MetadataPastPaper.sampleTitle

(a) Cold: reduces/prevents enzyme activity to stop organelle self-digestion. Isotonic: maintains same water potential to prevent osmotic water movement, preventing organelles from bursting or shrinking. Buffered: maintains constant pH to prevent protein/enzyme denaturation. (b) Homogenate is spun at a low speed first to pellet large debris and nuclei. The supernatant is poured off and spun at a higher speed. The resulting pellet contains the chloroplasts. (c) Convert image length to micrometres: \( 67.5\text{ mm} \times 1000 = 67,500\ \mu\text{m} \). Magnification = \( \text{Image size} / \text{Actual size} = 67,500\ \mu\text{m} / 4.5\ \mu\text{m} = 15,000 \). (d) The nucleolus is responsible for synthesizing ribosomal RNA (rRNA) and assembling ribosomes.

(a) 1 mark for cold: prevents enzyme activity; 1 mark for isotonic: prevents osmosis/bursting/shrinking of organelles; 1 mark for buffered: maintains pH to prevent enzyme denaturation. (b) 1 mark for centrifuging at a low speed first; 1 mark for separating supernatant from pellet; 1 mark for centrifuging supernatant at a higher speed to pellet chloroplasts. (c) 1 mark for correct unit conversion (67.5 mm to 67,500 micrometres); 1 mark for dividing image size by actual size; 0.7 marks for correct final answer of \times 15,000 (or 15000). (d) 1 mark for synthesizing rRNA; 1 mark for assembling/producing ribosomes.

(a) A saturated fatty acid has no carbon-carbon double bonds (\( C=C \)) in its hydrocarbon chain, whereas an unsaturated fatty acid contains at least one carbon-carbon double bond. (b) A peptide bond is formed via a condensation reaction between the amine group (\( -NH_2 \)) of one amino acid and the carboxyl group (\( -COOH \)) of another, releasing a molecule of water. (c) A triglyceride has three fatty acids attached to glycerol, making it highly hydrophobic and ideal for compact energy storage. A phospholipid has two fatty acids and a hydrophilic phosphate group attached to glycerol, allowing it to form a stable bilayer in cell membranes. (d) Add dilute hydrochloric acid to the sample and heat in a water bath to hydrolyse glycosidic bonds. Neutralise with sodium hydrogencarbonate. Add Benedict's reagent and heat again. A colour change from blue to green/yellow/orange/brick-red indicates a positive result.

(a) 1 mark for saturated having only single carbon-carbon bonds / no double bonds; 1 mark for unsaturated having at least one carbon-carbon double bond (\( C=C \)). (b) 1 mark for condensation reaction; 1 mark for reaction between amine group of one and carboxyl group of another; 0.7 marks for release of water. (c) 1 mark for triglyceride having 3 fatty acids vs phospholipid having 2 fatty acids and 1 phosphate; 1 mark for triglyceride being hydrophobic/non-polar for energy storage; 1 mark for phospholipid forming a bilayer due to hydrophilic head and hydrophobic tails. (d) 1 mark for heating with acid and neutralising with alkali; 1 mark for adding Benedict's reagent and heating; 1 mark for observing green/yellow/orange/brick-red precipitate.

(a) Sodium ions are actively transported out of the epithelial cells into the blood by the sodium-potassium pump. This maintains a lower concentration of sodium ions inside the epithelial cell than in the lumen of the ileum, establishing a concentration gradient. Sodium ions then diffuse down their concentration gradient into the epithelial cell via a co-transporter protein, carrying glucose molecules into the cell alongside them against their concentration gradient. (b) Change in mass = \( 3.22 - 3.68 = -0.46\text{ g} \). Percentage change = \( (-0.46 / 3.68) \times 100 = -12.5\% \). (c) Factor 1: Concentration gradient; a steeper gradient increases the rate of diffusion as more molecules move down the gradient per unit time. Factor 2: Surface area; a larger membrane surface area provides more space/pathways for diffusion, increasing the overall rate.

(a) 1 mark for sodium ions actively transported out of cell into blood; 1 mark for creating a concentration gradient of sodium ions (lower inside cell); 1 mark for sodium ions diffusing in via co-transporter protein; 1 mark for glucose being carried in against its concentration gradient. (b) 1 mark for calculating correct change in mass (\( -0.46\text{ g} \)); 1 mark for dividing by initial mass (\( 3.68\text{ g} \)) and multiplying by 100; 0.7 marks for final answer of \( -12.5\% \) (accept 12.5% loss/decrease). (c) 1 mark each for identifying two correct factors (e.g. concentration gradient, surface area, temperature, thickness of diffusion path); 1 mark each for correct explanation of how each factor affects the rate.

(a) In the induced-fit model, the active site is not initially fully complementary to the substrate. As the substrate binds, the active site changes shape slightly to fit tightly around the substrate, putting physical strain on its bonds and lowering the activation energy. This differs from the lock-and-key model, which proposes that the active site is rigid and perfectly complementary to the substrate before binding. (b) Malonate has a similar molecular structure to the substrate of succinate dehydrogenase. It binds to the active site, blocking substrate molecules from entering and reducing the number of enzyme-substrate complexes formed. This inhibition can be overcome by significantly increasing the concentration of the substrate, which increases the probability of substrate molecules colliding with active sites instead of the inhibitor. (c) A non-competitive inhibitor binds to an allosteric site on the enzyme. This changes the tertiary structure of the enzyme, altering the shape of the active site so it is no longer complementary to the substrate. Substrates can no longer bind, meaning increasing substrate concentration has no effect on the rate of reaction.

(a) 1 mark for active site being flexible/not fully complementary initially; 1 mark for active site changing shape upon binding; 1 mark for putting strain on substrate bonds to lower activation energy; 1 mark for stating lock-and-key model describes a rigid/pre-formed complementary active site. (b) 1 mark for inhibitor having similar structure to substrate; 1 mark for binding to active site / blocking substrate; 1 mark for reducing rate of enzyme-substrate complex formation; 0.7 marks for stating it is overcome by increasing substrate concentration. (c) 1 mark for binding to an allosteric site; 1 mark for changing the tertiary structure and shape of the active site; 1 mark for stating substrate can no longer bind / increasing substrate concentration has no effect.

(a) Species richness is the total number of different species present in a community or habitat. (b) First, find total population \( N \): \( N = 15 + 12 + 8 + 4 + 1 = 40 \). Calculate \( N(N - 1) \): \( 40 \times 39 = 1560 \). Calculate \( n(n - 1) \) for each species: Species A: \( 15 \times 14 = 210 \), Species B: \( 12 \times 11 = 132 \), Species C: \( 8 \times 7 = 56 \), Species D: \( 4 \times 3 = 12 \), Species E: \( 1 \times 0 = 0 \). Sum of \( n(n - 1) = 210 + 132 + 56 + 12 + 0 = 410 \). Calculate \( d \): \( d = 1560 / 410 = 3.804878... \approx 3.80 \). (c) A high index of diversity means there are many different species present with more complex food webs. If abiotic conditions change, some species are likely to have adaptations or alleles that allow them to survive. This ensures that even if some species decline, other species can fill their ecological niches and predators have alternative food sources, maintaining community stability.

(a) 2 marks for stating the number of different species in a community/habitat (1 mark if 'number of species' is mentioned without 'different' or 'in a community'). (b) 1.7 marks for calculating total \( N = 40 \) and \( N(N-1) = 1560 \); 2 marks for calculating individual \( n(n-1) \) values and correct sum of 410; 1 mark for correct final answer of 3.80 (allow 3.8). (c) 1 mark for complex food webs/many species; 1 mark for some species possessing alleles/adaptations to survive the abiotic change; 1 mark for alternative food sources for predators; 1 mark for ecosystem/community stability.

(a) Hemoglobin has a quaternary structure consisting of four polypeptide chains, each containing a prosthetic heme group with an iron ion (\( Fe^{2+} \)) that binds reversibly to an oxygen molecule. The binding of the first oxygen molecule alters the quaternary shape, making it easier for the remaining oxygen molecules to bind (cooperative binding). (b) An increased concentration of carbon dioxide leads to carbonic acid formation, lowering the pH. This low pH alters the shape of hemoglobin, decreasing its affinity for oxygen. Consequently, the oxygen dissociation curve shifts to the right, allowing oxygen to be unloaded more easily at respiring tissues. (c) The worm's oxygen dissociation curve would be shifted to the left compared to the human curve. This means its hemoglobin has a higher affinity for oxygen, allowing it to load/saturate with oxygen at the very low partial pressures of oxygen found in its mud habitat.

(a) 1 mark for four polypeptide chains each with a heme group / \( Fe^{2+} \); 1 mark for oxygen binding reversibly; 1 mark for cooperative binding / shape change on first oxygen binding. (b) 1 mark for CO2 lowering pH; 1 mark for low pH changing hemoglobin shape; 1 mark for reducing affinity for oxygen; 1 mark for shifting curve to the right to unload oxygen to tissues. (c) 1 mark for curve shifted to the left; 1 mark for higher affinity for oxygen; 1.7 marks for allowing oxygen loading at very low oxygen partial pressures (pO2) in the mud.

(a) During meiosis I, homologous chromosomes pair up to form bivalents. Non-sister chromatids wrap around each other, break at points called chiasmata, and swap equivalent portions of maternal and paternal chromatids. This results in recombinant chromatids with new combinations of maternal and paternal alleles. (b) Homologous pairs of chromosomes align randomly along the equator of the spindle in metaphase I. It is completely random which chromosome of each pair goes to which pole during anaphase I. This random separation produces many different combinations of maternal and paternal chromosomes in the resulting gametes. (c) (i) Since \( 2n = 16 \), the haploid number \( n = 8 \). The number of combinations is \( 2^n = 2^8 = 256 \). (ii) During non-disjunction, homologous chromosomes or sister chromatids fail to separate properly during anaphase. This results in some gametes containing an extra chromosome (\( n + 1 \)) and other gametes lacking a chromosome (\( n - 1 \)).

(a) 1 mark for homologous chromosomes pairing up to form bivalents; 1 mark for chromatids breaking and joining at chiasmata; 1 mark for exchanging alleles between non-sister chromatids; 1 mark for producing recombinant chromosomes / new combinations of alleles. (b) 1 mark for random alignment of homologous pairs along the equator of the cell; 1 mark for random separation of maternal/paternal chromosomes to opposite poles; 1 mark for resulting in new chromosome combinations in gametes. (c) (i) 1 mark for identifying the haploid number as \( n = 8 \); 0.7 marks for calculating \( 2^8 = 256 \). (ii) 1 mark for stating chromosomes/chromatids fail to separate; 1 mark for stating some gametes have one extra chromosome while others are missing a chromosome.

(a) 1. Assembling underwater prevents air from entering the xylem of the stem, which would break the continuous water column / cohesive pathway of water. 2. Sealing joints with petroleum jelly ensures the apparatus is completely airtight, preventing air leaks that would interfere with water uptake and bubble movement. (b) 1. Use an electric fan to generate different wind speeds (e.g., low and high fan settings or placing the fan at different distances). 2. Measure the distance the air bubble moves along the capillary tube in a fixed time period (e.g., 5 minutes). 3. Use the reservoir to return the air bubble to the starting point of the scale before changing conditions. 4. Control other environmental variables, such as light intensity and temperature, and repeat measurements to calculate a mean rate. (c) 1. The cell walls of the endodermis contain a waterproof Casparian strip made of suberin. 2. This blocks the apoplast (cell wall) pathway, forcing water and dissolved mineral ions to enter the cytoplasm through the selectively permeable cell surface membrane of the endodermal cells (symplast pathway), allowing selective control over ion absorption.

(a) Max 4 marks:
- 1 mark for stating underwater assembly prevents air entering the xylem.
- 1 mark for explaining this maintains the continuous water column / prevents break in cohesion.
- 1 mark for stating sealing prevents air leaks.
- 1 mark for explaining this ensures all water lost from the leaves causes bubble movement / maintains pressure gradient.

(b) Max 4 marks:
- 1 mark for describing how wind speed is varied (e.g., fan speed or distance).
- 1 mark for measuring the distance moved by the bubble in a set time (or time taken for a set distance).
- 1 mark for detail of resetting the bubble using the reservoir/tap.
- 1 mark for controlling a relevant variable (e.g., temperature, light intensity) or conducting repeats to calculate a mean.

(c) Max 2.7 marks:
- 1 mark for identifying the Casparian strip / suberin layer in endodermal cell walls.
- 1 mark for stating this blocks the apoplast pathway.
- 0.7 marks for explaining water/ions must enter the symplast pathway / pass through the cell membrane, allowing selective uptake.

(a) 1. Cut a small section (about 5 mm) of the root tip and place it in warm hydrochloric acid to break down the middle lamella, separating the cells. 2. Rinse the root tip in distilled water, place on a slide, and add a drop of acetic orcein or toluidine blue stain to bind to chromatin/chromosomes. 3. Place a coverslip over the specimen and press down firmly and vertically with a thumb. 4. Do not smear the coverslip sideways, as this would damage or roll the cells rather than spreading them into a single, thin layer. (b) 1. Formula: \( \text{Mitotic Index} = \frac{\text{Number of cells in mitosis}}{\text{Total number of cells}} \times 100 \). 2. Total mitotic cells = \( 18 + 7 + 4 + 3 = 32 \). 3. Total cells = \( 32 + 120 = 152 \). 4. Calculation: \( \frac{32}{152} \times 100 = 21.0526\% \). Rounding to 1 decimal place gives 21.1%. (c) 1. This region contains the apical meristem. 2. This is the only zone of the root where cells are actively dividing by mitosis. 3. Further up the root, cells are in the zone of elongation or differentiation and do not undergo mitosis.

(a) Max 4 marks:
- 1 mark for acid hydrolysis step (mentioning hydrochloric acid to separate cells/break middle lamella).
- 1 mark for naming an appropriate stain (e.g., acetic orcein, toluidine blue, Feulgen).
- 1 mark for pushing down vertically on the coverslip (squashing) to obtain a single cell layer thick.
- 1 mark for stating not to slide/smear the coverslip to prevent overlapping/damaged cells.

(b) Max 3.7 marks:
- 1 mark for showing correct sum of mitotic cells (32).
- 1 mark for showing correct sum of total cells (152).
- 1 mark for correct division and multiplication by 100.
- 0.7 marks for correct final answer to 1 decimal place (21.1%). (Allow 1 mark overall if formula is correct but minor arithmetic slip is made).

(c) Max 3 marks:
- 1 mark for identifying this as the meristematic region.
- 1 mark for stating that cell division / mitosis only occurs in this region.
- 1 mark for explaining that cells outside this zone are undergoing elongation or differentiation (not dividing).

(a) 1. Sodium-potassium pumps in the basolateral membrane of the epithelial cell actively transport sodium ions (\( \text{Na}^+ \)) out of the cell and into the blood. 2. This requires ATP and lowers the concentration of sodium ions inside the epithelial cell compared to the lumen. 3. Sodium ions move down their concentration gradient from the lumen into the epithelial cell via a co-transport (symporter) protein in the microvillus membrane. 4. As sodium ions pass through the symporter, they bring a glucose molecule with them against its concentration gradient. 4.7. Glucose then accumulates in the cell and moves down its concentration gradient into the blood via facilitated diffusion. (b) 1. Without ATP, active transport of sodium ions out of the epithelial cell stops. 2. The concentration gradient of sodium ions between the lumen and the cytoplasm of the epithelial cell is lost. 3. Therefore, co-transport of glucose into the cell stops, and glucose cannot be absorbed from the lumen. 4. Facilitated diffusion of glucose into the blood also ceases once existing cellular glucose is depleted. (c) 1. A high density of mitochondria to produce ATP for active transport of sodium ions. 2. Or the presence of many carrier/channel proteins in the membrane.

(a) Max 4.7 marks:
- 1 mark for active transport of sodium ions out of the cell into the blood via the Na+/K+ pump.
- 1 mark for establishing a low sodium ion concentration inside the cell (concentration gradient).
- 1 mark for sodium ions entering the cell down their gradient from the lumen.
- 1 mark for glucose entering alongside sodium ions via a co-transporter/symport protein (against its gradient).
- 0.7 marks for glucose leaving the epithelial cell into the blood by facilitated diffusion.

(b) Max 4 marks:
- 1 mark for stating DNP stops/reduces ATP synthesis.
- 1 mark for identifying that the sodium-potassium pump stops.
- 1 mark for explaining that the sodium ion gradient is lost/depleted.
- 1 mark for concluding that co-transport of glucose into the cell is prevented.

(c) Max 2 marks:
- 2 marks for stating 'high density of mitochondria' or 'large number of carrier proteins' (1 mark for name, 1 mark for explaining how it aids absorption).

(a) 1. Atrioventricular (AV) valves prevent the backflow of blood from the ventricles into the atria when the ventricles contract (systole). 2. They open when atrial pressure exceeds ventricular pressure, allowing ventricles to fill. 3. Semi-lunar (SL) valves prevent the backflow of blood from the aorta/pulmonary artery into the ventricles during ventricular relaxation (diastole). 4. They open when ventricular pressure exceeds arterial pressure during ventricular systole, allowing blood to leave the heart. (b) 1. At 0.15 s, since left ventricular pressure exceeds left atrial pressure, the left AV (bicuspid) valve closes. 2. This prevents blood flowing backwards into the atrium, beginning ventricular systole. 3. At 0.35 s, since left ventricular pressure exceeds aortic pressure, the aortic semi-lunar valve opens. 4. This allows blood to be actively ejected from the left ventricle into the aorta. (c) 1. Mouse: \( T_{\text{mouse}} = \frac{60 \text{ s}}{600} = 0.1 \text{ s} \). 2. Elephant: \( T_{\text{elephant}} = \frac{60 \text{ s}}{30} = 2.0 \text{ s} \). 3. 2.7. The small mouse has a high surface area to volume ratio, resulting in rapid heat loss. To maintain its body temperature, it must sustain a very high metabolic rate (respiration rate), demanding rapid delivery of oxygen and glucose via a fast heart rate.

(a) Max 4 marks:
- 1 mark for stating AV valves prevent backflow from ventricles to atria.
- 1 mark for stating AV valves open when atrial pressure > ventricular pressure (or close when ventricular pressure > atrial pressure).
- 1 mark for stating SL valves prevent backflow from arteries (aorta/pulmonary artery) to ventricles.
- 1 mark for stating SL valves open when ventricular pressure > arterial pressure (or close when arterial pressure > ventricular pressure).

(b) Max 4 marks:
- 1 mark for identifying closure of the AV valve at 0.15 s.
- 1 mark for explaining this prevents backflow of blood into the atrium.
- 1 mark for identifying opening of the SL valve at 0.35 s.
- 1 mark for explaining this allows blood to flow into the aorta.

(c) Max 2.7 marks:
- 1 mark for correct duration calculations (0.1 s for mouse and 2.0 s for elephant).
- 1 mark for linking high surface area to volume ratio in the mouse to rapid heat loss.
- 0.7 marks for explaining that high metabolic rate / respiration rate is needed to maintain body temperature, requiring fast delivery of oxygen/glucose.

(a) 1. Once inside, the viral enzyme reverse transcriptase synthesizes double-stranded viral DNA using the viral single-stranded RNA as a template. 2. This viral DNA is transported into the nucleus and integrated into the host cell's DNA using the enzyme integrase. 3. The host cell's RNA polymerase transcribes the integrated viral DNA into viral mRNA. 4. Host ribosomes translate viral mRNA into viral proteins (polyproteins), which are cleaved by protease, assembled with viral RNA, and bud out of the T-helper cell, wrapping themselves in the host cell membrane to form the envelope. (b) 1. Coat the bottom of a microtiter plate well with HIV antigen. 2. Add the patient's serum sample; if HIV antibodies are present, they bind to the HIV antigens. 3. Wash the wells to remove any unbound antibodies. 4. Add a secondary monoclonal antibody conjugated to an enzyme. This antibody specifically binds to the primary human antibody. 4.7. Wash the wells again to remove unbound secondary antibodies, then add the substrate for the enzyme. A color change indicates a positive result. (c) 1. HIV integrates into the host's genetic material and cannot be eliminated by the immune system, leading to a lifelong, persistent infection. 2. There is currently no effective vaccine for HIV, so antibodies cannot arise from vaccination, only from active infection.

(a) Max 4 marks:
- 1 mark for reverse transcriptase making DNA from RNA.
- 1 mark for viral DNA integrating into the host's DNA.
- 1 mark for transcription and translation using host cell machinery (ribosomes) to make viral proteins.
- 1 mark for assembly of new capsids and budding from the host cell membrane to acquire the envelope.

(b) Max 4.7 marks:
- 1 mark for coating wells with HIV antigen and adding patient serum.
- 1 mark for washing to remove unbound primary antibodies.
- 1 mark for adding enzyme-linked secondary antibody.
- 1 mark for washing again to remove unbound secondary antibody (crucial step to prevent false positives).
- 0.7 marks for adding substrate and detecting color change.

(c) Max 2 marks:
- 1 mark for stating HIV is a persistent infection that integrates into host DNA and cannot be cleared.
- 1 mark for noting that there is no vaccine for HIV, so antibodies cannot represent immunization.

(a) 1. The macrophage is attracted to the bacterium by chemical signals (chemotaxis). 2. Receptors on the macrophage membrane bind to foreign antigens on the bacterium. 3. The macrophage cell membrane extends around the bacterium (forming pseudopodia) and engulfs it into a vesicle called a phagosome. 4. Lysosomes fuse with the phagosome to form a phagolysosome, releasing hydrolytic enzymes (such as lysozyme) that digest and destroy the bacterium. (b) 1. Macrophages have a highly flexible cell membrane and many lysosomes, facilitating the engulfment and breakdown of diverse pathogens. 2. Plasma cells have an extensive rough endoplasmic reticulum (RER) and a well-developed Golgi apparatus. 3.7. This allows them to synthesize, package, and secrete massive quantities of protein antibodies (immunoglobulins). (c) 1. Following the digestion of the pathogen within the phagolysosome, the antigen molecules are not completely destroyed. 2. Specific antigen fragments (peptides) are bound to Major Histocompatibility Complex (MHC) class II proteins. 3. This MHC-antigen complex is transported via vesicles to the cell surface membrane and displayed, allowing T-helper cells to recognize the antigen.

(a) Max 4 marks:
- 1 mark for chemotaxis / attraction and binding of receptors to antigens.
- 1 mark for engulfment to form a phagosome (vesicle).
- 1 mark for fusion of lysosomes with the phagosome.
- 1 mark for hydrolytic/lytic enzymes (accept lysozymes) digesting/hydrolyzing the pathogen.

(b) Max 3.7 marks:
- 1 mark for explaining macrophage flexibility/lysosomes for engulfment and intracellular destruction.
- 1 mark for stating plasma cells have extensive RER and Golgi apparatus.
- 1.7 marks for linking RER/Golgi to the fast synthesis, modification, and secretion of antibodies (glycoproteins).

(c) Max 3 marks:
- 1 mark for stating that digested antigen fragments are salvaged/retained.
- 1 mark for binding of antigens to MHC proteins.
- 1 mark for movement of the vesicle to and fusion with the cell surface membrane to display the antigen.

(a) 1. Damage occurs to the endothelial lining of the coronary artery (due to high blood pressure, toxins from smoking, etc.). 2. Lipids (LDL cholesterol) and white blood cells (macrophages) accumulate in the artery wall beneath the endothelium. 3. Over time, these form a fatty deposit/fibrous plaque (atheroma) which narrows the lumen of the artery. 4. If the atheroma ruptures, it triggers the formation of a blood clot (thrombus). 4.7. The clot blocks the coronary artery, cutting off oxygen and glucose to the downstream myocardium, preventing aerobic respiration and causing cardiac muscle tissue to die (myocardial infarction). (b) 1. Correlation means there is an association or link between physical activity and CHD incidence (as physical activity increases, CHD decreases). 2. Causation means there is a direct biological mechanism by which high physical activity directly lowers CHD risk (e.g., by lowering blood pressure and strengthening cardiac muscle). 3. A correlative link could be indirect, where physically active people might just share another healthy habit (like not smoking). (c) 1. Age (older individuals have higher baseline CHD risk). 2. Diet or smoking status (smokers or those with high-fat diets have significantly increased risk of atheroma).

(a) Max 4.7 marks:
- 1 mark for damage to the endothelium of the coronary artery.
- 1 mark for accumulation of lipids/macrophages in the artery wall to form a plaque.
- 1 mark for plaque narrowing the lumen (restricting blood flow).
- 1 mark for rupture of the atheroma triggering a blood clot (thrombosis).
- 0.7 marks for blocking the artery, causing lack of oxygen/glucose, leading to localized cardiac muscle death.

(b) Max 3 marks:
- 1 mark for defining correlation (an association/link between two variables).
- 1 mark for defining causation (one variable directly causes the change in the other via a biological mechanism).
- 1 mark for explaining that physical activity might just correlate because active people smoke less, whereas the true cause is the biological effect of exercise on blood pressure/blood vessels.

(c) Max 3 marks:
- 1.5 marks for first correct confounding variable with brief reason (e.g., Age - risk increases naturally with age).
- 1.5 marks for second correct confounding variable with brief reason (e.g., Smoking - increases blood pressure and damages endothelium).

(a) Conditions required for the Hardy-Weinberg principle:
1. Large population size.
2. No mutation.
3. Random mating.
4. No immigration or emigration (isolated population).
5. No natural selection (all genotypes have equal reproductive success).

(b) Working:
Let \(q^2\) be the frequency of the homozygous recessive genotype (\(ww\)).
\(q^2 = 0.04\)
\(q = \sqrt{0.04} = 0.2\)
Since \(p + q = 1\):
\(p = 1 - 0.2 = 0.8\)
The frequency of heterozygotes is represented by \(2pq\):
\(2pq = 2 \times 0.8 \times 0.2 = 0.32\)
Percentage of heterozygotes = 32\%.

(c) The frequency of the recessive allele (\(w\)) will decrease. White-flowered plants (\(ww\)) are selected against, meaning they have a lower survival rate and lower reproductive success. They are less likely to pass on their alleles to the next generation. However, the allele will not be completely eliminated quickly because the recessive allele remains protected in heterozygous individuals.

(d) Heterozygotes (\(Ww\)) carry one copy of the recessive lethal allele but do not show the lethal phenotype because they have a functional dominant allele. They survive, reproduce, and pass the recessive allele on to 50\% of their offspring, keeping it in the gene pool.

(a) [3 marks]
- 1 mark for each correct condition mentioned (Max 3): Large population, No mutation, Random mating, No migration, No natural selection.

(b) [3 marks]
- 1 mark for correct calculation of \(q = 0.2\).
- 1 mark for correct calculation of \(p = 0.8\).
- 1 mark for correct calculation of \(2pq = 0.32\) or 32\%.

(c) [3 marks]
- 1 mark for stating that the frequency of the recessive allele decreases.
- 1 mark for explaining that white-flowered plants have lower reproductive success / are less likely to survive.
- 1 mark for stating fewer recessive alleles are passed on to the offspring.

(d) [2 marks]
- 1 mark for stating the recessive allele is carried / hidden in heterozygotes.
- 1 mark for explaining that heterozygotes have the normal dominant phenotype and are not selected against / survive to reproduce.

(a) Geographical isolation prevents gene flow between the two populations. Each island has different environmental conditions (such as food sources, predators, and climate), representing different biotic and abiotic selection pressures. Spontaneous mutations occur independently in both populations, introducing new alleles. Individuals with advantageous alleles are more likely to survive, reproduce, and pass these alleles on to their offspring (natural selection). Over many generations, the allele frequencies change independently in both populations, leading to significant genetic divergence and reproductive isolation.

(b) Scientists can collect individuals from both populations and attempt to interbreed them in a laboratory environment. If they are unable to produce offspring, or if the offspring produced are sterile/non-viable, then speciation has occurred and they are separate species.

(c) In small, isolated populations, chance events (rather than natural selection) can cause significant, unpredictable fluctuations in allele frequencies. Certain alleles may become fixed or completely lost from the population purely by chance. This random change can lead to rapid genetic divergence from the parent population, speeding up the process of speciation.

(a) [5 marks]
- 1 mark for stating geographical isolation prevents gene flow.
- 1 mark for noting different environmental conditions / selection pressures on each island.
- 1 mark for spontaneous mutations introducing new alleles.
- 1 mark for natural selection favoring advantageous alleles (increased survival and reproduction).
- 1 mark for change in allele frequencies over time leading to genetic divergence / reproductive isolation.

(b) [2 marks]
- 1 mark for attempting to interbreed individuals from both populations.
- 1 mark for checking if they can produce fertile/viable offspring.

(c) [4 marks]
- 1 mark for defining genetic drift as random / chance-driven changes in allele frequencies.
- 1 mark for stating that genetic drift is highly significant in small populations.
- 1 mark for explaining that some alleles can become fixed or lost purely by chance.
- 1 mark for stating this speeds up genetic divergence from the original population.

(a) When the light is switched off, the light-dependent reaction stops, so no ATP or reduced NADP is produced. Consequently, glycerate 3-phosphate (GP) cannot be reduced to triose phosphate (TP). However, carbon dioxide fixation continues initially because ribulose bisphosphate (RuBP) is still available to combine with \(\text{CO}_2\) to form GP, leading to a temporary accumulation of GP.

(b) RuBP continues to combine with \(\text{CO}_2\) to form GP, which consumes RuBP. Since the regeneration of RuBP from TP requires ATP (which is no longer produced by the light-dependent reaction), no new RuBP can be regenerated, causing its concentration to fall.

(c) Stroma of the chloroplast.

(d) In the Calvin cycle, carbon dioxide combines with RuBP to form two molecules of GP. ATP provides the energy and a phosphate group, while reduced NADP provides the hydrogen (and electrons) to reduce GP into triose phosphate (TP). Two molecules of TP out of every twelve are then used to synthesize hexose sugars like glucose, while the rest are used to regenerate RuBP.

(a) [3 marks]
- 1 mark for stating the light-dependent reaction stops, preventing production of ATP and reduced NADP.
- 1 mark for explaining GP cannot be reduced to TP.
- 1 mark for explaining that carbon fixation (RuBP + \(\text{CO}_2\)) continues, causing GP to accumulate.

(b) [3 marks]
- 1 mark for stating RuBP is consumed as it combines with carbon dioxide to form GP.
- 1 mark for noting ATP and reduced NADP are required to regenerate RuBP from TP.
- 1 mark for explaining that since these are absent in the dark, RuBP cannot be regenerated.

(c) [1 mark]
- 1 mark for stroma.

(d) [4 marks]
- 1 mark for stating carbon dioxide combines with RuBP.
- 1 mark for ATP providing energy / phosphate.
- 1 mark for reduced NADP providing hydrogen/electrons to reduce GP to TP.
- 1 mark for stating that TP molecules are converted into hexose sugars (allow '2 out of 12 TP' or '1/6th of TP').

(a) Oxygen acts as the final electron acceptor in the electron transport chain (ETC) on the inner mitochondrial membrane. It combines with electrons (at the end of the ETC) and protons (\(\text{H}^+\)) in the matrix to form water. Without oxygen, electrons cannot pass down the ETC, meaning reduced NAD and reduced FAD cannot be reoxidized. This depletes the pool of oxidized NAD and FAD in the matrix, which are essential coenzymes for the dehydrogenase enzymes in the Link reaction and the Krebs cycle, causing them to halt.

(b) Pyruvate undergoes decarboxylation to form ethanal, releasing carbon dioxide (catalyzed by pyruvate decarboxylase). Ethanal is then reduced to ethanol by alcohol dehydrogenase, using hydrogen from reduced NAD. This reoxidizes reduced NAD to NAD, allowing glycolysis to continue.

(c) The theoretical yield of aerobic respiration is approximately 30 to 32 ATP per glucose molecule, whereas anaerobic respiration in yeast yields only 2 ATP per glucose molecule (produced during glycolysis). The actual yield in aerobic respiration is lower because: some protons leak across the inner mitochondrial membrane (reducing the proton gradient), some ATP is actively transported out of the mitochondrial matrix into the cytoplasm (consuming energy), and some energy is lost as heat.

(a) [4 marks]
- 1 mark for stating oxygen acts as the terminal electron acceptor in the electron transport chain.
- 1 mark for stating oxygen combines with electrons and protons to form water.
- 1 mark for stating that without oxygen, reduced NAD/FAD cannot be reoxidized.
- 1 mark for explaining that a lack of oxidized NAD/FAD halts the Link reaction and Krebs cycle because they lack coenzymes.

(b) [3 marks]
- 1 mark for stating pyruvate is decarboxylated to form ethanal and carbon dioxide.
- 1 mark for stating ethanal is reduced to ethanol.
- 1 mark for stating this uses hydrogen from reduced NAD, reoxidizing it to NAD (allowing glycolysis to continue).

(c) [4 marks]
- 1 mark for stating aerobic respiration produces ~30-32 ATP per glucose and anaerobic produces 2 ATP.
- 1 mark for noting energy is lost as heat / proton leakage across the inner mitochondrial membrane.
- 1 mark for stating active transport is required to move pyruvate / ADP / ATP across mitochondrial membranes.
- 1 mark for stating some reduced NAD from glycolysis is used for other cellular processes.

(a) Gross Primary Productivity (GPP) is the total chemical energy stored in plant biomass, in a given area or volume, in a given time. Net Primary Productivity (NPP) is the chemical energy store remaining in plant biomass after respiratory losses (\(R\)) are subtracted (\(NPP = GPP - R\)). NPP represents the energy available for plant growth and reproduction, and the energy available to consumers at the next trophic level.

(b) \(NPP = GPP - R\)
\(NPP = 4.8 \times 10^4 - 2.9 \times 10^4 = 1.9 \times 10^4 \text{ kJ m}^{-2} \text{ yr}^{-1}\).

(c) Efficiency = \((\text{Energy in primary consumers} / \text{NPP of producers}) \times 100\)
\(\text{Efficiency} = \frac{1.9 \times 10^3}{1.9 \times 10^4} \times 100 = 10\%\).

(d) Primary consumers (herbivores) consume plant matter which contains a high proportion of indigestible, fibrous material, such as cellulose and lignin. Much of this energy is lost as waste in faeces. Secondary consumers (carnivores) feed on animal tissues which are highly digestible (mostly protein and lipids), meaning less energy is lost in faeces, resulting in higher assimilation efficiency.

(a) [3 marks]
- 1 mark for defining GPP as the total chemical energy stored in plant biomass in a given area/time.
- 1 mark for defining NPP as the chemical energy store in plant biomass after respiratory losses have been taken into account.
- 1 mark for stating the equation \(NPP = GPP - R\).

(b) [2 marks]
- 1 mark for correct calculation: \(1.9 \times 10^4\) (accept 19000).
- 1 mark for correct units: \(\text{kJ m}^{-2} \text{ yr}^{-1}\).

(c) [3 marks]
- 1 mark for dividing the energy of primary consumers by the calculated NPP: \(\frac{1.9 \times 10^3}{1.9 \times 10^4}\).
- 1 mark for multiplying by 100.
- 1 mark for correct answer of 10\% (must be to 2 significant figures, so accept 10 or 10.\%).

(d) [3 marks]
- 1 mark for stating plant material is difficult to digest / contains cellulose and lignin.
- 1 mark for stating herbivores lose more energy in faeces.
- 1 mark for noting carnivore diet is highly digestible / more easily absorbed.

(a) Autosomal linkage refers to two or more genes located on the same non-sex chromosome (autosome) that do not assort independently during gamete formation and tend to be inherited together.

(b) If the genes were not linked (independent assortment), we would expect a 1:1:1:1 ratio (approx. 250 of each phenotype). The high frequency of parental phenotypes (grey body, red eyes and black body, purple eyes) and the very low frequency of recombinant phenotypes (grey body, purple eyes and black body, red eyes) indicates that the genes are linked on the same chromosome. The recombinant phenotypes are present because crossing over occurred between non-sister chromatids of homologous chromosomes during prophase I of meiosis, exchanging alleles and creating new combinations.

(c) Recombination frequency (map distance) = \((\text{Number of recombinants} / \text{Total offspring}) \times 100\)
Total offspring = \(462 + 458 + 41 + 39 = 1000\)
Recombinants = \(41 + 39 = 80\)
Recombination frequency = \(\frac{80}{1000} \times 100 = 8.0\%\) (or 8.0 map units).

(d) The alleles are in coupling (cis) phase because the parental alleles were inherited together. The genotype of the F1 fly is written as \(\frac{GR}{gr}\) where \(G\) (grey) and \(R\) (red) are linked on one chromosome, and \(g\) (black) and \(r\) (purple) are linked on the homologous chromosome.

(a) [2 marks]
- 1 mark for stating the genes are located on the same chromosome.
- 1 mark for specifying they are on autosomes (non-sex chromosomes) / do not assort independently.

(b) [5 marks]
- 1 mark for stating that unlinked genes would yield a 1:1:1:1 ratio.
- 1 mark for pointing out that the high numbers of parental phenotypes indicate linkage.
- 1 mark for identifying the grey-purple and black-red flies as recombinants.
- 1 mark for explaining recombinants are produced by crossing over.
- 1 mark for specifying crossing over occurs during prophase I of meiosis.

(c) [2 marks]
- 1 mark for correct calculation of total offspring (1000) and recombinants (80).
- 1 mark for correct map distance: 8.0\% or 8 map units (accept 8).

(d) [2 marks]
- 1 mark for showing alleles G and R on one chromosome and alleles g and r on the homologous chromosome.
- 1 mark for indicating coupling / cis linkage (e.g., \(\frac{GR}{gr}\) or equivalent diagram).

(a) Saprobionts decompose organic waste, containing nitrogen (such as proteins, DNA, and urea), by secreting extracellular enzymes (extracellular digestion). This releases ammonium ions (\(\text{NH}_4^+\)) into the soil via ammonification. Nitrifying bacteria, such as *Nitrosomonas*, then oxidize ammonium ions into nitrite ions (\(\text{NO}_2^-\)), and another group of nitrifying bacteria, such as *Nitrobacter*, oxidize nitrite ions into nitrate ions (\(\text{NO}_3^-\)), which plants can easily absorb.

(b) Waterlogged soil has water-filled air spaces, excluding oxygen and creating anaerobic conditions. Without oxygen, root cells cannot perform aerobic respiration (specifically, oxidative phosphorylation in the electron transport chain). This severely reduces the production of ATP. Active transport of mineral ions against their concentration gradient into the root hair cells requires ATP. Therefore, the rate of mineral uptake by active transport decreases significantly.

(c) The nitrogen-fixing bacteria (such as *Rhizobium* in root nodules) reduce gaseous nitrogen (\(\text{N}_2\)) to ammonia, which is used to synthesize amino acids. This provides the leguminous host plant with a source of usable nitrogen for protein synthesis and growth. In return, the plant provides the bacteria with organic carbohydrates (sucrose/sugars) produced during photosynthesis, which the bacteria use as a respiratory substrate.

(a) [4 marks]
- 1 mark for saprobionts performing extracellular digestion/decomposition of nitrogenous organic matter (proteins/urea).
- 1 mark for releasing ammonium ions via ammonification.
- 1 mark for nitrifying bacteria oxidizing ammonium to nitrite.
- 1 mark for nitrifying bacteria oxidizing nitrite to nitrate.

(b) [4 marks]
- 1 mark for stating waterlogging causes anaerobic conditions / reduces oxygen concentration.
- 1 mark for noting root hair cells cannot perform aerobic respiration.
- 1 mark for explaining this leads to decreased ATP production.
- 1 mark for explaining that active transport of mineral ions requires ATP.

(c) [3 marks]
- 1 mark for stating bacteria convert nitrogen gas into ammonia / amino acids for the plant.
- 1 mark for stating the plant provides carbohydrates / photosynthetic products to the bacteria.
- 1 mark for identifying this as a mutualistic relationship where both organisms benefit.

(a) The sodium-potassium pump actively transports three sodium ions out of the axon for every two potassium ions transported in, using energy from ATP hydrolysis. This creates concentration gradients. The membrane is more permeable to potassium ions than sodium ions because there are more open potassium leakage channels, allowing potassium to diffuse out down its concentration gradient, leaving a net negative charge inside. (b) Repolarisation would be delayed or prevented because potassium ions cannot exit the axon down their electrochemical gradient to restore the negative resting potential. The membrane remains depolarised for a prolonged period. (c) Use a computerised reaction time test where the subject presses a button in response to a light (visual) or a sound (auditory). Control variables: age, caffeine intake, same dominant hand used, and ambient noise levels. Measure distance from sensory receptor (eye or ear) to brain, and brain to effector muscle. Calculate speed as estimated pathway distance divided by reaction time.

(a) 1. Active transport of 3 Na+ out and 2 K+ in via sodium-potassium pump (1 mark); 2. Membrane is more permeable to K+ due to open potassium leak channels (1 mark); 3. K+ diffuses out down concentration gradient, creating negative charge inside of \(-70\text{ mV}\) (1 mark). (b) 1. Repolarisation is prevented or delayed (1 mark); 2. Because voltage-gated K+ channels cannot open to allow K+ efflux (1 mark); 3. Axon membrane remains depolarised or cannot return to resting potential (1 mark). (c) 1. Method to measure reaction time to visual and auditory stimuli using a standardised timer / software (1 mark); 2. Identify two control variables (e.g., age of participants, sleep level, stimulus intensity) (1 mark); 3. Measure pathway distance from receptor to effector (1 mark); 4. Speed calculation: Distance divided by reaction time (1.7 marks).

(a) Unilateral light causes IAA to migrate from the light side to the shaded side of the shoot tip. IAA is then transported down the shaded side of the shoot, where it stimulates cell elongation by loosening cell walls. The greater elongation of cells on the shaded side causes the shoot to bend towards the light. (b) To make a 10-fold serial dilution: transfer 1 cm^3 of the 1000 mg dm^-3 stock solution to 9 cm^3 of distilled water to make 100 mg dm^-3. Repeat this process: transfer 1 cm^3 of 100 mg dm^-3 to 9 cm^3 of water to make 10 mg dm^-3, then 1 mg dm^-3, 0.1 mg dm^-3, and finally 0.01 mg dm^-3. (c) Use decapitated coleoptiles of the same age, species, and initial length. Place the agar blocks offset on one side of the decapitated coleoptile stem. Keep all setups in complete darkness to eliminate light phototropism. Use at least 5 replicates per concentration to calculate a mean and identify anomalies.

(a) 1. IAA moves / diffuses to the shaded side of the shoot (1 mark); 2. IAA stimulates cell elongation / expansion on the shaded side (1 mark); 3. Asymmetric growth causes the shoot to bend towards the light (1 mark). (b) 1. Dilution method: Serial dilution described (1 mark); 2. Correct volumes: 1 cm^3 solute to 9 cm^3 solvent / water (1 mark); 3. Range of 5 concentrations calculated correctly (1 mark); 4. Mixing thoroughly at each step (1 mark). (c) 1. Controls: age/species/length of coleoptiles and complete darkness (1.7 marks); 2. Experimental design: offset placement of agar blocks (1 mark); 3. Reliability: Repeat at least 5 times per concentration to calculate mean (1 mark).

(a) Adrenaline acts as the first messenger and binds to a specific transmembrane receptor on the liver cell membrane. This binding causes a conformational change that activates a G-protein, which in turn activates the enzyme adenylate cyclase. Adenylate cyclase converts ATP into cyclic AMP (cAMP), which acts as the second messenger. cAMP binds to and activates protein kinase A. (b) Activated protein kinase A initiates a cascade of enzymatic reactions that activates glycogen phosphorylase. This enzyme catalyses glycogenolysis (the breakdown of glycogen to glucose). Glucose is then transported out of the liver cells into the blood by facilitated diffusion, increasing blood glucose. (c) Prepare glucose solutions of known concentrations (e.g., 0, 2, 4, 6, 8, 10 mmol dm^-3). Heat each with a fixed volume of Benedict's reagent, filter the precipitate, and measure the absorbance or transmission of the remaining solution using a colorimeter. Plot a calibration curve of absorbance against glucose concentration. Test the unknown plasma sample in the same way, measure its absorbance, and use the curve to read off the glucose concentration.

(a) 1. Adrenaline binds to specific receptor on liver cell membrane (1 mark); 2. Activates G-protein, which activates adenylate cyclase (1 mark); 3. Adenylate cyclase converts ATP to cyclic AMP (cAMP) (1 mark); 4. cAMP activates protein kinase (1 mark). (b) 1. Protein kinase activates glycogen phosphorylase / enzyme cascade (1 mark); 2. Glygogenolysis converts glycogen to glucose (1 mark); 3. Glucose diffuses out of hepatocytes into blood via carrier proteins (1 mark). (c) 1. Produce known glucose standards and heat with Benedict's reagent (1 mark); 2. Use colorimeter to measure absorbance of standards and plot calibration curve (1.7 marks); 3. Test unknown sample under identical conditions and read concentration from curve (1 mark).

(a) Calcium ions 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 actin-myosin cross-bridges. ATP binds to the myosin head, causing it to detach from the actin filament. Hydrolysis of ATP by ATPase provides energy to reset the myosin head (cocking). The myosin head binds to a new site on actin, and the release of ADP and Pi causes the myosin head to pivot (power stroke), pulling the actin filament along the myosin filament. (b) Slow-twitch fibres are adapted for aerobic respiration; they contain more mitochondria, myoglobin, and a richer capillary supply to generate ATP efficiently over long periods. Fast-twitch fibres contain high concentrations of glycogen and phosphocreatine, relying on anaerobic glycolysis for rapid, short-term ATP generation. (c) Stain the muscle cross-section for an aerobic enzyme such as succinate dehydrogenase or cytochrome oxidase. Slow-twitch fibres will stain more intensely (darker) because they have high mitochondrial density. Count the number of stained and unstained fibres under a microscope and calculate the percentage of slow-twitch fibres.

(a) 1. Ca2+ binds to troponin, displacing tropomyosin (1 mark); 2. Exposes binding sites on actin, allowing cross-bridge formation (1 mark); 3. ATP binds to myosin head, causing detachment (1 mark); 4. ATP hydrolysis provides energy to cock / reset myosin head (1 mark); 5. Power stroke occurs as actin is pulled over myosin, releasing ADP/Pi (1 mark). (b) 1. Slow-twitch produces ATP aerobically, fast-twitch anaerobically (1 mark); 2. Slow-twitch has high mitochondrial density / myoglobin for oxidative phosphorylation (1 mark); 3. Fast-twitch has high glycogen / phosphocreatine for rapid glycolysis (1 mark). (c) 1. Use stain specific to mitochondrial enzymes / aerobic respiration (1 mark); 2. Identify slow-twitch as highly stained fibres and count total vs stained (1 mark); 3. Calculate percentage / proportion (0.7 marks).

(a) mRNA for the target protein is extracted from cells where the gene is highly expressed. Reverse transcriptase is added along with free DNA nucleotides to synthesise a single-stranded complementary DNA (cDNA) copy from the mRNA template. DNA polymerase is then used to make the cDNA double-stranded. This cDNA can then be amplified using PCR. (b) Restriction endonucleases are used to cut both the isolated cDNA and the plasmid vector at specific palindromic recognition sequences, creating matching single-stranded 'sticky ends'. The cDNA and cut plasmids are mixed together, allowing their complementary sticky ends to pair up via hydrogen bonding. DNA ligase is added to catalyse the formation of phosphodieser bonds, sealing the sugar-phosphate backbones. (c) PCR involves heating the mixture to 95 degrees Celsius to separate DNA strands (denaturation). Cool to 55 degrees Celsius to allow primers to bind (annealing). Heat to 72 degrees Celsius for Taq polymerase to synthesise complementary strands (extension). Primers are short single-stranded DNA sequences that bind to the start of the target sequence, providing a starting point for DNA polymerase to bind and begin synthesis.

(a) 1. Extract mRNA from cells actively expressing the gene (1 mark); 2. Use reverse transcriptase to synthesise single-stranded cDNA (1 mark); 3. Use DNA polymerase to make double-stranded DNA (1 mark); 4. Amplify cDNA using PCR (1 mark). (b) 1. Use same restriction endonuclease on plasmid and cDNA (1 mark); 2. To produce complementary sticky ends (1 mark); 3. Mix plasmid and gene to allow base pairing of sticky ends (1 mark); 4. Use DNA ligase to form phosphodieser bonds (1 mark). (c) 1. PCR steps: 95C (denaturing), 55C (annealing), 72C (extension) (1 mark); 2. Primers are short single-stranded DNA sequences (0.7 marks); 3. Primers bind to target sequence to allow DNA polymerase to attach / initiate synthesis (1 mark).

(a) Increased DNA methylation involves adding methyl groups to CpG islands in the promoter region of the gene. This prevents the binding of transcription factors. Decreased histone acetylation increases the positive charge on histones, increasing their attraction to the negatively charged DNA. This causes the chromatin to condense into heterochromatin, making the gene inaccessible to RNA polymerase. (b) Double-stranded siRNA is cleaved into short double-stranded fragments. One strand of the siRNA combines with an enzyme complex called the RNA-induced silencing complex (RISC). The single-stranded siRNA acts as a guide, guiding the RISC to the target mRNA by complementary base pairing. Once bound, the RISC enzyme cleaves the mRNA, preventing it from being translated into a protein. (c) siRNA therapies are highly specific because they rely on complementary base pairing (usually 20-25 base pairs) to target only one unique mRNA sequence (e.g., an oncogene). Traditional chemotherapy targets all rapidly dividing cells indiscriminately, leading to widespread side effects like hair loss and immune suppression.

(a) 1. DNA methylation: addition of methyl groups to CpG islands / promoter (1 mark); 2. Prevents transcription factor binding (1 mark); 3. Decreased histone acetylation increases positive charge on histones, causing DNA to wind tightly / form heterochromatin (1 mark); 4. RNA polymerase cannot access gene, preventing transcription (1 mark). (b) 1. siRNA associates with RNA-induced silencing complex (RISC) (1 mark); 2. One siRNA strand remains to act as a guide (1 mark); 3. siRNA binds to target mRNA via complementary base pairing (1 mark); 4. RISC cleaves/breaks down mRNA, preventing translation (1 mark). (c) 1. siRNA relies on highly specific complementary base pairing (1 mark); 2. Targets only a specific oncogene mRNA without affecting normal genes (1 mark); 3. Chemotherapy is non-specific / targets all rapidly dividing cells (0.7 marks).

(a) Increased cellular respiration during exercise produces more carbon dioxide, which dissolves in the blood to form carbonic acid, lowering blood pH. Chemoreceptors in the carotid arteries and aorta detect this drop in pH and send a higher frequency of nerve impulses to the cardiovascular centre in the medulla oblongata. The medulla oblongata increases the frequency of impulses along sympathetic neurones to the sinoatrial node (SAN). This causes the SAN to fire electrical impulses more frequently, increasing heart rate. (b) The sinoatrial node (SAN) generates a wave of electrical excitation that spreads across both atria, causing them to contract (atrial systole). Non-conductive tissue prevents this wave from spreading directly to the ventricles. The wave reaches the atrioventricular node (AVN), which delays the impulse to allow the atria to empty completely into the ventricles. The wave then travels down the Bundle of His to the apex, and up through the Purkyne fibres, causing ventricular contraction from the apex upwards. (c) Baroreceptors detect high blood pressure and send impulses to the medulla oblongata. The medulla sends impulses along parasympathetic (vagus) neurones to the SAN, releasing acetylcholine to decrease the frequency of electrical impulses generated by the SAN.

(a) 1. Respiration produces CO2, which dissolves to form carbonic acid / lowers pH (1 mark); 2. Chemoreceptors in carotid bodies / aortic bodies detect pH drop (1 mark); 3. Send more impulses to cardiovascular centre in medulla oblongata (1 mark); 4. Medulla sends impulses along sympathetic nerves (1 mark); 5. To SAN, increasing frequency of heartbeats (1 mark). (b) 1. SAN initiates electrical wave spreading across atria (1 mark); 2. Non-conducting tissue prevents direct spread to ventricles (1 mark); 3. AVN delays wave to allow ventricles to fill (1 mark); 4. Wave travels down Bundle of His and Purkyne fibres to cause contraction from the apex up (1 mark). (c) 1. Baroreceptors detect high BP and send impulses to medulla (1 mark); 2. Medulla sends parasympathetic / vagus impulses to SAN to decrease heart rate (0.7 marks).

For (a), the increase is from 1.85 to 11.20. Absolute increase = 11.20 - 1.85 = 9.35. Percentage increase = (9.35 / 1.85) * 100 = 505.4%. For (b), glycogen phosphorylase breaks down glycogen reserves to glucose-1-phosphate, increasing the pool of respiratory substrates available for glycolysis to meet high ATP demands anaerobically. For (c), lower pH (higher hydrogen ion concentration) denatures enzymes (like phosphofructokinase or ATPase) by disrupting ionic/hydrogen bonds in their tertiary structure, slowing metabolic pathways. H+ ions also compete with calcium ions for the binding sites on troponin, preventing tropomyosin from moving away from actin binding sites, reducing cross-bridge formation. For (d), ATP is produced by substrate-level phosphorylation in glycolysis. Pyruvate is reduced to lactate, which oxidizes NADH back to NAD+ to ensure glycolysis can continue.

(a) [2 marks]
- Correct calculation of absolute increase: 11.20 - 1.85 = 9.35 (1 mark)
- Correct calculation of percentage increase: (9.35 / 1.85) * 100 = 505.4% (accept 505% or 505.41%) (1 mark)

(b) [3 marks]
- Breaks down stored glycogen to glucose-1-phosphate / glucose (1 mark)
- Increases availability of substrate for glycolysis (1 mark)
- Allows rapid production of ATP to sustain high contraction frequencies under anaerobic conditions (1 mark)

(c) [5.5 marks]
- Lower pH represents a higher concentration of hydrogen ions (0.5 marks)
- Hydrogen ions disrupt ionic and hydrogen bonds within the tertiary structure of proteins / enzymes (1 mark)
- Denatures enzymes (e.g., ATPase or respiratory enzymes), reducing active site complementarity and reaction rate (1 mark)
- Hydrogen ions compete with calcium ions (1 mark)
- ...for specific binding sites on troponin (1 mark)
- Tropomyosin remains in position blocking the myosin-binding sites on actin, reducing cross-bridge formation (1 mark)

(d) [2 marks]
- Net yield of 2 ATP molecules is produced in glycolysis via substrate-level phosphorylation (1 mark)
- NADH reduces pyruvate to lactate, regenerating oxidized NAD+ to allow glycolysis to proceed (1 mark)

For (a), restriction endonucleases target and cut specific DNA sequences with palindromic symmetry to form sticky ends. Plasmid vectors are cut with the same restriction enzyme, creating complementary sticky ends. DNA ligase seals the phosphodiester backbone. For (b), expressing the toxin only upon wounding saves valuable metabolic energy and resources when the plant is not under attack, reducing negative selection pressure for resistance in pests. For (c), transcription factors move from the cytoplasm into the nucleus where they bind to specific promoter sequences on the DNA, recruiting RNA polymerase to begin transcription. For (d), reverse transcriptase converts isolated mRNA into cDNA, which is then amplified via PCR using primers specific to the Bt gene.

(a) [4 marks]
- Restriction endonucleases cut DNA at specific palindromic recognition sequences (1 mark)
- Producing sticky ends with exposed, single-stranded bases (1 mark)
- The plasmid vector is cut with the same restriction endonuclease to produce complementary sticky ends (1 mark)
- DNA ligase joins the gene to the plasmid by forming phosphodiester bonds (1 mark)

(b) [3 marks]
- Saves metabolic resources / energy (amino acids and ATP) when the plant is undamaged (1 mark)
- Reduces selection pressure on pest populations to develop resistance to the Bt toxin (1 mark)
- Minimizes exposure/toxicity to non-target herbivores or consumers eating healthy tissue (1 mark)

(c) [3.5 marks]
- Transcription factors move from the cytoplasm into the nucleus (0.5 marks)
- Bind to specific DNA base sequences / promoter regions of the target gene (1 mark)
- This binding recruits / allows RNA polymerase to bind to the template DNA strand (1 mark)
- Initiates the synthesis of pre-mRNA from the DNA template (1 mark)

(d) [2 marks]
- Extract mRNA from leaves and use reverse transcriptase to synthesize complementary DNA (cDNA) (1 mark)
- Perform PCR using primers specific to the Bt gene sequence to amplify cDNA, then analyze via gel electrophoresis (1 mark)

For (a), Cardiac Output (CO) = Stroke Volume (SV) x Heart Rate (HR). Converting CO during exercise to cm3: 24.75 dm3 min-1 = 24,750 cm3 min-1. SV = 24,750 / 165 = 150 cm3. For (b), increased CO2 leads to carbonic acid production, lowering blood pH. This is detected by chemoreceptors in the carotid and aortic bodies, sending impulses to the medulla oblongata, which increases the frequency of sympathetic nerve impulses to the sinoatrial node (SAN) to increase heart rate. For (c), high blood pressure stretches baroreceptors, sending more impulses to the medulla. The medulla then increases impulses via the parasympathetic system (vagus nerve), releasing acetylcholine at the SAN to lower heart rate and reduce blood pressure. For (d), mechanical pressure deforms lamellae, opening stretch-mediated Na+ channels, causing Na+ influx to generate a generator potential.

(a) [2 marks]
- Convert cardiac output to cm3: 24.75 * 1000 = 24,750 cm3 min-1 (1 mark)
- Stroke volume = 24,750 / 165 = 150 cm3 (1 mark) (Correct answer only = 2 marks)

(b) [5 marks]
- CO2 dissolves in blood forming carbonic acid, which lowers blood pH (1 mark)
- This pH drop is detected by chemoreceptors in the carotid bodies / aortic bodies (1 mark)
- Chemoreceptors send a higher frequency of nerve impulses to the medulla oblongata / cardiovascular center (1 mark)
- Medulla oblongata sends a higher frequency of impulses via the sympathetic nervous system (1 mark)
- ...to the sinoatrial node (SAN), increasing the rate of depolarization and heart rate (1 mark)

(c) [3.5 marks]
- High blood pressure stretches baroreceptors in carotid sinus / aorta (0.5 marks)
- Baroreceptors send higher frequency of impulses to the medulla oblongata (1 mark)
- Medulla increases impulses via parasympathetic nervous system / vagus nerve (1 mark)
- Secretion of acetylcholine at SAN decreases the heart rate, reducing blood pressure to normal (1 mark)

(d) [2 marks]
- Mechanical pressure deforms the lamellae, stretching the membrane of the sensory neurone ending (1 mark)
- Opens stretch-mediated sodium channels, causing influx of sodium ions (Na+) and producing a generator potential (1 mark)

For (a), Gross Primary Productivity (GPP) = 1.5% of 1,800,000 = 27,000 kJ m-2 yr-1. Net Primary Productivity (NPP) = GPP - Respiration (R). R = 60% of GPP = 16,200. NPP = 27,000 - 16,200 = 10,800 kJ m-2 yr-1 (or 40% of 27,000). For (b), water is split by photolysis into protons, electrons, and oxygen. The electrons replace those lost by photosystem II. The accumulation of protons in the thylakoid space builds a concentration gradient; protons flow back through ATP synthase to catalyze ATP generation. For (c), CO2 combines with RuBP (catalyzed by Rubisco) to form GP, which is then reduced to TP using ATP and NADPH. For (d), energy is lost because parts of the plants are indigestible (cellulose in feces) or uneaten, and significant energy is dissipated as heat during respiration.

(a) [3 marks]
- Correct calculation of GPP: 1.8 x 10^6 * 0.015 = 27,000 kJ m-2 yr-1 (1 mark)
- Subtracting 60% respiration loss or calculating 40% of GPP (1 mark)
- NPP = 10,800 kJ m-2 yr-1 (1 mark) (Correct answer only = 3 marks)

(b) [4 marks]
- Photolysis of water splits water into protons (H+), electrons (e-), and oxygen (O2) (1 mark)
- Electrons replace those excited and lost from chlorophyll / photosystem II (1 mark)
- Protons accumulate inside the thylakoid space, creating an electrochemical / proton gradient (1 mark)
- Protons diffuse down this gradient into the stroma via ATP synthase, driving photophosphorylation of ADP to ATP (1 mark)

(c) [3.5 marks]
- Carbon dioxide reacts with ribulose bisphosphate (RuBP), a 5C compound, catalyzed by Rubisco (1 mark)
- Produces two molecules of glycerate 3-phosphate (GP) (1 mark)
- GP is reduced to triose phosphate (TP) (0.5 marks)
- ...using reduced NADP (NADPH) and energy from the hydrolysis of ATP (1 mark)

(d) [2 marks]
- Some plant tissues (such as cellulose) are indigestible and lost as feces (1 mark)
- Not all parts of plants are eaten (roots/wood), and substantial energy is lost as heat via consumer respiration (1 mark)

For (a), the sodium-potassium pump actively transports 3 Na+ out of the axon for every 2 K+ inside. Potassium leak channels allow K+ to diffuse down its concentration gradient, leaving a negative resting potential of about -70mV. For (b), an action potential opens calcium channels. Ca2+ entry causes ACh vesicles to fuse with the presynaptic membrane and release ACh via exocytosis. ACh diffuses and binds postsynaptic receptors, opening ligand-gated Na+ channels and depolarizing the membrane. For (c), IAA moves to the shaded side of the shoot, stimulating cell elongation there, causing unilateral growth toward the light source. Tropisms in plants are much slower (hours/days vs milliseconds) and produce permanent, long-term architectural modifications compared to the transient effects of nervous responses.

(a) [3 marks]
- Active transport of 3 Na+ out and 2 K+ in via sodium-potassium pump, using ATP (1 mark)
- Axon membrane is more permeable to K+ than Na+ due to open potassium leak channels (1 mark)
- Net diffusion of K+ out of axon creates an electrochemical gradient with a resting potential of approx. -70 mV (1 mark)

(b) [5 marks]
- Depolarization of presynaptic membrane opens voltage-gated calcium channels, causing Ca2+ influx (1 mark)
- Calcium ions cause synaptic vesicles to fuse with the presynaptic membrane, releasing acetylcholine (ACh) by exocytosis (1 mark)
- ACh diffuses across the synaptic cleft (1 mark)
- ACh binds to specific receptors on ligand-gated sodium channels on the postsynaptic membrane (1 mark)
- Sodium channels open, Na+ diffuses in, depolarizing the postsynaptic membrane and initiating an action potential if threshold is reached (1 mark)

(c) [4.5 marks]
- Light causes IAA to migrate to the shaded side of the shoot tip (1 mark)
- IAA is transported down the shaded side of the shoot (0.5 marks)
- High concentration of IAA in shoots promotes cell elongation on the shaded side (1 mark)
- Unequal growth causes the shoot to bend towards the light source (1 mark)
- Contrast: Plant growth responses are much slower (hours/days) compared to nervous impulses (milliseconds) and cause long-lasting/permanent structural changes (1 mark)

For (a), glucagon binds to cell surface receptors on hepatocytes, activating adenylate cyclase. Adenylate cyclase converts ATP to cyclic AMP (cAMP), the second messenger. cAMP activates protein kinase A, triggering a cascade that stimulates glycogenolysis and gluconeogenesis. For (b), the heterozygous man has genotype Mm, and the healthy woman has mm. Gametes are M and m from the father, and m from the mother. The genetic cross yields a 1:1 ratio of Mm (affected) to mm (unaffected), giving a probability of 0.5 (or 50%). For (c), negative feedback detects a deviation from the set point and triggers corrective actions to return the parameter to the norm. Insulin reduces hyperglycemia, and glucagon raises hypoglycemia. For (d), the genetic code is universal; codons represent the same amino acids across species, enabling bacteria to translate human insulin mRNA correctly.

(a) [4 marks]
- Glucagon binds to specific, complementary receptors on liver cell membranes (1 mark)
- Activates the enzyme adenylate cyclase (1 mark)
- Adenylate cyclase converts ATP into cyclic AMP (cAMP) (1 mark)
- cAMP acts as a second messenger, activating protein kinase A, which triggers a cascade initiating glycogenolysis / gluconeogenesis (1 mark)

(b) [3.5 marks]
- Correct parent genotypes: Father = Mm, Mother = mm (1 mark)
- Correct gametes shown: Father (M and m), Mother (m) (0.5 marks)
- Correct offspring genotypes/phenotypes: Mm (MODY) and mm (healthy) in equal ratios (1 mark)
- Correct probability of child inheriting MODY = 0.5 / 50% / 1 in 2 (1 mark)

(c) [3 marks]
- Negative feedback is a regulatory mechanism where any change from a physiological set point triggers responses to reverse the change (1 mark)
- If blood glucose increases, insulin is secreted, stimulating glycogen synthesis to lower glucose levels back to normal (1 mark)
- If blood glucose drops, glucagon is secreted, stimulating glycogen breakdown to raise blood glucose back to normal (1 mark)

(d) [2 marks]
- The genetic code is universal (1 mark)
- The same mRNA codons code for the exact same amino acids in both human and bacterial cells, ensuring functional human protein synthesis (1 mark)