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An increase in blood plasma osmolarity (dehydration) stimulates osmoreceptors in the hypothalamus. This triggers the release of antidiuretic hormone (ADH) from the posterior pituitary gland. ADH binds to receptors on the cells of the collecting ducts in the kidney, leading to the translocation of aquaporin water channels to the apical membrane. This increases water reabsorption back into the blood, resulting in the production of a low volume of highly concentrated urine.

Award [1] for the correct answer (C).

Pure lipid bilayers are highly impermeable to large polar molecules like glucose. Therefore, glucose cannot easily pass through the bilayer without transport proteins. However, the membrane is semi-permeable and allows water molecules to pass slowly. Water will move down its concentration gradient (from a region of lower solute concentration, Side X, to higher solute concentration, Side Y) by osmosis.

Award [1] for the correct answer (B).

Solar radiation that reaches Earth is mostly short-wave radiation (UV and visible light). The Earth absorbs this radiation and re-emits it as longer wavelength radiation (infrared/heat). Greenhouse gases in the atmosphere absorb these longer wavelengths (infrared radiation) and then re-emit them in all directions, including back towards the Earth's surface, thereby trapping heat.

Award [1] for the correct answer (B).

Active loading of sucrose into phloem sieve tubes involves companion cells. Proton pumps (\(H^+\)-ATPases) use ATP to actively pump protons (\(H^+\)) out of the companion cells into the cell wall space (apoplast). This creates a high electrochemical gradient of protons. Sucrose is then co-transported into the companion cell-sieve tube complex along with the protons as they diffuse back in down their concentration gradient through a sucrose-proton cotransporter protein (symport).

Award [1] for the correct answer (A).

Competitive inhibitors bind to the active site of the enzyme, competing directly with the substrate. Because they can be outcompeted by very high concentrations of substrate, the maximum rate of reaction (\(V_{max}\)) remains the same. However, because more substrate is needed to achieve the same rate of reaction, the \(K_m\) increases.

Award [1] for the correct answer (B).

During the follicular phase, estrogen levels rise as the follicle develops. Once estrogen reaches a threshold high concentration, it exerts positive feedback on the hypothalamus and anterior pituitary, leading to a rapid surge in luteinizing hormone (LH). This LH surge triggers the release of the oocyte from the mature follicle (ovulation).

Award [1] for the correct answer (B).

Sweating is an effective cooling mechanism because water has a high latent heat of vaporization. To change state from liquid water (sweat) to water vapor, a significant amount of heat energy (thermal energy) must be absorbed from the skin to break the hydrogen bonds between water molecules. This removes heat from the body, cooling it down.

Award [1] for the correct answer (B).

Since the parents have blood groups A and B but produced a child with blood group O (genotype \(ii\)), both parents must be heterozygous. The mother's genotype is \(I^A i\) and the father's genotype is \(I^B i\). The cross (\(I^A i \times I^B i\)) yields the following possible offspring genotypes: \(I^A I^B\) (Blood group AB, 25%), \(I^A i\) (Blood group A, 25%), \(I^B i\) (Blood group B, 25%), and \(ii\) (Blood group O, 25%). Thus, the probability of having a child with blood group AB is 25%.

Award [1] for the correct answer (B).

The scenario describes a transport mechanism that requires a membrane protein (since it is inhibited by a competitive protein inhibitor), occurs along a concentration gradient (down its gradient), and does not require metabolic energy (no ATP required). This is the definition of facilitated diffusion. Simple diffusion does not require a membrane protein and would not be stopped by a competitive inhibitor of Protein X. Active transport mechanisms require ATP.

Award [1] mark for the correct option B.
- Reject A because simple diffusion does not require membrane proteins and would not be affected by a protein-specific inhibitor.
- Reject C and D because active transport processes require ATP to move solutes against their concentration gradient.

In end-product inhibition, the final product of the pathway (substance D) acts as an allosteric inhibitor of the first enzyme (Enzyme 1). When an excess of substance D is introduced, Enzyme 1 is inhibited, preventing the conversion of substrate A into intermediate B. Consequently, substance A will accumulate (its consumption ceases), while intermediates B and C will continue to be converted into downstream products by Enzyme 2 and Enzyme 3, causing their concentrations to decrease.

Award [1] mark for the correct option B.
- Reject A, C, and D because they do not correctly identify that the inhibition of Enzyme 1 stops the consumption of A (leading to accumulation) and prevents the generation of new B and C (leading to depletion of existing intermediates).

Consuming a hypertonic solution increases the solute concentration (osmolarity) of the blood. Osmoreceptors in the hypothalamus detect this high solute concentration and stimulate the posterior pituitary gland to secrete more antidiuretic hormone (ADH) into the bloodstream. ADH acts on the collecting duct of the kidney, making its cells more permeable to water (by promoting the insertion of aquaporin channels), allowing more water to be reabsorbed back into the body to dilute the blood.

Award [1] mark for the correct option A.
- Reject B and D because the hypothalamus detects high, not low, solute concentration under hypertonic conditions.
- Reject C because high solute concentration stimulates an increase, not a decrease, in ADH secretion.

According to the pressure-flow hypothesis, sucrose is actively loaded into the phloem sieve tubes at the source tissue. This active transport lowers the water potential (makes it more negative) inside the sieve tubes, drawing water from the adjacent xylem by osmosis. The influx of water generates a high hydrostatic pressure at the source. At the sink, sucrose is unloaded, raising the water potential, causing water to leave the sieve tubes and lowering the hydrostatic pressure. This hydrostatic pressure gradient drives mass flow from source to sink.

Award [1] mark for the correct option B.
- Reject A and C because hydrostatic pressure is higher at the source, not the sink.
- Reject D because sucrose is actively (not passively) loaded into the phloem sieve tubes at the source, which lowers (not raises) the water potential.

During the early to mid-follicular phase, low to moderate levels of estrogen exert negative feedback on the anterior pituitary gland and hypothalamus, inhibiting the secretion of FSH and LH. This keeps their levels relatively low. However, near the end of the follicular phase, estrogen levels rise and remain above a high threshold for a sustained period, switching the feedback to positive feedback. This triggers a massive surge in LH (the LH surge), which induces ovulation.

Award [1] mark for the correct option C.
- Reject A and B because estrogen initially exerts negative feedback, not positive feedback.
- Reject D because estrogen exerts positive, not negative, feedback immediately before ovulation to trigger the LH surge.

Plasma cells are differentiated B lymphocytes that produce a single, specific type of antibody. However, they are short-lived and cannot divide or survive long-term in laboratory culture. Tumor (myeloma) cells are immortal and can divide indefinitely in culture, but they do not produce the desired antibody. Fusing them creates a hybridoma cell that possesses both desired traits: the ability to produce the specific antibody and the ability to replicate indefinitely in culture.

Award [1] mark for the correct option B.
- Reject A because an individual plasma cell is clonal and produces only one type of antibody, not a variety.
- Reject C because tumor cells do not produce the required antibodies.
- Reject D because the fusion does not aim to induce mutations in the mouse genome.

Water-soluble hormones like epinephrine cannot pass through the hydrophobic lipid bilayer of the cell membrane. Consequently, they bind to external receptor proteins on the cell surface, which initiates an intracellular signal transduction cascade involving second messengers (such as cyclic AMP). Lipid-soluble steroid hormones like cortisol can easily diffuse across the plasma membrane and bind to intracellular receptor proteins in the cytoplasm or nucleus. The hormone-receptor complex then binds to DNA to directly regulate gene transcription.

Award [1] mark for the correct option B.
- Reject A because it reverses the receptor locations and signaling pathways of epinephrine and cortisol.
- Reject C because steroid hormones do not typically bind to cell surface receptors to open ion channels.
- Reject D because epinephrine cannot cross the plasma membrane by simple diffusion.

Since the genes are linked in a cis-coupling configuration (\(PL / pl\)), the alleles \(P\) and \(L\) are on one homologous chromosome, and \(p\) and \(l\) are on the other. The parental, non-recombinant gametes produced by the heterozygote are \(PL\) and \(pl\). If crossing over occurs, recombinant gametes (\(Pl\) and \(pL\)) are formed, but at a lower frequency than parental gametes. When test-crossed with a \(ppll\) plant (which only contributes \(pl\) gametes), the parental phenotypes (purple-long and red-round) will appear in high frequency, while the recombinant phenotypes (purple-round and red-long) will appear in lower frequency.

Award [1] mark for the correct option C.
- Reject A because a \(1:1:1:1\) ratio is only expected under independent assortment (unlinked genes).
- Reject B because crossing over does occur, meaning recombinant phenotypes will be present, albeit in smaller quantities.
- Reject D because recombinant phenotypes cannot be more frequent than parental phenotypes when genes are linked.

Carrier proteins used in facilitated diffusion (like the glucose transporter) undergo a conformational change to transport solute molecules down their concentration gradient without consuming ATP. Simple diffusion (A) does not use a protein. Active transport (C) requires ATP hydrolysis to power the conformational change. Aquaporins (D) are channel proteins that allow water to pass rapidly through a pore without undergoing cyclic conformational changes for each water molecule.

Award [1] for the correct answer (B).
- Award [0] for any other option.

Water deprivation leads to high blood osmolarity, stimulating the posterior pituitary to secrete ADH. ADH binds to receptors on the collecting duct cells, triggering the translocation of aquaporin water channels to the apical (luminal) membrane, which increases water reabsorption back into the blood, resulting in a low volume of concentrated urine.

Award [1] for the correct answer (C).
- Award [0] for any other option.

Companion cells use ATP-driven proton pumps to pump hydrogen ions (\(\text{H}^+\)) out of the cell. This creates an electrochemical gradient. The protons then diffuse back into the companion cell down their gradient through a sucrose-\(H^+\) symporter, co-transporting sucrose against its concentration gradient into the phloem.

Award [1] for the correct answer (B).
- Award [0] for any other option.

Because the inhibition can be completely overcome at very high substrate concentrations, the substrate outcompetes the inhibitor for the active site. This is a classic characteristic of a competitive inhibitor, which binds directly to the active site.

Award [1] for the correct answer (A).
- Award [0] for any other option.

In the late follicular phase, the developing follicle secretes high levels of estrogen. Once estrogen levels cross a threshold and remain high for about 36 hours, it switches from negative feedback to positive feedback, stimulating a massive surge of LH from the anterior pituitary, which triggers ovulation.

Award [1] for the correct answer (B).
- Award [0] for any other option.

The solar radiation that reaches the Earth's surface consists mainly of short-wave radiation (visible and UV light). The Earth absorbs this and re-emits it as long-wave radiation (infrared/heat). Greenhouse gases in the atmosphere absorb and re-emit this long-wave radiation, keeping the atmosphere warm.

Award [1] for the correct answer (B).
- Award [0] for any other option.

The man is blood group A but must have inherited the recessive allele \(i\) from his father (who was \(ii\)), so his genotype is \(I^A i\). Similarly, the woman is blood group B and must have inherited \(i\) from her mother, making her genotype \(I^B i\). A cross of \(I^A i \times I^B i\) yields a 1 in 4 (25%) chance of producing a child with genotype \(ii\) (blood group O).

Award [1] for the correct answer (B).
- Award [0] for any other option.

Epinephrine is a hydrophilic hormone that cannot cross the plasma membrane. It binds to an extracellular G-protein coupled receptor (GPCR) on the target cell membrane. This activates a G-protein, which in turn activates the enzyme adenylyl cyclase. Adenylyl cyclase catalyzes the synthesis of the second messenger cyclic AMP (cAMP) from ATP.

Award [1] for the correct answer (B).
- Award [0] for any other option.

ADH functions by stimulating the insertion of aquaporin channels into the apical membranes of collecting duct epithelial cells. With abnormally low ADH levels, fewer aquaporins are inserted (lower concentration), reducing water reabsorption. Consequently, a large volume of highly dilute urine is excreted.

Award 1 mark for the correct option (B).
- Reject A, C, D as they incorrect identify the aquaporin concentration change or the resulting urine concentration.

The sodium-potassium pump is an active transport mechanism that uses the energy from the hydrolysis of one ATP molecule to pump 3 sodium ions (\(\text{Na}^+\)) out of the cell against their concentration gradient, and 2 potassium ions (\(\text{K}^+\)) into the cell against their concentration gradient.

Award 1 mark for the correct option (A).
- Reject B because the directions of ion transport are reversed.
- Reject C and D because the pump requires active transport (ATP hydrolysis) rather than passive transport.

Non-competitive inhibitors bind to an allosteric site (a site other than the active site) of the enzyme. This binding changes the conformation of the active site, reducing the overall rate of product formation (lower \(V_{\max}\)) regardless of the amount of substrate present. Since the substrate can still bind to the active site, the affinity of the enzyme for the substrate (represented by \(K_m\)) remains unchanged.

Award 1 mark for the correct option (D).
- Reject A and B because competitive inhibitors increase \(K_m\) and do not alter \(V_{\max}\).
- Reject C because non-competitive inhibitors do not bind at the active site.

Active loading of sucrose into the phloem sieve tubes increases the solute concentration inside the sieve tube elements (decreasing the water potential). This causes water to flow from the high water potential of adjacent xylem vessels into the sieve tubes by osmosis, thereby generating high hydrostatic pressure.

Award 1 mark for the correct option (B).
- Reject A because increased solute concentration, not decreased, leads to osmosis of water.
- Reject C because water is not actively pumped directly.
- Reject D because water moves into, not out of, the phloem sieve tube elements.

Dissolved \(\text{CO}_2\) reacts with water to form carbonic acid, which dissociates into hydrogen ions (\(\text{H}^+\)) and bicarbonate ions (\(\text{HCO}_3^-\)). The excess hydrogen ions bind with carbonate ions (\(\text{CO}_3^{2-}\)) to form more bicarbonate. This reduces the concentration of free carbonate ions, making it harder for corals to absorb them and secrete the calcium carbonate (\(\text{CaCO}_3\)) skeleton.

Award 1 mark for the correct option (B).
- Reject A because the availability of carbonate ions decreases, not increases.
- Reject C because the hydrogen ion concentration increases (acidification), not decreases.
- Reject D because increased ocean acidification hinders rather than increases the precipitation of calcium carbonate.

A sudden surge in luteinizing hormone (LH) around day 14 of the menstrual cycle triggers ovulation (the release of the mature oocyte from the Graafian follicle) and stimulates the remaining follicular cells to develop into the corpus luteum.

Award 1 mark for the correct option (C).
- Reject A because the degeneration of the corpus luteum is caused by a drop in LH and hCG, not a surge.
- Reject B because FSH primarily stimulates the growth and maturation of primary follicles.
- Reject D because progesterone is secreted by the corpus luteum in the ovary, not the pituitary gland.

Mitochondria and chloroplasts share key characteristics with prokaryotic cells, which supports the theory that they evolved from once-free-living bacteria. Specifically, they contain their own 70S ribosomes (unlike eukaryotic 80S cytoplasmic ribosomes) and single circular DNA molecules (similar to bacterial chromosomes).

Award 1 mark for the correct option (C).
- Reject A because eukaryotes contain 80S ribosomes and linear DNA, while mitochondria/chloroplasts do not.
- Reject B because they are surrounded by a double membrane (outer from vesicle, inner from bacterium) and divide by binary-like fission, not mitosis.
- Reject D because although they can synthesize some of their own proteins, they still rely on nuclear genes for many vital proteins.

Steroid hormones are lipophilic (lipid-soluble) because they are derived from lipids (cholesterol), allowing them to pass directly through the hydrophobic core of the cell membrane's phospholipid bilayer. In contrast, peptide hormones are polar/hydrophilic proteins and cannot freely diffuse across the hydrophobic cell membrane, requiring them to bind to cell-surface receptors.

Award 1 mark for the correct option (B).
- Reject A because steroid hormones are hydrophobic, not hydrophilic.
- Reject C because molecular size is not the primary factor; charge/solubility is. Moreover, steroid hormones cross by passive diffusion, not endocytosis.
- Reject D because it does not explain the physical-chemical properties preventing/allowing membrane crossing.

When blood solute concentration increases (dehydration or hypertonic plasma), water leaves the osmoreceptor cells in the hypothalamus by osmosis. This causes the osmoreceptor cells to shrink. The shrinkage of these cells triggers nerve impulses to the posterior pituitary gland, stimulating the release of antidiuretic hormone (ADH) into the blood to increase water reabsorption in the collecting ducts of the kidneys.

[1 mark] awarded for the correct option (B).
- Reject other options because they state incorrect anatomical locations, incorrect receptor responses (swelling), or incorrect hormone release.

At the source, sucrose is actively loaded into the sieve tube elements. This high solute concentration draws water from the adjacent xylem into the phloem via osmosis. The influx of water increases the hydrostatic pressure at the source. At the sink, sucrose is unloaded, water leaves the sieve tube, and hydrostatic pressure decreases, generating a mass flow gradient.

[1 mark] awarded for the correct option (A).
- Reject options describing passive water movement as driving the loading, or using transpiration pull (which drives xylem transport, not phloem mass flow).

A non-competitive inhibitor binds to an allosteric site of the enzyme, changing the conformation of the active site so that substrates can no longer bind or be converted to products. This effectively reduces the concentration of active enzyme molecules, decreasing the maximum velocity (\(V_{max}\)). However, because the remaining uninhibited enzymes still have the same affinity for the substrate, the substrate concentration required to reach half of the new maximum rate (apparent \(K_m\)) remains unchanged.

[1 mark] awarded for the correct option (A).
- Reject options where \(V_{max}\) remains unchanged (which describes competitive inhibition) or where \(K_m\) changes.

The sodium-potassium pump moves three sodium ions (\(\text{Na}^+\)) out of the cell and two potassium ions (\(\text{K}^+\)) into the cell per cycle against their concentration gradients. This is an active transport process that requires the hydrolysis of ATP, which phosphorylates the pump to trigger a conformational change.

[1 mark] awarded for the correct option (A).
- Reject options with incorrect ion numbers (e.g., 2 Na+ out, 3 K+ in) or incorrect phosphorylation descriptions.

Starting from day 1 (menstruation), FSH peaks first to stimulate follicle recruitment. The growing follicle secretes estrogen, which peaks shortly before ovulation. High estrogen levels trigger a rapid surge in LH, which induces ovulation. Following ovulation, the corpus luteum secretes progesterone, causing it to peak during the luteal phase.

[1 mark] awarded for the correct option (B).
- Reject options that suggest progesterone peaks before ovulation, or that estrogen peaks before FSH.

During active (forced) exhalation, the external intercostal muscles relax, and the internal intercostal muscles contract antagonistically to pull the rib cage down and in. Simultaneously, the diaphragm relaxes and domes upwards, which decreases the volume of the thoracic cavity and increases air pressure, forcing air out of the lungs.

[1 mark] awarded for the correct option (B).
- Reject options that claim the external intercostals contract during exhalation or that the diaphragm flattens during exhalation (which occurs during inhalation).

Water has a high specific heat capacity, meaning it requires a relatively large amount of heat energy to raise its temperature. This is because many hydrogen bonds between water molecules must absorb energy and break before the molecules can move faster (increasing temperature). This property helps aquatic ecosystems and multicellular organisms maintain a relatively stable temperature.

[1 mark] awarded for the correct option (B).
- Reject options mentioning covalent bonds within water molecules as the basis for thermal buffering, or referring to vaporization (which relates to cooling mechanisms rather than temperature stability of the bulk water).

Since the genes are linked on the same autosome and in coupling phase (\(AB / ab\)), the parental gametes produced will be \(AB\) and \(ab\). When crossed with a homozygous recessive tester (\(ab / ab\)), the offspring will display a significantly higher proportion of parental phenotypes (\(AB\) and \(ab\)). Recombinant phenotypes (\(Ab\) and \(aB\)) will only appear due to crossing over during prophase I of meiosis, resulting in a low frequency of these phenotypes.

[1 mark] awarded for the correct option (C).
- Reject A because a 1:1:1:1 ratio is only expected with independent assortment (unlinked genes).
- Reject B because 9:3:3:1 is a dihybrid self-cross ratio, not a test cross.
- Reject D because crossing over still occurs (even at a low frequency), meaning some recombinants will be present.

(a) Percentage increase = \( \frac{1.8 - 1.2}{1.2} \times 100\% = 50\% \). (b) High relative humidity means the air is highly saturated with water vapor, reducing the concentration gradient of water vapor between the skin and the surrounding air. This significantly slows down the rate of sweat evaporation. Since the evaporation of sweat (latent heat of vaporization) is the primary physiological mechanism for heat loss during exercise, less metabolic heat is dissipated, leading to greater heat retention and a steeper rise in core body temperature. (c) Thermoreceptors in the skin and the hypothalamus detect the increase in blood and body temperature. The hypothalamus acts as the control center and integrates this information, sending electrical impulses via autonomic motor neurons to effectors. It triggers the vasodilation of skin arterioles to increase radiative and convective heat loss, and stimulates sweat glands to secrete sweat to lower body temperature.

(a) Award [1] for showing correct working (e.g., \( 0.6 / 1.2 \) or similar fraction) and [1] for the correct final percentage value of 50%. (b) Award [1] for explaining that high humidity reduces the sweat evaporation rate due to a lower concentration/vapor gradient, [1] for identifying that sweat evaporation is the main cooling/heat dissipation mechanism, and [1] for concluding that metabolic heat is retained, causing a higher core temperature. (c) Award [1] for detection of temperature change by thermoreceptors/hypothalamus, [1] for transmission of signals via autonomic nerves/pathways to effectors, and [1] for correct description of an effector response (vasodilation of arterioles or activation of sweat glands).

(a) Difference = \( 4.2 - 1.1 = 3.1\text{ mmol m}^{-2}\text{ s}^{-1} \). (b) Similarities: Both the wild-type and the phot1 mutant show an increase in transpiration rate when moved from complete darkness to blue light. Differences: The increase in transpiration rate is much greater in wild-type plants (an increase of \( 3.4\text{ mmol m}^{-2}\text{ s}^{-1} \)) compared to the phot1 mutant (an increase of only \( 0.4\text{ mmol m}^{-2}\text{ s}^{-1} \)). The mutant maintains a very low rate under blue light, almost identical to its baseline dark level, whereas the wild-type responds robustly. (c) Blue light is detected by phototropin-1 photoreceptors in guard cells. This activation stimulates proton pumps (\( \text{H}^+\text{-ATPase} \)) in the guard cell plasma membrane to actively pump hydrogen ions out of the cytoplasm. The resulting hyperpolarization/electrochemical gradient drives the inward movement of potassium ions (\( \text{K}^+ \)) through voltage-gated ion channels. The accumulation of potassium ions and associated anions lowers the solute/water potential inside the guard cells, causing water to enter by osmosis. This increases turgor pressure, causing the guard cells to swell and bow outward, opening the stomatal pore.

(a) Award [1] for 3.1 (accept with or without units; reject any other values). (b) Award [1] for similarity (both increase), and up to [2] for differences (such as quantifying the rate differences or stating that the wild-type has a significantly larger magnitude of response compared to the mutant). (c) Award [1] for blue light detection by phototropins, [1] for activation of proton pumps moving H+ out of the guard cell, [1] for inward movement of potassium ions (K+), and [1] for water entry by osmosis lowering water potential, increasing cell turgor to open the stomata.

(a) There is a positive, non-linear relationship between sea surface temperature anomaly and coral bleaching. From Week 0 to Week 4, when the anomaly is small (up to \( +0.5^\circ\text{C} \)), bleaching remains very low (2% to 5%). However, once the anomaly exceeds \( +1.0^\circ\text{C} \) (Weeks 8 to 12), the rate of coral bleaching accelerates dramatically/exponentially to reach 76%. (b) Rate of increase = \( \frac{\text{Bleaching at Week 12} - \text{Bleaching at Week 4}}{12 - 4\text{ weeks}} = \frac{76.0\% - 5.0\%}{8} = \frac{71.0\%}{8} = 8.875\% \) per week (accept 8.9% or 8.88% per week). (c) Corals exist in a mutualistic symbiotic relationship with photosynthetic zooxanthellae (algae) residing inside their tissues. Elevated ocean temperatures cause physiological stress to these algae, leading to the hyperproduction of toxic reactive oxygen species. In response, the coral host expels the algae, losing its pigment and primary energy source (bleaching). Rising greenhouse gases (like \( \text{CO}_2 \)) in the atmosphere trap outgoing longwave radiation, driving global warming and increasing sea surface temperatures, leading to prolonged marine heatwaves that trigger these bleaching events.

(a) Award [1] for describing a positive correlation/relationship and [1] for identifying that it is non-linear or that bleaching increases rapidly/exponentially above a threshold anomaly (approx. 1°C). (b) Award [1] for correct substitution in the rate formula (\( (76 - 5) / (12 - 4) \)) and [1] for the correct final value of 8.875% per week (or 8.9% per week) with correct units. (c) Award [1] for explaining the symbiotic/mutualistic relationship with zooxanthellae, [1] for noting that thermal stress causes algae to become damaged/expelled, [1] for explaining that atmospheric greenhouse gases trap solar heat/infrared radiation, and [1] for linking this global warming directly to elevated sea surface temperatures.

(a) Percentage inhibition = \( \frac{\text{Rate without} - \text{Rate with}}{\text{Rate without}} \times 100 = \frac{50 - 15}{50} \times 100 = \frac{35}{50} \times 100 = 70\% \). (b) Inhibitor X acts as a competitive inhibitor, meaning it has a complementary structure to lactose and binds reversibly to the active site of lactase. At low substrate concentrations (5 mM), there is less competition, allowing the inhibitor to bind to a large proportion of the active sites, severely reducing the rate of reaction. At very high substrate concentrations (100 mM), the substrate molecules vastly outnumber the inhibitor molecules, making it highly probable that a substrate molecule will bind to the active site rather than an inhibitor molecule, allowing the reaction rate to reach near maximum capacity (Vmax). (c) Factors that must be controlled are: 1. Temperature (since kinetic energy affects collision rates and denaturation), 2. pH (as extreme changes alter active site charges and conformation), 3. Enzyme (lactase) concentration/volume (to keep the number of active sites constant across trials).

(a) Award [1] for showing correct working (e.g., \( 35 / 50 \)) and [1] for the correct final value of 70%. (b) Award [1] for identifying that competitive inhibitors have a similar shape to the substrate and bind to the active site, [1] for explaining that low substrate concentration fails to outcompete the inhibitor, and [1] for explaining that high substrate concentration outcompetes the inhibitor, saturating active sites and restoring Vmax. (c) Award [1] for each correct factor listed up to a maximum of [3]: Temperature, pH, and Enzyme concentration (or enzyme volume/source). Do not accept 'substrate concentration' or 'inhibitor concentration'.

(a) Carrier-mediated transport / facilitated diffusion / active cotransport. Reason: The uptake rate plateaus because transport proteins (such as SGLT1 or GLUT2) become fully saturated when all substrate-binding sites are occupied. (b) At 50 mM, glucose transport is operating at its maximum rate (Vmax = 85 nmol min⁻¹ mg⁻¹) because all glucose transporter proteins are fully saturated. Conversely, fructose transport (mediated by different proteins like GLUT5) has not reached saturation at 50 mM (it is far below its Vmax), or is occurring via a higher capacity mechanism/diffusion, so its rate (40 nmol min⁻¹ mg⁻¹) continues to rise linearly. (c) The sodium-potassium pump (\( \text{Na}^+/\text{K}^+\text{ ATPase} \)) on the basolateral membrane of the enterocytes uses ATP to actively pump \( \text{Na}^+ \) ions out of the cytoplasm and into the blood. This maintains an extremely low intracellular concentration of \( \text{Na}^+ \). This low internal concentration establishes a steep concentration/electrochemical gradient for \( \text{Na}^+ \) between the intestinal lumen and the cell interior. Sodium-glucose cotransporter proteins (SGLT1) on the apical membrane then use this downward sodium gradient to actively pull glucose into the cell against its own concentration gradient (secondary active transport).

(a) Award [1] for suggesting carrier-mediated transport/facilitated diffusion/active cotransport, and [1] for explaining that the plateau is due to the saturation of carrier proteins. (b) Award [1] for stating that glucose transport proteins are fully saturated / at Vmax, and [1] for explaining that fructose transporters are not yet saturated / have a higher Km / capacity at 50 mM. (c) Award [1] for stating that the basolateral Na+/K+ pump uses active transport/ATP to pump Na+ out, [1] for explaining that this maintains a low intracellular Na+ concentration, [1] for noting that this creates a steep Na+ gradient across the apical membrane, and [1] for explaining that SGLT1 cotransports glucose inward alongside Na+ using this gradient.

Part (a):
Negative feedback is a fundamental homeostatic control mechanism. When a change in a physiological variable (such as blood glucose) occurs, the system responds by reversing the direction of that change to return the variable back to its normal range or set point. Receptors detect a deviation from the set point, a coordinator processes the signal, and effectors carry out a corrective response. In blood glucose regulation, if concentration rises above the norm (e.g., after eating), negative feedback triggers insulin release to lower it. If concentration falls below the norm (e.g., during exercise or fasting), negative feedback triggers glucagon release to raise it.

Part (b):
When blood glucose levels rise above the set point, sensory mechanisms detect this change:
1. High blood glucose levels are detected by the beta (\(\beta\)) cells located in the islets of Langerhans within the pancreas.
2. This detection stimulates the beta cells to secrete the hormone insulin into the bloodstream.
3. Insulin travels via the circulatory system to target tissues, primarily the liver and skeletal muscle cells.
4. Insulin binds to specific protein receptors on the cell surface membrane of these target cells.
5. This binding triggers the mobilization of glucose transporter proteins (GLUT4) to the plasma membrane, increasing the rate of facilitated diffusion of glucose into the cells.
6. Within liver cells (hepatocytes), insulin activates enzymes that convert glucose into insoluble glycogen (a process called glycogenesis), which is stored.
7. This uptake and conversion remove glucose from the blood, returning blood glucose levels back down to the normal set point.

Part (c):
Type I and Type II diabetes are both characterized by an inability to regulate blood glucose effectively, but they differ significantly:
1. Cause: Type I is an autoimmune condition where the body's immune system mistakenly destroys the insulin-producing beta cells of the pancreas, leading to little or no insulin production. Type II is characterized by insulin resistance, where target cells (such as liver and muscle cells) fail to respond appropriately to normal or even elevated levels of insulin because their insulin receptors are unresponsive.
2. Onset and Risk Factors: Type I usually develops during childhood or adolescence and is primarily linked to genetic predisposition combined with an environmental trigger (such as a viral infection). Type II typically develops in adulthood (though increasingly in younger people) and is strongly associated with lifestyle factors, including obesity, physical inactivity, poor diet, and genetic factors.
3. Treatment: Type I must be managed with daily external insulin administration (via injections or an insulin pump) combined with regular blood glucose monitoring. Type II is primarily managed through lifestyle interventions such as dietary modifications (reducing simple carbohydrate intake), weight loss, and regular physical exercise. It may also be treated with oral medications that improve insulin sensitivity (e.g., metformin) or, in later stages, insulin therapy.

Part (a) [Max 4 marks]:
- Award [1] for defining negative feedback as a mechanism that reverses any deviation from a set point to restore homeostatic balance.
- Award [1] for stating that receptors detect a change and trigger a response that counteracts the stimulus.
- Award [1] for applying to glucose: state that an increase in blood glucose triggers mechanisms to lower it, and a decrease triggers mechanisms to raise it.
- Award [1] for mentioning the roles of opposing hormones (insulin and glucagon) in maintaining this balance.

Part (b) [Max 5 marks]:
- Award [1] for stating that high blood glucose is detected by beta (\(\beta\)) cells in the islets of Langerhans / pancreas.
- Award [1] for stating that pancreatic beta cells secrete insulin into the bloodstream.
- Award [1] for stating that insulin binds to receptors on the membrane of target cells (such as hepatocytes/liver cells or muscle cells).
- Award [1] for stating that insulin increases glucose uptake by target cells (via glucose transporters/GLUT4).
- Award [1] for stating that insulin stimulates glycogenesis / the conversion of glucose into glycogen for storage in the liver/muscle cells.
- Award [1] for explaining that this overall process reduces the concentration of glucose in the blood back to the set point.

Part (c) [Max 6 marks]:
- Award [1] for stating that Type I is caused by autoimmune destruction of beta (\(\beta\)) cells, leading to a lack of insulin production.
- Award [1] for stating that Type II is caused by insulin resistance / decreased sensitivity of target cell receptors to insulin.
- Award [1] for stating that Type I usually develops in childhood/youth, whereas Type II usually develops in adulthood.
- Award [1] for stating that Type II is strongly linked to lifestyle risk factors such as obesity, physical inactivity, and diets high in sugars/fats, whereas Type I is linked to genetics/environmental triggers.
- Award [1] for stating that Type I requires daily insulin therapy (injections/pumps) and regular blood glucose monitoring.
- Award [1] for stating that Type II is managed primarily through lifestyle adjustments (diet, exercise, weight loss) and may use oral drugs or insulin as needed.

Quality of Communication [1 mark]:
- Award [1] if the response is well-structured, exhibits logical flow between parts, and uses precise biological vocabulary (e.g., distinguishing glucagon from glycogen; correctly referencing beta cells and receptor sensitivity) consistently.

Part (a):
Enzymes are globular proteins that act as biological catalysts. They increase the rate of chemical reactions by lowering the activation energy (\(E_a\)) required for the reaction to proceed. Substrate molecules bind to a specific region on the enzyme known as the active site, forming an enzyme-substrate complex. According to the induced fit model, when the substrate binds, the enzyme's active site undergoes a conformational change to fit the substrate more tightly. This binding stresses and destabilizes chemical bonds in the substrate, reducing the energy threshold needed to convert reactants into products.

Part (b):
Competitive and non-competitive inhibitors differ in how and where they interact with an enzyme:
1. Binding site: Competitive inhibitors have a structure similar to the substrate and bind directly to the active site. Non-competitive inhibitors have a structure different from the substrate and bind to an alternative site on the enzyme known as the allosteric site.
2. Mechanism: Competitive inhibitors compete directly with the substrate for the active site, physically blocking substrate binding. Non-competitive inhibitors, when bound to the allosteric site, induce a conformational change in the enzyme's structure, modifying the shape of the active site so that the substrate can no longer bind or the reaction cannot be catalyzed.
3. Effect of substrate concentration: The effect of a competitive inhibitor can be overcome by increasing the substrate concentration, as the substrate is more likely to outcompete the inhibitor for the active site; the maximum rate of reaction (\(V_{\max}\)) can still be reached. In contrast, increasing the substrate concentration has no effect on non-competitive inhibition, because the substrate and inhibitor do not compete for the same site; the maximum rate of reaction (\(V_{\max}\)) is significantly reduced.
4. Examples: An example of a competitive inhibitor is malonate (which competes with succinate for succinate dehydrogenase). An example of a non-competitive inhibitor is cyanide (which binds to the allosteric site of cytochrome oxidase).

Part (c):
Metabolic pathways consist of chains or cycles of enzyme-catalyzed steps. End-product inhibition is a form of negative feedback control where the final product of a pathway inhibits an enzyme that catalyzes an early step in that same pathway.
1. When the concentration of the end-product is high, it binds to the allosteric site of the first enzyme in the pathway.
2. This binding causes a conformational change in the enzyme's active site, temporarily inactivating it.
3. Consequently, the production of intermediate metabolites stops, halting the entire pathway and preventing the wasteful overproduction of the end-product.
4. When the cell consumes the end-product and its concentration drops, the product molecules detach from the allosteric sites of the enzymes, restoring their activity and allowing the pathway to resume.
5. A classic example of this is the pathway that converts the amino acid threonine into the amino acid isoleucine in five steps. The first enzyme in this pathway is threonine deaminase. When isoleucine (the end-product) accumulates, it binds to the allosteric site of threonine deaminase, inhibiting the enzyme and shutting down its own synthesis until levels fall.

Part (a) [Max 3 marks]:
- Award [1] for explaining that enzymes lower the activation energy (\(E_a\)) of a reaction.
- Award [1] for stating that the substrate binds to the specific active site on the enzyme to form an enzyme-substrate complex.
- Award [1] for referencing the induced fit model, explaining that conformational changes strain/weaken bonds in the substrate to facilitate the reaction.

Part (b) [Max 6 marks]:
- Award [1] for stating that competitive inhibitors are structurally similar to the substrate, whereas non-competitive inhibitors are structurally different.
- Award [1] for stating that competitive inhibitors bind to the active site, whereas non-competitive inhibitors bind to the allosteric site.
- Award [1] for stating that competitive inhibitors block the active site directly, whereas non-competitive inhibitors change the shape of the active site when they bind.
- Award [1] for stating that the effect of competitive inhibitors can be overcome by increasing substrate concentration, whereas the effect of non-competitive inhibitors cannot be overcome.
- Award [1] for stating that competitive inhibitors allow \(V_{\max}\) to be reached (at high substrate concentrations), whereas non-competitive inhibitors reduce \(V_{\max}\).
- Award [1] for providing a valid example of competitive inhibition (e.g., malonate inhibiting succinate dehydrogenase) and/or non-competitive inhibition (e.g., cyanide inhibiting cytochrome oxidase).

Part (c) [Max 6 marks]:
- Award [1] for defining a metabolic pathway as a chain/cycle of enzyme-controlled steps.
- Award [1] for stating that the end-product of the pathway acts as an inhibitor of the first (or an early) enzyme in the pathway.
- Award [1] for explaining that this is a form of negative feedback control.
- Award [1] for explaining that the end-product binds to an allosteric site, altering the active site's shape and halting the pathway when product concentration is high.
- Award [1] for explaining that when product levels fall, the inhibitor dissociates, restarting the pathway.
- Award [1] for describing a specific example: the pathway converting threonine to isoleucine, where isoleucine inhibits threonine deaminase.

Quality of Communication [1 mark]:
- Award [1] if the candidate's response is structured logically, uses appropriate terminology (such as active site, allosteric site, conformative/conformational change, negative feedback, threonine/isoleucine) correctly, and presents ideas clearly in continuous prose.

1. The independent variable is the parameter manipulated by the investigator, which is the substrate (p-nitrophenyl phosphate) concentration. 2. Competitive inhibitors bind to the active site and can be outcompeted by high concentrations of substrate. Therefore, the maximum rate of reaction (\(V_{\max}\)) is unaffected, but a higher concentration of substrate is required to reach half of this maximum rate (increased \(K_m\)). 3. At very high substrate concentrations, every enzyme molecule has its active site occupied by a substrate molecule. The enzyme is working at its maximum capacity (saturated), meaning any further increase in substrate concentration has no effect on the rate.

1. Substrate concentration / concentration of p-nitrophenyl phosphate [1 mark]. 2. Competitive inhibition [1 mark] AND justification: the maximum rate (\(V_{\max}\)) remains unchanged OR the substrate concentration needed to reach half-maximum velocity (\(K_m\)) increases / more substrate is needed to overcome inhibition [1 mark]. 3. Saturated active sites / all active sites occupied [1 mark] AND enzyme concentration becomes the limiting factor [1 mark].

1. The darkness control ensures that any color change observed is due to the light-dependent reactions of photosynthesis and not an independent chemical breakdown of DCPIP or a dark reaction. 2. From 0 to 1500 lux, there is a positive correlation where light intensity is the limiting factor; increasing light intensity directly increases the rate of reaction. 3. Photoactivation of photosystem II by light absorption leads to the excitation of electrons. These excited electrons are passed along an electron transport chain. In this cell-free Hill reaction setup, DCPIP acts as an artificial electron acceptor, intercepting the electrons and becoming reduced (changing color from blue to colorless).

1. To serve as a control / to prove that the reduction of DCPIP is dependent on light [1 mark]. 2. Rate of reduction increases as light intensity increases / positive correlation [1 mark] (Do not accept just 'it increases' without referencing the independent variable relationship). 3. Any three from: Light energy/photons excite electrons in chlorophyll / photosystem II [1 mark]; Photolysis of water splits water to supply replacement electrons [1 mark]; Electrons flow through an electron transport chain [1 mark]; DCPIP accepts these electrons and is reduced (changing from blue to colorless) in place of NADP+ [1 mark] (Maximum 3 marks).

1. In osmosis experiments, the surface area to volume ratio affects the rate of water diffusion, and the source tissue must have identical initial osmolarity. 2. Percentage change standardizes the results because it is virtually impossible to cut potato cylinders with exactly the same starting mass. 3. An isotonic solution has the same solute concentration as the cytoplasm, meaning there is no net movement of water by osmosis. This corresponds to a 0% change in mass. Looking at the data, the mass change crosses 0% between the \(0.2\text{ mol dm}^{-3}\) solution (\(+7\%\)) and the \(0.4\text{ mol dm}^{-3}\) solution (\(-2\%\)), which suggests an isotonic point of roughly \(0.35\text{ mol dm}^{-3}\).

1. Award [1 mark] for each of any two controlled variables: surface area-to-volume ratio / dimensions / length / diameter of cylinders; source/variety/age of potato tissue; temperature of the solutions; duration of immersion (60 minutes); volume of sucrose solution used. 2. To account for differences in the initial mass of the potato cylinders [1 mark]. 3. A concentration between \(0.30\text{ mol dm}^{-3}\) and \(0.40\text{ mol dm}^{-3}\) (accept \(0.35\text{ mol dm}^{-3}\) or \(0.36\text{ mol dm}^{-3}\)) [1 mark]; This is the concentration where there is no net osmosis / no mass change / 0% change in mass [1 mark].

(a) The sight or smell of food triggers nervous impulses in the brain (specifically the medulla oblongata). These impulses travel via the vagus nerve (parasympathetic system) directly to the stomach, stimulating the gastric glands to secrete gastric juice. (b) When food enters the stomach, physical distension of the stomach wall is detected by stretch receptors. This stimulates the G-cells in the stomach mucosa to release the hormone gastrin into the bloodstream. Gastrin travels via circulation back to the gastric glands, where it stimulates a sustained release of hydrochloric acid and pepsinogen. (c) Pepsinogen (or pepsin) and mucus (or intrinsic factor).

(a) Max 2 marks: Award 1 mark for mentioning the involvement of the brain / medulla oblongata / sensory inputs. Award 1 mark for identifying that signals travel via the vagus nerve to the stomach to initiate secretion. (b) Max 2 marks: Award 1 mark for stating that physical stretch / distension triggers gastrin release. Award 1 mark for explaining that gastrin travels in blood to stimulate gastric glands / acid production. (c) Max 2 marks: Award 1 mark for each correct component named: pepsinogen/pepsin (1), mucus (1), intrinsic factor (1).

(a) The arrival of an action potential depolarizes the pre-synaptic membrane, which triggers the opening of voltage-gated calcium channels. Calcium ions (\(\text{Ca}^{2+}\)) rapidly diffuse into the pre-synaptic neuron. This influx of calcium causes synaptic vesicles filled with neurotransmitter molecules to move toward and fuse with the pre-synaptic membrane, releasing the neurotransmitters into the synaptic cleft via exocytosis. (b) Excitatory neurotransmitters open sodium channels on the post-synaptic membrane, causing an influx of positive sodium ions that depolarizes the membrane (creating an excitatory post-synaptic potential or EPSP) and moving it closer to the threshold required to fire an action potential. Inhibitory neurotransmitters open chloride or potassium channels, causing chloride influx or potassium efflux, which hyperpolarizes the membrane (creating an inhibitory post-synaptic potential or IPSP) and moves it further from the threshold, making an action potential less likely to fire.

(a) Max 3 marks: Award 1 mark for depolarization opening voltage-gated calcium channels. Award 1 mark for calcium ion influx into the pre-synaptic neuron. Award 1 mark for vesicle fusion and exocytosis of neurotransmitters. (b) Max 3 marks: Award 1 mark for excitatory causing depolarization / EPSP / opening sodium channels. Award 1 mark for inhibitory causing hyperpolarization / IPSP / opening chloride or potassium channels. Award 1 mark for contrasting the likelihood of generating an action potential (excitatory increases, inhibitory decreases).

(a) In situ conservation involves the preservation of ecosystems and natural habitats and the maintenance and recovery of viable populations of species in their natural surroundings (e.g., national parks, nature reserves, or marine sanctuaries). Ex situ conservation involves the preservation of components of biological diversity outside their natural habitats (e.g., zoos, botanical gardens, seed banks, or captive breeding facilities). (b) Advantages of in situ conservation include: 1. It allows species to continue evolving in response to natural selective pressures in their native environment. 2. It preserves the broader ecological interactions, protecting other species and the habitat as a whole. 3. It is generally more cost-effective for maintaining large populations and avoids problems of captive behavior, loss of wild survival skills, or inbreeding depression.

(a) Max 3 marks: Award 1 mark for defining in situ as conservation in natural habitats with a correct example. Award 1 mark for defining ex situ as conservation outside natural habitats with a correct example. Award 1 mark for a clear comparative distinction. (b) Max 3 marks: Award 1 mark for each distinct advantage discussed, up to a maximum of 3 marks (e.g., permits natural evolution, maintains ecological relationships, preserves habitat, avoids captive breeding issues like behavioral anomalies or inbreeding, more cost-effective).

(a) The guide RNA (gRNA) is engineered to contain a sequence of nucleotides that is complementary to the specific target DNA sequence in the genome, allowing it to locate and bind to the target site via complementary base pairing. The Cas9 protein is an endonuclease enzyme that acts as a pair of molecular scissors; guided by the gRNA, it binds to the DNA target and introduces a double-stranded break at that precise location. (b) To knock out a gene, the CRISPR-Cas9 system is designed to cut within the coding sequence of the target gene. After Cas9 creates the double-stranded break, the cell's natural DNA repair mechanism (non-homologous end joining) attempts to repair the cut. Because this repair pathway is highly error-prone, it frequently introduces insertions or deletions (indels) at the cut site. These mutations disrupt the reading frame (causing frameshift mutations) or introduce premature stop codons, preventing the synthesis of a functional protein, thereby knocking out the gene's function.

(a) Max 3 marks: Award 1 mark for gRNA guiding the complex via complementary base pairing to the target DNA. Award 1 mark for Cas9 functioning as an endonuclease enzyme. Award 1 mark for Cas9 making a double-stranded break at the target site. (b) Max 3 marks: Award 1 mark for targeting a cut within the coding region of the target gene. Award 1 mark for explaining that the cell's error-prone repair mechanism (non-homologous end joining) repairs the break. Award 1 mark for stating that insertions/deletions (indels) create frameshifts/premature stop codons, leading to a non-functional protein (knockout).

(a) During intense muscular exercise, active tissues consume oxygen rapidly and produce large amounts of carbon dioxide and lactic acid. This raises partial pressure of carbon dioxide and lowers local pH (increasing hydrogen ion concentration). The Bohr shift describes how these conditions decrease hemoglobin's affinity for oxygen, causing the oxygen dissociation curve to shift to the right. This is physiologically significant because it ensures that hemoglobin releases (unloads) more oxygen to the highly active, respiring muscle tissues where oxygen is urgently needed. (b) Carbon dioxide is transported in three main ways: 1. About 7-10% is dissolved directly in the blood plasma. 2. About 20-23% binds to the amino groups of hemoglobin to form carbaminohemoglobin. 3. The majority (about 70%) is transported as hydrogencarbonate ions (\(\text{HCO}_3^-\)). Carbon dioxide diffuses into red blood cells where the enzyme carbonic anhydrase catalyzes its reaction with water to form carbonic acid, which then dissociates into hydrogen ions and hydrogencarbonate ions. The hydrogencarbonate ions then diffuse out into the plasma.

(a) Max 3 marks: Award 1 mark for stating that exercise increases carbon dioxide/lactic acid levels, lowering local pH. Award 1 mark for explaining that lowered pH/increased carbon dioxide decreases hemoglobin's affinity for oxygen (shifting the curve to the right). Award 1 mark for linking this to increased oxygen unloading/release to respiring tissues. (b) Max 3 marks: Award 1 mark for mentioning transport dissolved in plasma AND bound to hemoglobin (carbaminohemoglobin). Award 1 mark for identifying that the majority is transported as hydrogencarbonate ions (\(\text{HCO}_3^-\)). Award 1 mark for explaining the role of carbonic anhydrase in converting carbon dioxide and water to carbonic acid which dissociates into hydrogen ions and hydrogencarbonate ions.