2025 HKDSE Biology Answers & Marking Scheme

(a) Polysaccharide antigens are T-independent antigens, meaning they cannot be processed and presented on MHC class II molecules of antigen-presenting cells (APCs) to helper T cells. By linking them to a carrier protein, APCs take up the conjugate, process the protein, and present the peptide fragments to helper T cells, stimulating their activation and cytokine release.
(b) Upon secondary infection, memory B and T cells already present in the body recognize the pathogen's antigens immediately. Memory B cells rapidly proliferate and differentiate into antibody-secreting plasma cells, producing a much higher concentration of specific antibodies in a shorter time compared to the primary response.

(a)
- Correctly states that T cell receptors can only recognize peptide antigens presented by antigen-presenting cells (or polysaccharides alone do not activate helper T cells) (1 mark)
- Explains that the carrier protein is processed and presented to helper T cells, triggering their activation (1 mark)

(b)
- States that memory B and T cells persist in the blood/lymphatic system after vaccination (1 mark)
- Explains that upon reinfection, memory B cells recognize the antigen immediately and differentiate into plasma cells without delay (1 mark)
- Mentions that a much larger quantity of antibodies is produced rapidly to clear the pathogen (1 mark)

(a) Active transport requires metabolic energy (ATP). Since glucose uptake dropped by 90% when ATP synthesis was blocked, its primary transport mechanism is active transport. In contrast, facilitated diffusion is a passive process that does not require ATP. Since fructose uptake was unaffected, its primary transport mechanism is facilitated diffusion.
(b) Active transport of glucose utilizes energy derived from ATP hydrolysis, converting ATP to ADP. When this transport is inhibited, ATP consumption decreases, leading to a high cellular ATP-to-ADP ratio. This high ratio feedback-inhibits key enzymes in cellular respiration (such as those in glycolysis and the Krebs cycle), slowing down the electron transport chain where oxygen acts as the final electron acceptor, thus decreasing oxygen consumption.

(a)
- Identifies glucose transport as active transport because it is highly dependent on ATP/metabolic energy (1 mark)
- Identifies fructose transport as facilitated diffusion because it is independent of metabolic energy (1 mark)
- Explains that active transport moves molecules against a concentration gradient using ATP, while facilitated diffusion moves them down a gradient via carrier/channel proteins without ATP (1 mark)

(b)
- States that less active transport results in decreased ATP hydrolysis to ADP (1 mark)
- Explains that a higher ATP/ADP ratio feedback-inhibits cellular respiration / electron transport chain, reducing the consumption of oxygen as the final electron acceptor (1 mark)

(a) High sucrose concentration serves two main purposes: it provides an abundant carbon and energy source for rapid biomass growth of Aspergillus niger, and it maintains a high rate of glycolysis. This high glycolytic rate ensures a continuous, rapid supply of pyruvate and acetyl-CoA, which are the direct precursors for citric acid synthesis.
(b) In the Krebs cycle, citric acid is converted to isocitric acid by the enzyme aconitase, which strictly requires iron ions (\(Fe^{2+}\)/\(Fe^{3+}\)) as a cofactor to function. When iron is limited in the medium, aconitase becomes inactive. As a result, the metabolic pathway of the Krebs cycle is blocked at the very first step after citric acid formation, leading to the accumulation of citric acid within the fungal cells, which is subsequently excreted.

(a)
- States that excess sucrose provides an abundant carbon/energy source for fungal growth and respiration (1 mark)
- Explains that it maintains a high glycolytic rate to yield sufficient precursors (acetyl-CoA/pyruvate) for citric acid synthesis (1 mark)

(b)
- States that aconitase requires iron ions as a crucial cofactor for its catalytic activity (1 mark)
- Explains that iron deficiency inactivates aconitase, blocking the conversion of citric acid to isocitric acid (1 mark)
- Concludes that citric acid cannot be further metabolized in the Krebs cycle, resulting in its accumulation and excretion (1 mark)

(a) A patient with central diabetes insipidus will produce a very large volume of highly dilute urine. In a healthy individual, ADH acts on the cells of the collecting ducts, increasing their permeability to water by inserting aquaporins. Due to the lack of ADH in this patient, the collecting ducts remain impermeable to water, and less water is reabsorbed from the filtrate back into the blood capillaries, leaving more water in the urine.
(b) Desmopressin (or any ADH analogue) can be administered. It acts as a synthetic replacement for ADH, binding to the V2 receptors on the basolateral membrane of the collecting duct cells. This triggers the insertion of water channels (aquaporins) into the apical membrane, restoring water reabsorption by osmosis and returning urine output to normal levels.

(a)
- Identifies that the patient produces a large volume of dilute urine (1 mark)
- Explains that the absence of ADH results in decreased water permeability of the collecting ducts/distal convoluted tubules (1 mark)
- Explains that less water is reabsorbed by osmosis back into the blood, leading to high water loss in urine (1 mark)

(b)
- Suggests Desmopressin / synthetic ADH analogue (1 mark)
- Explains that it mimics endogenous ADH to increase the water permeability of the collecting ducts, promoting water reabsorption and reducing urine volume (1 mark)

(a) The binding of ABA to guard cell receptors triggers signaling pathways that stimulate the efflux of potassium ions (\(K^+\)) out of the guard cells into the surrounding epidermal cells. This loss of solute increases the water potential inside the guard cells. Consequently, water leaves the guard cells by osmosis down the water potential gradient. The guard cells lose turgidity, become flaccid, and the stomatal pore closes.
(b) When stomata are closed, the diffusion of carbon dioxide (\(CO_2\)), a raw material for photosynthesis, into the leaf is severely restricted. This leads to a substantial decline in the rate of photosynthesis. Over time, the rate of carbon fixation becomes lower than the rate of respiration (which continues to consume organic matter), resulting in a net loss of organic substances and thus a reduction in the plant's dry mass.

(a)
- States that ABA causes the efflux/outflow of potassium ions (\(K^+\)) from guard cells (1 mark)
- Explains that the loss of potassium ions increases the water potential inside the guard cells, causing water to leave by osmosis (1 mark)
- Explains that guard cells lose turgor / become flaccid, which leads to the closure of the stomatal pore (1 mark)

(b)
- States that closed stomata block or greatly restrict the diffusion of carbon dioxide into the leaves (1 mark)
- Explains that the rate of photosynthesis drops below the rate of respiration, causing a net loss of organic matter/dry mass (1 mark)

(a) Independent variable: Surface area-to-volume ratio of the agar blocks. Dependent variable: Depth or rate of diffusion of the acid. (b) The acid (HCl) diffuses from a region of higher concentration (beaker) to lower concentration (agar). The acid neutralizes the sodium hydroxide, lowering the pH and causing the phenolphthalein indicator to turn from pink to colorless. (c) (i) Diffusion is driven by the concentration gradient and kinetic energy of particles, which are the same across all treatments since the concentration of HCl and temperature are kept constant. (ii) Total volume of the 1 cm cube = \(10\text{ mm} \times 10\text{ mm} \times 10\text{ mm} = 1000\text{ mm}^3\). The acid penetrated 2 mm on all sides, leaving a central pink cube of side length \(10\text{ mm} - 2\text{ mm} - 2\text{ mm} = 6\text{ mm}\). Volume of remaining pink region = \(6\text{ mm} \times 6\text{ mm} \times 6\text{ mm} = 216\text{ mm}^3\). Percentage remaining pink = \((216 / 1000) \times 100\% = 21.6\%\). (iii) Large organisms have small surface area-to-volume ratios, making simple diffusion too slow to reach internal cells over long distances.

(a) Independent variable: SA/Vol ratio / cube size (1 mark). Dependent variable: Diffusion rate / depth of acid penetration (1 mark). (b) HCl diffuses into the agar (1 mark). Neutralization of NaOH decreases pH, changing phenolphthalein from pink to colorless (1 mark). (c) (i) Diffusion rate is determined by intrinsic factors (concentration gradient, temperature) which are kept constant (1 mark); thus, the speed of acid penetration is the same (1 mark). (ii) Volume of total cube = \(1000\text{ mm}^3\) (0.5 mark). Dimensions of remaining pink cube = \(6\text{ mm} \times 6\text{ mm} \times 6\text{ mm} = 216\text{ mm}^3\) (1 mark). Percentage = \(21.6\%\) (0.7 mark). (iii) Large multicellular organisms have a small SA/Vol ratio, meaning diffusion distance is too long for external substances to reach deeply located tissues in time (1 mark).

(a) Cutting the shoot under water prevents air locks in the xylem, ensuring continuous water column cohesion. Sealing joints with grease ensures no air leaks, maintaining the suction tension generated by transpiration. (b) (i) Wind sweeps away the humid boundary layer of water vapor near the stomatal openings. This increases the concentration gradient of water vapor between the sub-stomatal cavity and the dry external atmosphere, accelerating diffusion (transpiration). (ii) The potometer measures the rate of water absorption. Not all absorbed water is lost via transpiration; some water is utilized for cellular metabolism (photosynthesis) or stored to maintain cell turgidity.

(a) Precaution 1: Cut the shoot stem under water (1 mark). Reason: Prevents air bubbles from blocking xylem vessels / maintains continuous water column (1.1 marks).
Precaution 2: Seal joints with petroleum jelly (1 mark). Reason: Prevents air leakage into the system, which would disrupt negative pressure / prevent bubble movement (1.1 marks).
(b) (i) Wind sweeps away water vapor near the leaf surface (1 mark). This increases/maintains a steeper water vapor concentration gradient between the substomatal cavity and the atmosphere (1 mark). Therefore, more water diffuses out through stomata per unit time, leading to higher water uptake and faster bubble movement (1 mark).
(ii) Not all water absorbed is transpired (1 mark). Some water is used in metabolic processes (e.g., photosynthesis) or to maintain cell turgidity/growth (1 mark).

(a) Run a transect line perpendicular to the water line from the low tide to the high tide mark. Place quadrats at fixed intervals (e.g., every 2 m) along the line to systematically count the gastropods. (b) (i) Desiccation (or temperature fluctuation/wave action). As you move from low to high tide, exposure to air increases. Species A cannot tolerate prolonged desiccation, restricting it to lower regions. (ii) Species B can tolerate desiccation better, so it survives in the upper shore. It is restricted to the upper shore because Species A, which is a stronger competitor for space, excludes it from the lower shore where physical conditions are favorable for both. (c) Quadrats assume stationary organisms. Mobile animals can move away during counting or hide. An alternative is the mark-release-recapture method (Lincoln index).

(a) Lay a transect line perpendicular to the tide marks (from low to high tide) (1 mark). Place quadrats at regular/fixed intervals along the transect line (1 mark). Identify and count Species A and B in each quadrat (1 mark). (b) (i) Abiotic factor: Desiccation / duration of exposure to air (1 mark). Explanation: Near the high tide mark, exposure to air is longer; Species A loses water rapidly and dies, limiting its distribution to lower zones (1.2 marks). (ii) Species A is limited to the lower shore due to physical constraints (desiccation) (1 mark). Species B, being stress-tolerant, survives in the upper shore, escaping competitive exclusion by Species A (1 mark). (c) Limitation: Highly mobile animals may flee or be counted multiple times / inaccurate counts (1 mark). Alternative: Mark-release-recapture method (1 mark).

(a) Liquid paraffin is less dense than water and insoluble in it, forming a barrier that prevents atmospheric oxygen from dissolving into the yeast suspension, thus enforcing anaerobic conditions. (b) (i) Glucose is the primary respiratory substrate. An increase in concentration up to 10% increases substrate availability, resulting in a higher rate of enzyme-substrate complexes forming and faster carbon dioxide production. (ii) A 25% glucose solution has a very low water potential. Water leaves the yeast cells via osmosis down the water potential gradient, causing cellular dehydration (plasmolysis) which decreases the activity of intracellular enzymes required for respiration. (c) A control setup consists of replacing the active yeast suspension with dead yeast (boiled and cooled) or distilled water, while keeping all other variables (glucose concentration, temperature, setup) identical.

(a) To exclude oxygen / prevent oxygen from dissolving into the mixture (1 mark). This ensures anaerobic conditions for yeast fermentation (1 mark). (b) (i) Glucose is the respiratory substrate (1 mark). Higher concentration of substrate increases the frequency of collisions between substrate molecules and respiratory enzymes, accelerating fermentation and producing more gas (1.2 marks). (ii) 25% glucose solution has a lower water potential than yeast cytoplasm (1 mark). Water leaves yeast cells by osmosis (1 mark), causing plasmolysis / dehydration which denatures/inhibits enzymes and slows down metabolic activities (1 mark). (c) Use dead (boiled and cooled) yeast suspension (1 mark) with the same glucose concentrations under identical conditions (1 mark).

(a) The primary response (Day 0 injection) features a lag phase and a lower concentration of antibodies because naive B cells must be selected and clonal expansion takes time. The secondary response (Day 28 injection) is faster, stronger, and more prolonged because memory B cells generated during the primary response are activated directly and rapidly differentiate into antibody-secreting plasma cells. (b) On Day 35, Group 2 shows a primary response to Antigen Y (first exposure on Day 28), yielding low antibody concentrations. Group 1 shows a secondary response to Antigen X (second exposure), yielding high antibody concentrations. This demonstrates that immunological memory is antigen-specific; exposure to Antigen X does not confer memory against Antigen Y. (c) Age, gender, and genetic background are confounding variables that affect immune system responsiveness. Using genetically identical mice of the same age and gender ensures a fair comparison, ensuring that the measured antibody production represents only the effect of the treatment.

(a) Description: First injection leads to a slow, low, and short-lived antibody response, while the second injection leads to a rapid, high, and long-lasting antibody response (1 mark). Explanation: First injection triggers a primary response, producing memory B cells (1 mark). Second injection triggers a secondary response where memory cells recognize Antigen X immediately (1 mark), rapidly dividing and differentiating into plasma cells to produce large amounts of antibodies (1 mark).
(b) Comparison: Antibody level against Y in Group 2 is lower than against X in Group 1 (1 mark). Explanation: Injection of Y in Group 2 is a primary response (first exposure) (1 mark), which requires time for antigen presentation and clonal selection (1.2 marks), whereas Group 1 is exhibiting a secondary response to X.
(c) To ensure that any variations in antibody production are due to the treatment (antigens injected) and not differences in individual immunity (1 mark); different ages, genders, or genetics would affect the baseline immune capability and introduce bias (1 mark).

**Similarities:**
Both natural selection and artificial selection are mechanisms of evolutionary change that operate on the existing genetic variation within a population. Both processes lead to changes in allele frequencies over successive generations by favoring individuals with specific heritable phenotypes, allowing them to reproduce and pass on their genes.

**Differences:**
1. **Driving Forces:**
- Natural selection is driven by natural environmental factors, including biotic pressures (such as predation, disease, and competition) and abiotic pressures (such as climate change and habitat characteristics).
- In contrast, artificial selection is driven by human intervention, where humans deliberately choose which individuals are allowed to breed based on desired characteristics.

2. **Criteria for Selecting Traits:**
- In natural selection, traits are selected based on how well they enhance the organism's survival and reproductive success (evolutionary fitness) in its natural habitat.
- In artificial selection, traits are selected based on their utility, economic value, or aesthetic appeal to humans. These traits often have no survival advantage or may even be disadvantageous to the organism in the wild.

3. **Impact on Genetic Diversity:**
- Natural selection tends to maintain a level of genetic variation that allows the population to adapt to changing environmental conditions.
- Artificial selection typically leads to a rapid and drastic reduction in genetic diversity because humans often breed closely related individuals (inbreeding) or use cloning to propagate a few highly desirable phenotypes.

4. **Impact on Long-term Survival:**
- Under natural selection, the population becomes better adapted to its natural environment, which increases its long-term viability and survival in the wild.
- Under artificial selection, the population often becomes highly dependent on human care for survival. The selected traits often reduce their fitness and survival capacity if they were returned to the wild.

**Content (max 9 marks):**
*Similarities (max 1 mark):*
- Both involve selection based on existing genetic variation / lead to changes in allele frequencies over generations. (1)

*Driving Force (max 2 marks):*
- Natural selection is driven by environmental factors / selection pressures (biotic and abiotic). (1)
- Artificial selection is driven by human preference / choice / intervention. (1)

*Criteria for Selecting Traits (max 2 marks):*
- Under natural selection, traits that enhance survival and reproductive success (evolutionary fitness) in the wild are selected. (1)
- Under artificial selection, traits that are desirable or useful to humans are selected (even if they disadvantage the organism). (1)

*Impact on Genetic Diversity (max 2 marks):*
- Natural selection preserves genetic diversity to allow adaptation to changing environments. (1)
- Artificial selection rapidly and drastically reduces genetic diversity / increases homozygosity due to inbreeding or selective breeding of few individuals. (1)

*Impact on Long-term Survival (max 2 marks):*
- Natural selection increases the adaptability and long-term survival of the population in its natural habitat. (1)
- Artificial selection often reduces the fitness / survival of the population in the wild, making them dependent on human support. (1)

**Communication (max 2 marks):**
- **2 marks**: Systematic and logical comparison, clearly distinguishing between similarities and differences under structured themes (driving force, traits, diversity, survival). Free of major biological misconceptions, with fluent expression.
- **1 mark**: Shows some structure, but the comparison is disorganized or some parts are presented as isolated descriptions without active comparison.
- **0 marks**: Disorganized presentation with major conceptual gaps or very poor clarity.