2022 HKDSE Biology Practice Paper with Answers

When the water potential of the external solution equals that of the cell sap, there is no net movement of water into or out of the cells by osmosis. Consequently, the percentage change in mass is zero. The cell wall does not exert maximum pressure (which occurs in hypotonic solution) and plasmolysis does not occur.

A (1 mark)

In arid environments, plants minimize transpirational water loss by having a very low overall stomatal density and confining stomata strictly to the lower epidermis (or sunken pits) to reduce evaporation caused by direct sunlight and wind. Species S has the lowest total number of stomata (15 per mm²) with zero stomata on the upper epidermis, making it best adapted to dry conditions.

D (1 mark)

At the arterial end, hydrostatic pressure (4.3 kPa) exceeds oncotic pressure (3.3 kPa), producing a net filtration pressure outward into the tissue spaces (formation of tissue fluid). At the venous end, oncotic pressure (3.3 kPa) exceeds hydrostatic pressure (1.6 kPa), causing net reabsorption of fluid into the capillary.

A (1 mark)

Since unaffected parents gave birth to an affected child, the condition must be caused by a recessive allele, so (1) is correct. If it were X-linked recessive, the affected daughter (XᵃXᵃ) must have inherited an Xᵃ from her father, who would have the genotype XᵃY and thus be affected; since the father is unaffected, the gene must be autosomal, so (2) is incorrect. The unaffected son has a 2/3 probability of being a carrier (Aa) and a 1/3 probability of being homozygous dominant (AA), so he does not necessarily have to be a carrier, making (3) incorrect.

A (1 mark)

A double-stranded DNA segment of 240 base pairs contains a total of \(240 \times 2 = 480\) bases. Since adenine makes up 30%, thymine also makes up 30%. Therefore, adenine and thymine together account for 60% of all bases. The remaining 40% consists equally of guanine and cytosine, so cytosine makes up 20% of the total bases. Total number of cytosine bases \(= 480 \times 0.20 = 96\).

C (1 mark)

During mitosis and meiosis II (specifically anaphase and anaphase II), centromeres divide and sister chromatids separate toward opposite poles. Pairing of homologous chromosomes (synapsis) and crossing over occur only in prophase I of meiosis. Independent assortment occurs in metaphase I / anaphase I of meiosis.

C (1 mark)

Ecological efficiency from primary consumers to secondary consumers is calculated as \(\frac{\text{Energy in secondary consumers}}{\text{Energy in primary consumers}} \times 100\% = \frac{316.8}{3520} \times 100\% = 9.0\%\).

B (1 mark)

In a cold environment, the sympathetic nervous system triggers vasoconstriction of arterioles supplying superficial skin capillaries, diverting blood flow through arteriovenous shunt vessels away from the skin surface to decrease heat loss by radiation, conduction, and convection. Additionally, contraction of erector pili muscles causes hairs to stand erect, which in animals traps an insulating layer of air.

C (1 mark)

In conditions 2 and 3, aerobic respiration is inhibited, preventing ATP synthesis via oxidative phosphorylation. The uptake of ion X drops to negligible levels under these conditions, showing that its transport requires metabolic energy (ATP) against a concentration gradient, which is characteristic of active transport ((1) is correct). In contrast, the absorption of substance Y remains high when aerobic respiration is blocked, indicating that its transport does not depend on ATP produced by aerobic respiration ((2) is correct). Substance Y could enter through simple diffusion or facilitated diffusion via channel/carrier proteins down a concentration gradient; the data do not prove it enters by simple diffusion only ((3) is incorrect). Therefore, only (1) and (2) are correct.

A (1 mark): (1) and (2) only. Deduct no marks for incorrect choices.

A double-stranded DNA segment of 1200 base pairs contains a total of \(1200 \times 2 = 2400\) nitrogenous bases. According to Chargaff's rules for double-stranded DNA, the percentage of cytosine (C) equals the percentage of guanine (G), so \(\text{C} = 28\%\). Together, \(\text{G} + \text{C} = 28\% + 28\% = 56\%\). The remaining bases are adenine (A) and thymine (T), which make up \(100\% - 56\% = 44\%\). Since \(\text{A} = \text{T}\), the percentage of thymine is \(44\% \div 2 = 22\%\). The number of thymine bases is \(2400 \times 0.22 = 528\).

B (1 mark): 528.

(1) Stomata are predominantly located on the lower epidermis; coating them with petroleum jelly blocks stoma pores, drastically reducing transpiration pull and water uptake. (2) Increasing humidity decreases the water vapour concentration gradient between the internal air spaces and the atmosphere, reducing transpiration rate and water uptake. (3) Enclosing the shoot inside a transparent bag traps transpired water vapour, causing local relative humidity to reach near-saturation, which sharply decreases transpiration and water uptake. Thus, all three statements are correct.

D (1 mark): (1), (2) and (3).

A is incorrect because arteries do not have pocket valves along their length; valves are found in veins to prevent backflow under low pressure. B is correct because capillaries form an extensive branching network having the greatest total cross-sectional area of all vessel types. C is incorrect because arteries (P) have much thicker, more muscular and elastic walls than veins (R) to withstand high blood pressure. D is incorrect because the rate of blood flow (velocity) is slowest in capillaries (Q) due to their large total cross-sectional area, not in veins.

B (1 mark): Vessel Q has the highest total cross-sectional area among the three vessels.

(1) Since unaffected parents produce an affected offspring, the allele causing the disorder must be recessive and masked in the heterozygous parents. (2) If the disorder were X-linked recessive, an affected female (\(X^a X^a\)) must inherit an affected X chromosome (\(X^a\)) from her father, making the father affected (\(X^a Y\)). Since the father is phenotypically normal, the gene cannot be on the X chromosome and must be on an autosome. (3) Since it is an autosomal recessive condition (genotype \(aa\)), both unaffected parents must carry the recessive allele and are thus heterozygous carriers (\(Aa\)). Therefore, (1), (2), and (3) are all correct.

D (1 mark): (1), (2) and (3).

Phytoplankton have a very high rate of reproduction and cell turnover (short generation time). Even though their standing biomass at any one instant is small, their high rate of primary productivity generates sufficient energy over time to support the larger standing biomass of longer-lived zooplankton. Energy transfer is never 100% efficient due to metabolic heat loss and unconsumed parts. Phytoplankton are autotrophs, not heterotrophs.

A (1 mark): Phytoplankton have a very high turnover rate and rapid reproduction.

Rapid intake and absorption of pure water into the bloodstream dilutes the blood plasma, increasing its water potential (making it less negative / lower osmotic pressure). This increase is detected by osmoreceptors in the hypothalamus, which decreases the release of antidiuretic hormone (ADH) from the posterior pituitary gland. Lower ADH levels reduce the permeability of the collecting ducts to water, resulting in less water reabsorption and the production of a larger volume of dilute urine.

B (1 mark): The water potential of the blood plasma increases.

According to Darwinian natural selection, genetic variations arise randomly before exposure to an environmental pressure. Pre-existing resistant mutant individuals had a selective survival advantage when the insecticide was applied. They survived and reproduced, passing their resistance alleles to the next generation. Over successive generations, the frequency of resistance alleles increased in the population. Insecticides do not induce directed mutations or promote individual acquired physiological adaptations that are inherited.

C (1 mark): Pests with pre-existing resistance alleles survived and passed these alleles to their offspring.

The rate of uptake of substance P is directly proportional to its external concentration without showing saturation, and does not require ATP (unaffected by cyanide). Thus, substance P crosses the membrane by simple diffusion. The uptake of substance Q shows saturation kinetics (plateaus at high concentration, indicating carrier proteins are involved) and is dependent on ATP production (inhibited by cyanide), which indicates active transport.

A (1 mark)

When light is turned off, the light-dependent reactions stop producing ATP and NADPH. Consequently, the reduction of GP (3-C compound) to triose phosphate (GALP) decreases, causing GP to accumulate. Meanwhile, without ATP, RuBP cannot be regenerated from triose phosphate, but the remaining RuBP continues to react with available \(\text{CO}_2\) to form GP, leading to a rapid decrease in the concentration of RuBP (5-C compound).

B (1 mark)

An affected mother having all affected sons and unaffected daughters (when father is unaffected) strongly points to an X-linked recessive trait. If the mother is \(X^a X^a\) and the father is \(X^A Y\), all sons receive \(X^a\) from the mother and \(Y\) from the father, so 100% of sons are affected (\(X^a Y\)). All daughters receive \(X^A\) from the father, so 100% of daughters are unaffected carriers (\(X^A X^a\)). When an affected male (\(X^a Y\)) mates with a homozygous normal female (\(X^A X^A\)), all children will be normal phenotype (daughters: \(X^A X^a\), sons: \(X^A Y\)). Thus, X-linked recessive is fully consistent.

C (1 mark)

Two hours after a carbohydrate-rich meal, glucose is actively absorbed from the small intestine, making the glucose concentration in the hepatic portal vein higher than in the hepatic vein (as excess glucose is converted to glycogen in the liver). Deamination of excess amino acids takes place in the liver, so urea is produced by the liver and released into the hepatic vein, making the urea concentration higher in the hepatic vein than in the hepatic portal vein. Both vessels carry deoxygenated blood, so oxygen level is relatively low in both, but hepatic vein is not higher in oxygen.

D (1 mark)

During inhalation, external intercostal muscles contract (pulling ribs up and out) and the diaphragm contracts and flattens, increasing thoracic volume and decreasing alveolar pressure below atmospheric pressure so air flows in. Statement (4) is incorrect because elastic recoil occurs during passive exhalation to decrease lung volume.

B (1 mark)

Phytoplankton (trophic level 1): \(4.8 \times 10^5\text{ kJ m}^{-2}\text{ year}^{-1}\).
Zooplankton (trophic level 2): \(4.8 \times 10^5 \times 0.10 = 4.8 \times 10^4\text{ kJ m}^{-2}\text{ year}^{-1}\).
Small fish (trophic level 3): \(4.8 \times 10^4 \times 0.10 = 4.8 \times 10^3\text{ kJ m}^{-2}\text{ year}^{-1}\).
Large predatory fish (trophic level 4): \(4.8 \times 10^3 \times 0.10 = 4.8 \times 10^2\text{ kJ m}^{-2}\text{ year}^{-1} = 480\text{ kJ m}^{-2}\text{ year}^{-1}\).

C (1 mark)

Focusing on a distant object (the whiteboard) requires accommodation for distance: the circular ciliary muscles relax, increasing the tension on the suspensory ligaments (they become taut), which pulls on the elastic eye lens, causing it to become thinner/flatter (decreasing its curvature and reducing its refractive power). Therefore, statements (1), (2), and (3) are all correct.

D (1 mark)

The mRNA transcribed from the original coding sequence is 5'-AUG CCG GCU UAC UGA-3', which codes for Met - Pro - Ala - Tyr (4 amino acids before the stop codon). With the T to C mutation at the 10th base, the coding strand is 5'-ATG CCG GCT CAC TGA-3', producing mRNA 5'-AUG CCG GCU CAC UGA-3'. This codes for Met - Pro - Ala - His (also 4 amino acids). Thus, the length of the polypeptide remains unchanged (4 amino acids), but tyrosine is replaced by histidine at the 4th position.

B (1 mark)

The rate of uptake shows saturation kinetics (reaches a maximum rate / plateau as concentration increases), which indicates the involvement of membrane transport proteins (carriers or channels). Since the metabolic inhibitor (preventing ATP synthesis) has no effect on the rate, the process does not require cellular energy (ATP). Therefore, substance X is transported across the cell membrane by facilitated diffusion.

B (1 mark) - Correct identification of facilitated diffusion based on carrier saturation and lack of ATP requirement.

During fasting, the liver undergoes glycogenolysis and gluconeogenesis to release glucose into the general circulation, so blood exiting the liver via the hepatic vein has the highest glucose concentration. Urea is filtered out of the blood and excreted by the kidneys, so blood leaving the kidney via the renal vein has the lowest concentration of urea.

B (1 mark) - Hepatic vein carries released glucose from glycogen breakdown during fasting; renal vein carries blood after urea removal by the kidneys.

All three statements are correct. (1) A potometer measures the rate of water absorption by the cut shoot by following bubble movement. (2) Not all absorbed water is lost through transpiration; a small fraction (around 1-2%) is used for maintaining turgidity, growth, and metabolic processes such as photosynthesis. (3) Petroleum jelly (Vaseline) ensures an airtight seal at connections.

D (1 mark) - All statements (1), (2), and (3) are scientifically accurate regarding potometer operation.

Since the first child has blood group O (genotype ii), each parent must carry the recessive allele i. Therefore, the father's genotype is \(I^A i\) and the mother's genotype is \(I^B i\). The possible genotypes of their offspring are \(I^A I^B\) (group AB), \(I^A i\) (group A), \(I^B i\) (group B), and \(i i\) (group O), each with a probability of \(\frac{1}{4}\). The probability of having a boy is \(\frac{1}{2}\). Thus, the probability of having a boy with blood group AB is \(\frac{1}{4} \times \frac{1}{2} = \frac{1}{8}\).

B (1 mark) - Correct determination of parental genotypes and calculation: (1/4 for AB) * (1/2 for boy) = 1/8.

The finch populations were separated by geographic barriers (different islands in an archipelago), preventing gene flow between them. This geographic isolation led to allopatric speciation. Under different environmental conditions and food availability, different beak phenotypes had different survival advantages, driving speciation via natural selection.

A (1 mark) - Allopatric speciation is caused by geographic barriers, and the adaptation to food sources is driven by natural selection.

Consuming a concentrated salt solution lowers the blood water potential (increases blood solute concentration). (1) Osmoreceptors in the hypothalamus detect this decrease in blood water potential and send impulses to the pituitary gland. (2) The posterior pituitary gland releases more ADH into the bloodstream. (3) ADH increases the water permeability of the epithelial cells in the collecting duct (and distal convoluted tubule), allowing more water to be reabsorbed by osmosis into the surrounding capillaries, producing a smaller volume of concentrated urine.

D (1 mark) - All three events (1), (2), and (3) are standard steps in the negative feedback regulation of osmoregulation.

1200 base pairs equal \(1200 \times 2 = 2400\) total bases. According to Chargaff's rules, the percentage of guanine equals the percentage of cytosine, so \(\%\text{G} = \%\text{C} = 28\%\). Together, \(\text{C} + \text{G} = 56\%\). Therefore, \(\text{A} + \text{T} = 100\% - 56\% = 44\%\), and \(\%\text{A} = 22\%\). The total number of adenine bases is \(2400 \times 0.22 = 528\).

B (1 mark) - Correct calculation: 2400 total bases * 22% A = 528 bases.

All three statements are correct. (1) In human skeletal muscles, anaerobic respiration follows lactic acid fermentation (glucose -> lactic acid), which produces no carbon dioxide. (2) Because lactic acid still contains a large amount of chemical energy in its bonds, only a net gain of 2 ATP molecules is produced per glucose molecule (compared to ~30-32 ATP in aerobic respiration). (3) Glycolysis and the subsequent reduction of pyruvate to lactic acid occur exclusively in the cytoplasm and do not involve mitochondria.

D (1 mark) - All statements (1), (2), and (3) accurately describe lactic acid fermentation in human skeletal muscles.

A segment containing 1200 base pairs has a total of \(1200 \times 2 = 2400\) bases. According to Chargaff's rule in double-stranded DNA, the percentage of guanine (G) equals that of cytosine (C), so \(\%\text{G} = \%\text{C} = 28\%\). Together, G and C make up \(28\% + 28\% = 56\%\) of the total bases. The remaining \(100\% - 56\% = 44\%\) is shared equally between adenine (A) and thymine (T), meaning \(\%\text{A} = \%\text{T} = 22\%\). Therefore, the number of thymine bases is \(2400 \times 22\% = 528\).

B (1 mark)

At the light compensation point, the rate of photosynthetic carbon dioxide uptake and oxygen release exactly matches the rate of respiratory carbon dioxide release and oxygen uptake. Therefore, (1) is correct, and (2) is correct as there is no net exchange of oxygen. Statement (3) is incorrect because photosynthesis is actively occurring at a rate sufficient to balance cellular respiration, so photophosphorylation does not stop.

A (1 mark)

Vessels leaving the ventricles (arteries) have high blood pressure, while vessels returning to the atria (veins) have low blood pressure. Vessel Q has high blood pressure and carries deoxygenated (low oxygen) blood, which is the pulmonary artery (pumping blood from the right ventricle to the lungs). Vessel R has low blood pressure and carries oxygenated (high oxygen) blood, which is the pulmonary vein (returning blood from the lungs to the left atrium).

A (1 mark)

Since the couple has a child with blood group O (genotype \(ii\)), both parents must carry the recessive allele \(i\). Thus, the man's genotype is \(I^A i\) and the woman's genotype is \(I^B i\). The possible offspring genotypes are \(I^A I^B\) (group AB), \(I^A i\) (group A), \(I^B i\) (group B), and \(ii\) (group O), each with a probability of \(\frac{1}{4}\). The probability of having a child with blood group AB is \(\frac{1}{4}\). The probability of the child being a boy is \(\frac{1}{2}\). Therefore, the overall probability is \(\frac{1}{4} \times \frac{1}{2} = \frac{1}{8}\).

C (1 mark)

(a) Cutting the stem underwater prevents air from entering the exposed cut end of the xylem vessels, which would otherwise break the continuous water column and prevent water uptake.

(b) Over the 20-minute period, the bubble in Set-up Q moved 66 mm (average rate of 3.3 mm min\(^{-1}\)), while in Set-up P it moved 30 mm (average rate of 1.5 mm min\(^{-1}\)). Thus, the rate of water uptake/transpiration in Set-up Q is more than double that of Set-up P.

(c) In still air, water vapor evaporating from the stomata accumulates in the layer of still air immediately surrounding the leaf surface, decreasing the diffusion gradient. Wind sweeps away this humid layer of air, maintaining a steep concentration gradient of water vapor from the intercellular air spaces to the atmosphere, thereby accelerating diffusion/evaporation.

(d) 1. The apparatus must be completely airtight (e.g., sealed with petroleum jelly) so that water movement is driven purely by transpiration and not external leaks.
2. The leafy shoot surface/foliage should be dry at the start so that water evaporation is not hindered by external liquid droplets.

(a) Prevent air from entering the xylem vessels / maintain continuous water column (1)

(b) Correct comparison stating that Set-up Q shows a higher rate of movement / transpiration rate than Set-up P (1)
Supporting data / calculation of rates (e.g. 3.3 mm/min vs 1.5 mm/min or total distance 66 mm vs 30 mm) (1)

(c) Wind blows away / removes water vapor accumulated around the leaf surface / outer stomatal region (1)
Maintains a steeper / higher concentration gradient of water vapor between inner leaf spaces and ambient air (1)

(d) Any two valid precautions (1 mark each, max 2):
- Ensure all joints / connection points are airtight / smear with petroleum jelly (1)
- Ensure leaves are dry before the experiment starts (1)
- Introduce the bubble only once and let the plant equilibrate before taking readings (1)

(a) Independent variable: Temperature of the water bath.
Dependent variable: Absorbance / intensity of red pigment leaked into the solution.

(b) Cutting damages cells at the cut edges, releasing intracellular pigment. Washing ensures that any pigment measured during the experiment is solely due to temperature-induced permeability changes.

(c) At 30 °C and 45 °C, the phospholipid bilayer and embedded transport proteins remain intact and selectively permeable, preventing large betalain molecules from diffusing out. At 60 °C and above, membrane proteins denature, and the thermal motion of phospholipid molecules increases drastically, disrupting bilayer stability and forming pores. This loss of selective permeability allows betalain stored in the vacuole to diffuse out into the surrounding solution.

(d) The vacuole is bounded by a single selectively permeable tonoplast that contains specific transport systems to retain solutes inside.

(a) Independent variable: Temperature (1)
Dependent variable: Absorbance / intensity / concentration of red pigment in the surrounding solution (1)

(b) To wash away pigments released from cells broken/damaged during cutting/preparation (1)

(c) At lower temperatures (30-45 °C), cell membrane / tonoplast remains intact / selectively permeable (1)
At higher temperatures (60 °C and above), membrane proteins denature (1)
Phospholipids gain kinetic energy / bilayer structure is disrupted / becomes permeable, allowing betalain to diffuse out (1)

(d) Bound by a selectively permeable membrane / tonoplast / possesses a large central volume (1)

(a) The father is affected, so his genotype is \(X^D Y\). The mother is normal and recessive, so her genotype is \(X^d X^d\).

(b)
Parental phenotypes: Affected father \(\times\) Normal mother
Parental genotypes: \(X^D Y \times X^d X^d\)
Gametes: \(X^D, Y\) and \(X^d\)
Offspring genotypes:
- \(X^D X^d\) (affected female)
- \(X^d Y\) (normal male)
Phenotypic ratio: 1 affected female : 1 normal male (or all daughters affected, all sons normal).

(c) The daughter's genotype is \(X^D X^d\) and the normal man is \(X^d Y\).
Gametes from mother: \(X^D\) (1/2), \(X^d\) (1/2).
Gametes from father: \(X^d\) (1/2), \(Y\) (1/2).
Possible child genotypes: \(X^D X^d\) (affected female, 1/4), \(X^d X^d\) (normal female, 1/4), \(X^D Y\) (affected male, 1/4), \(X^d Y\) (normal male, 1/4).
Therefore, the probability of having an affected boy (\(X^D Y\)) is 1/4 (25% or 0.25).

(a) Father: \(X^D Y\) (1)
Mother: \(X^d X^d\) (1)

(b) Genetic diagram:
- Correct gametes shown: \(X^D, Y\) and \(X^d\) (1)
- Correct offspring genotypes: \(X^D X^d\) and \(X^d Y\) (1)
- Correct phenotype corresponding to genotypes: all daughters are affected and all sons are normal (1)

(c) Probability: 1/4 or 25% or 0.25 (1)
Explanation: The mother has a 50% chance of passing \(X^D\) allele and the father provides the Y chromosome with a 50% chance (\(0.5 \times 0.5 = 0.25\)) (1)

(a) 1. Microvilli on the epithelial cells greatly increase the surface area for diffusion/absorption.
2. The intestinal epithelium is only one-cell thick, providing a very short diffusion pathway.
3. Each villus contains a central lacteal (lymph vessel), into which reformed triglycerides (chylomicrons) enter, maintaining a continuous concentration gradient.

(b) 1. Deamination: Excess amino acids are broken down in the liver where the amino group is removed to form ammonia and converted to urea.
2. Protein synthesis: Amino acids are assembled into functional proteins such as plasma proteins (albumin, fibrinogen, globulins) or cellular proteins in the liver.

(c) The damaged liver fails to synthesise sufficient plasma proteins (such as albumin), resulting in a lower concentration of plasma proteins in the blood. This decreases the colloid osmotic/water potential gradient pulling water back into venous capillaries, so less tissue fluid is reabsorbed into the capillaries, leading to fluid retention in tissues (oedema).

(a) Any three structural features linked to absorption (1 mark each, max 3):
- Epithelium is one-cell thick / thin lining, providing a short diffusion distance (1)
- Microvilli on epithelial cells provide a large surface area for absorption (1)
- Presence of lacteal inside each villus to transport fats away and maintain a concentration gradient (1)

(b) Process 1: Deamination (1) - excess amino acids have amino groups removed to produce urea (1/2)
Process 2: Protein synthesis / assimilation (1) - amino acids are incorporated into plasma proteins / liver enzymes / cellular proteins (1/2)
(Award 1.5 marks per process correctly named and explained, total 3 marks)

(c) Liver damage decreases the production/synthesis of plasma proteins (e.g., albumin) (1)
Lowers blood colloid osmotic pressure / higher water potential in blood capillaries, reducing reabsorption of tissue fluid / more tissue fluid remains in interstitial spaces (1)

(a) Thermoreceptors/cold receptors in the skin detect the decrease in ambient temperature and generate action potentials/nerve impulses. These impulses are transmitted along sensory neurones to the thermoregulatory centre located in the hypothalamus.

(b) Nerve impulses from the hypothalamus stimulate sympathetic motor neurones causing the smooth muscle in the walls of superficial skin arterioles to contract (vasoconstriction). This reduces blood flow through the capillary networks near the surface of the skin, diverting blood to deeper core vessels. Consequently, less heat is lost from the blood to the surrounding air via radiation, convection, and conduction.

(c) Shivering involves rapid, involuntary rhythmic contractions of skeletal muscles. Muscle contractions require ATP generated through cellular respiration, a process that is not 100% efficient and releases substantial thermal energy (heat) as a byproduct to warm the body.

(a) Cold receptors in skin detect temperature decrease and generate nerve impulses (1)
Impulses transmitted via sensory neurones to the hypothalamus / thermoregulatory centre (1)

(b) Smooth muscles of skin arterioles contract / diameter decreases (vasoconstriction) (1)
Blood flow to superficial capillaries / skin surface is reduced / diverted to deeper tissues (1)
Reduces heat loss from the body surface via radiation / conduction / convection (1)

(c) Involuntary rapid contraction of skeletal muscles (1)
Increases rate of cellular respiration / ATP breakdown which releases heat / thermal energy (1)

(a) Percentage \(= \frac{\text{GPP}}{\text{Incident solar}} \times 100\% = \frac{18\,000}{1\,500\,000} \times 100\% = 1.2\%\).

(b) \(\text{NPP} = \text{GPP} - R = 18\,000 - 10\,800 = 7\,200\text{ kJ m}^{-2} \text{year}^{-1}\).

(c) \(\text{Assimilation efficiency} = \frac{\text{Energy assimilated}}{\text{Energy consumed}} \times 100\% = \frac{720}{2\,400} \times 100\% = 30\%\).

(d) 1. A large proportion of assimilated energy is lost as heat during cellular respiration to support metabolic activities and movement.
2. Some energy remains locked in unconsumed structures (bones, fur) or is lost in excretory products (urine) and egested materials (faeces).

(e) According to the laws of thermodynamics, energy transfers between trophic levels are inefficient. At each trophic level, energy is irreversibly lost to the environment as heat through respiration. Therefore, the total energy content at a lower trophic level is always greater than that at any successive higher trophic level, making an inverted energy pyramid physically impossible.

(a) \(\frac{18\,000}{1\,500\,000} \times 100\% = 1.2\%\) (1)

(b) \(18\,000 - 10\,800 = 7\,200\text{ kJ m}^{-2} \text{year}^{-1}\) (unit required) (1)

(c) \(\frac{720}{2\,400} \times 100\% = 30\%\) (1)

(d) Any two valid points (1 mark each, max 2):
- Energy lost as heat through respiration / metabolic activities / movement (1)
- Some biomass is not eaten / unconsumed by predators (e.g. bones, horns) (1)
- Some consumed matter is indigestible and lost in faeces / egestion / excretion (1)

(e) Energy is continuously lost as heat through respiration at each trophic level (1)
Energy cannot be recycled / total energy entering a trophic level is always less than the level below it (1)

(a) Stage A is Metaphase (chromosomes lined up at equatorial plate).
Stage B is Anaphase (sister chromatids pulled apart to opposite spindle poles).
Stage C is Telophase (chromosomes uncoil, nuclear envelopes reform, cell plate appears).

(b) Spindle fibres attach to kinetochores/centromeres of each chromosome during metaphase. During anaphase, motor proteins and microtubule depolymerisation cause the spindle fibres to contract/shorten, exerting tension that splits the centromeres and pulls the separated sister chromatids to opposite poles.

(c) In plant cells, vesicles fuse to form a cell plate that develops into the middle lamella and new cellulose cell wall. In animal cells, a contractile ring of actin microfilaments pinches the plasma membrane inward, forming a cleavage furrow.

(d) The root tip contains the apical meristematic zone, which has a high rate of active cell division (mitosis) for plant growth.

(a) Stage A: Metaphase (1)
Stage B: Anaphase (1)
Stage C: Telophase (1)

(b) Spindle fibers attach to centromeres of chromosomes (1)
Spindle fibers shorten / contract to separate sister chromatids and pull them to opposite poles (1)

(c) In plant cells, a cell plate forms to divide the cytoplasm, whereas in animal cells, a cleavage furrow / constriction of membrane forms (1)

(d) The root tip contains meristematic tissue / actively dividing cells for root elongation (1)

(a) An antigen is a foreign macromolecule (often a protein or polysaccharide on pathogen surfaces) that triggers an immune response. An antibody is a Y-shaped immunoglobulin protein produced by activated B cells (plasma cells) in response to a specific antigen, designed to bind specifically to that antigen.

(b)(i) The lag time (latent period) is longer in the primary response (several days) and much shorter/almost immediate in the secondary response.
(b)(ii) The peak concentration of antibodies is significantly higher in the secondary response compared to the primary response.

(c) During the primary response, specific memory B cells (and memory T cells) are formed and persist in lymphoid tissues. When re-exposed to the antigen on Day 28, these memory B cells recognise the antigen immediately, rapidly divide by clonal expansion, and differentiate into a large population of plasma cells, which secrete vast amounts of antibodies in a short time.

(d) Antibodies possess antigen-binding sites with a specific tertiary structure complementary to the epitopes of the original spike protein. If the gene coding for the viral spike glycoprotein mutates, the spatial conformation of the antigen changes. The existing antibodies can no longer bind complementarily to the mutated spike protein, failing to neutralise the virus.

(a) Antigen: A foreign substance/molecule that triggers an immune response / antibody production (1)
Antibody: A specific protein / immunoglobulin produced by plasma cells that binds to a specific antigen (1)

(b)(i) Secondary response has a shorter lag time / responds faster than primary response (1)
(b)(ii) Secondary response produces a higher peak concentration / titer of antibodies (1)

(c) Memory B cells produced during the first vaccination persist in the body (1)
Upon second exposure, memory B cells rapidly proliferate and differentiate into plasma cells that secrete large amounts of antibodies (1)

(d) Mutation changes the shape / conformation / amino acid sequence of the spike glycoprotein (1)
Existing antibodies cannot bind complementarily to the altered antigen / cannot neutralize the variant (1)

(a) The transpiration rate increases with increasing wind speed from 0 to 4 m s^-1, and then remains relatively constant / levels off between 4 and 6 m s^-1.

(b) In still air (0 m s^-1), water vapour diffuses out of stomata and accumulates near the leaf surface, forming a layer of humid air. Increased wind speed carries away this layer of humid air, maintaining a steep concentration gradient of water vapour between the sub-stomatal air space and the external air. This increases the rate of evaporation and diffusion of water vapour out of the leaves.

(c) Sunken stomata are recessed within pits on the leaf surface. This pit shelters the moist air from being easily blown away by moving air/wind. As a result, a microclimate of high humidity is retained within the pit, preventing the steepening of the water vapour concentration gradient even when external wind speed increases.

(d) 1. Wrap the pot and soil surface completely with a waterproof plastic bag tied around the base of the stem to prevent water loss via evaporation from the soil. 2. Ensure no water spills on the pot or foliage during the experimental period.

(a) Transpiration rate increases as wind speed increases, then levels off / plateaus (1 mark)

(b)
- Wind blows away / removes water vapour accumulated around the leaf surface / stomata (1 mark)
- Steepens the water vapour concentration gradient between the internal leaf tissue/air space and the surrounding atmosphere, promoting diffusion (1 mark)

(c)
- Sunken pits trap a layer of humid / still / stagnant air directly over the stomata (1 mark)
- This layer is protected from being swept away by the wind, reducing the water vapour concentration gradient / lowering water loss (1 mark)

(d) Any two of the following (1 mark each, max 2 marks):
- Wrap the pot and soil tightly with a waterproof / polythene bag (to prevent evaporation from soil)
- Seal the opening around the stem with petroleum jelly / string
- Keep the outer pot dry / wipe off any spilled water before each weighing

(a)(i) Substance X is bile / bile salts.
(ii) Bile salts lower the surface tension of large fat globules and break them down mechanically into smaller droplets (emulsification). This greatly increases the surface-area-to-volume ratio of the substrate, allowing pancreatic lipase to digest the triglycerides much more rapidly.

(b) Enzyme Y is pancreatic lipase, produced by the pancreas. Other digestive enzymes produced by the pancreas include pancreatic amylase, trypsin (or trypsinogen / peptidases), and nucleases.

(c)(i) Vessel Z is the lacteal (central lymphatic vessel of the villus).
(ii) It belongs to the lymphatic system. Chylomicrons are large lipoprotein complexes that cannot penetrate the narrow junctions / basement membrane of blood capillaries, whereas lacteal endothelia possess larger intercellular gaps / overlapping endothelial flaps that permit large particles like chylomicrons to enter.

(a)
(i) Bile salts / bile (1 mark)
(ii)
- Emulsifies large fat globules / droplets into tiny droplets (1 mark)
- Increases the total surface area for enzyme action / lipase action (1 mark)

(b)
- Pancreas (1 mark)
- Pancreatic amylase / trypsin / pancreatic protease / nuclease (1 mark)

(c)
(i) Lacteal / lymphatic vessel (1 mark)
(ii) Lymphatic system AND chylomicrons are too large to enter blood capillaries / lacteal walls have higher permeability / larger gaps to allow large lipid particles to enter (1 mark)

Structural adaptations for reducing water loss:
- Thick waxy cuticle: Present on the epidermal surfaces to provide a waterproof barrier, substantially reducing cuticular transpiration.
- Sunken stomata / hairs (trichomes) / rolled leaves: Stomata located in pits, covered by epidermal hairs, or situated inside rolled leaves trap a layer of moist, humid air next to the leaf surface. This reduces air movement across stomata and decreases the water vapour concentration gradient between the leaf interior and the atmosphere, thereby reducing transpiration rate.
- Reduced leaf surface area: Modification of leaves into spines, needles, or small scales decreases the surface area-to-volume ratio exposed to sunlight and dry air, minimizing water loss.
- Deep or widespread root systems: Extensive deep roots tap water from deep underground water tables, or widely spread shallow root networks rapidly absorb moisture from light rainfall.

Physiological mechanism of stomatal closure during drought:
- When soil water availability is low, plant roots and leaves experience water stress, triggering the synthesis/release of abscisic acid (ABA).
- This causes an efflux of potassium ions and other solutes from guard cells, increasing their water potential.
- Water moves out of the guard cells into surrounding epidermal cells by osmosis, causing guard cells to lose turgidity and become flaccid, leading to the closure of the stomatal pore.
- This effectively cuts down stomatal transpiration and helps the plant maintain internal turgor pressure and prevent wilting.

Trade-offs of stomatal closure on plant growth and transport:
- Reduction in photosynthesis and growth: Stomatal closure significantly restricts the diffusion of carbon dioxide into the sub-stomatal cavity and mesophyll cells. Carbon fixation in the Calvin cycle is inhibited, resulting in a decreased rate of photosynthesis and lower synthesis of organic nutrients/sugars, which stunts plant growth.
- Reduced transport of minerals: Transpiration pull is the main driving force for the upward transport of water and dissolved inorganic mineral ions in the xylem. Stomatal closure lowers the transpiration pull, reducing the supply of essential mineral ions to shoots and leaves.
- Impaired evaporative cooling: Transpiration provides evaporative cooling to prevent leaf overheating. Prolonged stomatal closure under intense sunlight may lead to heat stress and enzyme denaturation in leaf tissues.

Content Marks (max. 9 marks):

Structural adaptations of xerophytic plants (max. 4 marks):
- Thick waxy cuticle on the epidermis acts as an impermeable barrier to reduce cuticular transpiration. (1)
- Sunken stomata in pits / presence of epidermal hairs / rolled leaves trap a microclimate of humid air around stomata. (1)
- This reduces the water vapour concentration gradient between the sub-stomatal air space and the external air, reducing the transpiration rate. (1)
- Modification of leaves into spines / needles / reduced leaf surface area decreases the surface area available for transpiration. (1)
- Deep tap roots to reach underground water table / extensive shallow root systems to absorb scarce rainfall quickly. (1)

Physiological mechanism of stomatal closure during water stress (max. 3 marks):
- Water deficit triggers the release of the plant hormone abscisic acid (ABA) / leads to the loss of solutes (such as \(\text{K}^+\)) from guard cells. (1)
- Water leaves guard cells by osmosis, causing guard cells to become flaccid / lose turgor. (1)
- Guard cells change shape and close the stomatal pore, greatly reducing water loss through stomata to maintain tissue turgor and prevent desiccation/wilting. (1)

Trade-offs on plant growth and transport (max. 3 marks):
- Stomatal closure restricts the entry / diffusion of carbon dioxide into the mesophyll. (1)
- Rate of photosynthesis / Calvin cycle decreases, reducing organic nutrient production and inhibiting plant growth. (1)
- Reduced transpiration diminishes transpiration pull, which slows down the upward transport of water and mineral ions in the xylem / reduces evaporative cooling, increasing susceptibility to thermal stress. (1)

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Effective Communication (0–3 marks):

| Mark | Clarity of expression and relevance to the question | Logical and systematic presentation |
| :---: | :--- | :--- |
| 3 | Answers are easy to understand. They are fluent, showing good command of language. There is no or little irrelevant material. | Answers are well structured, showing coherence of thought and organisation of ideas. |
| 2 | Language used is understandable but there is some inappropriate use of words. A little irrelevant material is included but does not mar the overall answer. | Answers are organised, but there is some repetition of ideas. |
| 1 | Markers have to use some time and effort in understanding the answer(s). Irrelevant material obscures some minor ideas. | Answers are a bit disorganised, but paragraphing is evident. Repetition is noticeable. |
| 0 | Language used is incomprehensible. Irrelevant material buries the major ideas required by the question. | Ideas are not coherent or systematic. Candidates show no attempt to organise thoughts. |

(a) (i) Sweating causes loss of water, leading to an increase in blood solute concentration / decrease in blood water potential. Osmoreceptors in the hypothalamus detect the drop in blood water potential and stimulate the posterior pituitary gland to release more antidiuretic hormone (ADH) into the bloodstream. ADH increases the permeability of the collecting ducts and distal convoluted tubules to water. As a result, more water is reabsorbed by osmosis into the surrounding capillaries, resulting in a smaller volume of concentrated urine.

(ii) Desert animals possess a longer loop of Henle that extends deeper into the renal medulla. This creates a steeper osmotic gradient in the medullary interstitial fluid (higher concentration of solutes in the inner medulla). Consequently, a greater proportion of water can be reabsorbed from the collecting duct by osmosis, allowing them to produce hypertonic urine and conserve water.

(iii) Rapid ingestion of large volumes of water lowers blood solute concentration / increases plasma water potential. Water enters red blood cells by osmosis down the water potential gradient, causing them to swell and potentially lyse (haemolyse). Meanwhile, the sudden absorption of water expands the blood volume, which increases venous return and cardiac output, leading to an immediate transient increase in blood pressure.

(b) (i) \( \text{CO}_2 \) level is a much more potent stimulus. In Condition 2, increasing \( \text{CO}_2 \) from 0.04% to 4.0% caused a large increase in ventilation rate from 6.0 to 22.5 L min\(^{-1}\) (nearly fourfold increase). In Condition 3, decreasing \( \text{O}_2 \) substantially from 21% to 12% caused only a minor increase in ventilation rate from 6.0 to 8.2 L min\(^{-1}\).

(ii) Elevated blood \( \text{CO}_2 \) reacts with water to form carbonic acid, which dissociates and lowers the pH of blood and cerebrospinal fluid (increasing \( \text{H}^+ \) concentration). This is detected by central chemoreceptors in the medulla oblongata (and peripheral chemoreceptors in carotid and aortic bodies). The respiratory centre in the medulla generates more frequent nerve impulses sent via motor nerves (intercostal nerves and phrenic nerve) to the respiratory muscles (external intercostal muscles and diaphragm), increasing the rate and depth of breathing.

(iii) Hyperventilation removes large amounts of \( \text{CO}_2 \) from the blood without significantly increasing \( \text{O}_2 \) saturation (as haemoglobin is already saturated). The urge to breathe is driven primarily by high arterial \( \text{CO}_2 \). Because initial \( \text{CO}_2 \) is very low, the threshold to trigger breathing is not reached even while \( \text{O}_2 \) is continuously consumed. Arterial \( \text{O}_2 \) drops critically low before \( \text{CO}_2 \) rises enough to stimulate breathing, leading to brain hypoxia and blackout without prior warning.

(a) (i)
- Decrease in blood water potential / increase in blood osmolarity detected by osmoreceptors in hypothalamus (1)
- Hypothalamus stimulates pituitary gland / posterior pituitary to release more ADH into blood (1)
- ADH increases water permeability of collecting duct / distal convoluted tubule (1)
- More water is reabsorbed by osmosis into blood, producing a smaller volume of concentrated urine (1)

(ii)
- Longer loop of Henle / extending deeper into medulla (1)
- Establishes a steeper / higher solute concentration gradient in medullary interstitial fluid (1)
- Enables greater water reabsorption from collecting ducts by osmosis (1)

(iii)
- Water enters red blood cells by osmosis / down water potential gradient, causing swelling / lysis (haemolysis) (1)
- Blood volume increases (1)
- Blood pressure increases (1)

(b) (i)
- \( \text{CO}_2 \) level is the more potent stimulus (1)
- When \( \text{CO}_2 \) increased to 4%, ventilation rate increased significantly by 16.5 L min\(^{-1}\) / by ~275% (1)
- When \( \text{O}_2 \) dropped substantially to 12%, ventilation rate only increased slightly by 2.2 L min\(^{-1}\) / by ~37% (1)

(ii)
- High \( \text{CO}_2 \) lowers pH / increases \( [\text{H}^+] \) in blood / cerebrospinal fluid (1)
- Detected by central chemoreceptors in the medulla / peripheral chemoreceptors in carotid and aortic bodies (1)
- Respiratory centre in medulla sends more nerve impulses (1)
- Via motor nerves to diaphragm and external intercostal muscles to increase breathing rate and depth (1)

(iii)
- Hyperventilation significantly reduces blood \( \text{CO}_2 \) level without adding extra \( \text{O}_2 \) (1)
- The respiratory centre is not stimulated to breathe as \( \text{CO}_2 \) remains below threshold (1)
- Blood \( \text{O}_2 \) level drops to dangerously low levels before \( \text{CO}_2 \) rises enough to trigger breathing, causing blackout / brain hypoxia (1)

(a) (i) The restriction endonuclease cuts both the plasmid vector at the multiple cloning site and the hGH cDNA fragments at specific recognition sequences, generating complementary sticky ends. DNA ligase then catalyses the formation of phosphodiester bonds between the sugar-phosphate backbones of the vector and cDNA, sealing the cDNA into the plasmid.

(ii) The transformed E. coli are grown on agar containing ampicillin and X-gal. Non-transformed bacteria lack the \( amp^R \) gene and are killed by ampicillin. Bacteria with intact non-recombinant plasmids have an intact \( lacZ \) gene, producing functional \( \beta \)-galactosidase that hydrolyses X-gal to form blue colonies. Bacteria containing recombinant plasmids have the \( lacZ \) gene disrupted by the inserted cDNA (insertional inactivation), so they cannot produce functional \( \beta \)-galactosidase and form white colonies. Therefore, white colonies growing on ampicillin plates contain the recombinant plasmid.

(iii) Genomic human DNA contains non-coding introns as well as coding exons. Bacterial cells lack RNA splicing machinery (spliceosomes) to remove introns from pre-mRNA transcripts. As a result, transcription of genomic DNA in bacteria would lead to incorrect mRNA and non-functional proteins. In contrast, cDNA is synthesized by reverse transcription of mature mRNA and contains only coding exons without introns.

(b) (i) First, the target beta-carotene biosynthesis genes and an appropriate plant promoter are inserted into the T-DNA region of the Ti plasmid of Agrobacterium tumefaciens. The recombinant A. tumefaciens is then co-cultivated with cultured rice plant cells/tissue (callus). The bacteria infect the plant cells and transfer the T-DNA carrying the target genes into the plant genome. Transformed plant cells are selected using selective media and regenerated into complete transgenic rice plants by plant tissue culture.

(ii) (1) Heating to \( 95^\circ\text{C} \) breaks hydrogen bonds between complementary strands to denature double-stranded DNA into single strands. Cooling to \( 55^\circ\text{C} \) allows specific DNA primers to anneal/bind complementarily to the single-stranded DNA templates.
(2) Taq polymerase is heat-stable/thermostable and will not denature at the high temperatures (\( 95^\circ\text{C} \)) used during the denaturation step of repeated PCR cycles.

(iii) Environmental concern: Potential gene flow/cross-pollination of the transgene to wild rice relatives, possibly creating invasive weeds or reducing wild biodiversity. Socio-economic concern: High cost of patented GM seeds creating economic dependence of local farmers on multinational biotechnology companies / consumer resistance due to lack of public acceptance.

(a) (i)
- Restriction endonuclease cuts plasmid and target cDNA at specific recognition sites (1)
- To produce complementary sticky ends (1)
- DNA ligase joins / forms phosphodiester bonds between plasmid and cDNA fragments (1)

(ii)
- Ampicillin in agar kills untransformed bacteria (only transformed cells survive) (1)
- Insertion of cDNA disrupts the \( lacZ \) gene (insertional inactivation) (1)
- Non-recombinant cells have functional \( \beta \)-galactosidase and turn X-gal blue (1)
- Recombinant cells cannot break down X-gal and form white colonies (1)

(iii)
- Human genomic DNA contains introns (non-coding sequences) (1)
- Bacteria lack RNA splicing mechanisms / cannot remove introns (1)
- cDNA is synthesized from mature mRNA which contains only exons / coding sequences (1)

(b) (i)
- Insert target genes into the T-DNA / Ti plasmid of Agrobacterium tumefaciens (1)
- Co-cultivate / infect rice plant cells (callus) with recombinant A. tumefaciens (1)
- A. tumefaciens transfers T-DNA containing target genes into the rice plant genome (1)
- Regenerate transgenic rice plants from transformed cells via tissue culture (1)

(ii) (1)
- \( 95^\circ\text{C} \): Denature double-stranded DNA / break hydrogen bonds to yield single strands (1)
- \( 55^\circ\text{C} \): Allow primers to anneal / bind to single-stranded template DNA (1)

(ii) (2)
- Taq polymerase is thermostable / resistant to denaturation at high temperatures (1)

(iii)
- Environmental concern (any one): Gene transfer / gene flow to wild relatives via cross-pollination; reduction in genetic diversity of native rice varieties; potential unforeseen impact on non-target organisms (1)
- Socio-economic concern (any one): Dependence of smallholder farmers on biotechnology companies for expensive patented seeds; trade barriers or market rejection of GM food; ethical concerns over patenting living organisms (1)
- Clarity and coherence of presentation of both points (1)