2025 HKDSE Biology Practice Paper with Answers

(a) The potometer measures water uptake by the leafy shoot. It is assumed that the rate of water uptake is approximately equal to the rate of transpiration, as only a very small fraction (usually <1-2%) of absorbed water is retained for metabolic processes (e.g., photosynthesis) or maintaining turgidity.

(b) In still air, water vapor accumulates immediately outside the stomatal pores, forming a humid boundary layer. Increased air movement sweeps away this humid layer, reducing the ambient relative humidity right at the leaf surface. This steepens the concentration gradient of water vapor between the moist intercellular air spaces inside the leaf and the external air, accelerating the diffusion of water vapor (transpiration) and hence water absorption.

(c) Stomatal opening is an active process: guard cells utilize ATP to actively pump ions (notably \(\text{K}^+\)) into their cytoplasm/vacuoles, lowering their water potential below that of adjacent epidermal cells. Water then enters guard cells down the water potential gradient by osmosis, generating turgor pressure that bends the cells open. Blocking ATP synthesis stops active solute accumulation, preventing guard cell turgidity and leading to stomatal closure, which drastically decreases transpiration.

(a) Water uptake by the shoot is equal to / directly proportional to the rate of water loss via transpiration (or: water consumed in metabolism / photosynthesis is negligible). (1 mark)

(b) Wind / air movement removes the humid boundary layer / water vapor around the leaf surface. (1 mark)
This maintains / steepens the water vapor concentration gradient between the internal leaf air space and the outside atmosphere (leading to faster diffusion/transpiration). (1 mark)

(c) Active accumulation / transport of ions (such as \(\text{K}^+\)) into guard cells requires ATP / energy. (1 mark)
Without ATP, guard cells cannot lower their water potential to draw in water by osmosis / cannot become turgid, leading to stomatal closure (thus reducing transpiration). (1 mark)

(a) The epithelial cells possess microvilli forming a brush border, which massively increases the surface area for diffusion and active transport. Alternatively, having numerous transport/carrier proteins or abundant mitochondria for supplying ATP for active transport is also valid.

(b) Amino acids are absorbed into epithelial cells and exit across the basolateral membrane into the rich network of blood capillaries inside the villus, proceeding directly to the liver via the hepatic portal vein. In contrast, fatty acids are re-synthesized into triglycerides and packaged into chylomicrons inside epithelial cells, which are then released by exocytosis into the central lacteal (lymphatic capillary) rather than blood capillaries.

(c) Fat-soluble vitamins require lipid digestion and emulsification (micelle formation) to be presented to the microvillus membrane for absorption. Insufficient secretion of pancreatic lipase (and colipase) results in incomplete lipid hydrolysis; unhydrolyzed lipids pass through the gut unabsorbed, carrying fat-soluble vitamins along into the feces.

(a) Presence of microvilli (brush border) to increase surface area for absorption / abundant mitochondria to provide ATP for active transport / presence of carrier proteins in the membrane. (Any 1, 1 mark)

(b) Amino acids enter blood capillaries / hepatic portal system, (1 mark)
whereas fatty acids / chylomicrons enter lacteals / lymphatic vessels. (1 mark)

(c) Pancreatic lipase secretion is deficient, so dietary fats / lipids cannot be properly digested / broken down. (1 mark)
Fat-soluble vitamins cannot be incorporated into micelles / remain trapped in undigested fats, thus failing to be absorbed by epithelial cells. (1 mark)

(a) Energy transfer between trophic levels is typically only about 10% efficient due to multiple losses:
1. Zooplankton use a significant portion of their assimilated energy for cellular respiration to maintain metabolic activities and locomotion, releasing energy as heat which cannot be utilized by consumers.
2. Not all individual zooplankton or their body parts are captured and ingested by small fish.
3. Some ingested material is indigestible and lost from the trophic transfer as feces (egestion) or excretory products.

(b) This phenomenon is biomagnification (or bioaccumulation along the food chain). Because Pollutant Z is persistent (non-biodegradable) and fat-soluble, it is retained within the tissues of organisms rather than being metabolized or excreted. Each osprey eats many small fish, each of which has eaten numerous zooplankton, leading to progressive concentration of the chemical at each successive trophic level, reaching the highest concentration in apex predators.

(a) Any THREE of the following for 1 mark each (max 3 marks):
- Energy is lost as heat during cellular respiration / metabolic activities of zooplankton.
- Not all zooplankton organisms are eaten / ingested by small fish (uneaten biomass).
- Some consumed biomass is indigestible / lost as feces / egested.
- Energy is lost in excretory wastes.

(b) Identification of biomagnification / bioaccumulation: Pollutant Z cannot be broken down / metabolized / excreted (is non-biodegradable and fat-soluble). (1 mark)
Top predators (osprey) consume large amounts of contaminated biomass from lower trophic levels, leading to accumulation at higher concentrations at the top of the food chain. (1 mark)

(a) The coding strand has the sequence \(5'\text{-ATG GCT TTT CGA-3'}\). Since mRNA corresponds to the coding strand with U replacing T, the mRNA sequence is \(5'\text{-AUG GCU UUU CGA-3'}\).
The first codon is \(5'\text{-AUG-3'}\), so its complementary tRNA anticodon is \(3'\text{-UAC-5'}\) (or \(5'\text{-CAU-3'}\)).
The second codon is \(5'\text{-GCU-3'}\), so its complementary tRNA anticodon is \(3'\text{-CGA-5'}\) (or \(5'\text{-AGC-3'}\)).

(b) 1. The replacement of phenylalanine by leucine alters the specific sequence of amino acids (primary structure).
2. This affects internal bonding/interactions (e.g., hydrophobic interactions, ionic bonds, or Van der Waals forces) during folding, leading to an altered three-dimensional / tertiary conformation of the polypeptide chain.
3. If this alteration distorts the active site, the substrate can no longer fit / bind into the active site (loss of structural complementarity), preventing the formation of enzyme-substrate complexes and destroying catalytic function.

(a) mRNA sequence: \(5'\text{-AUG GCU UUU CGA-3'}\) (1 mark)
tRNA anticodons: \(\text{UAC}\) and \(\text{CGA}\) (accept written \(3'\to 5'\) or \(5'\to 3'\) if correctly paired). (1 mark)

(b) The substitution alters the amino acid sequence / primary structure of the enzyme. (1 mark)
This disrupts intramolecular interactions / bonding (e.g. hydrophobic interactions / hydrogen bonds), altering the 3D / tertiary conformation of the protein. (1 mark)
The shape of the active site is changed so that it is no longer complementary to the substrate / cannot form enzyme-substrate complexes. (1 mark)

(a) Osmoreceptors in the hypothalamus monitor the osmolarity / water potential of circulating blood.

(b) 1. When blood water potential falls, the hypothalamus stimulates the posterior pituitary gland to release more ADH into the bloodstream.
2. ADH travels to the kidneys and binds to target epithelial cells of the distal convoluted tubules and collecting ducts.
3. This increases the water permeability of these tubular walls (facilitated by the insertion of aquaporin water channels into the cell membranes).
4. As the filtrate passes through the hypertonic renal medulla, a greater volume of water is reabsorbed by osmosis into the medullary tissue fluid and peritubular capillaries, resulting in concentrated, low-volume urine.

(c) Because a large amount of water is reabsorbed from the collecting duct while urea is not actively reabsorbed to the same extent, the urea remaining in the tubular fluid becomes highly concentrated in the final urine.

(a) Osmoreceptors in the hypothalamus. (1 mark)

(b) Posterior pituitary releases more ADH into the blood. (1 mark)
ADH increases the permeability of collecting duct / distal convoluted tubule walls to water. (1 mark)
More water is reabsorbed from filtrate into the blood / medulla capillaries by osmosis. (1 mark)

(c) The concentration of urea in the urine increases (due to reduced urine volume / greater water reabsorption). (1 mark)

(a) Sodium hydrogencarbonate releases dissolved hydrogencarbonate ions / carbon dioxide in aqueous solution, which acts as the essential inorganic carbon source / raw material for the light-independent reactions of photosynthesis.

(b) Air is trapped within the intercellular air spaces of the spongy mesophyll layer in normal leaves. Vacuum creates negative pressure that draws this gas out, replacing it with the higher-density liquid medium; this increases the overall buoyant density of the leaf disc relative to water, causing it to sink. Upon illumination, photosynthesis occurs and oxygen gas is generated via the photolysis of water in the thylakoid membranes. As oxygen gas accumulates in the intercellular air spaces, the overall density of the leaf disc decreases until it becomes less dense than the liquid, causing the disc to rise to the surface.

(c) The waxy cuticle on the leaf surface is hydrophobic and resists water entry; dilute detergent acts as a surfactant / wetting agent that reduces surface tension, allowing the aqueous medium to penetrate stomata and enter mesophyll spaces.

(d)(i) Beaker A is placed closer to the lamp, receiving higher photon flux / light intensity. Chlorophyll pigments absorb more light energy per unit time, resulting in higher rates of light absorption, ATP/NADPH generation, and water photolysis, thereby generating oxygen bubbles at a faster rate.

(ii) Any two of:
- Source / age / physiological state of the spinach leaf from which discs were punched.
- Concentration of sodium hydrogencarbonate solution in both beakers.
- Surface area / diameter of leaf discs.
- Wavelength / spectrum of light used.

(a) Provides dissolved carbon dioxide / \(\text{CO}_2\) / carbon source for photosynthesis (1)

(b)
- Vacuum pulls air out of intercellular spaces / spongy mesophyll and replaces it with solution, increasing disc density (causing them to sink) (1)
- Illumination drives photosynthesis / photolysis of water producing oxygen (1)
- Accumulation of oxygen gas inside leaf air spaces reduces disc density / increases buoyancy (causing them to float) (1)

(c) Reduces surface tension / breaks down hydrophobic cuticle barrier to enable infiltration of solution (1)

(d)(i)
- Higher light intensity provides more light energy absorbed by chlorophyll (1)
- Higher rate of photolysis / light-dependent reactions leads to faster oxygen production (1)

(d)(ii) Any TWO (1 mark each, max 2):
- Concentration / batch of \(\text{NaHCO}_3\) solution (1)
- Size / diameter / thickness of leaf discs (1)
- Physiological condition / age / position of leaves used (1)
- Light wavelength / colour of light source (1)

(a) Because it is impossible to cut all potato cylinders to the exact same initial mass, calculating percentage change normalises the differences and allows meaningful comparison between treatments.

(b) Cylinders are gently rolled on a clean filter paper/paper towel to remove adhering surface solution. If unremoved, excess external solution clinging to the cylinder surface would be weighed, leading to an overestimation of cylinder mass and inconsistent errors.

(c) By interpolation of data, zero percentage change in mass occurs between \(0.4\text{ M}\) (\(+1.0\%\)) and \(0.6\text{ M}\) (\(-5.8\%\)), approximately at \(0.42\text{ - }0.44\text{ M}\). At this concentration, the water potential of the sucrose solution equals the internal water potential of the tuber cells, resulting in no net water movement by osmosis.

(d) The \(0.8\text{ M}\) sucrose solution has a lower water potential (is hypertonic) compared to the cell sap of the potato cells. Water moves out of the vacuole and cytoplasm across the selectively permeable cell membrane into the external solution via osmosis down the water potential gradient, leading to decreased cell volume and mass.

(e) Boiling exposes the tissue to high temperatures, denaturing membrane proteins and disrupting phospholipid bilayers, thereby destroying cell membrane selective permeability. Cell contents / solutes leak out and cells lose the ability to establish an osmotic gradient, so little to no water is drawn in, or mass decreases due to solute loss.

(a) Eliminates bias/error caused by slight variations in initial mass / allows fair comparison across treatments (1)

(b)
- Removing residual/adhering fluid from the tissue surface (1)
- Prevents overestimating final mass / ensures only water taken up by tissue is measured (1)

(c)
- Concentration range: \(0.42\text{ M}\) to \(0.44\text{ M}\) (accept \(0.41\text{ - }0.45\text{ M}\)) (1)
- At this point, percentage mass change is \(0\%\) / no net osmotic movement of water / solution is isotonic to cell sap (1)

(d)
- Water potential of external solution is lower than that of potato cell sap (1)
- Water leaves cells by osmosis across selectively permeable membrane down the water potential gradient (1)

(e)
- Percentage mass gain would decrease drastically / become negligible / show mass loss (1)
- High temperature denatures membrane proteins / destroys membrane selective permeability, causing leakage of solutes (1)

(a) Lipids present in milk are hydrolysed by lipase into glycerol and free fatty acids. The accumulation of fatty acids produces \(\text{H}^+\) ions which neutralise the alkaline sodium carbonate, lowering the pH below the threshold of phenolphthalein (\(\text{pH} < 8.2\)), causing the pink colour to decolourise.

(b) Tube W decolourised four times faster than Tube X. Bile salts do not contain enzymes but act as emulsifying agents. They break down large lipid globules into microscopic lipid droplets (emulsification). This greatly increases the surface area-to-volume ratio of lipids accessible to water-soluble lipase, speeding up the rate of enzymatic hydrolysis into fatty acids.

(c) Tube Y serves as a negative control to confirm that lipid hydrolysis is catalysed specifically by active enzymatic action rather than non-biological breakdown. High temperature in boiling denatures lipase by disrupting tertiary structure and altering the active site shape, abolishing catalytic activity.

(d) Adding distilled water maintains an equal total reaction volume (\(7.0\text{ cm}^3\)) and constant initial reactant concentrations across all tubes, ensuring a fair test.

(e) Judging the visual disappearance of pink colour is subjective and susceptible to human error. An improvement is using a digital pH probe connected to a data logger to continuously record pH changes over time.

(a)
- Hydrolysis of triglycerides/lipids produces fatty acids (1)
- Fatty acids lower the pH / neutralise alkaline medium below phenolphthalein endpoint (1)

(b)
- Tube W reacted faster / took less time to decolourise than Tube X (1)
- Bile salts emulsify large fat globules into tiny droplets (1)
- This increases total surface area for lipase to bind and catalyse digestion (1)

(c)
- Acts as a control to demonstrate that lipid digestion requires active/functional enzyme (1)
- Boiling denatures lipase active site so no fatty acids are formed to lower pH (1)

(d) To keep total volume constant / ensure concentration of reactants is equal across all tubes (1)

(e)
- Limitation: Subjective visual judgment of endpoint / endpoint color change is gradual (0.5)
- Improvement: Use a pH meter / data logger / spectrophotometer to track rate quantitatively (0.5)

(a) Key aseptic techniques include:
1. Flamin the neck of culture tubes / sterilising forceps by dipping in ethanol and flaming.
2. Working adjacent to a Bunsen burner flame to maintain a sterile convection updraft / lifting Petri dish lids at an angle rather than fully removing them.

(b) In school laboratories, incubation at \(37^\circ\text{C}\) (human body core temperature) is avoided because it promotes the selective proliferation of potential human pathogenic bacteria, posing safety risks.

(c) The paper disc itself has a diameter of \(6.0\text{ mm}\). A value of \(6.0\text{ mm}\) indicates zero zone of clearance beyond the disc edge, confirming that sterile distilled water has no antibacterial effect.

(d) Bioactive antimicrobial chemicals in garlic (e.g., allicin) diffuse radially into the agar gel, establishing a concentration gradient. Where the chemical concentration exceeds the minimum inhibitory concentration, bacterial binary fission / cell division is prevented, resulting in a clear zone devoid of bacterial colonies.

(e) The claim is invalid because:
1. The test was conducted only in vitro on a single Gram-negative bacterium (E. coli); efficacy may differ against other bacteria (e.g., Gram-positive species).
2. Inside the human body, active substances may be degraded by stomach acid/enzymes, poorly absorbed into the bloodstream, or metabolized before reaching target sites.

(f) Take a sterile loop or swab from the clear zone, streak it onto a fresh agar plate lacking garlic extract, and incubate at \(30^\circ\text{C}\). If bacterial colonies grow, the substance is bacteriostatic; if no growth occurs, it is bactericidal.

(a) Any TWO (1 mark each, max 2):
- Sterilising forceps / loop in a Bunsen flame before use (1)
- Opening Petri dish lid minimally / at an angle during transfer (1)
- Working near a lit Bunsen burner / wiping bench with disinfectant (1)

(b) Prevents the growth of human pathogens / microorganisms adapted to human body temperature (1)

(c) \(6.0\text{ mm}\) represents the diameter of the paper disc itself / indicates no antibacterial inhibition (1)

(d)
- Antimicrobial chemicals diffuse outward through the agar medium (1)
- High chemical concentration inhibits bacterial reproduction / kills bacteria, preventing colony formation (1)

(e) Any TWO (1 mark each, max 2):
- Only tested on E. coli / cannot generalise to other bacterial species (1)
- In vitro conditions do not reflect complex physiological environments in the human body / compounds may be digested or diluted (1)

(f) Swab the clear zone and inoculate onto fresh nutrient agar without garlic; observe if bacterial growth resumes (growth = bacteriostatic, no growth = bactericidal) (1)

(a) Transpiration generates negative pressure (tension) in xylem vessels. Cutting the stem in air would draw air into the cut xylem vessels, causing cavitation / air embolism which breaks the cohesive water column and prevents continuous water transport. Cutting under water ensures that only water enters the xylem lumen.

(b) Under still air (Condition 1), water evaporating through stomata forms a stagnant, humid boundary layer around the leaf surface. The fan in Condition 2 blows away this boundary layer, decreasing external water vapour concentration and steepening the concentration gradient of water vapour between the sub-stomatal air spaces and the surrounding air. This increases the rate of transpiration and subsequent transpirational pull.

(c) The enclosed humid bag creates an atmosphere saturated with water vapour (high relative humidity). This drastically flattens/reduces the water vapour concentration gradient between the internal leaf air spaces and the ambient air, slowing down transpiration and reducing water uptake.

(d) Residual movement is due to small amounts of water utilized in cellular metabolism (e.g. photolysis in photosynthesis) or slight cuticular transpiration / expansion of plant tissues.

(e) No. The potometer directly measures the rate of water absorption/uptake by the shoot, not transpiration directly. A small fraction (approx. \(1\text{ - }2\%\)) of the absorbed water is retained by plant cells for maintaining turgidity, cellular expansion, and biochemical reactions (such as photosynthesis), so water uptake slightly exceeds water loss via transpiration.

(a)
- Transpiration tension draws air into xylem vessels if cut in air (1)
- Prevents air locks / cavitation that disrupts continuous water column in xylem (1)

(b)
- Wind removes humid boundary layer of air around leaf surface (1)
- Steepens water vapour concentration gradient between intercellular spaces and outside air, increasing transpiration pull (1)

(c)
- High ambient humidity inside the bag (1)
- Reduces water vapour concentration gradient across stomata, lowering transpiration rate (1)

(d) Water consumed in metabolic processes / photosynthesis / maintaining cell turgor (1)

(e)
- No (1)
- Some absorbed water is retained for turgidity / used in metabolic reactions (e.g., photosynthesis) rather than lost by evaporation (1)

An exemplary essay should address the following key points:

1. Differences in Starting Conditions:
- Primary succession begins on newly formed, completely barren terrain where no living organisms or pre-existing soil exist (e.g., bare rock, cooled lava flows, newly formed sand dunes).
- Secondary succession occurs in an area where an existing community has been disrupted or cleared by disturbances (e.g., forest fire, flood, abandoned farmland), but pre-existing soil, seed bank, and underground vegetative structures remain.

2. Differences in Pioneer Species:
- In primary succession, pioneer species must be highly tolerant of extreme abiotic stress (e.g., lichens, mosses) capable of surviving without developed soil and capable of weathering rock.
- In secondary succession, pioneer species are typically fast-growing, opportunistic, sun-loving herbaceous plants or grasses that germinate quickly from the remaining soil seed bank or wind-dispersed seeds.

3. Soil Development and Rate of Succession:
- Primary succession involves a lengthy process of pedogenesis (soil formation); death and decomposition of pioneer species gradually accumulate organic matter (humus) over hundreds to thousands of years, resulting in a very slow rate of change.
- Secondary succession progresses much faster because the soil layer and nutrients are already present, allowing rapid colonization and faster canopy formation.

4. Climax Community:
- Both successions eventually progress towards a stable, complex climax community (e.g., mature woodland/forest) characterized by high biodiversity, complex food webs, and balanced nutrient cycling, provided environmental conditions remain stable.

5. Human Disruption of Ecological Succession:
- Deforestation / clear-cutting / urbanization removes biomass, resets the succession to an earlier stage, or permanently replaces natural communities with artificial ecosystems.
- Agricultural practices (such as regular plowing, monoculture planting, and grazing by livestock) maintain succession at a deflected/arrested stage (plagioclimax), preventing the establishment of climax shrubland or forest.
- Introduction of invasive alien species may outcompete native species, altering the trajectory and biodiversity of succession.

Content (max. 9 marks):

Starting Conditions:
- Primary succession starts on bare ground/substrate with no pre-existing soil/organic matter (1)
- Secondary succession starts in an area where existing vegetation was removed/disturbed, but soil/seed bank is already present (1)

Pioneer Species:
- Pioneer species in primary succession (e.g., lichens, mosses) can survive harsh conditions/grow on bare rock (1)
- Pioneer species in secondary succession (e.g., fast-growing weeds/grasses) quickly colonize the pre-existing soil (1)

Soil Development and Rate:
- Primary succession involves soil formation through weathering and organic matter accumulation, thus it takes a very long time / proceeds very slowly (1)
- Secondary succession proceeds much faster as fertile soil with nutrients is already available (1)

End-point / Climax:
- Both pathways transition through gradual replacement of species until reaching a stable climax community with high biodiversity / complex food webs (1)

Human Interruption (any 2 points, 1 mark each, max 2 marks):
- Regular mowing / plowing / grazing prevents woody plant establishment, maintaining the community at an early/deflected stage (plagioclimax) (1)
- Deforestation / urbanization destroys established vegetation, forcing succession to restart or preventing recovery (1)
- Introduction of invasive alien species outcompetes native colonizers, permanently altering the community structure/trajectory (1)

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Communication (max. 2 marks):
- 2 marks: Answer is well-structured, coherent, logically sequenced with appropriate biological terminology, addressing all required aspects of comparison and human impact.
- 1 mark: Answer is understandable and covers most aspects, but shows minor logical gaps, slight disorganization, or omissions.
- 0 marks: Irrelevant, highly fragmented, or completely disorganized answer.