2023 HKDSE 生物 模拟试题及答案详解

Dialysis tubing is selectively permeable, allowing small molecules such as iodine and water to pass through while preventing large macromolecules such as starch from diffusing out. As iodine molecules diffuse into the tubing, they react with starch to turn the inside solution blue-black. Starch cannot exit into the beaker, so the solution in the beaker remains yellowish-brown (the color of dilute iodine).

A (1 mark) - Correct identification of selective permeability of dialysis tubing and the resulting localized iodine-starch reaction.

Transpiration causes loss of water, reducing the total mass of the set-up. Shoot 4 has all stomata open, losing the most water (lightest final mass). Shoot 1 has lower stomata open (where most stomata are located), so it loses more water than Shoot 2 (which has lower stomata blocked). Shoot 3 has all stomata blocked and loses the least water (heaviest final mass). Therefore, the order from lightest to heaviest is Shoot 4 < Shoot 1 < Shoot 2 < Shoot 3.

A (1 mark) - Correct deduction of relative transpiration rates based on stomatal distribution and coatings.

(1) is correct: Glucose absorbed in the ileum enters the hepatic portal vein, and excess glucose is converted to glycogen and stored in the liver, meaning glucose concentration is higher in the hepatic portal vein than the hepatic vein.
(2) is incorrect: The liver produces urea through deamination of excess amino acids, so the blood leaving via the hepatic vein contains a higher concentration of urea than the blood entering via the hepatic portal vein.
(3) is incorrect: Hepatic vein carries deoxygenated blood leaving the liver after the liver tissues consume oxygen for cellular respiration, so its oxygen concentration is lower than or similar to that of blood entering the liver.

A (1 mark) - Only statement (1) is correct.

Glycolytic enzymes are located in the cytoplasm, so isolated intact mitochondria cannot metabolise glucose, starch, or glycogen directly. Pyruvate is the substrate that is transported across the mitochondrial membrane to enter the link reaction and Krebs cycle, supplying electrons to the electron transport chain where oxygen acts as the terminal electron acceptor.

B (1 mark) - Mitochondria utilize pyruvate, not glucose or polysaccharides.

Energy transfer across successive trophic levels is inefficient due to respiratory heat loss, uneaten parts, and undigested materials. Therefore, the apex predator (hawk) receives only a minute fraction of the chemical energy originally fixed by the primary producers (grass). Energy flow is unidirectional and cannot be recycled into photosynthesis.

B (1 mark) - Correct understanding of energy dissipation and trophic efficiency.

The woman inherits \(X^b\) from her affected father, making her a carrier with genotype \(X^B X^b\). The father has normal vision with genotype \(X^B Y\). The possible offspring genotypes are \(X^B X^B\) (normal female, 0.25), \(X^B X^b\) (carrier female, 0.25), \(X^B Y\) (normal male, 0.25), and \(X^b Y\) (colour blind male, 0.25). Thus, the probability of having a colour blind boy is 0.25 (or 1/4).

B (1 mark) - Probability is calculated as \(\frac{1}{2} \text{ (carrier mother passing } X^b) \times \frac{1}{2} \text{ (father passing } Y) = 0.25\).

The primary response leads to the production of long-lived memory B cells. Upon a second exposure to the same antigen in the booster, memory B cells recognize the antigen immediately, proliferating and differentiating into antibody-secreting plasma cells much faster and in larger numbers than naive B cells in the primary response.

B (1 mark) - Secondary immune response is mediated by rapid activation and proliferation of memory B cells.

The knee-jerk reflex is a monosynaptic stretch reflex: stretching of the patellar tendon stimulates stretch receptors in the quadriceps muscle, which generate impulses in sensory neurones. These sensory neurones synapse directly with motor neurones in the spinal cord without relay neurones (interneurones) in the monosynaptic pathway, triggering contraction of the extensor muscle.

A (1 mark) - Correct description of the monosynaptic pathway of the knee-jerk reflex.

High light intensity stimulates stomatal opening, increasing the diffusion of water vapor. Low relative humidity steepens the water vapor concentration gradient between the internal air spaces of the leaf and the surrounding atmosphere. High wind speed removes the humid boundary layer around the stomata, maintaining a steep concentration gradient. Together, these conditions maximize the rate of transpiration, leading to the fastest rate of water uptake and hence the fastest movement of the air bubble.

A (1 mark)

At each trophic level, a large proportion of the ingested energy is lost as heat through cellular respiration, excreted as metabolic wastes, or remains uneaten/undigested in dead tissues and faeces. Therefore, only around 10% is converted into new biomass and made available to the next trophic level.

C (1 mark)

Cross: BbTt × BbTt.
For fur colour: Probability of brown fur (bb) = 1/4.
For tail length: Probability of short tail (TT or Tt) = 3/4.
Since the genes assort independently, the combined probability is \( \frac{1}{4} \times \frac{3}{4} = \frac{3}{16} \).

B (1 mark)

Digestion products of lipids (fatty acids and glycerol) enter epithelial cells, where they re-assemble into triglycerides and form chylomicrons, which are then exocytosed into the central lacteal (lymphatic vessel). In contrast, water-soluble nutrients such as glucose and amino acids are directly absorbed into the blood capillaries.

B (1 mark)

In the thylakoid membranes during the light-dependent stage, light energy absorbed by chlorophyll excites electrons, driving photolysis of water into protons, electrons, and oxygen gas, while producing ATP and reduced NADP (NADPH). Carbon dioxide fixation occurs in the stroma during the Calvin cycle (light-independent reactions).

D (1 mark)

When there is no change in length, there is no net movement of water into or out of the potato cells by osmosis. This indicates that the water potential of the 0.4 M sucrose solution equals the average water potential of the potato cell sap.

A (1 mark)

When an action potential arrives at the synaptic knob (axon terminal), voltage-gated calcium channels open, causing neurotransmitter vesicles to fuse with the presynaptic membrane and release neurotransmitters by exocytosis. The neurotransmitters diffuse across the synaptic cleft and bind to specific receptor proteins on the postsynaptic membrane.

C (1 mark)

Due to the presence of memory B cells and memory T cells formed during the primary response (vaccination), the secondary immune response upon subsequent encounter with the live virus has a shorter lag period, produces a much higher peak concentration of specific antibodies, and maintains high antibody levels for a longer duration.

B (1 mark)

The rate of transpiration increases with higher light intensity (which stimulates stomatal opening), higher temperature (which increases the kinetic energy and evaporation rate of water molecules), increased air movement/wind speed (which removes the humid boundary layer near stomata), and lower relative humidity (which steepens the water vapour concentration gradient). High light intensity, high wind speed, and low relative humidity maximize the transpiration pull and thus water uptake.

B (1 mark)

Energy transfer across trophic levels with \(10\%\) efficiency:
- Trophic level 1 (Phytoplankton): \(10\,000\text{ kJ m}^{-2}\text{ year}^{-1}\)
- Trophic level 2 (Zooplankton): \(10\,000 \times 10\% = 1\,000\text{ kJ m}^{-2}\text{ year}^{-1}\)
- Trophic level 3 (Small fish): \(1\,000 \times 10\% = 100\text{ kJ m}^{-2}\text{ year}^{-1}\)
- Trophic level 4 (Tuna): \(100 \times 10\% = 10\text{ kJ m}^{-2}\text{ year}^{-1}\).
Therefore, the energy transferred to the tuna is \(10\text{ kJ m}^{-2}\text{ year}^{-1}\).

A (1 mark)

Since the child is of blood group O (genotype \(I^O I^O\)), both parents must carry the allele \(I^O\). Thus, the father's genotype is \(I^A I^O\) and the mother's is \(I^B I^O\).
The possible genotypes of their offspring are \(I^A I^B\) (Group AB), \(I^A I^O\) (Group A), \(I^B I^O\) (Group B), and \(I^O I^O\) (Group O), each with a probability of \(\frac{1}{4}\).
The probability of the child being a boy is \(\frac{1}{2}\).
Therefore, the combined probability of having a boy with blood group AB is \(\frac{1}{4} \times \frac{1}{2} = \frac{1}{8}\) (or \(0.125\) / \(12.5\%\)).

C (1 mark)

During ventricular systole, the ventricles contract, raising intraventricular pressure above atrial pressure. This causes the atrioventricular (bicuspid and tricuspid) valves to close to prevent backflow into the atria. When ventricular pressure exceeds arterial pressure, the semilunar valves open, and blood is pumped into the aorta and pulmonary artery.

D (1 mark)

The isotonic concentration corresponds to zero percentage change in mass. The mass increases by \(+7\%\) at \(0.2\text{ M}\) (water enters cells by osmosis) and decreases by \(-3\%\) at \(0.4\text{ M}\) (water leaves cells by osmosis). Therefore, zero change in mass occurs between \(0.2\text{ M}\) and \(0.4\text{ M}\), approximately around \(0.34\text{ M}\).

B (1 mark)

In \(\text{G}_1\), the cell has \(2n = 8\) chromosomes and \(4\text{ a.u.}\) of DNA. After S phase, DNA replicates so each chromosome consists of two sister chromatids; the cell has 8 chromosomes and \(8\text{ a.u.}\) of DNA. During Meiosis I, homologous chromosomes separate into two daughter cells. Each daughter cell now contains \(n = 4\) chromosomes, each with two chromatids, giving a DNA content of \(8 / 2 = 4\text{ a.u.}\)

C (1 mark)

When an action potential arrives at the axon terminal, voltage-gated calcium channels open, causing an influx of \(\text{Ca}^{2+}\) ions into the presynaptic terminal. This triggers the exocytosis of synaptic vesicles containing neurotransmitters into the synaptic cleft. The neurotransmitter then diffuses across the cleft and binds to specific receptor proteins on the postsynaptic membrane.

A (1 mark)

Statement (1) is correct: Active immunity stimulates the host's immune system to produce antibodies and memory cells, providing long-term protection.
Statement (2) is correct: Passive immunity provides antibodies directly (e.g. maternal antibodies across the placenta or antivenom injection).
Statement (3) is incorrect: Passive immunity provides immediate protection because ready-made antibodies are introduced, whereas active immunity requires a lag period for clonal selection and expansion of lymphocytes.

B (1 mark)

High humidity reduces the water vapour concentration gradient between the intercellular air spaces and the surrounding atmosphere, resulting in the lowest transpiration rate (R). Moving air (windy condition in Q) removes water vapour accumulated near the stomata, steepening the diffusion gradient and increasing transpiration rate more significantly than still air at room temperature (P). Elevated temperature (S) increases the kinetic energy and evaporation rate of water molecules, but high wind speed (Q) typically maintains a very steep concentration gradient. Between the tested conditions, high humidity gives the lowest rate (R), followed by normal room condition (P), elevated temperature (S), and windy condition (Q) or S/Q depending on gradient steepness, so R < P < S < Q is the correct sequence.

B (1 mark)

Process 1 is nitrogen fixation (carried out by nitrogen-fixing bacteria). Process 2 and Process 3 are stages of nitrification: ammonium to nitrite (by Nitrosomonas) and nitrite to nitrate (by Nitrobacter), both of which are nitrifying bacteria requiring aerobic conditions (oxygen is used to oxidise nitrogen). Process 4 is denitrification, carried out by denitrifying bacteria under anaerobic conditions. Therefore, only option C is correct.

C (1 mark)

In a dihybrid cross of two heterozygous individuals (GgRr × GgRr), the expected phenotypic ratio among the offspring is 9 (grey, red) : 3 (grey, sepia) : 3 (black, red) : 1 (black, sepia). The fraction of offspring exhibiting grey body and sepia eyes is 3/16. Thus, expected number = 320 × (3/16) = 60.

A (1 mark)

During ventricular systole, the ventricles contract, causing ventricular pressure to rise above atrial pressure. This causes the atrioventricular (bicuspid and tricuspid) valves to close to prevent backflow into the atria (statement 1 is correct; statement 3 is incorrect as ventricular pressure is higher). When the ventricular pressure exceeds the pressure in the aorta and pulmonary artery, the semi-lunar valves are forced open (statement 2 is correct).

A (1 mark)

At 0.3 M sucrose solution, there is no net movement of water, meaning the water potential (and effective solute concentration) of the potato cells is equal to that of the 0.3 M sucrose solution. Option A is incorrect because the cylinder gained mass in 0.1 M, meaning the solution had higher water potential than the tissue. Option B is incorrect because dynamic equilibrium occurs—water molecules enter and leave at equal rates. Option D is incorrect because water moves by osmosis (passive transport), not active transport.

C (1 mark)

Oxidative phosphorylation, which takes place across the inner mitochondrial membrane (cristae) via the electron transport chain and ATP synthase, produces the vast majority of ATP molecules (approx. 28-34 ATP) during aerobic respiration.

D (1 mark)

During the secondary immune response, specific memory B cells formed during the primary response recognize the familiar antigens quickly, proliferate rapidly, and differentiate into plasma cells that secrete high titres of specific antibodies. Phagocytes do not form memory cells (they are part of non-specific immunity).

A (1 mark)

The withdrawal reflex is a spinal reflex. Sensory impulses enter the spinal cord through the dorsal root (1), pass through interneurones, and exit via motor neurones through the ventral root to stimulate effector muscles (2). The reflex action occurs automatically without conscious decision-making; the impulses sent to the cerebral cortex to perceive pain arrive after or simultaneously with the withdrawal, not before the withdrawal movement begins (statement 3 is incorrect).

A (1 mark)

In Twig S (both surfaces coated), the water loss was 2.2 mm (mainly cuticular transpiration or loss via the stem). In Twig Q (upper surface coated), the remaining water loss was 15.6 mm, indicating water loss from the lower surface is approximately 15.6 - 2.2 = 13.4 mm. In Twig R (lower surface coated), the water loss was 4.8 mm, indicating water loss from the upper surface is approximately 4.8 - 2.2 = 2.6 mm. Therefore, significantly more water is lost through the lower epidermis than through the upper epidermis.

Award 1 mark for the correct answer B.

The efficiency of energy transfer from primary consumers to secondary consumers is calculated as \(\frac{\text{Energy in secondary consumers}}{\text{Energy in primary consumers}} \times 100\% = \frac{316.8}{3520} \times 100\% = 9.0\%\). Energy pyramids are never inverted because energy is always lost at each trophic level. Decomposers receive organic matter and energy from all trophic levels. The number of trophic levels is limited because energy is lost (not accumulated) between trophic levels.

Award 1 mark for the correct answer A.

The phenotypic ratio of the offspring is approximately 9 : 3 : 3 : 1, which reveals that both parents are dihybrid heterozygotes (\(BbRr\)). A brown straight individual has the genotype \(bbrr\). When \(BbRr\) is crossed with \(bbrr\), the possible offspring genotypes are \(BbRr\), \(Bbrr\), \(bbRr\), and \(bbrr\) in a 1 : 1 : 1 : 1 ratio. The only genotype homozygous for both traits is \(bbrr\), which has a probability of \(1/4 = 0.25\).

Award 1 mark for the correct answer B.

As blood flows through kidney tissues, kidney cells carry out cellular respiration, consuming oxygen and producing carbon dioxide. Consequently, the blood leaving via the renal vein has a lower oxygen concentration and a higher carbon dioxide concentration compared to the blood entering via the renal artery. Urea concentration decreases in the renal vein due to ultrafiltration and excretion in urine.

Award 1 mark for the correct answer B.

(a) Petroleum jelly forms an impermeable layer over the cuticle and stomata, preventing water vapour from evaporating through that surface.
(b) Terrestrial dicotyledonous leaves typically have many more stomata on their lower epidermis than on their upper epidermis. Shoot B only has its upper surface blocked, leaving the stomata-rich lower surface exposed, whereas Shoot C has its lower surface blocked. Hence, Shoot B will lose water at a higher rate.
(c) The upper epidermis is covered by a thick, transparent waxy cuticle which is impermeable to water, thereby reducing non-stomatal cuticular transpiration.

(a) To seal / block stomata (or prevent transpiration/water loss) on that leaf surface (1)

(b)
- Shoot B (1)
- The lower epidermis has more stomata / higher stomatal density than the upper epidermis in terrestrial dicots (1)

(c) Any one of the following:
- Presence of a (thick) waxy cuticle (1)
- Few or no stomata on the upper epidermis (1)

(a) Digested lipids (fatty acids and glycerol) enter the lacteal / lymphatic vessel inside the core of the villus.
(b) Microvilli are microscopic cellular membrane protrusions on the apical surface of intestinal epithelial cells. They dramatically increase the surface area available for transport proteins and simple diffusion, thereby increasing the rate of nutrient absorption into the blood/lymph.
(c) The hepatic portal vein drains nutrient-rich venous blood from the gastrointestinal tract directly to the liver for metabolic processing.

(a) Lacteal / lymphatic vessel (1)

(b)
- Increases surface area of the epithelial cell membrane (1)
- Increases the rate of diffusion / active transport of nutrients (1)

(c) Hepatic portal vein (1)

(a) Percentage efficiency = \( \frac{480}{4800} \times 100\% = 10\% \).
(b) Energy is lost because primary consumers do not ingest all parts of the producers (e.g. roots), not all ingested material is digested/absorbed (egested as faeces), and energy is lost as metabolic heat during respiration.
(c) Because approximately 90% of energy is lost at each trophic step, the amount of energy remaining after 4-5 trophic transfers is too small to sustain another viable top-predator population.

(a) \( 10\% \) (1)

(b) Any two of the following (1 mark each, max 2):
- Not all parts of producers are eaten / consumed by herbivores (1)
- Some consumed material is indigestible and egested in faeces (1)
- Energy is lost as heat through respiration / metabolic activities of producers/consumers (1)

(c) Progressive energy loss at each trophic level leaves insufficient energy to support another viable population / trophic level (1)

(a) Cutting and assembling underwater prevents air from entering the xylem vessels / stem (1 mark), ensuring a continuous water column is maintained for transpiration pull (1 mark).

(b)(i) Volume in 30 min = \(120\text{ mm} \times 0.8\text{ mm}^2 = 96\text{ mm}^3\).
Rate per hour = \(96\text{ mm}^3 \times 2 = 192\text{ mm}^3\text{ h}^{-1}\) (2 marks).

(ii) In windy conditions (Condition Y), moving air blows away the accumulated water vapour around the leaf surface / stomata (1 mark). This steepens the concentration gradient of water vapour between the intercellular air spaces and the surrounding atmosphere (1 mark), resulting in a faster rate of diffusion / evaporation of water vapour out of the stomata (1 mark).

(c) Some water absorbed is retained/used by plant cells for photosynthesis or maintaining cell turgor / growth (1 mark).

(d) Xylem / xylem vessels (1 mark).

(a)
- Prevent air bubbles from entering the xylem vessels / lumen (1)
- Maintain a continuous / unbroken water column / transpiration pull (1)

(b)(i)
- Correct calculation of volume: \(120 \times 0.8 = 96\text{ mm}^3\) (1)
- Correct hourly rate: \(192\text{ mm}^3\text{ h}^{-1}\) [correct unit required] (1)

(b)(ii)
- Wind removes the layer of humid air / water vapour around the leaf surface (1)
- Increases / steepens the water vapour concentration gradient between the substomatal space and the atmosphere (1)
- Increases the rate of transpiration / evaporation, pulling water up faster (1)

(c)
- A small fraction of water is consumed in photosynthesis / metabolic reactions / cell expansion / turgidity (1)

(d)
- Xylem (1)

(a) The presence of circular folds / folded mucosal lining (1 mark) and finger-like projections called villi / microvilli on epithelial cells (1 mark).

(b) Glucose and amino acids are absorbed directly into the blood capillaries within villi and transported to the liver via the hepatic portal vein (1 mark). Fatty acids and glycerol diffuse into epithelial cells, recombine into lipids/chylomicrons, and enter the lacteals / lymphatic vessels (1 mark), eventually entering the bloodstream via the thoracic duct / subclavian vein (1 mark).

(c)(i) Removal of the ileum drastically reduces the total surface area and number of carrier proteins available for nutrient absorption (1 mark), meaning digested nutrients pass out unabsorbed, leading to energy deficit and tissue breakdown (1 mark).

(ii) Reduced absorption of solutes in the small intestine leaves a higher concentration of unabsorbed nutrients in the lumen, lowering the water potential of the intestinal contents (1 mark). Less water is reabsorbed by osmosis (or water is drawn into the lumen), resulting in excess water in faeces (1 mark).

(a)
- Presence of folds / villi (1)
- Microvilli on the surface of epithelial cells (1)

(b)
- Glucose and amino acids are absorbed into blood capillaries (1)
- Transported to the liver via the hepatic portal vein (1)
- Fatty acids/glycerol (or fats/chylomicrons) enter lacteals / lymphatic system before entering the bloodstream (1)

(c)(i)
- Reduced total surface area / contact time for absorption of nutrients (1)
- Less glucose/amino acids/lipids absorbed into the blood, leading to catabolism of body reserves / weight loss (1)

(c)(ii)
- High solute concentration in lumen lowers water potential of intestinal contents (1)
- Osmotic reabsorption of water is reduced / water moves into lumen by osmosis, producing watery faeces (1)

(a) Energy is continually lost as heat during respiration and unconsumed/undigested organic matter at each successive trophic level, so available energy must strictly decrease along the food chain (2 marks). In contrast, a pyramid of numbers only counts individuals regardless of size; one single large producer (e.g. a large tree) can support many small primary consumers (e.g. insects), leading to an inverted number pyramid (1 mark).

(b)(i) Trophic efficiency = \(\frac{216}{1800} \times 100\% = 12\%\) (1 mark).

(ii) Energy lost through cellular respiration of grasshoppers as metabolic heat (1 mark); energy lost in undigested materials / faeces / egestion (1 mark).

(c) Hawks (1 mark); Biomagnification / biological magnification (1 mark).

(d) They act as decomposers, breaking down dead organic matter into simple inorganic nutrients (e.g. nitrates, phosphates) to be recycled for plant uptake (1 mark).

(a)
- Energy is lost at each trophic level via respiration / metabolic heat / excretion (1)
- Only ~10% of energy is passed to the next level, so energy content must decrease progressively (1)
- Number pyramid ignores biomass/size of individuals; one large producer can support numerous small consumers (1)

(b)(i)
- \(\frac{216}{1800} \times 100\% = 12\%\) (1)

(b)(ii) Any two:
- Heat lost during cellular respiration (1)
- Unconsumed parts / uneaten body parts of grasshoppers (1)
- Undigested food / faeces / egested waste (1)

(c)
- Hawk / top carnivore (1)
- Biomagnification / bioaccumulation along food chain (1)

(d)
- Decompose dead organic matter to recycle inorganic nutrients / minerals to the soil (1)

(a)(i) Genotype: \(RrSs\) (1 mark). Phenotype: Red flower and smooth seed (1 mark).
(ii) 9 red smooth : 3 red wrinkled : 3 white smooth : 1 white wrinkled (1 mark).

(b)(i) A cross between an individual of unknown or heterozygous genotype with a homozygous recessive individual (1 mark).
(ii) In a dihybrid test cross with independent assortment, the theoretical phenotypic ratio is 1 : 1 : 1 : 1 (1 mark). The observed ratio of 204 : 198 : 202 : 196 is approximately 1 : 1 : 1 : 1 (1 mark), which confirms that the alleles of the two genes segregate and assort independently during gamete formation (1 mark).

(c) The parental combinations were \(Rs\) and \(rS\). With complete linkage and no crossing over, the \(F_1\) produces only \(Rs\) and \(rS\) gametes (1 mark). Test-crossing with \(rs\) gives a 1 : 1 ratio of Red wrinkled : White smooth (1 mark).

(a)(i)
- Genotype: \(RrSs\) (1)
- Phenotype: Red flower, smooth seed (1)

(a)(ii)
- 9 : 3 : 3 : 1 (1)

(b)(i)
- Mating an individual with a homozygous recessive individual to determine genotype / test linkage (1)

(b)(ii)
- Expected ratio for independent assortment in a test cross is 1 : 1 : 1 : 1 (1)
- Observed counts (204 : 198 : 202 : 196) match closely with ~1 : 1 : 1 : 1 / equal proportions (1)
- Demonstrates that the two genes assort independently on different chromosomes (1)

(c)
- Only parental phenotypes appear / recombinant phenotypes absent (1)
- Phenotypic ratio: 1 Red wrinkled : 1 White smooth (1)

(a) Count the number of oxygen bubbles released from the cut stem per unit time (or measure the volume of gas collected using a gas syringe over a fixed time) (2 marks).

(b)(i) Light intensity is no longer the limiting factor (1 mark). Another factor, such as carbon dioxide concentration or temperature, has become the limiting factor for enzyme-catalyzed reactions in the Calvin cycle (1 mark).

(ii) Increase the concentration of \(\text{NaHCO}_3\) / dissolved \(\text{CO}_2\) in the water (1 mark). More carbon dioxide provides more substrate for carbon fixation by Rubisco, increasing the overall rate of photosynthesis (1 mark).

(c) ATP provides energy (1 mark), and NADPH provides reducing power / hydrogen atoms (1 mark) to reduce glycerate 3-phosphate (GP / 3-PGA) to triose phosphate (TP).

(d) Stroma of the chloroplast (1 mark).

(a)
- Counting number of oxygen bubbles produced per unit time / measuring volume of oxygen gas collected with syringe (1)
- Fixed duration / time period specified (1)

(b)(i)
- Light intensity is no longer the limiting factor (1)
- Another factor (e.g. \(\text{CO}_2\) availability / temperature / enzyme activity) becomes limiting (1)

(b)(ii)
- Increase \(\text{NaHCO}_3\) concentration / temperature (within optimal range) (1)
- Explanation: \(\text{CO}_2\) is essential substrate for carbon fixation / higher temperature increases kinetic energy of enzymes (1)

(c)
- ATP provides chemical energy for phosphorylation / conversion of GP to triose phosphate (1)
- NADPH provides hydrogen / electrons for reduction of 3-PGA to TP (1)

(d)
- Stroma (1) [Reject: matrix / cytoplasm / thylakoid]

(a) Vasodilation occurs: smooth muscles in the walls of arterioles near the skin surface relax (1 mark), increasing the diameter/lumen of the arterioles so more blood flows to superficial capillary networks near the skin surface (1 mark). Heat is lost more rapidly to the surroundings by radiation, convection, and conduction (1 mark).

(b)(i) Lactic acid releases hydrogen ions (\(\text{H}^+\)), decreasing blood pH / making blood more acidic (1 mark).

(ii) The drop in pH / increase in \(\text{H}^+\) is detected by central chemoreceptors in the medulla oblongata and peripheral chemoreceptors in the carotid and aortic bodies (1 mark). Nerve impulses are sent at higher frequency to the respiratory centre in the medulla oblongata (1 mark). The respiratory centre sends more nerve impulses via the phrenic and intercostal nerves to the diaphragm and external intercostal muscles, increasing the rate and depth of contraction (1 mark).

(c) Essential enzymes in metabolic pathways become denatured, leading to heat stroke / cell death / organ failure (1 mark).

(a)
- Smooth muscle of arterioles relaxes / arterioles dilate (vasodilation) (1)
- More blood diverted / flows through superficial capillaries in the skin dermis (1)
- Facilitates heat loss to surroundings by radiation / convection (1)

(b)(i)
- Lowers blood pH / increases acidity (1)

(b)(ii)
- Chemoreceptors (carotid/aortic bodies or medulla) detect fall in blood pH (1)
- Impulses transmitted to respiratory centre in medulla oblongata (1)
- Increased nerve impulses sent to intercostal muscles and diaphragm to contract faster and more forcefully (1)

(c)
- Denaturation of enzymes / loss of cell function / death (1)

(a) Helper T cells / \(\text{CD4}^+\) T lymphocytes (1 mark).

(b) Helper T cells play a central role in activating both B lymphocytes (for antibody production) and cytotoxic T cells (for destroying infected host cells) (1 mark). As helper T cell numbers decline drastically, antibody production and cell-mediated immunity are severely impaired (1 mark). The body can no longer mount an effective immune response against common pathogens, leading to opportunistic infections (1 mark).

(c)(i) Active immunity: antibodies are produced by the individual's own plasma cells / B cells; Passive immunity: antibodies are received from an external source (e.g. maternal transfer or injection of antiserum) (1 mark).
(ii) Active immunity gives long-term protection due to memory cells; passive immunity gives short-term / temporary protection as no memory cells are formed (1 mark).

(d) HIV is an RNA virus with high mutation rates during reverse transcription (1 mark). This alters the antigenic structure / surface glycoproteins of the virus rapidly, so antibodies and memory cells produced by a specific vaccine fail to recognize mutated strains (1 mark).

(a)
- Helper T cells / \(\text{CD4}^+\) T lymphocytes (1)

(b)
- Helper T cells secrete cytokines / activate B cells and cytotoxic T cells (1)
- Destruction of helper T cells impairs both humoral and cell-mediated immunity (1)
- Pathogens cannot be eliminated efficiently by the weakened immune system (1)

(c)(i)
- Active: produced by own body / plasma cells; Passive: received from outside / preformed antibodies (1)

(c)(ii)
- Active: long-lasting (due to memory cells); Passive: short-lived / temporary (antibodies degraded, no memory cells) (1)

(d)
- High mutation rate in viral RNA / surface glycoproteins (1)
- Antigenic variation / antibodies produced by vaccine cannot bind to mutated viral antigens (1)

### Structural adaptations for photosynthesis and transpiration pull (max. 5 marks)
- Broad and flat lamina / leaf blade: Provides a large surface area for absorbing sunlight and taking up carbon dioxide.
- Thin leaf lamina / transparent cuticle and epidermis: Allows light to penetrate easily to the photosynthetic cells below and shortens diffusion distance for gases.
- Palisade mesophyll: Densely packed, vertically elongated cells containing abundant chloroplasts located right beneath the upper epidermis to maximize light capture for photosynthesis.
- Spongy mesophyll with large intercellular air spaces: Allows rapid diffusion and circulation of gases (\(\text{CO}_2\) and \(\text{O}_2\)) for photosynthesis, and provides a large, moist internal surface area for water to evaporate into air spaces.
- Presence of stomata: Facilitates the entry of \(\text{CO}_2\) required for carbon fixation in photosynthesis; the diffusion of water vapour out of stomata creates a water potential gradient between the leaf and atmosphere, generating the transpiration pull.
- Vascular bundles (xylem): Branch extensively throughout the leaf, supplying water and dissolved mineral ions to photosynthesizing mesophyll cells and sustaining the continuous water column for the transpiration stream.

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### Physiological mechanisms to reduce water loss under hot, dry conditions (max. 3 marks)
- Stomatal closure: When the rate of transpiration exceeds the rate of water absorption by roots, guard cells lose water and turgidity, causing the stomatal pore to close (or narrow), which drastically reduces further loss of water vapour through transpiration.
- Wilting / leaf rolling or drooping: Loss of turgor in leaf cells causes leaves to droop or curl, which reduces the surface area directly exposed to sunlight and dry air, lowering the leaf temperature and transpiration rate.

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### Effective Communication (3 marks)
- 3 marks: Answers are well-structured, showing coherence of thought and clear logical presentation of biological concepts with appropriate scientific terminology.
- 2 marks: Answers are understandable and organized, though with minor repetition or slight lack of fluency.
- 1 mark: Fragmented points, noticeable disorganization, or difficult to follow.
- 0 marks: Irrelevant material or incomprehensible answer.

Concept for mark award:

1. Structural adaptations of the leaf for photosynthesis and transpiration pull (max. 5 marks):
* Broad / flat leaf blade provides a large surface area for sunlight absorption / \(\text{CO}_2\) uptake (1)
* Thin leaf / transparent cuticle and epidermis allows light penetration / shortens gas diffusion distance (1)
* Palisade mesophyll cells packed with chloroplasts near the upper surface to maximize light absorption for photosynthesis (1)
* Spongy mesophyll with loose arrangement / large intercellular air spaces facilitates gas diffusion / provides large moist surface area for evaporation of water (1)
* Stomata allow diffusion of \(\text{CO}_2\) into the leaf, and evaporation/diffusion of water vapour creates transpiration pull (1)
* Extensive xylem network / veins supply water to mesophyll cells for photosynthesis and maintain continuous transpiration stream (1)

2. Mechanisms to reduce water loss under hot and dry conditions (max. 3 marks):
* Transpiration exceeding water absorption leads to water deficit / loss of turgor in guard cells (1)
* Guard cells become flaccid leading to stomatal closure / reduction in stomatal aperture, significantly decreasing transpiration (1)
* Wilting / drooping / curling of leaves reduces surface area exposed to direct sunlight / dry air, decreasing heat absorption and water loss (1)

(Content marks: max. 8 marks)

3. Effective communication (0–3 marks):
* 3 marks: Answer is easy to understand, well structured, shows fluent and coherent presentation of ideas with accurate biological terminology.
* 2 marks: Answer is understandable with clear ideas, but contains some repetition or slight organizational flaws.
* 1 mark: Ideas are disorganized, paragraphing is poor, or scientific terminology is lacking.
* 0 marks: Incomprehensible or fails to address the prompt.

(a)(i) From 0.0 hr to 1.0 hr, the urine production rate of Subject M increased dramatically from 1.2 mL min^-1 to a peak of 11.5 mL min^-1, while the urine solute concentration dropped sharply from 850 mOsm L^-1 to 95 mOsm L^-1.

(a)(ii) Drinking pure distilled water caused water to be absorbed into the bloodstream, lowering the osmolarity / solute concentration (or raising the water potential) of the blood plasma. Osmoreceptors in the hypothalamus detected this decrease in blood osmolarity (increase in water potential) and stimulated the posterior pituitary gland to decrease / inhibit the secretion of antidiuretic hormone (ADH) into the blood. With a lower circulating level of ADH, the permeability of the epithelial cells in the distal convoluted tubules and collecting ducts of the kidneys to water decreased (fewer aquaporins). Consequently, less water was reabsorbed from the filtrate into the peritubular capillaries, resulting in the excretion of a large volume of dilute urine (diuresis).

(a)(iii) The 0.9% NaCl solution consumed by Subject N is isotonic to blood plasma. Therefore, its absorption expanded the extracellular fluid volume without altering the blood osmolarity / water potential. Because blood osmolarity remained unchanged, osmoreceptors in the hypothalamus were not stimulated to decrease ADH secretion; hence, ADH levels and renal water reabsorption remained relatively stable.

(a)(iv) The urine production rate would remain low / would not increase significantly. This is because synthetic ADH would keep the permeability of the collecting ducts and distal convoluted tubules high, causing continued high reabsorption of water back into the blood despite the lowered plasma osmolarity.

(b)(i) Hypothalamus.

(b)(ii) When core body temperature rises, central thermoreceptors in the hypothalamus (and peripheral thermoreceptors in the skin) detect the elevated blood temperature and send nerve impulses to the heat loss centre in the hypothalamus. The heat loss centre sends motor nerve impulses via the autonomic (sympathetic) nervous system to cause vasodilation of arterioles supplying the superficial capillary beds of the skin (and relaxation of precapillary sphincters / reduction of arteriovenous shunt flow), allowing more warm blood to flow near the skin surface for heat dissipation via radiation, conduction, and convection.

(b)(iii) High relative humidity reduces the water vapor concentration gradient between the wet skin surface and the surrounding air, which significantly lowers the rate of evaporation of sweat. Because latent heat of vaporisation cannot be efficiently removed from the skin, heat accumulates in the body, which can lead to hyperthermia / heat exhaustion / heat stroke / organ failure.

(b)(iv) Redistribution of blood flow ensures that active skeletal muscles receive a dramatically increased supply of oxygen and glucose to sustain a high rate of aerobic respiration for ATP generation during muscle contractions, while rapidly removing metabolic waste products like lactic acid, carbon dioxide, and excess heat. Vasoconstriction in the digestive tract allows cardiac output to be conserved and preferentially directed to working muscles without causing an unsustainable drop in systemic blood pressure.

(a)(i) [2 marks]
• Urine production rate increases markedly / reaches peak (from 1.2 to 11.5 mL min^-1) (1)
• Urine solute concentration decreases markedly (from 850 to 95 mOsm L^-1) (1)

(a)(ii) [4 marks]
• Ingestion of water increases blood water potential / decreases blood osmolarity (1)
• Detected by osmoreceptors in the hypothalamus (1)
• Posterior pituitary releases less / inhibits secretion of ADH (1)
• Permeability of collecting duct / distal convoluted tubule to water decreases, resulting in less water reabsorption (1)

(a)(iii) [2 marks]
• Isotonic saline has the same osmolarity / water potential as plasma, so blood osmolarity remains unchanged (1)
• Osmoreceptors are not stimulated / ADH secretion is not suppressed, so water reabsorption remains unchanged (1)

(a)(iv) [2 marks]
• Urine output remains low / does not surge (1)
• Circulating ADH keeps collecting ducts highly permeable to water, maintaining high water reabsorption (1)

(b)(i) [1 mark]
• Hypothalamus (1)

(b)(ii) [3 marks]
• Thermoreceptors detect temperature rise and send nerve impulses to the heat loss centre of the hypothalamus (1)
• Hypothalamus sends impulses along autonomic nerves to skin arterioles (1)
• Arterioles supplying superficial skin capillaries dilate / vasodilation occurs, increasing blood flow to skin surface (1)

(b)(iii) [3 marks]
• Smaller water vapor concentration gradient between skin and air (1)
• Lowers rate of sweat evaporation, reducing removal of latent heat (1)
• Danger: heat exhaustion / heat stroke / dangerously high core temperature / hyperthermia (1)

(b)(iv) [3 marks]
• Delivers more oxygen and glucose to skeletal muscles (1)
• Supports elevated aerobic cellular respiration / ATP production for muscle contraction (1)
• Accelerates removal of metabolic wastes (CO₂ / lactate) / excess heat from working muscles (1)

(a)(i) Both EcoRI and HindIII should be used simultaneously (double digestion). Using two distinct restriction enzymes that produce two different non-compatible sticky ends ensures that the Factor IX cDNA fragment can only ligate into the vector in one specific, correct orientation (directional cloning). Furthermore, it prevents the digested plasmid vector from undergoing self-ligation without the insert.

(a)(ii) DNA ligase.

(a)(iii) Transformation mixtures are first plated on medium containing ampicillin. Only bacteria that have taken up a plasmid (whether intact or recombinant) can survive because the plasmid carries the intact amp^R gene (selection for transformants). Colonies from this master plate are then replica-plated onto a medium containing tetracycline. Because the Factor IX cDNA was inserted into the multiple cloning site inside the tet^R gene, insertional inactivation occurs, disrupting the tet^R gene. Thus, colonies harboring the recombinant plasmid can grow on ampicillin plates but CANNOT grow on tetracycline plates.

(a)(iv) Bacteria are prokaryotes and lack the cellular machinery (endoplasmic reticulum and Golgi apparatus) for proper post-translational modifications, such as correct glycosylation, gamma-carboxylation (essential for Factor IX), or proper protein folding and disulfide bond formation.

(b)(i)
1. Denaturation (at approx. 94–95 °C): High temperature breaks the hydrogen bonds between complementary strands of template DNA, separating them into single strands.
2. Annealing (at approx. 50–60 °C): Lower temperature allows specific DNA primers to bind / hybridize to their complementary sequences on the single-stranded template DNA flanking the target STR region.
3. Extension / Elongation (at approx. 72 °C): Thermostable DNA polymerase (e.g. Taq polymerase) synthesises the new complementary DNA strands by adding free deoxynucleotide triphosphates (dNTPs) extending from the primers in the 5' to 3' direction.

(b)(ii) Alleged Father 2 (AF2) is the biological father.
Explanation:
• For Locus 1: The child has alleles (14, 16). The mother contributes the allele '14', so the paternal allele must be '16'. Both AF1 (14, 16) and AF2 (12, 16) possess allele 16.
• For Locus 2: The child has alleles (8, 10). The mother contributes allele '8', so the paternal allele must be '10'. AF1 has (10, 12) and AF2 has (8, 10); both possess allele 10.
• For Locus 3: The child has alleles (15, 18). The mother has (15, 17) and contributes allele '15', meaning the child must have inherited the paternal allele '18'. AF1 has alleles (17, 19) and lacks allele 18, which completely excludes AF1. AF2 has alleles (18, 20) and possesses the required allele 18. Therefore, AF2 is the biological father.

(b)(iii) A single STR locus has limited allele variation in the general human population; two unrelated individuals could easily share the same alleles at one locus by random coincidence. Analysing multiple independent STR loci drastically reduces the probability of a coincidental match (random match probability), making the power of discrimination extremely high (approaching 1 in billions).

(b)(iv) Violation of genetic privacy / potential misuse of genetic data by third parties (insurers, employers) / stigmatisation and discrimination / potential risk of unauthorized surveillance or data breaches.

(a)(i) [3 marks]
• EcoRI and HindIII (1)
• Generates two distinct / non-complementary sticky ends (1)
• Ensures directional insertion of the cDNA / prevents vector self-ligation (1)

(a)(ii) [1 mark]
• DNA ligase (1)

(a)(iii) [4 marks]
• Cells plated on ampicillin medium select for transformed bacteria containing plasmids (1)
• Insertion of target gene into MCS disrupts / inactivates the tet^R gene / insertional inactivation (1)
• Recombinant colonies grow on ampicillin agar (1)
• Recombinant colonies fail to grow / die on tetracycline agar (1)

(a)(iv) [2 marks]
• Bacteria lack post-translational modification mechanisms (e.g. glycosylation / carboxylation / phosphorylation) (1)
• Bacteria fail to fold eukaryotic proteins correctly into their native tertiary / quaternary structure (1)

(b)(i) [3 marks]
• Denaturation (90–95 °C): separates double-stranded DNA into single strands / breaks hydrogen bonds (1)
• Annealing (50–65 °C): primers bind / hybridise specifically to flanking regions of template DNA (1)
• Extension (70–75 °C): Taq polymerase synthesises complementary DNA strand using dNTPs (1)

(b)(ii) [4 marks]
• Identification that child receives one allele from mother and one from biological father at each locus (1)
• Locus 3 paternal allele required is 18 (1)
• AF1 lacks allele 18 / possesses 17 and 19, hence AF1 is excluded (1)
• AF2 possesses allele 18 (and matches alleles 16 at locus 1 and 10 at locus 2), confirming AF2 as biological father (1)

(b)(iii) [2 marks]
• Different individuals may share the same allele profile at one locus purely by chance / coincidence (1)
• Testing multiple loci multiplies individual match probabilities together, dramatically decreasing the random match probability / providing near-unique identification (1)

(b)(iv) [1 mark]
• Privacy intrusion / potential misuse of sensitive genetic information / discrimination / risk of false matches from contamination (1)