2023 HKDSE Biology Answers & Marking Scheme

Active transport of sucrose into sieve tubes (phloem loading) and the uptake of mineral ions against their concentration gradient from a dilute soil solution both require ATP. The movement of water across the root cortex occurs passively by osmosis down a water potential gradient.

1 mark for selecting B. Correctly identifies active and passive processes in plants.

During the light-dependent reactions of photosynthesis, photolysis of water splits water molecules into hydrogen ions, electrons, and oxygen gas. Therefore, all oxygen gas released during photosynthesis originates from water, meaning the oxygen gas will contain \(^{18}\text{O}\).

1 mark for B. Demonstrates understanding that the oxygen released in photosynthesis comes entirely from water.

In daylight, potassium ions (\(\text{K}^+\)) are actively transported into guard cells, which lowers (not increases) their water potential. Water then enters the guard cells by osmosis down the water potential gradient, causing them to become turgid and bow outward, opening the pore.

1 mark for B. Correctly identifies the ionic and osmotic changes during stomatal opening.

Statement I is correct because energy is lost at each trophic level as heat and during transfer, making energy pyramids always upright. Statement II is incorrect because a single large producer (e.g., a tree) can support many primary consumers, creating an inverted or spindle-shaped pyramid of numbers. Statement III is correct because rapid turnover rates of phytoplankton in oceans can result in a smaller standing crop biomass than that of the zooplankton feeding on them.

1 mark for B. Correctly identifies the properties and exceptions of various ecological pyramids.

Waterlogged soils lack oxygen, creating anaerobic conditions. Denitrifying bacteria thrive in anaerobic conditions and convert soil nitrates into gaseous nitrogen (\(\text{N}_2\)), which escapes into the atmosphere. This process reduces the level of nitrates available to plants, lowering soil fertility.

1 mark for C. Understands the anaerobic nature of denitrifying bacteria and its ecological impact.

Let the yellow allele be \(Y\) and the grey allele be \(y\). Since yellow is dominant, yellow mice must carry at least one \(Y\) allele. In a cross between two yellow mice (\(Yy \times Yy\)), the expected genetic ratio is \(1 YY : 2 Yy : 1 yy\). If the homozygous dominant condition (\(YY\)) is lethal, the embryo dies before birth, resulting in a phenotypic ratio of 2 yellow (\(Yy\)) to 1 grey (\(yy\)).

1 mark for C. Correctly interprets the lethal allele hypothesis based on the 2:1 inheritance ratio.

In independent assortment, a test cross of \(AaBb \times aabb\) should yield approximately equal proportions (1:1:1:1) of the four phenotypes. The significant deviation, with high numbers of parental phenotypes (AB and ab) and lower numbers of recombinant phenotypes (Ab and aB), indicates that the genes \(A\) and \(B\) are linked on the same chromosome, and the recombinant types arose from crossing over during meiosis.

1 mark for B. Correctly identifies gene linkage and crossing over based on phenotypic ratios.

Pepsin is a protease produced by gastric glands in the stomach wall, and it functions in the stomach to hydrolyze proteins into polypeptides. Bile is produced in the liver (only stored in the gall bladder). Pancreatic amylase hydrolyzes starch to maltose (not glucose). Lipase hydrolyzes lipids into fatty acids and glycerol, whereas bile emulsifies lipids.

1 mark for C. Correctly identifies the production site and specific function of human digestive enzymes.

In red blood cells, carbon dioxide is converted to carbonic acid by carbonic anhydrase (I), which dissociates into hydrogen ions and hydrogencarbonate ions (\(\text{HCO}_3^-\)). \(\text{HCO}_3^-\)) diffuses out into the plasma down its concentration gradient (II). To balance this loss of negative charge, chloride ions (\(\text{Cl}^-\)) diffuse into the red blood cells (III), a phenomenon known as the 'chloride shift'.

1 mark for D. Demonstrates complete understanding of carbon dioxide transport mechanisms and the chloride shift.

In capillaries, the capillary wall is permeable to small solutes like water, ions, and glucose, but impermeable to large plasma proteins and blood cells. Therefore, blood plasma contains a significantly higher concentration of proteins than tissue fluid. Red blood cells stay inside capillaries and are not found in tissue fluid. Hydrostatic pressure of blood plasma is higher than that of tissue fluid at the arteriole end, causing filtration.

1 mark for B. Correctly identifies differences in composition and pressure between blood plasma and tissue fluid.

Let U be the stomatal transpiration from the upper epidermis, L be that from the lower epidermis, and C be the cuticular transpiration. From treatment (4), the cuticular transpiration C = 1 mm/min. From treatment (2), upper coated leaving lower and cuticle: L + C = 8, so L = 8 - 1 = 7 mm/min. From treatment (3), lower coated leaving upper and cuticle: U + C = 3, so U = 3 - 1 = 2 mm/min. Therefore, the ratio of stomatal transpiration of upper to lower epidermis is U : L = 2:7.

Award 1 mark for the correct option B. Correct analysis shows that upper stomatal transpiration is 2 mm/min and lower stomatal transpiration is 7 mm/min.

Phosphate ions are absorbed by root hair cells via active transport, which requires ATP from aerobic respiration. (1) Bubbling nitrogen gas deprives the root cells of oxygen, inhibiting aerobic respiration. (2) A respiratory inhibitor directly blocks ATP synthesis. Both will significantly decrease active transport. (3) Increasing light intensity increases transpiration and photosynthesis but does not decrease active transport of ions. Thus, only (1) and (2) are correct.

Award 1 mark for option A. Statements (1) and (2) describe conditions that limit ATP production, which is necessary for active transport of ions.

Ring-barking removes the phloem, blocking the translocation of organic nutrients (sugars) from the leaves to the roots. Since roots cannot photosynthesize, they rely on translocated sugars for cellular respiration to produce ATP. Over several weeks, the lack of sugars inhibits respiration, leading to a shortage of ATP, which reduces the active transport of mineral ions by the roots. Option A is incorrect because swelling occurs above the ring. Option B is incorrect because accumulation of sugars decreases water potential. Option D is incorrect as root damage eventually reduces water uptake and transpiration.

Award 1 mark for option C. Ring-barking blocks translocation to the roots, starving them of respiratory substrates, reducing active transport.

Phytoplankton (primary producers) have a very high rate of reproduction and turnover. Although their standing crop biomass at any single moment is small, they produce biomass rapidly enough to support a larger biomass of zooplankton (primary consumers). The pyramid of energy must always be upright because energy is lost as heat and through waste at each trophic level.

Award 1 mark for option A. The high turnover rate of producers allows a small standing crop to support a larger consumer biomass.

Process W is nitrogen fixation, which can be done by free-living or symbiotic bacteria, or lightning. Process X is nitrification, which is carried out by aerobic nitrifying bacteria and is inhibited in waterlogged soils. Process Y is denitrification, which is performed by anaerobic denitrifying bacteria in anaerobic environments (like waterlogged soils). Thus, option C is correct.

Award 1 mark for option C. Denitrification (Process Y) is an anaerobic process carried out by denitrifying bacteria.

During primary succession, pioneer species weather the rock and die, adding organic matter. Over time, soil depth increases, which allows larger plants to grow, leading to an increase in species diversity and biomass. This increases food web complexity and community stability. Thus, option B is correct.

Award 1 mark for option B. Succession leads to increased soil depth, species richness, biomass, and community stability.

Since unaffected parents have affected children, the disease-causing allele must be recessive. If the allele were X-linked recessive, the affected daughter must inherit one recessive allele from her father, which means her father would also carry the recessive allele and express the disease. However, the father is unaffected. Therefore, the allele cannot be on the X chromosome and must be autosomal recessive.

Award 1 mark for option B. An affected daughter from unaffected parents rules out both dominant inheritance and X-linked recessive inheritance.

The parent plant has the genotype AaBb, which exhibits the dominant phenotypes (round and yellow seeds). For the offspring to have the same phenotype, it must have at least one dominant allele for both genes (genotype A_B_). Since the genes assort independently: (1) Probability of expressing the dominant phenotype for shape (A_) = \frac{3}{4}; (2) Probability of expressing the dominant phenotype for color (B_) = \frac{3}{4}. Therefore, the probability of having both dominant phenotypes is \frac{3}{4} \times \frac{3}{4} = \frac{9}{16}.

Award 1 mark for option C. Calculation: \frac{3}{4} \times \frac{3}{4} = \frac{9}{16}.

Fatty acids and glycerol are lipid-soluble products of lipid digestion. They can easily dissolve in and diffuse across the phospholipid bilayer of the microvilli membrane by passive simple diffusion. Glucose, amino acids, and sodium ions are polar or charged, and require specific transport proteins, often involving active transport or facilitated diffusion.

Award 1 mark for option C. Lipid-soluble molecules like fatty acids diffuse directly across the cell membrane without requiring ATP.

Two hours after a carbohydrate-rich meal, a large amount of glucose is absorbed from the small intestine into the hepatic portal vein, resulting in a high glucose concentration. As this blood passes through the liver, insulin stimulates liver cells to convert excess glucose into glycogen for storage. Therefore, the blood leaving the liver via the hepatic vein has a lower, regulated glucose concentration. Urea is produced in the liver, so the hepatic vein has a higher urea concentration than the hepatic portal vein.

Award 1 mark for option B. After a meal, the liver removes glucose from the hepatic portal blood, making hepatic vein glucose lower.

The symplastic pathway involves the movement of water from cell to cell via the cytoplasm and plasmodesmata. The apoplastic pathway involves water movement through cell walls and intercellular spaces. The Casparian strip blocks the apoplastic pathway, forcing water to enter the symplast. Water movement is passive.

1 mark for identifying that symplastic transport involves plasmodesmata (C is correct). Reject options associated with apoplastic pathway (A, B) or active transport of water (D).

In marine ecosystems, the primary producers (phytoplankton) are highly productive but have a very small standing crop (biomass) at any one time because they are consumed rapidly by zooplankton. This results in an inverted pyramid of biomass. Pyramids of energy can never be inverted because energy is lost at each trophic level.

1 mark for selecting B. Option A is incorrect because energy pyramids are never inverted. Option C is incorrect because biomass and number pyramids often differ. Option D is incorrect as energy is lost as heat.

If a healthy couple has an affected daughter: (1) and (2) are ruled out because dominant disorders require at least one parent to be affected. (3) is also ruled out because an affected female (XaXa) must inherit one recessive allele from her father, which would make the father affected (XaY). Since her father is healthy, X-linked recessive is ruled out.

1 mark for D. (1), (2), and (3) are all logically impossible for healthy parents with an affected daughter.

Pancreatic lipase breaks down lipids into fatty acids and glycerol. The production of fatty acids increases the concentration of hydrogen ions, lowering the pH. Lipase works best at its optimal temperature of 37 °C and in the presence of bile salts, which emulsify lipids to increase the surface area for enzyme action. Therefore, Tube 1 has the fastest rate of reaction and the fastest decrease in pH.

1 mark for A. Tube 2 has denatured enzyme, Tube 3 lacks bile salts, and Tube 4 is at a temperature that deactivates enzyme kinetics, so they all have slower or no pH change.

Drinking a large volume of water dilutes the blood, lowering the blood osmotic pressure. The osmoreceptors in the hypothalamus detect this change and stimulate the pituitary gland to secrete less ADH. With lower ADH concentration in the blood, the cells of the collecting ducts become less permeable to water, reducing water reabsorption and resulting in a large volume of dilute urine.

1 mark for B. Correctly matching the decrease in blood osmotic pressure to decreased ADH and reduced permeability of collecting ducts. Reject A (incorrect physiological response), C (water moves passively), and D (GFR is stable).

Photolysis of water splits water molecules into oxygen, protons (\(\text{H}^+\)), and electrons. These electrons and protons are accepted by \(\text{NADP}^+\) to form \(\text{NADPH}\), which is then used in the light-independent reactions (Calvin cycle) to reduce 3-phosphoglycerate (3-PGA) to triose phosphate.

1 mark for B. Oxygen is not used as an electron acceptor in the Calvin cycle (A). Carbon dioxide fixation is mediated by RuBP, not free electrons (C). Respiration, not photosynthesis, forms water from oxygen and protons (D).

The diploid number is \(2n = 8\). During meiosis I, homologous chromosomes separate, so each daughter cell gets \(n = 4\) chromosomes. At metaphase II, these 4 chromosomes align at the equator. Since the sister chromatids have not yet separated, each chromosome consists of two sister chromatids. Therefore, there are 4 chromosomes and \(4 \times 2 = 8\) DNA molecules in a single cell at metaphase II.

1 mark for B. Option A describes the state after anaphase II / telophase II. Options C and D contain incorrect chromosome counts for haploid metaphase II.

An injection of ready-made antibodies (immunoglobulins) provides passive artificial immunity. This type of immunity provides immediate protection because the antibodies are already present to neutralize the toxin, but it is temporary because the injected antibodies break down over time and no memory cells are produced. Therefore, a subsequent exposure will not trigger a rapid secondary response.

1 mark for B. Active immunity is incorrect because no host immune response or memory cells are generated (reject A, C, D).

A single base deletion results in a frameshift mutation. From the point of deletion onwards, the triplet reading frame of the mRNA is shifted by one base. This changes all subsequent codons, leading to a completely different sequence of amino acids in the polypeptide and potentially altering the position of the stop codon.

1 mark for C. Frame-shift mutations alter all downstream codons, not just one (reject A). Translation continues until a stop codon is met (reject B). Transcription and translation can still occur, but the product is altered (reject D).

Transpiration is driven by the water vapor concentration gradient between the air spaces inside the leaf and the external air. Low relative humidity of external air decreases its water vapor potential, increasing the gradient. High wind speed sweeps away the water vapor boundary layer around the leaves, keeping the external humidity near the stomata low. High temperature increases the rate of evaporation of water from the mesophyll cell walls. Thus, low humidity, high wind speed, and high temperature maximize transpiration.

1 mark for C. High humidity reduces the concentration gradient (reject A, D). Low wind speed and low temperature reduce evaporation and diffusion rates (reject B).

(1) is correct because most stomata of dicotyledonous plants are located on the lower epidermis; smearing vaseline on the lower surface blocks the stomata, significantly reducing transpiration and thus water uptake. (2) is correct because enclosing the shoot in a bag with water increases relative humidity, reducing the water vapor concentration gradient between the leaf interior and the air, which decreases transpiration and water uptake. (3) is incorrect because wind increases the rate of transpiration by removing water vapor from around the leaves, which would speed up the movement of the air bubble.

1 mark for option A.

\(^{14}\text{CO}_2\) from the atmosphere enters the leaf through stomata into the air spaces, then diffuses into mesophyll cells where photosynthesis occurs to synthesize radioactive sugars. These organic nutrients are loaded into the phloem sieve tubes and translocated downwards to the roots. In the roots, the sugars are stored as starch in the cortical cells of the cortex.

1 mark for option A.

Process 2 is nitrification, converting ammonium compounds into nitrates. This aerobic process is carried out by nitrifying bacteria which require oxygen. Process 1 is decomposition/ammonification by decomposers. Process 3 is denitrification carried out by denitrifying bacteria under anaerobic conditions.

1 mark for option B.

(1) is correct: Normal parents having an affected offspring indicates the disease allele is recessive. (2) is correct: If the disease were X-linked recessive, the affected daughter must have an affected father, but her father is normal, hence it must be autosomal. (3) is correct: Since the affected daughter is homozygous recessive (aa) and she has an affected son (aa), the son must inherit one recessive allele from his mother and the other from his normal father, making the normal father heterozygous (Aa).

1 mark for option D.

Amino acids are hydrophilic and absorbed into the blood capillaries of villi in the ileum. They travel through the hepatic portal vein to the liver, and then via the hepatic vein to the inferior vena cava and into the heart. Glucose follows this same capillary-blood pathway, not the lacteal pathway. Fatty acids and glycerol are absorbed into the lacteals and transported via the lymphatic system to the subclavian vein.

1 mark for option B.

The yellow-brown result in Test 1 indicates starch is absent. The blue result in Test 2 indicates reducing sugars are absent. The purple color in Test 3 indicates proteins are present. The brick-red precipitate in Test 4 indicates non-reducing sugars (e.g., sucrose) are present because acid hydrolysis breaks them down into reducing sugars, which then react with Benedict's reagent. Therefore, the sample contains non-reducing sugar and protein.

1 mark for option C.

Prediction: The rate of water uptake decreases. Explanation: Active transport is responsible for absorbing mineral ions against their concentration gradient into the root hair cells. With active transport inhibited, the accumulation of mineral ions in the root cells decreases, leading to a higher water potential (less negative) inside the root cells. Consequently, the water potential gradient between the soil solution and the root cells is reduced, leading to a slower rate of water entry into the root cells by osmosis.

- Prediction: Rate of water uptake decreases / drops. [1 mark] - Explanation: Without active transport, root hair cells cannot absorb mineral ions against the concentration gradient / accumulate mineral ions. [1 mark]; This results in a decrease in the solute concentration inside the root hair cells / an increase in water potential inside the root cells. [1 mark]; The water potential gradient between the soil and the root cells decreases, resulting in less water being absorbed by osmosis. [1 mark]

1. Mutualistic relationship: Legumes have root nodules containing nitrogen-fixing bacteria (e.g., Rhizobium). 2. Nitrogen fixation: These bacteria fix atmospheric nitrogen gas into nitrogenous compounds (such as ammonium ions). 3. Enrichment of soil: When legumes shed leaves or their roots die and decompose, these nitrogen compounds are released into the soil, increasing soil nitrogen content. 4. Corn growth: Corn plants absorb the released nitrogen (as nitrates) from the soil and use it to synthesize essential molecules like proteins and chlorophyll, enhancing their growth.

- Legumes contain nitrogen-fixing bacteria / Rhizobium in their root nodules. [1 mark] - These bacteria convert/fix atmospheric nitrogen gas into nitrogenous compounds (e.g. ammonium / ammonia). [1 mark] - Decomposed parts of legumes release these nitrogenous compounds into the soil, increasing soil nitrogen content / nitrates. [1 mark] - Corn plants absorb these nitrogenous nutrients (nitrates) to synthesize proteins / nucleic acids / chlorophyll for growth. [1 mark]

(a) Let \(X^B\) represent the dominant allele for normal vision and \(X^b\) represent the recessive allele for red-green color blindness. The father (Mr. Chan) has normal vision, so his genotype must be \(X^B Y\). He passes his Y chromosome to his son. The mother (Mrs. Chan) has normal vision but must carry the recessive allele to pass it to her son. Thus, her genotype is \(X^B X^b\). The son inherits the Y chromosome from the father and the \(X^b\) chromosome from the mother, resulting in genotype \(X^b Y\), which phenotypically presents as color blindness. (b) For daughters, they must receive the \(X^B\) chromosome from the father (genotype \(X^B Y\)). From the mother (genotype \(X^B X^b\)), a daughter has a 50% chance of receiving \(X^B\) (genotype \(X^B X^B\), non-carrier) and a 50% chance of receiving \(X^b\) (genotype \(X^B X^b\), carrier). Thus, the probability that the next daughter is a carrier is 0.5 (or 50%).

(a) - Define alleles: Let \(X^B\) be the allele for normal vision and \(X^b\) be the allele for red-green color blindness. [1 mark] - State genotypes: Mr. Chan's genotype is \(X^B Y\) and Mrs. Chan's genotype is \(X^B X^b\). [1 mark] - Explain inheritance: The son inherits the Y chromosome from the father and the \(X^b\) allele from the mother, resulting in genotype \(X^b Y\) (color-blind). [1 mark] (b) - 0.5 / 50% / 1/2. [1 mark] (Note: Reject 25% because the question specifies 'their next daughter', meaning the pool of offspring is restricted to females).

(a) Cut the leafy shoot under water to prevent air bubbles from entering the xylem. Assemble the potometer completely under water to ensure no air leaks. Apply petroleum jelly to all joints to make the apparatus completely airtight.

(b) Wind sweeps away the accumulated water vapor around the stomata. This maintains a steep concentration gradient of water vapor between the intercellular spaces and the surrounding air, increasing the rate of transpiration and thus water uptake.

(c) The effect of wind speed on water uptake would be less pronounced. At 80% relative humidity, the surrounding air already has a very high water vapor concentration. Even if the boundary layer is removed, the concentration gradient remains narrow, limiting transpiration.

(a) Cut stem under water [1]; Assemble apparatus under water [1]; Apply petroleum jelly to joints to ensure airtightness [1].
(b) Wind removes water vapor around leaves [1]; Steepens concentration gradient of water vapor between leaf interior and air [1]; Increases transpiration rate, which increases water uptake [1].
(c) Less pronounced increase [1]; High humidity means surrounding air already has high water vapor concentration [1]; Concentration gradient remains narrow even with wind [1].

(a) Nymphaea leaves float on water, meaning the lower epidermis is in direct contact with water, where gas diffusion is extremely slow. Stomata on the upper epidermis allow direct contact with air for efficient gas exchange. For Lilium, having stomata on the lower epidermis shields them from direct sunlight, reducing water loss through transpiration while still allowing gas exchange.

(b) The rate of photosynthesis would decrease significantly. Carbon dioxide diffuses much slower in water than in air. Under submerged conditions, the availability of dissolved carbon dioxide becomes extremely low, acting as a limiting factor.

(c) Floating aquatic leaves typically have large intercellular air spaces or aerenchyma in the spongy mesophyll to facilitate rapid gas diffusion within the leaf and provide buoyancy.

(a) Nymphaea lower epidermis in contact with water [1]; Gas diffusion in water is slow, so stomata must be on upper epidermis for air exposure [1]; Lilium lower epidermis is shaded from direct sunlight [1]; Reduces transpiration water loss while permitting gas exchange [1].
(b) Photosynthesis rate decreases significantly [1]; Carbon dioxide diffuses much slower in water than in air [1]; Dissolved carbon dioxide becomes a limiting factor [1].
(c) Large intercellular air spaces / aerenchyma [1]; Facilitates gas diffusion / transport within the leaf [1].

(a) Area A undergoes primary succession, while Area B undergoes secondary succession. The key difference is that primary succession begins in an environment completely devoid of soil and organic matter, whereas secondary succession occurs in an area where soil and organic matter are already present.

(b) Lichens and mosses secrete organic acids that chemically weather the bare rock into smaller mineral particles. When these pioneer organisms die and decompose, their organic remains (humus) mix with the mineral particles to form simple soil, increasing its water-holding capacity and nutrient availability for herbaceous plants.

(c) A climax community has higher species diversity, more complex food webs (which provide higher ecological stability), and a much larger total biomass dominated by long-lived, shade-tolerant trees compared to the early stages.

(a) Area A: Primary succession AND Area B: Secondary succession [1]; Primary succession starts on bare rock/lava with no soil [1]; Secondary succession starts where soil is already present [1].
(b) Pioneer species secrete acids to weather rock into minerals [1]; Decomposed pioneer organic matter (humus) mixes with minerals to form soil [1]; Soil increases water retention / nutrient levels, allowing larger plants to grow [1].
(c) High species richness / biodiversity [1]; Complex food webs / higher ecological stability [1]; Large biomass dominated by climax species (e.g., trees) [1].

(a) Ecological efficiency = (Energy in primary consumers / Energy in primary producers) * 100% = (7.2 * 10^4 / 8.4 * 10^5) * 100% = 8.57% (or 8.6%).

(b) Energy is lost because: 1. Some parts of the primary consumers are not eaten by secondary consumers (e.g., bones, fur). 2. Some consumed biomass is indigestible and lost as feces (egestion). 3. Much of the energy is lost as heat during cellular respiration for life processes like movement and metabolic maintenance.

(c) The primary producers in aquatic systems (phytoplankton) have a very high rate of reproduction and turnover (short life span). They produce biomass at a rapid rate (high productivity) which quickly supports the primary consumers, even though their standing crop (biomass at any one instant) is small.

(a) Correct calculation setup: (7.2 * 10^4) / (8.4 * 10^5) * 100% [1]; Correct answer (8.57% or 8.6%) [1].
(b) Any three of: Uneaten parts (e.g., bones, teeth) [1]; Egested feces / undigested food [1]; Heat loss during respiration for metabolic processes [1]; Excretory products (e.g., urea) [1].
(c) Phytoplankton have high reproduction and turnover rate / short lifespan [1]; High productivity allows rapid replenishment of biomass [1]; Can support larger consumer standing crop despite small producer standing crop [1].

(a) The disorder is dominant. II-1 and II-4 are both affected, but they have an unaffected daughter, III-1. If the allele were recessive, both affected parents would be homozygous recessive and could only have affected offspring. Therefore, II-1 and II-4 must be heterozygous, and the unaffected III-1 inherited the recessive normal allele from both parents.

(b) The gene is autosomal. II-4 is an affected male who has an unaffected daughter, III-1. If the disorder were X-linked dominant, the affected father would pass his X chromosome carrying the dominant allele to all his daughters, making them all affected. Since III-1 is unaffected, the gene cannot be X-linked.

(c) Since II-3's father (I-1) is unaffected (homozygous recessive, aa), II-3 must have inherited a recessive allele from his father. Thus, II-3's genotype must be heterozygous (Aa). If he marries an unaffected woman (aa), the cross is Aa x aa. The probability of their first child inheriting the disorder (Aa) is 50% (or 0.5).

(a) Dominant allele [1]; Affected parents II-1 and II-4 produce unaffected daughter III-1 [1]; If recessive, both parents would be homozygous and could only produce affected offspring [1]; Parents must be heterozygous and passed normal recessive alleles to III-1 [1].
(b) Autosomal [1]; Affected father II-4 has unaffected daughter III-1 [1]; If X-linked dominant, affected father must pass dominant allele to all daughters, making III-1 affected [1].
(c) Genotype of II-3 is heterozygous (Aa) because father I-1 is unaffected (aa) [1]; Probability of affected child (Aa) is 0.5 (or 50%) [1].

(a) Genotype of F1: TtRr. Phenotype of F1: Tall stem, red fruit.

(b) (i) The phenotypic ratio would be 1 tall, yellow fruit : 1 tall, red fruit : 1 dwarf, yellow fruit : 1 dwarf, red fruit.

(ii) The two genes are linked (located on the same chromosome). Specifically, T and r are linked on one chromosome, and t and R are linked on the homologous chromosome, tending to be inherited together. Recombinant phenotypes only arise when crossing over occurs between these two gene loci, which happens at a low frequency.

(c) Crossing over (during prophase I of meiosis).

(a) Phenotype: Tall stem, red fruit [1]; Genotype: TtRr [1].
(b) (i) 1 tall, yellow : 1 tall, red : 1 dwarf, yellow : 1 dwarf, red [2] (Deduct 1 mark if phenotypes are omitted or incorrect).
(ii) The two genes are linked / on the same chromosome [1]; T is linked with r, and t is linked with R [1]; Recombinants only form due to crossing over, which occurs at low frequency [1].
(c) Crossing over [1].

(a) Bile salts in bile normally emulsify large lipid droplets into smaller droplets, increasing the surface area-to-volume ratio for lipase to act on. With a blocked bile duct, bile cannot enter the duodenum, leading to poor emulsification and a significantly slower rate of lipid digestion.

(b) Pale feces occur because bile contains bile pigments (bilirubin) which are converted by intestinal bacteria into the brown compounds that color normal feces; without bile, these pigments are absent. Greasy and foul-smelling feces occur because undigested lipids cannot be absorbed and are excreted in the feces, where they are decomposed by gut bacteria.

(c) Structural adaptations of the ileum include: 1. It is very long, providing ample surface area and time for absorption. 2. Its inner wall is folded with numerous villi and microvilli, greatly increasing the surface area. 3. The epithelium of the villi is only one cell thick, reducing diffusion distance. 4. Villi contain a rich network of blood capillaries and a central lacteal to carry away nutrients rapidly, maintaining a steep concentration gradient.

(a) Bile contains bile salts which emulsify lipids into smaller droplets [1]; Increases surface area-to-volume ratio for lipase action [1]; Blocked duct prevents bile entry, decreasing lipid digestion rate [1].
(b) Pale color: Absence of bile pigments (bilirubin) which normally color feces brown [1.5]; Greasy/smelly: Undigested/unabsorbed lipids excreted in feces and decomposed by bacteria [1.5].
(c) Any three of: Very long (increases time/area) [1]; Inner wall folded with villi and microvilli (increases surface area) [1]; Epithelium is one-cell thick (shortens diffusion distance) [1]; Rich capillary network and lacteal (maintains steep concentration gradient) [1].

The essay should be divided into two main parts: water transport and water conservation under hot and dry conditions.\n\n1. Water Transport (4 marks):\n- Xylem vessels are continuous, hollow tubes with no protoplasts or end walls, reducing resistance to water flow.\n- Vessel walls are thickened with lignin, preventing collapse under the tension created by transpiration pull.\n- Pits allow lateral water transport to adjacent cells.\n- Highly branched leaf veins distribute water to all mesophyll cells.\n\n2. Water Conservation / Adaptation (5 marks):\n- A thick, waterproof waxy cuticle on leaves minimizes water loss via non-stomatal evaporation.\n- Stomata are located primarily on the lower epidermis, reducing exposure to direct sunlight and wind.\n- Guard cells close stomata under dry conditions (water stress) to limit further water loss.\n- The stem is covered by a waterproof bark/cork layer to prevent desiccation of inner tissues.\n- Lenticels restrict gas exchange and water loss to small, localized areas.\n\n3. Communication (2 marks):\n- Awarded for clear, logical structure and accurate biological terminology.

Content (max 9 marks):\n- Xylem vessels as continuous hollow tubes / no end walls, reducing resistance to water flow (1 mark)\n- Lignified vessel walls provide support / prevent collapse under high tension (1 mark)\n- Pits allow lateral transport of water (1 mark)\n- Leaf veins branch extensively to distribute water to mesophyll cells (1 mark)\n- Waxy cuticle on leaf epidermis reduces non-stomatal evaporation (1 mark)\n- Stomata on lower epidermis are shaded, reducing transpiration rate (1 mark)\n- Guard cells close stomata under water stress to prevent further water loss (1 mark)\n- Waterproof bark/cork layer on stem prevents desiccation of inner tissues (1 mark)\n- Lenticels restrict gas exchange / water loss to localized areas compared to bare stem (1 mark)\n\nCommunication Skills (max 2 marks):\n- 2 marks: Clear, systematic exposition showing logical flow; appropriate terminology used; no major spelling/grammatical errors.\n- 1 mark: Structured answer but with minor logical gaps or minor misuse of terminology.\n- 0 marks: Highly disorganized or irrelevant answer.

1. Water Transport (4 marks): Xylem vessels are continuous, hollow tubes with no protoplasts or end walls, reducing resistance to water flow. Vessel walls are thickened with lignin, preventing collapse under the tension created by transpiration pull. Pits allow lateral water transport to adjacent cells. Highly branched leaf veins distribute water to all mesophyll cells.\n\n2. Water Conservation / Adaptation (5 marks): A thick, waterproof waxy cuticle on leaves minimizes water loss via non-stomatal evaporation. Stomata are located primarily on the lower epidermis, reducing exposure to direct sunlight and wind. Guard cells close stomata under dry conditions to limit further water loss. The stem is covered by a waterproof bark/cork layer to prevent desiccation. Lenticels restrict gas exchange and water loss to small, localized areas.\n\n3. Communication (2 marks): Awarded for clear, logical structure and accurate biological terminology.

Content (max 9 marks):\n- Xylem vessels as continuous hollow tubes / no end walls, reducing resistance to water flow (1 mark)\n- Lignified vessel walls provide support / prevent collapse under high tension (1 mark)\n- Pits allow lateral transport of water (1 mark)\n- Leaf veins branch extensively to distribute water to mesophyll cells (1 mark)\n- Waxy cuticle on leaf epidermis reduces non-stomatal evaporation (1 mark)\n- Stomata on lower epidermis are shaded, reducing transpiration rate (1 mark)\n- Guard cells close stomata under water stress to prevent further water loss (1 mark)\n- Waterproof bark/cork layer on stem prevents desiccation of inner tissues (1 mark)\n- Lenticels restrict gas exchange / water loss to localized areas compared to bare stem (1 mark)\n\nCommunication Skills (max 2 marks):\n- 2 marks: Clear, systematic exposition showing logical flow; appropriate terminology used; no major spelling/grammatical errors.\n- 1 mark: Structured answer but with minor logical gaps or minor misuse of terminology.\n- 0 marks: Highly disorganized or irrelevant answer.

(a) In Condition Y, the high relative humidity (85%) reduces the water vapor pressure gradient between the skin and the surrounding air. Consequently, the rate of sweat evaporation is significantly lower. Since less heat is lost from the body as latent heat of vaporization, core body temperature rises faster.\n(b) Sweat is hypotonic compared to plasma, meaning it has a lower solute concentration than blood. Prolonged sweating thus removes more water than solutes from the blood, resulting in an increase in blood osmolarity. This increase in blood osmolarity is detected by osmoreceptors in the hypothalamus, which stimulates the pituitary gland to secrete more antidiuretic hormone (ADH) into the blood. ADH travels to the kidneys and increases the water permeability of the cells in the collecting ducts and distal convoluted tubules. As a result, more water is reabsorbed from the glomerular filtrate back into the surrounding blood capillaries. The body produces a smaller volume of concentrated urine, conserving water and restoring blood osmolarity.\n(c)(i) Drinking a large volume of pure water rapidly dilutes the blood, causing a sharp drop in blood osmolarity. This change is detected by osmoreceptors, which inhibit the secretion of ADH from the pituitary gland. Without ADH, the permeability of the collecting ducts to water decreases dramatically, leading to less water reabsorption. Consequently, the excess water is rapidly excreted as a large volume of dilute urine, preventing the full restoration of body fluid volume.\n(c)(ii) Isotonic sports drinks contain sodium ions, which help maintain the osmolarity of the blood. Drinking them prevents a rapid drop in blood osmolarity, thereby avoiding the sudden inhibition of ADH secretion. This allows the kidneys to continue reabsorbing water, ensuring the ingested fluid is retained in the body to fully restore fluid balance.

Part (a) (Max 4 marks):\n- Condition Y has higher relative humidity / smaller water vapor pressure gradient between skin and air (1)\n- Lower rate of sweat evaporation in Condition Y (1)\n- Less latent heat of vaporization is lost from the body surface (1)\n- Leads to faster accumulation of metabolic heat and rise in core temperature (1)\n\nPart (b) (Max 8 marks):\n- Sweat is hypotonic to blood / contains less solute per unit volume (1)\n- Sweating causes relatively more water loss than solute loss, increasing blood osmolarity (1)\n- Osmoreceptors in the hypothalamus detect the increase in blood osmolarity (1)\n- Pituitary gland is stimulated to secrete more ADH into the circulation (1)\n- ADH increases the water permeability of collecting ducts / distal convoluted tubules (1)\n- More water is reabsorbed from renal filtrate into blood capillaries (1)\n- Smaller volume of concentrated urine is excreted (1)\n- Normal blood osmolarity is restored via negative feedback (1)\n\nPart (c)(i) (Max 5 marks):\n- Pure water absorption rapidly dilutes blood plasma / lowers blood osmolarity (1)\n- Detected by osmoreceptors, leading to inhibition of ADH secretion (1)\n- Decreased water permeability of collecting ducts (1)\n- Less water is reabsorbed, resulting in rapid excretion of a large volume of dilute urine (1)\n- Pure water is lost before it can be effectively integrated into body fluids (1)\n\nPart (c)(ii) (Max 3 marks):\n- Isotonic sports drinks contain sodium ions to maintain blood osmolarity / prevent rapid dilution (1)\n- ADH secretion is not abruptly shut down / remains at a level that allows water reabsorption (1)\n- Absorbed water is retained in the extracellular fluid, achieving effective rehydration (1)

(a)(i) Habitat fragmentation divides the dolphin population into smaller, isolated subpopulations. This isolation prevents gene flow between different groups. Over generations, the probability of inbreeding within each small subpopulation increases, which leads to a decrease in genetic variation and the accumulation of harmful recessive alleles, reducing their genetic diversity and adaptability to environmental changes.\n(a)(ii) Reclamation activities stir up sediment, causing high water turbidity. This blocks sunlight penetration, reducing the rate of photosynthesis in marine plants and phytoplankton. Additionally, suspended sediment can clog the gills of fish or filter-feeding marine organisms, leading to suffocation and death.\n(b)(i) The data illustrates biomagnification (or bioaccumulation). PCBs are persistent organic pollutants that are lipophilic and cannot be easily broken down or excreted. Croaker fish accumulate PCBs in their body tissues when they ingest contaminated seawater and plankton over time. Since Chinese White Dolphins are apex predators, they consume many croaker fish, transferring and highly concentrating PCBs at this higher trophic level, leading to a much higher concentration in dolphins than in their prey.\n(b)(ii) Microplastics have a much higher surface-area-to-volume ratio compared to larger plastic debris, allowing them to adsorb and concentrate higher amounts of toxic organic pollutants like PCBs from surrounding water. Furthermore, their small size makes them easily ingestible by low-trophic-level marine organisms, introducing these chemical toxins directly into the base of the marine food web.\n(c)(i) The Marine Park regulates human activities, such as restricting high-speed vessel traffic, which reduces physical injuries from collisions and minimizes acoustic noise pollution that disrupts dolphin communication and echolocation. It also bans or controls commercial fishing, which increases prey abundance (e.g., croaker fish) and protects breeding areas, supporting dolphin reproduction and population recovery.\n(c)(ii) 1. Deploying artificial reefs to create complex habitats and nurseries for marine organisms. 2. Conducting regular clean-ups of marine litter and ghost nets to prevent entanglement of marine life.

Part (a)(i) (Max 4 marks):\n- Separation of dolphins into smaller, isolated subpopulations (1)\n- Prevents or restricts gene flow/migration between subpopulations (1)\n- Promotes inbreeding within the small subpopulations (1)\n- Leads to reduction in genetic variation / increased homozygosity / accumulation of harmful recessive traits over generations (1)\n\nPart (a)(ii) (Max 3 marks):\n- Stirring up of seabed sediment / high turbidity (1)\n- Reduced light penetration reduces photosynthesis in benthic plants / phytoplankton (1)\n- Clogging of gills of fish / respiratory organs of filter-feeders, leading to suffocation (1)\n[Accept alternative valid physical impacts, e.g., alterations in coastal water currents causing erosion or siltation of surrounding habitats (1)]\n\nPart (b)(i) (Max 4 marks):\n- Concept of biomagnification / bioaccumulation identified (1)\n- PCBs are chemically stable / non-biodegradable / lipophilic, meaning they cannot be excreted or broken down (1)\n- Level increases from seawater to croaker fish due to continuous intake and accumulation (1)\n- Dolphin, as the top predator, consumes large numbers of contaminated fish, further concentrating PCBs at the highest trophic level (1)\n\nPart (b)(ii) (Max 3 marks):\n- Microplastics have a higher surface-area-to-volume ratio than large plastics (1)\n- Adsorb and concentrate organic pollutants (PCBs) at a higher density (1)\n- Small size matches the prey size of primary consumers/plankton, facilitating entry into the food chain (1)\n\nPart (c)(i) (Max 4 marks):\n- Restriction of marine vessel traffic/speed decreases collision risk (1)\n- Reduces underwater acoustic noise, which interferes with dolphin echolocation/communication (1)\n- Ban/restrictions on commercial fishing increases food/prey availability for dolphins (1)\n- Protects crucial habitats (such as nursery and feeding grounds) from physical disturbance (1)\n\nPart (c)(ii) (Max 2 marks):\n- Deployment of artificial reefs (1)\n- Marine litter/ghost net clearance programs (1)\n- Species monitoring / ecological research (1)\n- Public educational tours/campaigns (1)\n(Any two, 1 mark each)