2021 HKDSE Biology Practice Paper with Answers

To ensure a fair test when investigating the effect of light intensity, the independent variable is the light intensity (which can be varied by changing the distance between the light source and the aquatic plant). All other factors affecting photosynthesis, such as temperature (1) and carbon dioxide availability/sodium hydrogencarbonate concentration (2), must be kept constant as controlled variables. Distance (3) is the means of changing the independent variable, so it should not be kept constant.

A (1 mark): (1) and (2) only

During normal quiet exhalation (a passive process), the external intercostal muscles relax, causing the rib cage to move downwards and inwards. The diaphragm relaxes and returns to its dome shape. This decreases the thoracic volume, leading to an increase in lung pressure above atmospheric pressure, pushing air out of the lungs.

B (1 mark): The external intercostal muscles relax and the diaphragm relaxes.

According to Chargaff's rules for double-stranded DNA, the percentage of cytosine (C) equals the percentage of guanine (G), so G = 28%. Together, C + G = 56%. The remaining bases are adenine (A) and thymine (T), which make up 100% - 56% = 44%. Since A = T, the percentage of adenine is 44% / 2 = 22%.

C (1 mark): 22%

Crossing over occurs between non-sister chromatids of homologous chromosomes during prophase I of meiosis. Sister chromatids separate during anaphase II (not anaphase I, where homologous chromosomes separate). DNA replication occurs during interphase prior to meiosis, not between meiosis I and II. Meiosis yields four genetically non-identical haploid gametes.

D (1 mark): Crossing over between homologous chromosomes occurs during prophase I.

The 1% sucrose solution in the beaker has a higher water potential than the 10% sucrose solution inside the dialysis tubing bag. Therefore, water moves by osmosis down the water potential gradient into the dialysis tubing bag, resulting in an increase in the volume of the bag.

B (1 mark): Net movement of water: Into the bag; Change in volume of the bag: Increases

Energy is lost at each trophic level mainly due to heat released during cellular respiration, unconsumed parts of organisms, and undigested waste/excretion. Hence, only around 10% of the energy is transferred to the next trophic level.

C (1 mark): Energy is lost as heat through respiration and in uneaten or undigested materials.

Untreated sewage provides rich organic matter that promotes the rapid growth and multiplication of aerobic decomposing bacteria. These bacteria consume dissolved oxygen rapidly during aerobic respiration, causing a sharp drop in the dissolved oxygen level in the pond water, which can lead to the suffocation and death of aquatic animals such as fish.

A (1 mark): A sharp decrease in the concentration of dissolved oxygen in the water.

Upon re-encountering the same antigen in a secondary immune response, memory B cells rapidly differentiate into plasma cells that produce a large quantity of specific antibodies in a shorter time period. Thus, the secondary response produces antibodies faster (1), reaches a higher peak concentration (2), and involves the activation of memory B cells (3).

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

Companion cells contain abundant mitochondria to produce ATP required for active loading of sugars (e.g. sucrose) into sieve tube elements against a concentration gradient. Translocation can occur bidirectionally (upwards or downwards) depending on source and sink locations, and mass flow is driven by hydrostatic pressure gradients generated by osmotic water uptake, not transpiration pull.

B (1 mark)

The light-dependent stage occurs in the thylakoid membranes and generates ATP and reduced NADP (NADPH) via photophosphorylation and photolysis of water. These energy-carrying molecules and reducing power are subsequently consumed in the light-independent stage (Calvin cycle) occurring in the stroma to reduce carbon dioxide into carbohydrates (such as triose phosphate).

C (1 mark)

The hepatic portal vein carries nutrient-rich blood from the capillary beds of the digestive tract (such as the small intestine) directly to the liver. The pulmonary artery carries deoxygenated blood while the pulmonary vein carries oxygenated blood; contraction of the right ventricle pumps deoxygenated blood to the lungs via the pulmonary trunk/artery; and the bicuspid (mitral) valve is located between the left atrium and left ventricle.

B (1 mark)

A double-stranded DNA of 2000 base pairs has a total of \(2000 \times 2 = 4000\) bases. Since adenine (A) makes up 28%, thymine (T) also makes up 28%. Together, \(A + T = 56\%\). Therefore, \(G + C = 100\% - 56\% = 44\%\). Since cytosine (C) and guanine (G) are present in equal amounts, \(C = 22\%\). The number of cytosine bases is \(4000 \times 0.22 = 880\).

C (1 mark)

The cross is a dihybrid test cross: \(BbEe \times bbee\). The \(BbEe\) parent produces gametes \(BE\), \(Be\), \(bE\), and \(be\) in equal proportions (1:1:1:1), while the \(bbee\) parent produces only \(be\) gametes. This results in four phenotypic classes (black short-eared, black long-eared, white short-eared, white long-eared) in an expected ratio of 1:1:1:1.

B (1 mark)

Statement (1) is correct as the ecological efficiency between trophic levels is generally around 10%. Statement (2) is correct because phytoplankton have a high turnover rate and rapid reproduction, resulting in a small standing crop biomass supporting a larger zooplankton biomass at any specific moment. Statement (3) is incorrect because non-biodegradable fat-soluble toxins undergo biomagnification, reaching the highest concentration in top predators (large fish).

B (1 mark)

Upon primary infection, memory B cells and memory T cells specific to the pathogen's antigens are formed and remain in lymphoid tissues. When exposed to the same pathogen again, these memory cells rapidly proliferate and differentiate, generating a much faster and greater secondary immune response with high antibody titers before symptoms develop.

C (1 mark)

At \(0.4\text{ M}\) sucrose, the final length of the cylinder remained \(50\text{ mm}\) (no net change in length), indicating that there is no net movement of water into or out of the cells by osmosis. Hence, the water potential of the potato cell sap is equivalent to that of the \(0.4\text{ M}\) sucrose solution.

A (1 mark)

Palisade mesophyll cells are located near the upper epidermis where light intensity is highest. They contain the highest density of chloroplasts per cell compared to spongy mesophyll cells, epidermal cells (which contain no chloroplasts except guard cells), and vascular tissues. Therefore, under bright sunlight, the palisade mesophyll layer has the highest rate of photosynthesis and the highest rate of carbon dioxide consumption.

A (1 mark)

A base substitution that results in a codon specifying the same amino acid is known as a silent (or synonymous) mutation. The primary structure (amino acid sequence) of the resulting polypeptide remains identical to the original protein, so its 3D conformation and function are unchanged. The mRNA transcribed will still carry the altered nucleotide (CGC instead of CGA), so the base sequence of mRNA is changed, but the translated protein is unaffected.

C (1 mark)

When there is no net change in mass of the potato tissue, the rate of water entering the cells equals the rate of water leaving the cells. This means the sucrose solution is isotonic to the cell sap, so the water potential of the external sucrose solution is equal to the average water potential of the potato cells.

B (1 mark)

Pepsin acts in the stomach in an acidic medium (pH 1.5–2.0). Salivary amylase acts in the mouth/buccal cavity in a neutral/slightly alkaline medium (pH 6.8–7.0). Pancreatic lipase works in the small intestine (duodenum) in an alkaline medium (pH 7.5–8.5). Trypsin acts in the small intestine in an alkaline environment (pH 7.5–8.5). Therefore, option D is correct.

D (1 mark)

Let \( X^B \) be normal allele and \( X^b \) be colour-blind allele. The father is \( X^B Y \). The mother must be a carrier \( X^B X^b \) since she had a colour-blind son (\( X^b Y \)).
Possible offspring from \( X^B X^b \times X^B Y \):
- \( X^B X^B \): normal daughter (\( 1/4 \))
- \( X^B X^b \): carrier daughter (\( 1/4 \))
- \( X^B Y \): normal son (\( 1/4 \))
- \( X^b Y \): colour-blind son (\( 1/4 \))
Thus, the probability that the next child is a carrier daughter is \( 1/4 \) (or 25%).

B (1 mark)

Oxygen acts as the terminal electron acceptor in the electron transport chain (oxidative phosphorylation) at the inner mitochondrial membrane. It accepts electrons and protons to form water. Glycolysis, the link reaction, and the Krebs cycle do not directly use molecular oxygen, although the link reaction and Krebs cycle require aerobic conditions to regenerate oxidized coenzymes via the electron transport chain.

C (1 mark)

In a pyramid of energy, energy is lost at each successive trophic level (typically around 90% loss, leaving ~10% transfer efficiency). Thus, producers must contain the most energy (Q = 12000), followed by primary consumers (S = 1200), secondary consumers (P = 120), and tertiary consumers (R = 12). The food chain is therefore \( Q \rightarrow S \rightarrow P \rightarrow R \).

A (1 mark)

For accommodation to near vision, the ciliary muscles contract. This releases tension on the suspensory ligaments (they become slackened). As a result, the elastic lens recoils and becomes more convex (thicker/more curved), increasing its refractive power to focus diverging light rays from the approaching object onto the retina.

B (1 mark)

The lower epidermis is shaded from direct sunlight and shielded from strong air currents, which reduces excessive water loss by transpiration while still allowing gas exchange to occur.

B (1 mark)

Dialysis tubing is selectively permeable, allowing small molecules like glucose to diffuse through into the surrounding water, which will give a brick-red precipitate with Benedict's reagent upon heating. Starch molecules are too large to pass through the membrane pores, so starch remains inside the tubing and will not be detected in the beaker water.

C (1 mark)

According to Chargaff's rules for double-stranded DNA, % Guanine = % Cytosine = 22%. Together, G + C = 44%. The remaining bases (A + T) make up 100% - 44% = 56%. Since % Adenine = % Thymine, % Adenine = 56% / 2 = 28%.

B (1 mark)

Energy flow through an ecosystem is unidirectional (non-cyclic) because energy dissipated as metabolic heat cannot be reused by producers. Thus statement (2) is incorrect. Statements (1) and (3) are correct biological principles describing trophic efficiency and food chain length.

B (1 mark)

Blood entering the kidneys via the renal arteries undergoes ultrafiltration and selective reabsorption, during which a large portion of urea is removed and excreted into urine. Therefore, blood leaving the kidneys via the renal veins has the lowest concentration of urea.

D (1 mark)

In a dihybrid cross of BbSs x BbSs with independent assortment: Probability of white coat (bb) = 1/4. Probability of short hair (SS or Ss) = 3/4. Therefore, probability of white coat and short hair = (1/4) * (3/4) = 3/16.

B (1 mark)

Glycolysis breaks down glucose into pyruvate in the cytoplasm without producing carbon dioxide. The link reaction converts pyruvate into acetyl-CoA, releasing one molecule of carbon dioxide per pyruvate. The Krebs cycle involves decarboxylation reactions releasing carbon dioxide. Therefore, only (2) and (3) directly produce carbon dioxide.

C (1 mark)

A spinal reflex is automatic, rapid, and involuntary, occurring without conscious intervention via a neural pathway through the spinal cord. Sensory neurones carry impulses to the CNS, not motor neurones. Impulses only reach the cerebral cortex after or simultaneously via ascending pathways, not before the effector reacts.

C (1 mark)

In darkness, stomata close, substantially reducing stomatal transpiration. Still air prevents the removal of the humid boundary layer around the leaves, and high relative humidity minimizes the water vapour concentration gradient between the internal leaf air spaces and the surrounding atmosphere. Together, these conditions minimize both cuticular and stomatal transpiration, leading to the slowest rate of water uptake.

C (1 mark)

Trypsin is synthesized and secreted by the pancreas as the inactive precursor trypsinogen. This prevents autolysis (self-digestion) of pancreatic tissue. It is later activated to trypsin in the duodenum by the enzyme enterokinase. Option A is incorrect because bile contains bile salts but no digestive enzymes. Option C is incorrect because monosaccharides and amino acids are absorbed directly into blood capillaries, not lacteals (which absorb fatty acids and glycerol). Option D is incorrect because peristalsis is an involuntary process.

B (1 mark)

The mRNA sequence is complementary to the template strand, meaning it has the same base sequence as the coding strand, with uracil (U) replacing thymine (T). Thus, the mRNA is 5'- AUG CCG AAU UAA -3'.

During translation:
- Codon 1 (AUG) codes for Methionine.
- Codon 2 (CCG) codes for Proline.
- Codon 3 (AAU) codes for Asparagine.
- Codon 4 (UAA) is a stop codon that terminates translation and does not code for any amino acid.

Hence, the resulting peptide consists of 3 amino acids.

A (1 mark)

Energy flows non-cyclically through ecosystems. At each trophic level, a large proportion of energy (typically ~90%) is lost to the environment as metabolic heat during cellular respiration, or remains in uneaten/undigested biomass. Therefore, available energy decreases continuously along a food chain. Option A is incorrect as energy enters as light and is converted to chemical energy. Option B is incorrect as transfer efficiency is typically only 10% (loss is 90%). Option C is incorrect because heat energy radiated to the atmosphere cannot be trapped or reused by producers.

D (1 mark)

(a) Even with both stomatal surfaces sealed, a small amount of cuticular transpiration can occur through the epidermis/stem, or water is consumed in cellular metabolism/photosynthesis.
(b) Comparing X (upper coated, 42 mm) with W (control, 48 mm), transpiration only decreased by 6 mm (12.5%). Comparing Y (lower coated, 10 mm) with W (48 mm), transpiration dropped by 38 mm (79.2%). This large difference demonstrates that stomata are far more numerous on the lower epidermis than on the upper epidermis.
(c) Increased air movement blows away water vapour accumulating near the stomata on the leaf surface. This lowers the relative humidity of the microenvironment around the leaf, increasing the water vapour concentration gradient between the intercellular spaces and the surrounding atmosphere. Consequently, transpiration rate rises, increasing the rate of water absorption and causing the bubble to move further.

(a) Water loss through the cuticle / stem lenticels OR water used in photosynthesis / maintaining cell turgidity (1 mark)

(b)
- Stomata are mainly / mostly located on the lower epidermis / surface (1 mark)
- Coating the upper surface resulted in only a small reduction in water uptake / transpiration (from 48 mm to 42 mm) (1 mark)
- Coating the lower surface caused a large / significant reduction in water uptake (from 48 mm to 10 mm) (1 mark)

(c)
- The distance moved by the bubble would increase / be greater than 48 mm (1 mark)
- Wind blows away water vapour from the leaf surface / prevents accumulation of a humid boundary layer (1 mark)
- This maintains / increases the steepness of the water vapour concentration gradient between the inside of the leaf and the external air (1 mark)

(a) Long-chain fatty acids and glycerol recombine into triglycerides inside epithelial cells and form chylomicrons, which enter lacteals.
(b) 1. The apical surface has folding in the form of microvilli, vastly increasing the surface area for diffusion and active transport proteins. 2. High density of mitochondria produces ATP to support secondary active transport of glucose and amino acids against concentration gradients.
(c) (i) Ileum resection reduces the available absorptive surface area. Unabsorbed nutrients remain in the intestinal lumen, lowering lumen water potential, which prevents water reabsorption by osmosis and causes diarrhea. The lack of absorbed carbohydrates, lipids, and amino acids causes progressive weight loss.
(ii) Eating small, frequent meals prevents saturation of the remaining digestive capacity; consuming predigested or elemental liquid formulas ensures efficient absorption.

(a) Fatty acids / glycerol / lipids / triglycerides / fat-soluble vitamins (1 mark)

(b) Any two of the following (1 mark each, max 2):
- Presence of microvilli on the cell surface to increase the surface area for absorption
- Abundant mitochondria to provide ATP / energy for active transport of nutrients
- Thin epithelial layer (one-cell thick) to provide a short diffusion distance
- Presence of specific transport proteins / carrier proteins on the membrane

(c)
(i)
- Reduced total surface area reduces absorption of nutrients, causing weight loss (1 mark)
- Unabsorbed nutrients decrease the water potential of intestinal contents, drawing/retaining water in the lumen by osmosis, resulting in diarrhea (1 mark)
(ii)
- Eat smaller and more frequent meals (1 mark)
- Choose easily absorbable food / nutrient-dense liquid diet / reduce high-fat intake (1 mark)

(a) A single adenine (A) is replaced by thymine (T) in the DNA sequence, which is a base substitution mutation.
(b) The coding strand sequence GAG transcribes to mRNA codon GAG, which codes for glutamic acid. The mutant sequence GTG transcribes to GUG, which codes for valine.
(c) Glutamic acid possesses a charged, polar side chain, whereas valine has a non-polar, hydrophobic side chain. This substitution disrupts normal intra-molecular bonding (hydrogen bonding/ionic attractions) during protein folding, altering the tertiary structure (3D shape) of the polypeptide, which destroys its normal functional properties (e.g., oxygen-carrying ability or enzyme catalysis).
(d) Because the genetic code has degeneracy, different triplets can specify the same amino acid (for instance, GAA and GAG both code for glutamic acid). If a substitution changes a base but results in a triplet that codes for the exact same amino acid, no change in protein primary structure occurs.

(a) Base substitution / point mutation / nucleotide replacement (1 mark)

(b)
- 7th codon changes from GAG to GUG (1 mark)
- Glutamic acid (Glu) is replaced by valine (Val) (1 mark)

(c)
- Change in primary structure / amino acid sequence (1 mark)
- Affects the folding / interactions (ionic bonds / hydrophobic interactions) in polypeptide chain (1 mark)
- Alters the three-dimensional / tertiary conformation of the protein, disabling its function (1 mark)

(d)
- The genetic code is degenerate / redundant (1 mark)
- Different codons / triplets can specify / code for the same amino acid (1 mark)

(a) Domestic sewage contains a high concentration of dissolved organic nutrients (carbohydrates, proteins, fats). Aerobic bacteria utilize these substrates for growth and multiplication, resulting in a population surge.
(b) The rapidly expanding bacterial population carries out vigorous aerobic cellular respiration to break down the organic waste, consuming large quantities of dissolved oxygen from the water faster than it can dissolve from the atmosphere.
(c) The breakdown of organic nitrogen and phosphorus by bacteria (ammonification and subsequent nitrification) releases large quantities of soluble inorganic ions ( ext{NO}_3^-, ext{PO}_4^{3-}). These act as limiting plant fertilizers, promoting explosive algal cell division (algal bloom).
(d) Constructing municipal sewage treatment works to process wastewater before discharge or implementing septic tank regulations prevents raw sewage release.

(a)
- Population density of aerobic decomposers increases (1 mark)
- Sewage provides abundant organic matter / nutrients / food source for bacterial growth and reproduction (1 mark)

(b)
- Bacteria undergo rapid aerobic respiration (1 mark)
- They consume large amounts of dissolved oxygen at a rate higher than re-oxygenation (1 mark)

(c)
- Decomposers break down organic waste into inorganic mineral nutrients / nitrates / phosphates (1 mark)
- High concentrations of these nutrients promote rapid multiplication / growth of algae (1 mark)

(d) Secondary/tertiary sewage treatment / building sewage treatment plants / strict enforcement of pollution control laws (1 mark)

(a) Stage P represents Metaphase I (homologous pairs on metaphase plate) and Stage R represents Anaphase I (separation of homologous chromosomes).
(b) 1. Crossing over / genetic recombination occurs at prophase I where homologous non-sister chromatids exchange alleles. 2. Independent assortment and random segregation of maternal and paternal homologous chromosomes at metaphase I / anaphase I generate diverse allele combinations.
(c) Reduction division produces haploid gametes ( ext{n}). Upon fertilization, the union of two gametes restores the diploid state ( ext{2n}). Without this halving, the chromosome count would double in each successive generation, causing catastrophic genetic instability.
(d) Failure of homologous chromosomes to separate in anaphase I causes both members of a pair to enter one daughter cell. Following meiosis II, the two cells derived from this daughter cell will each possess ext{n} + 1 chromosomes, while the two cells from the other daughter cell will each have ext{n} - 1 chromosomes.

(a) Stages P and R (1 mark)

(b) Any two of the following (1 mark each, max 2):
- Crossing over / exchange of genetic material between non-sister chromatids of homologous chromosomes (in Prophase I)
- Independent assortment / random orientation of homologous pairs along the equator (in Metaphase I)

(c)
- Produces haploid gametes / gametes with half the chromosome number (1 mark)
- Restores the diploid number upon fertilisation / maintains a constant chromosome number from generation to generation (1 mark)

(d)
- None of the gametes will have the normal chromosome number / all 4 gametes are abnormal (1 mark)
- Two gametes will have (n + 1) chromosomes and two gametes will have (n - 1) chromosomes (1 mark)

(a) When an action potential reaches the axon terminal, the membrane depolarizes, opening voltage-gated ext{Ca}^{2+} channels. Calcium ions diffuse rapidly into the presynaptic bulb down their concentration gradient, stimulating synaptic vesicles to move toward, fuse with the presynaptic membrane, and release acetylcholine via exocytosis.
(b) Acetylcholine molecules diffuse across the narrow synaptic cleft and bind specifically to ligand-gated receptors on the postsynaptic motor end plate. This opens cation/sodium channels, allowing ext{Na}^+ influx, producing an end-plate potential that exceeds threshold to generate a muscle action potential.
(c) (i) Acetylcholinesterase hydrolyses acetylcholine into choline and acetic acid, preventing continuous activation of receptors.
(ii) When inhibited, acetylcholine remains bound to receptors on the sarcolemma. This causes continuous influx of ext{Na}^+ and sustained generation of action potentials, causing uncontrollable, repetitive muscle contractions (spasms/cramping) and eventual exhaustion/flaccid paralysis due to receptor desensitization.
(d) Unidirectional transmission occurs because synaptic vesicles containing neurotransmitters exist solely in the presynaptic terminal, whereas functional complementary receptors are located exclusively on the postsynaptic membrane.

(a)
- Influx of calcium ions ( ext{Ca}^{2+}) into the presynaptic terminal through voltage-gated channels (1 mark)
- Synaptic vesicles fuse with the presynaptic membrane to release acetylcholine by exocytosis (1 mark)

(b)
- Acetylcholine diffuses across the cleft and binds to receptors on the postsynaptic membrane (1 mark)
- Opens sodium channels, causing influx of sodium ions ( ext{Na}^{+}) / depolarisation of the postsynaptic membrane (1 mark)

(c)
(i) To hydrolyse / break down acetylcholine to stop continuous stimulation (1 mark)
(ii)
- Acetylcholine accumulates in the synaptic cleft / is not broken down (1 mark)
- Causes continuous / prolonged stimulation of muscle receptors leading to continuous muscle contraction / spasms / convulsions (1 mark)

(d) Synaptic vesicles containing neurotransmitter are only located in the presynaptic neurone, and receptors are only present on the postsynaptic membrane (1 mark)

(a) Administration of an antigen (vaccine) that triggers the body's own immune system to generate antibodies and memory cells is classified as artificial active immunity.
(b) (i) Primary response shows a lag phase of ~5-7 days, whereas the secondary response shows an immediate rise within 1-2 days.
(ii) In the secondary response, antibodies accumulate at a much steeper rate, reaching a peak concentration (80 units) that is 8 times higher than that of the primary response (10 units).
(c) During primary exposure, naive B-lymphocytes undergo clonal selection, producing effector plasma cells and long-lived memory B cells. Upon booster vaccination, memory B cells recognize the antigen immediately without extensive lag time, proliferating rapidly by mitosis into numerous plasma cells that produce vast amounts of antibodies.
(d) Antibody molecules feature antigen-binding sites with a specific 3D conformation that only recognizes and binds to epitopes complementary in shape. The surface proteins (hemagglutinin/neuraminidase) of the influenza virus have completely different structures from the bacterial vaccine antigen.

(a) Artificial active immunity (1 mark)

(b)
(i) Shorter lag period / faster onset of antibody production in secondary response (1 mark)
(ii) Faster rate of production and higher maximum / peak antibody level in secondary response (1 mark)

(c)
- Presence of memory B cells formed during the primary response (1 mark)
- Memory cells rapidly recognise the specific antigen upon second exposure (1 mark)
- Quickly proliferate and differentiate into a large number of plasma cells to secrete abundant antibodies (1 mark)

(d) Antibodies are antigen-specific / antigen-binding sites only fit the complementary shape of the bacterial antigen, not the viral antigens of influenza (1 mark)

(a) Osmoreceptors situated in the hypothalamus detect changes in blood osmolarity/water potential. The hypothalamus acts as the coordinating centre, modulating ADH release via the posterior pituitary gland.
(b) Ingestion and absorption of 1000 mL water raises blood water potential (decreases osmolarity). Osmoreceptors in the hypothalamus swell slightly and reduce nerve impulses sent to the posterior pituitary, leading to decreased secretion of antidiuretic hormone (ADH) into the bloodstream. In the kidneys, lower circulating ADH reduces the insertion of aquaporins into the membranes of the distal convoluted tubule and collecting duct cells, decreasing their permeability to water. Consequently, less water is reabsorbed into the medullary capillaries, leaving more water in the tubule lumen, which is excreted as a large volume of dilute urine (diuresis).
(c) Isotonic saline contains a solute concentration identical to that of blood plasma. Thus, drinking it expands the extracellular fluid volume without changing the water potential of the blood. Since blood water potential remains stable, osmoreceptors are not stimulated to decrease ADH secretion significantly, producing a much smaller increase in urine output compared to distilled water.

(a)
- Receptor: Osmoreceptors (in the hypothalamus) (1 mark)
- Control centre: Hypothalamus / pituitary gland (1 mark)

(b)
- Drinking pure water raises / increases the water potential of blood (1 mark)
- Detected by osmoreceptors, causing the posterior pituitary to release less ADH / decrease ADH secretion (1 mark)
- Permeability of the collecting duct / distal convoluted tubule to water decreases (1 mark)
- Less water is reabsorbed into the blood by osmosis, resulting in a larger volume of dilute urine (1 mark)

(c)
- The increase in urine volume will be much less / slower (1 mark)
- Isotonic solution does not change the water potential of blood / does not suppress ADH secretion (1 mark)

(a) A potometer measures the rate of water uptake, which is assumed to be almost identical to the rate of water loss via transpiration (a very small fraction is used in metabolism/photosynthesis and cell turgidity).
(b) Moving air removes the humid boundary layer around the stomata, maintaining a steep water vapour diffusion gradient between the internal leaf airspace and external atmosphere. This accelerates transpiration pull, thereby increasing the rate of water uptake.
(c) Xerophytic adaptations that limit water loss include thick cuticles, lower stomatal density, sunken stomata, and epidermal hairs that trap still, moist air.
(d) Transpiration causes water loss at the leaves, creating a pulling force (tension) that draws water up the xylem vessels via continuous cohesive and adhesive forces (the cohesion-tension mechanism).

(a) The rate of water uptake is equal to the rate of water loss / transpiration (1 mark).

(b)
- Wind removes the boundary layer of water vapour / humid air around the leaf surface / stomata (1 mark).
- This maintains a steeper water vapour concentration / diffusion gradient between the leaf interior and the surrounding air, increasing the rate of transpiration / water uptake (1 mark).

(c) Any two structural features (1 mark each, max 2 marks):
- Thicker waxy cuticle on the epidermis
- Sunken stomata / stomata in pits / grooves
- Presence of epidermal hairs / trichomes
- Lower stomatal density / fewer stomata on upper or lower epidermis
(Do not accept physiological features like 'stomata close during day' unless explicitly asked for behaviour/mechanism).

(d)
- Evaporation of water from the mesophyll cells creates a transpiration pull / tension / negative pressure in the leaf xylem (1 mark).
- Cohesion between water molecules (and/or adhesion of water to xylem walls) pulls water upward as a continuous water column (1 mark).

(a) Starch is digested into maltose by amylase. Maltose is a disaccharide (reducing sugar) and is small enough to pass through the pores of the dialysis tubing into the surrounding water. Hence, when heated with Benedict's reagent, a brick-red precipitate forms.
(b) Protein molecules are too large to pass through the differentially permeable visking tubing, and no protease is present to digest them into smaller amino acids. Thus, protein does not appear in the beaker liquid.
(c) The real small intestine has many specialized adaptations, including folded mucosal walls, villi, and microvilli to increase surface area, rich capillary supplies to maintain diffusion gradients, and active transport mechanisms.

(a)
- Observation: Forms a brick-red / orange / yellow / green precipitate (1 mark).
- Explanation: Amylase digests / breaks down starch into maltose (a reducing sugar) (1 mark).
- Maltose molecules are small enough to pass / diffuse through the pores of the dialysis tubing into the surrounding water (1 mark).

(b)
- Protein molecules are too large to diffuse across / pass through the pores of the dialysis tubing (1 mark).
- No protease / protein-digesting enzyme is present to hydrolyse protein into smaller diffusible molecules / amino acids (1 mark).

(c)
- Correct structural feature named (1 mark), for example:
* Presence of folds / villi / microvilli
* Rich blood capillary network / continuous blood flow
* Epithelium is one cell thick
* Presence of transport proteins / active transport mechanisms
- Explanation of how it facilitates absorption (1 mark):
* Increases total surface area for absorption (for villi/microvilli/folds)
* Rapidly removes absorbed nutrients to maintain a steep concentration gradient (for capillary network/blood flow)
* Shortens the diffusion distance (for one-cell-thick epithelium)
* Allows absorption against a concentration gradient (for active transport)

### Absorption of Water and Mineral Ions into Root Xylem
1. Absorption by root hair cells:
- Root hair cells have long, slender extensions that greatly increase the surface area for the absorption of water and mineral ions.
- Mineral ions are absorbed from the soil solution into root hair cells primarily by active transport against a concentration gradient (requiring energy/ATP produced by cellular respiration), or by diffusion when the concentration in soil is higher.
- The active accumulation of mineral ions in the cytoplasm and vacuole of root hair cells lowers their water potential below that of the surrounding soil water.
- Water enters the root hair cells by osmosis down a water potential gradient across the selectively permeable cell membrane.

2. Movement across the root cortex to the xylem:
- Water and dissolved mineral ions move across the cortical cells towards the vascular cylinder along a water potential gradient via the apoplast pathway (through cell walls and intercellular spaces) and the symplast pathway (through cytoplasm and plasmodesmata).
- At the endodermis, the impermeable Casparian strip in the cell walls blocks the apoplast pathway, forcing water and dissolved ions through the selectively permeable membranes of the endodermal cells (symplast pathway), allowing selective uptake before entering the root xylem vessels.

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### Mechanism of Upward Water Transport in the Xylem
1. Transpiration pull (Cohesion-Tension Theory):
- Water constantly evaporates from the wet surfaces of spongy mesophyll cells into the substomatal air spaces and diffuses out through open stomata into the atmosphere (transpiration).
- Evaporation of water lowers the water potential of mesophyll cells, drawing water from neighboring cells and eventually pulling water out of the xylem vessels in the leaf veins, generating a strong negative pressure (suction force / transpiration pull).
- Because of cohesion (hydrogen bonding between adjacent water molecules), water molecules stick together to maintain an unbroken, continuous water column inside xylem vessels.
- Adhesion between polar water molecules and the hydrophilic cellulose/lignified walls of xylem vessels helps support the water column against gravity.

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### Physiological Importance of Water Transport to the Plant
1. Transport of dissolved mineral nutrients:
- Upward water flow distributes vital mineral ions (e.g., nitrates for amino acid/protein synthesis, magnesium for chlorophyll synthesis) from the roots to aerial photosynthetic organs.
2. Photosynthesis and metabolic reactions:
- Water serves as an essential raw material (source of electrons and protons via photolysis) in photosynthesis and acts as a universal solvent/medium for biochemical reactions.
3. Maintaining cell turgidity and mechanical support:
- Water uptake maintains turgor pressure in plant cells, keeping herbaceous tissues, leaves, and stems erect and mechanically supported, preventing wilting and keeping leaves properly oriented to capture sunlight.
4. Evaporative cooling:
- Transpiration absorbs latent heat of vaporization from the leaves, preventing leaves from overheating under direct sunlight and protecting heat-sensitive metabolic enzymes from denaturation.

### Content (max. 8 marks)

#### 1. Uptake of water and mineral ions into the root xylem (max. 3 marks)
- Root hairs provide a large surface area for absorption. (1)
- Mineral ions are absorbed actively by active transport (using ATP) / diffusion into root cells. (1)
- Uptake of mineral ions lowers the water potential in root cells, allowing water to enter root hair cells by osmosis. (1)
- Water and mineral ions move across the cortex via apoplast and symplast pathways to the xylem vessels / Casparian strip directs water into the symplast pathway. (1)

#### 2. Driving forces of upward water movement in xylem (max. 3 marks)
- Evaporation of water from mesophyll cells through stomata (transpiration) generates a tension / transpiration pull. (1)
- Cohesion between water molecules creates a continuous water column in the xylem vessels. (1)
- Adhesion of water molecules to the walls of xylem vessels prevents the column from breaking / supports the water column. (1)

#### 3. Physiological importance of water transport (max. 3 marks)
- Transport of dissolved mineral ions from roots to aerial parts for growth/metabolism. (1)
- Provides water as a raw material for photosynthesis / biochemical solvent. (1)
- Maintains turgor pressure in cells to provide mechanical support for non-woody tissues / prevent wilting. (1)
- Evaporative cooling effect of transpiration prevents overheating / protects enzymes from denaturation. (1)

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### Clarity of Expression and Logical Presentation (max. 3 marks)
- 3 marks: Answers are comprehensive, well-structured, logically developed with accurate use of biological terminology throughout.
- 2 marks: Relevant information is clearly conveyed with minor structural defects or minor inaccuracies in terminology.
- 1 mark: Disorganised presentation with limited relevance or frequent misuse of biological terms.
- 0 marks: Incoherent or irrelevant answer.

(a) (i) Ingesting pure water increases the blood water potential (decreases blood osmolarity). Osmoreceptors in the hypothalamus detect this change and stimulate the posterior pituitary gland to release less antidiuretic hormone (ADH) into the blood. The decrease in ADH concentration lowers the water permeability of the collecting duct and distal convoluted tubule cells. Consequently, less water is reabsorbed back into the bloodstream by osmosis, leading to the production of a large volume of dilute urine (higher urine flow rate and lower urine osmolarity).

(ii) Ingestion of hypertonic saline lowers the blood water potential (increases blood osmolarity). This deviation is detected by hypothalamic osmoreceptors, which send impulses to stimulate the posterior pituitary gland to secrete more ADH. ADH increases the permeability of the collecting ducts and distal convoluted tubules to water. As a result, more water is reabsorbed from the filtrate into the medullary capillaries, bringing blood water potential back to normal and resulting in a low volume of highly concentrated urine (urine osmolarity reaching 1250 mOsm/L at 2 h).

(iii) The presence of the long loop of Henle (which establishes a high solute concentration gradient in the renal medulla by countercurrent multiplier mechanism).

(iv) Marine birds possess specialised salt glands (nasal glands) that can excrete highly concentrated sodium chloride solution, whereas human kidneys cannot produce urine concentrated enough to remove the excess salt without net water loss.

(b) (i) FSH stimulates the growth, recruitment, and development of multiple ovarian follicles simultaneously.

(ii) High concentrations of estrogen stimulate the anterior pituitary gland to release a surge of luteinizing hormone (LH surge) (and FSH), which triggers ovulation and the formation of the corpus luteum.

(iii) An untimely endogenous LH surge would trigger premature ovulation inside the body, causing the oocytes to be released into the oviduct where they cannot be collected/harvested by the doctor.

(iv) hCG shares a very similar molecular/chemical structure and the same receptor with LH; therefore, hCG can bind to and activate LH receptors on granulosa/theca cells to induce final oocyte maturation.

(v) Massive fluid shift from blood vessels into peritoneal/abdominal cavity (causing ascites/abdominal distension/hypovolemia/blood clotting risk).

(a) (i) Water intake increases blood water potential / decreases blood solute concentration (1); detected by osmoreceptors in hypothalamus, causing posterior pituitary to release less ADH (1); permeability of collecting duct / distal convoluted tubule to water decreases (1); less water reabsorbed into blood, resulting in higher flow rate and lower osmolarity (1).
(ii) Increased blood osmolarity / decreased blood water potential detected by hypothalamus (1); increases secretion of ADH from posterior pituitary gland (1); increased water reabsorption from collecting duct / DCT (1); negative feedback restores blood water potential / produces highly concentrated urine (1).
(iii) Long loops of Henle / hypertonic renal medulla / countercurrent multiplier system (1).
(iv) Marine birds have salt-excreting nasal glands capable of secreting hypertonic salt solution (1).

(b) (i) Stimulates multiple follicles to grow / mature simultaneously (1).
(ii) High estrogen level stimulates anterior pituitary gland (1) to secrete a surge of LH / trigger LH surge (1).
(iii) Prevents premature / uncontrolled ovulation (1); so that intact mature eggs can be retrieved surgically before release (1).
(iv) hCG has a similar structure to LH / binds to LH receptors (1); triggers final oocyte maturation / mimics pre-ovulatory LH surge (1).
(v) Ovarian enlargement / ascites (fluid accumulation in abdomen) / hypovolaemia / increased risk of thrombosis (any 1, 1 mark).

(a) (i) Genomic human DNA contains non-coding introns in addition to exons. Bacteria lack the RNA splicing machinery (spliceosomes) to remove introns from pre-mRNA transcripts. Therefore, cDNA (which is synthesized from mature mRNA and contains only coding exons) must be used so that bacteria can correctly translate the human protein.

(ii) Reverse transcriptase.

(iii) E. coli cells are plated onto nutrient agar containing ampicillin and X-gal. Only transformed bacteria containing the plasmid with the ampicillin resistance gene can survive and form colonies (non-transformed bacteria are killed). In non-recombinant plasmids, the lacZ gene is intact and produces beta-galactosidase, which cleaves X-gal to produce blue colonies. In recombinant plasmids, insertion of the cDNA into the MCS disrupts the lacZ gene (insertional inactivation), preventing beta-galactosidase production, so recombinant colonies appear white.

(iv) A strong promoter provides a high affinity binding site for bacterial RNA polymerase, driving high rates of transcription of the inserted cDNA into mRNA.

(v) Potential release of antibiotic-resistant transgenic bacteria into the environment / risk of horizontal gene transfer of resistance genes to pathogenic strains (or requirement for strict containment and sterilization before disposal).

(b) (i) Primers bind/anneal specifically to the complementary sequences flanking the target DNA region, providing a free 3'-OH group. Thermostable DNA polymerase (e.g., Taq polymerase) synthesizes the complementary DNA strand by adding free deoxyribonucleoside triphosphates (dNTPs) to the 3' end of the primer along the template strand in the 5' to 3' direction.

(ii) Suspect 2 matches the crime scene sample. At Locus Alpha, the crime scene sample has alleles (12, 15), matching Suspect 2 (12, 15), whereas Suspect 1 is homozygous (12, 12). At Locus Beta, both the crime scene sample and Suspect 2 have alleles (8, 11).

(iii) The lawyer's argument is valid. Testing only two loci gives a relatively high probability that another unrelated individual in the population might share the same genotype by chance. In standard forensic profiling, at least 13 to 20 independent STR loci are analyzed to reduce the match probability to an astronomically low level.

(iv) (1) DNA fragments migrate from the negative electrode (cathode) toward the positive electrode (anode). This occurs because DNA has a negatively charged sugar-phosphate backbone due to phosphate groups.
(2) An allele with 8 repeats is shorter / has a smaller molecular size than an allele with 15 repeats. Smaller DNA fragments experience less physical resistance/retardation when passing through the porous agarose gel matrix, so they migrate faster.

(a) (i) Human genomic DNA contains introns (1); bacteria cannot perform post-transcriptional splicing / remove introns (1).
(ii) Reverse transcriptase (1).
(iii) Ampicillin selects for successfully transformed bacteria / kills untransformed bacteria (1); insertion of cDNA causes insertional inactivation of the lacZ gene (1); non-recombinant colonies express beta-galactosidase and turn blue (1); recombinant colonies cannot hydrolyse X-gal and appear white (1).
(iv) High affinity for RNA polymerase to ensure high rate of transcription (1).
(v) Containment / risk of releasing antibiotic resistance genes into the natural environment (1).

(b) (i) Primers provide a starting point / free 3'-OH end for DNA polymerase (1); primers bind specifically to complementary sequences flanking target region (1); DNA polymerase adds complementary dNTPs to extend the new DNA strand (1).
(ii) Suspect 2 (1); Alleles at both Locus Alpha (12, 15) and Locus Beta (8, 11) are identical to the crime scene sample, while Suspect 1 has a different genotype (12, 12) at Locus Alpha (1).
(iii) Valid: two loci are insufficient because random coincidence probability in the population is relatively high (1); more STR loci (e.g. 13-20 loci) must be examined to achieve high statistical confidence (1).
(iv) (1) Towards the positive pole / anode (1); because the phosphate groups on the DNA backbone carry net negative charges (1).
(2) 8-repeat fragment is shorter / has lower molecular mass (1); experiences less resistance / passes through gel pores more easily (1).