, preventing it from being translated by ribosomes, effectively silencing the gene.

Part (c):
- Oncogenes are mutated proto-oncogenes that are overexpressed, leading to uncontrolled cell division (tumours).
- Therapeutic siRNA can be designed with a base sequence that is perfectly complementary to the mRNA transcribed from the specific oncogene.
- Introducing this siRNA into cancer cells would target and destroy the oncogene mRNA, preventing the translation of the stimulatory proteins and thus halting uncontrolled cell division.

Part (a) [4 marks]:
- 1 mark: Oestrogen diffuses through the phospholipid bilayer of the cell membrane.
- 1 mark: Binds to a specific receptor on an inactive transcription factor.
- 1 mark: Alters the tertiary structure of the transcription factor / activates the DNA-binding site.
- 1 mark: Complex enters the nucleus, binds to promoter region of DNA, stimulating RNA polymerase / transcription.

Part (b) [4.7 marks]:
- 1 mark: Double-stranded RNA cleaved into siRNA (by Dicer).
- 1 mark: One strand of siRNA associates with the RISC enzyme complex.
- 1 mark: Single-stranded siRNA guides RISC to target mRNA by complementary base pairing.
- 1 mark: RISC cleaves / cuts the target mRNA.
- 0.7 mark: Cleaved mRNA is degraded, preventing translation.

Part (c) [2 marks]:
- 1 mark: siRNA designed to be complementary to the specific oncogene mRNA.
- 1 mark: Destroys oncogene mRNA, preventing translation of the stimulatory protein / halting uncontrolled mitosis.

(a) Proton pumps in the companion cell membrane actively pump hydrogen ions out of the companion cells into the apoplast/cell wall using energy from ATP. This creates an electrochemical gradient of hydrogen ions. Hydrogen ions then diffuse back into the companion cells down their concentration gradient via a co-transporter protein, bringing sucrose molecules along with them against their concentration gradient. Sucrose then diffuses into the sieve tube elements through plasmodesmata.

(b) The accumulation of sucrose in the sieve tube element lowers its water potential. Water moves from the xylem into the sieve tube element down a water potential gradient by osmosis, which increases the hydrostatic pressure inside the sieve tube at the source. At the sink, sucrose is unloaded, increasing the water potential inside the phloem, causing water to leave by osmosis and lowering the hydrostatic pressure. This creates a hydrostatic pressure gradient from source to sink, driving mass flow.

(c) Both treatments will significantly decrease or stop the translocation of sucrose. Cooling the stem to 3°C reduces the kinetic energy of enzymes involved in respiration, decreasing ATP production in companion cells. DNP directly inhibits ATP synthesis in mitochondria. Because translocation depends on active loading of sucrose (which requires ATP), a lack of ATP stops the active transport of hydrogen ions, stopping sucrose loading and halting mass flow.

(a) [Max 4 marks]
1. Active transport of \(\text{H}^+\) ions out of companion cell into cell wall/apoplast using ATP (1)
2. Establishes a \(\text{H}^+\) concentration/electrochemical gradient (1)
3. \(\text{H}^+\) ions diffuse back into companion cell down gradient via co-transporter protein (1)
4. Sucrose is co-transported against its concentration gradient (1)
5. Sucrose diffuses into sieve tube elements via plasmodesmata (1)

(b) [Max 4.5 marks]
1. Solutes/sucrose lower water potential in sieve tube (1)
2. Water enters by osmosis from xylem (1)
3. This increases hydrostatic pressure at source (1)
4. Sucrose unloaded at sink, water leaves phloem, lowering pressure at sink (1)
5. Hydrostatic pressure gradient created, causing mass flow from source to sink (0.5)

(c) [Max 4 marks]
1. Both treatments (cooling and DNP) stop/reduce translocation (1)
2. Cooling reduces enzyme activity/respiration rate, reducing ATP production (1)
3. DNP prevents ATP synthesis during oxidative phosphorylation (1)
4. No ATP means active loading of sucrose cannot occur, removing the pressure gradient (1)

(a) Light absorption excites electrons in chlorophyll / photosystem II (photoionization), causing them to leave the chlorophyll molecule. These high-energy electrons pass down an electron transport chain in the thylakoid membrane, releasing energy at each carrier. This energy is used to pump protons (\(\text{H}^+\)) from the stroma into the thylakoid lumen, creating a proton gradient. Protons diffuse back into the stroma through ATP synthase (chemiosmosis), which drives the synthesis of ATP from ADP and \(\text{P}_i\). Photolysis of water splits water molecules into protons, electrons, and oxygen, replacing the electrons lost by photosystem II. Finally, electrons from photosystem I and protons from the stroma are accepted by \(\text{NADP}^+\) with the help of NADP reductase to produce reduced NADP.

(b) Low \(\text{CO}_2\) concentration slows down the Calvin cycle because less RuBP is carboxylated to form GP, leading to less TP being formed and less regeneration of RuBP. This means that less ATP and reduced NADP are consumed in the reduction of GP. As a result, there is a shortage of ADP, \(\text{P}_i\), and \(\text{NADP}^+\) returning to the thylakoid membrane. Without these substrates to act as electron and phosphate acceptors, the light-dependent reaction cannot continue at its maximum rate.

(c) DCMU blocks the flow of electrons from PSII, which halts the electron transport chain and prevents the generation of the proton gradient needed for ATP synthesis. It also prevents the accumulation of electrons at PSI, meaning \(\text{NADP}^+\) cannot be reduced to NADPH. Because both ATP and NADPH are required to reduce GP to TP and to regenerate RuBP in the light-independent reaction, their absence halts the Calvin cycle, preventing the plant from fixing and taking up \(\text{CO}_2\).

(a) [Max 6 marks]
1. Light excites electrons in chlorophyll / PSII (photoionization) (1)
2. Electrons pass down electron transport chain (1)
3. Energy released is used to pump \(\text{H}^+\) ions into the thylakoid lumen (1)
4. \(\text{H}^+\) ions diffuse back into stroma through ATP synthase, producing ATP (1)
5. Photolysis of water produces protons, electrons, and oxygen, replacing lost electrons (1)
6. \(\text{NADP}^+\) is reduced to NADPH using electrons from PSI and protons (1)

(b) [Max 3.5 marks]
1. Less \(\text{CO}_2\) means less RuBP carboxylation / less GP and TP formed (1)
2. Therefore, less ATP and reduced NADP are used in the Calvin cycle (1)
3. Shortage of ADP, \(\text{P}_i\), and \(\text{NADP}^+\) returning to the light-dependent stage (1)
4. Light-dependent reaction slows as these acceptors are limiting (0.5)

(c) [Max 3 marks]
1. DCMU stops electron flow from PSII, preventing ATP synthesis (1)
2. Stops reduction of \(\text{NADP}^+\) to NADPH (1)
3. No ATP/NADPH means GP cannot be converted to TP / RuBP cannot be regenerated, halting \(\text{CO}_2\) fixation (1)

(a) Adrenaline acts as a first messenger. It binds to specific, complementary receptors on the cell-surface membrane of target cells (such as hepatocytes). This binding causes a conformational change that activates the transmembrane enzyme adenylate cyclase. Active adenylate cyclase converts ATP into cyclic AMP (cAMP), which acts as the second messenger. cAMP binds to and activates protein kinase A. This enzyme initiates a cascade of enzyme activations that ultimately activates glycogen phosphorylase, catalyzing glycogenolysis (the breakdown of glycogen to glucose), which is then released into the blood.

(b) Insulin binds to specific glycoprotein receptors on cell surface membranes (such as muscle and liver cells). This binding stimulates vesicles containing glucose transporter proteins (GLUT4) to fuse with the cell surface membrane, increasing the number of carrier proteins and making the membrane more permeable to glucose. It also activates intracellular enzymes that catalyze glycogenesis (converting glucose to glycogen) and increases the rate of cellular respiration. This keeps the intracellular glucose concentration low, maintaining a steep concentration gradient for glucose uptake.

(c) Uncontrolled Type 1 diabetes results in extremely high blood glucose levels (hyperglycemia). Consequently, the concentration of glucose in the glomerular filtrate in the kidneys is very high. This high concentration exceeds the capacity of the co-transporter proteins in the proximal convoluted tubule (PCT) responsible for selective reabsorption. Therefore, not all glucose can be reabsorbed back into the blood, and the excess glucose remains in the tubule fluid and is excreted in the urine.

(a) [Max 5 marks]
1. Adrenaline binds to specific receptor on membrane (1)
2. Activates adenylate cyclase (1)
3. Adenylate cyclase converts ATP to cAMP (second messenger) (1)
4. cAMP activates protein kinase A (1)
5. Leads to activation of glycogen phosphorylase / glycogenolysis (1)
6. Glucose released into blood (1)

(b) [Max 4.5 marks]
1. Insulin binds to specific receptors on target cells (1)
2. Causes vesicles containing glucose transporter proteins (GLUT4) to fuse with membrane (1)
3. Increases permeability of membrane to glucose (1)
4. Activates enzymes for glycogenesis / increases respiration of glucose (1)
5. Maintains steep concentration gradient for glucose diffusion into cell (0.5)

(c) [Max 3 marks]
1. High blood glucose concentration leads to high glucose concentration in glomerular filtrate (1)
2. Exceeds renal threshold / capacity of co-transporter proteins in PCT (1)
3. Glucose is not fully reabsorbed and remains in filtrate to be excreted (1)

(a) Glycolysis begins with the phosphorylation of glucose using phosphate groups from two ATP molecules to form fructose 1,6-bisphosphate. This makes the molecule more reactive and prevents it from leaving the cell. Fructose 1,6-bisphosphate is then split into two molecules of triose phosphate (TP). Each triose phosphate molecule is oxidized to pyruvate. This oxidation involves the removal of hydrogen, which is transferred to NAD+ to form reduced NAD (NADH). Additionally, four molecules of ATP are produced by substrate-level phosphorylation, giving a net yield of two ATP molecules.

(b) Oxygen acts as the terminal electron acceptor at the end of the electron transport chain (ETC) in oxidative phosphorylation. It combines with protons (H+) and electrons to form water: \(\frac{1}{2}\text{O}_2 + 2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2\text{O}\). If oxygen is absent, there is no terminal electron acceptor. This causes the ETC to stop because electrons can no longer be passed from carrier to carrier. As a result, NADH and FADH2 cannot transfer their electrons and remain reduced. Without the ETC operating, no proton gradient is established across the inner mitochondrial membrane, meaning ATP synthesis via chemiosmosis stops completely. The Link reaction and Krebs cycle also stop as NAD and FAD are not regenerated.

(c) Accumulation of lactic acid leads to its dissociation into lactate and hydrogen ions (H+), which lowers the pH of the muscle cell sarcoplasm (making it more acidic). This low pH alters the tertiary structure and ionic bonds of key cellular proteins. This affects enzymes involved in ATP production and cross-bridge cycling, and alters the shape of calcium-binding proteins (like troponin) or calcium channels on the sarcoplasmic reticulum. As a result, fewer calcium ions bind to troponin, fewer myosin-actin cross-bridges form, and ATP hydrolysis by myosin ATPase slows, reducing the force and speed of muscle contraction.

(a) [Max 4 marks]
1. Glucose phosphorylated using ATP to form fructose 1,6-bisphosphate (1)
2. Split into two triose phosphate (TP) molecules (1)
3. TP oxidized to pyruvate (1)
4. NAD+ reduced to NADH and 4 ATP produced by substrate-level phosphorylation (net gain of 2 ATP) (1)

(b) [Max 5.5 marks]
1. Oxygen is terminal electron acceptor (1)
2. Combines with H+ and electrons to form water / \(\frac{1}{2}\text{O}_2 + 2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2\text{O}\) (1)
3. In absence of oxygen, electron transport chain stops (1)
4. NADH and FADH2 remain reduced / cannot be oxidized (1)
5. No proton gradient established, stopping ATP synthesis via chemiosmosis / oxidative phosphorylation (1)
6. Krebs cycle / Link reaction stop due to lack of oxidized NAD/FAD (0.5)

(c) [Max 3 marks]
1. Lactic acid releases H+ ions, lowering sarcoplasmic pH (1)
2. Low pH denatures / alters tertiary structure of proteins (enzymes / troponin / calcium channels) (1)
3. Reduces calcium binding to troponin / reduces myosin ATPase activity, decreasing cross-bridge formation and contraction strength (1)

(a) During physical activity, the rate of cellular respiration increases, producing more carbon dioxide (CO2). This CO2 dissolves in blood plasma, forming carbonic acid, which lowers blood pH. Chemoreceptors in the walls of the carotid arteries and the aorta detect this decrease in pH. They send more frequent nerve impulses along sensory neurones to the cardiovascular center in the medulla oblongata. The medulla oblongata responds by sending more frequent nerve impulses along the sympathetic nervous system to the sinoatrial node (SAN). This causes the SAN to increase the frequency of electrical impulses it generates, leading to a faster heart rate.

(b) A sudden increase in blood pressure is detected by baroreceptors (pressure receptors) located in the carotid sinuses and aortic arch. These baroreceptors send more frequent nerve impulses to the cardiovascular center in the medulla oblongata. The medulla oblongata sends more frequent impulses along the parasympathetic (vagus) nervous system to the SAN, while reducing sympathetic activity. The parasympathetic nerve endings release acetylcholine, which binds to receptors on the SAN, decreasing the rate of electrical impulses generated. This slows the heart rate, reducing cardiac output and lowering blood pressure back to normal.

(c) First, calculate the exercise stroke volume:
\(\text{Cardiac Output} = \text{Heart Rate} \times \text{Stroke Volume}\)
\(22,400\text{ mL/min} = 160\text{ bpm} \times \text{Stroke Volume}\)
\(\text{Exercise Stroke Volume} = \frac{22,400}{160} = 140\text{ mL}\).
Next, calculate the percentage increase from resting stroke volume (70 mL):
\(\text{Increase in Stroke Volume} = 140 - 70 = 70\text{ mL}\)
\(\text{Percentage Increase} = \frac{70}{70} \times 100\% = 100\%\).

(a) [Max 6 marks]
1. Increased respiration releases more CO2, lowering blood pH (1)
2. Detected by chemoreceptors in carotid arteries/aorta (1)
3. More frequent impulses sent to medulla oblongata / cardiovascular center (1)
4. Medulla sends more frequent impulses along sympathetic nerves (1)
5. Releases noradrenaline at the SAN (1)
6. SAN increases frequency of electrical impulses, increasing heart rate (1)

(b) [Max 4 marks]
1. High blood pressure detected by baroreceptors in carotid sinuses/aorta (1)
2. More frequent impulses sent to medulla oblongata (1)
3. Medulla sends more frequent impulses along parasympathetic (vagus) nerves (1)
4. Acetylcholine released at SAN, decreasing frequency of SAN impulses and lowering heart rate (1)

(c) [Max 2.5 marks]
1. Calculates exercise stroke volume: \(22,400 / 160 = 140\text{ mL}\) (1)
2. Calculates difference: \(140 - 70 = 70\text{ mL}\) (0.5)
3. Calculates percentage increase: \((70 / 70) \times 100 = 100\%\) (1)
(Correct final answer of 100% with no working scores 2.5 marks)

(a) Phagocytosis begins with chemotaxis, where the macrophage is attracted to the bacterium by chemical signals (e.g., bacterial toxins or cytokines). Receptors on the macrophage membrane bind to complementary antigens on the surface of the bacterium. The macrophage then extends pseudopodia around the bacterium to engulf it, enclosing it within a membrane-bound vesicle called a phagosome. Lysosomes containing hydrolytic enzymes (such as lysozymes) fuse with the phagosome to form a phagolysosome. The enzymes are released onto the bacterium, digesting and hydrolyzing its cell wall and other molecules.

(b) T-helper cells (Th cells) are activated when their specific T-cell receptor binds to a complementary antigen presented on an antigen-presenting cell (APC), such as a macrophage or B cell. Once activated, the Th cell undergoes clonal expansion by mitosis. The activated Th cells release cytokines (chemical signaling molecules). In the cellular response, these cytokines stimulate cytotoxic T cells to divide and actively kill infected host cells. In the humoral response, cytokines stimulate antigen-sensitized B cells to undergo clonal expansion, differentiating into plasma cells (which produce and secrete large amounts of monoclonal antibodies) and memory B cells.

(c) A secondary immune response occurs when the immune system encounters the same antigen for a second time. It is much faster, produces a much higher concentration of antibodies, and lasts for a longer duration compared to the primary response. This is because memory B and T cells produced during the primary response persist in the blood. Upon re-exposure, memory B cells rapidly divide and differentiate into plasma cells, bypassing the time-consuming steps of antigen presentation and clonal selection. Vaccination introduces a dead, weakened, or harmless antigen into the body. This safely triggers the primary immune response, leading to the formation of specific memory B and T cells without causing disease, providing long-term immunity.

(a) [Max 4 marks]
1. Macrophage attracted to bacterium by chemical signals / chemotaxis (1)
2. Receptors on macrophage bind to antigens on bacterium (1)
3. Membrane engulfs bacterium to form a phagosome (1)
4. Lysosomes fuse with phagosome (forming phagolysosome) (1)
5. Lysozymes / hydrolytic enzymes digest / hydrolyze the bacterium (1)

(b) [Max 4.5 marks]
1. Th cells activated by binding to complementary antigen on APC (1)
2. Activated Th cells undergo clonal expansion / mitosis (1)
3. Th cells secrete cytokines (1)
4. Cytokines stimulate cytotoxic T cells to divide and destroy infected cells (cellular response) (1)
5. Cytokines stimulate B cells to undergo clonal expansion and differentiate into plasma cells / memory cells (humoral response) (0.5)

(c) [Max 4 marks]
1. Secondary response is faster, has higher antibody concentration, and lasts longer than primary response (1)
2. Memory B/T cells persist from the primary response (1)
3. Upon re-exposure, memory cells rapidly divide and differentiate into plasma cells (1)
4. Vaccination introduces harmless antigens to stimulate primary response and produce memory cells safely (1)