2021 HKDSE Biology Answers & Marking Scheme

During the 10 to 20 minute period, the rate of transpiration (water loss) exceeds the rate of water uptake (water absorption). Therefore, (1) leaf cells experience a net loss of water, so their water content decreases. (2) As water is pulled out of the xylem faster than it is replaced, the negative pressure or tension within the xylem vessels increases. (3) is incorrect because the rate of water loss (transpiration) is higher, not lower, than the rate of water absorption.

1 mark for correct option B. No partial marks.

At 500 m, although the bacterial population has decreased from its peak at 100 m, it is still very high (60 units). The rate of oxygen consumption by these bacteria for aerobic decomposition of organic matter still exceeds the rate of oxygen replenishment from the air and the newly growing algae. Therefore, the net dissolved oxygen level continues to fall to its lowest point.

1 mark for correct option B. No partial marks.

1. Amino acids are the digested products of proteins. They are absorbed into the blood capillary network of the villi by active transport and facilitated diffusion. Therefore, blood leaving the villus (Vessel Y) has a higher concentration of amino acids than blood entering (Vessel X). 2. The epithelial and other cells in the villi are metabolically active (e.g., performing active transport of nutrients) and carry out aerobic respiration. This consumes oxygen and produces carbon dioxide. Therefore, blood leaving the villus (Vessel Y) has a lower concentration of oxygen and a higher concentration of carbon dioxide than blood entering (Vessel X).

1 mark for correct option B. No partial marks.

For (1): Since the DNA is double-stranded with 30% A, then T = 30%. Therefore, G + C = 40%, meaning G = 20% and C = 20%. The ratio of A to G in the newly synthesized strands must match the original template composition, which is 30% : 20% = 3:2. Statement (1) is correct. For (2) and (3): The original non-radioactive DNA has 2 strands. After 2 rounds of semi-conservative replication, there are 4 DNA molecules (8 strands in total). The 2 original strands are distributed into 2 different DNA molecules, each paired with a newly synthesized radioactive strand. The remaining 2 DNA molecules are composed entirely of radioactive strands. Thus, 2 out of 4 (50%) DNA molecules are hybrid (one non-radioactive and one radioactive strand), and all 4 DNA molecules contain at least one radioactive strand. Statements (2) and (3) are correct.

1 mark for correct option D. No partial marks.

When the light is switched off: 1. The light-dependent reactions stop, so ATP and NADPH production ceases, and their concentrations decrease. 2. In the Calvin cycle, the reduction of GP to triose phosphate requires ATP and NADPH. Since these are depleted, GP cannot be reduced and therefore accumulates, leading to an increase in GP concentration. 3. The regeneration of RuBP from triose phosphate also requires ATP. Because regeneration stops but RuBP continues to combine with available CO2 to form GP (which does not directly require light), the concentration of RuBP decreases rapidly.

1 mark for correct option A. No partial marks.

Persistent organic pollutants (POPs) undergo biomagnification. Because they are non-biodegradable and insoluble in water (but highly fat-soluble), they cannot be easily excreted. Organisms at higher trophic levels must consume large amounts of biomass from lower trophic levels to obtain energy, accumulating and concentrating the POP in their tissues. Thus, the concentration increases up the food chain. Based on the concentration: P (0.05) -> S (0.45) -> R (1.80) -> Q (25.40). Option B is correct.

1 mark for correct option B. No partial marks.

Between t = 0 and t = 2 seconds, alveolar pressure is lower than atmospheric pressure, which drives air into the lungs (inhalation). Inhalation is an active process. It is brought about by: 1. The contraction of external intercostal muscles, which pulls the ribcage upwards and outwards. 2. The contraction of the diaphragm muscle, which flattens the diaphragm. These movements increase the volume of the thoracic cavity, lowering the pressure inside the lungs below atmospheric pressure, causing air to flow in.

1 mark for correct option B. No partial marks.

1. Transcribe the original template DNA sequence into mRNA: Template: 3'- T A C (1-3) G G T (4-6) C A T (7-9) T C A (10-12) A C T (13-15) -5'. mRNA: 5'- A U G (1-3) C C A (4-6) G U A (7-9) A G U (10-12) U G A (13-15) -3'. Codon 3 is GUA (Valine). 2. With the mutation, the 8th base from the 3' end of the template (which is A) is replaced by T. Mutated template: 3'- T A C G G T C T T T C A A C T -5'. 3. Transcribe the mutated template: Mutated mRNA: 5'- A U G C C A G A A A G U U G A -3'. Mutated codon 3 is GAA (Glutamic acid). Therefore, the amino acid valine in the original polypeptide is replaced by glutamic acid. This is a missense mutation.

1 mark for correct option B. No partial marks.

- Statement (1) is correct: The presence of memory B cells from the primary response allows for rapid recognition of the antigen during the secondary response, leading to immediate clonal expansion and differentiation into plasma cells. - Statement (2) is correct: The secondary response produces a significantly higher peak concentration of antibodies, and these antibodies persist in the bloodstream for a much longer time to clear the antigen. - Statement (3) is incorrect: Plasma cells (which are differentiated B-lymphocytes), not T-lymphocytes (like killer T cells), are responsible for producing and secreting antibodies.

1 mark for correct option B. No partial marks.

- Option A is incorrect: Human muscle cells perform lactic acid fermentation, which does not release carbon dioxide, whereas yeast cells perform ethanol fermentation, which releases carbon dioxide. - Option B is incorrect: The net ATP yield of anaerobic respiration for both human muscle cells and yeast is only 2 ATP per molecule of glucose (produced during glycolysis). - Option C is correct: The end product of lactic acid fermentation in muscle is lactic acid (lactate), while the end products of fermentation in yeast are ethanol and carbon dioxide. - Option D is incorrect: Anaerobic respiration (including glycolysis and fermentation steps) occurs entirely within the cytoplasm (cytosol) for both types of cells; it does not involve the mitochondria.

1 mark for correct option C. No partial marks.

High light intensity causes stomata to open wider. Increased air movement removes water vapour near the stomata, maintaining a steep water vapour concentration gradient. Elevated ambient temperature increases the rate of evaporation of water. High relative humidity, however, decreases the concentration gradient of water vapour between the leaf interior and the atmosphere, thereby decreasing the transpiration rate. Thus, (1), (3), and (4) promote transpiration.

Award 1 mark for the correct option B. No mark is given for other options.

During photosynthesis, the oxygen gas released originates entirely from the photolysis of water. Since the water provided contains normal oxygen (\(^{16}\text{O}\)), the released oxygen gas will not contain \(^{18}\text{O}\). On the other hand, carbon dioxide is fixed during the light-independent reaction to produce triose phosphate (and subsequently sugars/starch), meaning the carbon dioxide's oxygen (\(^{18}\text{O}\)) will be incorporated into triose phosphate.

Award 1 mark for the correct option B. No mark is given for other options.

When domestic sewage rich in organic matter is discharged, aerobic decomposers (bacteria) multiply rapidly as they feed on the organic matter. This microbial activity consumes dissolved oxygen rapidly, leading to a decrease (not increase) in dissolved oxygen (DO) levels. At the same time, the biological oxygen demand (BOD) increases significantly because of the high organic load and high bacterial activity. Therefore, only (1) is correct.

Award 1 mark for the correct option A. No mark is given for other options.

Since the pesticide is non-biodegradable, it cannot be broken down, and because it is fat-soluble, it accumulates in the fatty tissues of organisms (statement 2 is correct). As energy is lost at each trophic level, consumers must ingest a large amount of biomass from the lower trophic level, causing the toxin concentration to increase up the food chain (biomagnification), meaning fish-eating birds will have the highest concentration (statement 1 is correct). However, due to energy loss, the total biomass must decrease at higher trophic levels, so the biomass of fish-eating birds is much lower than that of phytoplankton (statement 3 is incorrect).

Award 1 mark for the correct option B. No mark is given for other options.

The ileum is the primary site for the absorption of digested food nutrients (such as glucose, amino acids, fatty acids, and glycerol) and also plays a major role in absorbing water from the alimentary canal. Removing the ileum will severely impair nutrient and water absorption (statements 2 and 3 are correct). However, most digestive enzymes in the small intestine are secreted by the pancreas (pancreatic juice), and the ileum itself does not secrete major digestive enzymes into the lumen (enzymes are mainly bound to the microvillar membranes), so enzyme secretion is not the primary function lost (statement 1 is incorrect).

Award 1 mark for the correct option D. No mark is given for other options.

Two hours after a meal, digestion and absorption are highly active. Glucose and amino acids absorbed from the small intestine enter the hepatic portal vein, making their concentrations higher than in the hepatic vein (statements 1 and 2 are correct), as the liver regulates blood glucose by storing excess as glycogen and uses/deaminates excess amino acids. Since the liver is the site of deamination where excess amino acids are broken down into urea, the blood leaving the liver via the hepatic vein contains a higher concentration of urea than the blood entering it (statement 3 is correct).

Award 1 mark for the correct option D. No mark is given for other options.

During transcription, RNA polymerase reads the template DNA strand in the 3' to 5' direction to synthesize the mRNA strand in the 5' to 3' direction. The bases on mRNA are complementary to those on the template DNA strand, with Adenine (A) pairing with Uracil (U) instead of Thymine (T). Thus:
- 3'-T pairs with 5'-A
- A pairs with U
- C pairs with G
- G pairs with C
- G pairs with C
- C pairs with G
- T pairs with A
- T pairs with A
- 5'-A pairs with 3'-U
This gives `5'- A U G C C G A A U -3'`.

Award 1 mark for the correct option A. No mark is given for other options.

A mutation that changes an amino acid codon to a stop codon (nonsense mutation) causes translation to terminate prematurely, resulting in a significantly truncated (shorter) polypeptide. Options A, C, and D are missense mutations that only change a single amino acid in the polypeptide without affecting its length.

Award 1 mark for the correct option B. No mark is given for other options.

At the source (e.g., photosynthesizing leaves), sucrose is actively loaded into the sieve tubes (statement 1 is correct). This lowers the water potential inside the sieve tubes, causing water to enter from the adjacent xylem by osmosis. The influx of water increases the hydrostatic pressure at the source, driving the bulk flow of sap toward the sink (statement 2 is correct). The transport of organic substances in phloem (translocation) can occur both upward (e.g., to flowers, fruits, or young leaves) and downward (e.g., to roots), depending on the location of sources and sinks (statement 3 is incorrect).

Award 1 mark for the correct option B. No mark is given for other options.

Clearing forests (deforestation) releases stored carbon as carbon dioxide (\(\text{CO}_2\)) when trees are burned or decomposed, and also reduces carbon dioxide uptake via photosynthesis (statement 1 is correct). Extensive use of nitrogen-based chemical fertilizers leads to the release of nitrous oxide (\(\text{N}_2\text{O}\)) by soil microbes, which is a potent greenhouse gas (statement 2 is correct). Combustion of coal in power stations directly releases large amounts of carbon dioxide (statement 3 is correct). Thus, all three activities contribute to the increase of greenhouse gases.

Award 1 mark for the correct option D. No mark is given for other options.

When a shoot is moved from a dark and humid environment to a bright and windy environment, stomata open rapidly and the concentration gradient of water vapor between the leaf and air increases greatly, leading to a sudden surge in transpiration. However, the cohesive force and tension take time to develop in the xylem vessels, so the increase in water absorption lags behind transpiration. Thus, during this transition, the rate of water absorption is significantly lower than the transpiration rate.

Award 1 mark for the correct option (A). No marks for incorrect options.

By comparing Condition 3 and Condition 4, only the \(\text{CO}_2\) concentration is increased (from 0.03% to 0.12%), while light intensity and temperature remain constant. This change leads to an increase in the rate of \(\text{CO}_2\) uptake from 30 to 45. Therefore, in Condition 3, the limiting factor is \(\text{CO}_2\) concentration. (In Condition 1, increasing \(\text{CO}_2\) to 0.12% in Condition 2 does not change the rate, but increasing light to 500 in Condition 3 does, meaning light is the limiting factor in Condition 1).

Award 1 mark for the correct option (C). No marks for incorrect options.

Aerobic decomposers break down the organic matter in sewage through aerobic respiration, consuming dissolved oxygen and causing its levels to decrease (1). During decomposition, organic nitrogenous compounds are broken down into ammonium ions, which are subsequently oxidized by nitrifying bacteria into nitrites and nitrates, increasing inorganic nutrient concentrations (2). Further downstream, as mineral nutrients become rich and the water recovers, algae absorb these nutrients and multiply rapidly, increasing the algal population (3). Therefore, all three statements are correct.

Award 1 mark for the correct option (D). No marks for incorrect options.

Trawling involves dragging heavily weighted nets along the seabed, which physically destroys delicate benthic habitats like coral reefs and sea grass beds (1 is a primary benefit). Trawling is also highly non-selective, capturing massive amounts of non-target species and juveniles as by-catch (2 is a primary benefit). While fishing activities can generate plastic debris (like lost nets), the ban on trawling is not a direct measure aimed at reducing the input of plastic pollutants into the sea (3 is not a primary benefit). Therefore, only (1) and (2) are correct.

Award 1 mark for the correct option (A). No marks for incorrect options.

Bile salts only perform physical digestion (emulsification) to increase the surface area of lipids, not chemical digestion (A is incorrect). Fatty acids and monoglycerides are lipid-soluble and enter the epithelial cells of the villi by simple diffusion, not active transport (B is incorrect). Once inside the epithelial cells, they are recombined into triglycerides in the smooth endoplasmic reticulum (C is correct). These triglycerides are then packaged into chylomicrons and released into the lacteals (lymphatic capillaries) rather than directly into the blood capillaries (D is incorrect).

Award 1 mark for the correct option (C). No marks for incorrect options.

During inhalation, the external intercostal muscles and diaphragm muscles contract (1 is correct). This muscle contraction increases the volume of the thoracic cavity and the lungs. An increase in lung volume causes the air pressure inside the lungs (intra-alveolar pressure) to decrease below atmospheric pressure, not increase (2 is incorrect). Since atmospheric pressure is now higher than the intra-alveolar pressure, air flows from the atmosphere into the lungs down the pressure gradient (3 is correct).

Award 1 mark for the correct option (B). No marks for incorrect options.

When the pressure in the left ventricle exceeds the pressure in the left atrium, it pushes the bicuspid valve closed to prevent the backflow of blood into the atrium. At the same time, because the ventricular pressure is still lower than the pressure in the aorta, it cannot push the semilunar valves open, so they remain closed. This phase corresponds to the isovolumetric ventricular contraction where both valves are closed.

Award 1 mark for the correct option (B). No marks for incorrect options.

1. The DNA molecule has 3000 base pairs, which means it contains a total of \(3000 \times 2 = 6000\) bases.\n2. Since adenine (A) is \(22\%\), according to Chargaff's rules, thymine (T) is also \(22\%\). Thus, A + T = \(44\%\).\n3. This leaves cytosine (C) and guanine (G) to make up the remaining percentage: C + G = \(100\% - 44\% = 56\%\).\n4. Since G = C, guanine (G) makes up \(56\% / 2 = 28\%\) of the total bases.\n5. The total number of G bases is \(6000 \times 28\% = 1680\).

Award 1 mark for the correct option (D). No marks for incorrect options.

1. The template DNA strand is read in the 3' to 5' direction. The third triplet is 3'-TTA-5'.\n2. During transcription, complementary base pairing produces the mRNA codon: 5'-AAU-3'.\n3. During translation, the tRNA anticodon binds to the mRNA codon in an antiparallel, complementary manner. The complementary sequence to 5'-AAU-3' is 3'-UUA-5'.\n4. Reading this tRNA anticodon in the 5' to 3' direction yields 5'-AUU-3'.

Award 1 mark for the correct option (C). No marks for incorrect options.

In a forest ecosystem, the primary producers are often very large trees, whereas in a grassland they are tiny grasses. A single massive tree has a huge biomass and can support a very large number of small herbivorous insects (primary consumers). Therefore, fewer individuals are needed at the producer level than at the primary consumer level, leading to an inverted pyramid of numbers. The size of individual organisms, rather than energy transfer efficiency or photosynthetic rates, is the key determinant of the shape of the pyramid of numbers.

Award 1 mark for the correct option (A). No marks for incorrect options.

All three statements are correct:
(1) In Setup S, enclosing the shoot in a plastic bag traps transpirational water vapor, increasing the relative humidity around the leaf. This reduces the concentration gradient of water vapor between the leaf and the atmosphere, reducing the transpiration rate compared to Setup P (P > S).
(2) In Setup Q, the higher temperature (\(35^\circ\text{C}\) vs \(25^\circ\text{C}\) in P) increases the kinetic energy of water molecules, leading to faster evaporation from the mesophyll cell walls into the air spaces, thus increasing transpiration rate (Q > P).
(3) In Setup R, moving air (wind) sweeps away the water vapor accumulating around the stomata, preventing the formation of a humid boundary layer and maintaining a steep water vapor concentration gradient, thus increasing transpiration rate compared to still air (R > P).

Award 1 mark for the correct option D. No marks are awarded for incorrect options.

C is correct because chemical X is a persistent organic pollutant that cannot be metabolized or excreted. As it is transferred along the food chain, its concentration increases at each successive trophic level due to biomagnification. Since large fish occupy a higher trophic level than benthic invertebrates, they accumulate a higher concentration of chemical X.
A is incorrect because biomagnification leads to the highest concentration of chemical X in top predators (sea birds).
B is incorrect because although producers have the highest total biomass, they have the lowest concentration of the pollutant per unit body mass as it has not undergone biomagnification.
D is incorrect because persistent organic pollutants are resistant to biodegradation.

Award 1 mark for the correct option C. No marks are awarded for incorrect options.

(1) is correct: Vessel P is the hepatic portal vein and Vessel Q is the hepatic vein. Two hours after a protein-rich meal, amino acids are actively absorbed from the small intestine into the hepatic portal vein (P). As blood passes through the liver, excess amino acids are deaminated or utilized, so the concentration of amino acids in the hepatic vein (Q) is lower.
(2) is correct: Urea is produced in the liver during deamination, so blood leaving the liver via the hepatic vein (Q) has a high urea concentration. Conversely, urea is filtered out of the blood in the kidneys, so blood leaving the kidneys via the renal vein (S) has a very low urea concentration. Thus, the concentration of urea in Q is higher than in S.
(3) is incorrect: Vessel R is the renal artery (carrying oxygenated blood), while Vessel S is the renal vein (carrying deoxygenated blood). The kidney cells consume oxygen for respiration, so the concentration of oxygen in R is higher than in S.

Award 1 mark for the correct option A. No marks are awarded for incorrect options.

The template strand is written in the 3' to 5' direction: `3'-CAG-GAC-GTC-CTG-5'`.
The complementary mRNA strand is synthesized in the 5' to 3' direction using complementary base-pairing (A-U, T-A, C-G, G-C):
`5'-GUC-CUG-CAG-GAC-3'`
Now we read the codons from 5' to 3':
- 1st codon: `GUC` -> Valine (Val)
- 2nd codon: `CUG` -> Leucine (Leu)
- 3rd codon: `CAG` -> Glutamine (Gln)
- 4th codon: `GAC` -> Aspartic acid (Asp)
Thus, the correct polypeptide sequence is Val - Leu - Gln - Asp.

Award 1 mark for the correct option A. No marks are awarded for incorrect options.

- (1) is correct: The supplied leaf acts as a 'source' of photosynthates. The radioactive sucrose synthesized during photosynthesis is translocated via the phloem to various 'sinks', which include both downward sinks (roots) and upward sinks (growing shoot tips and young leaves).
- (2) is correct: Organic nutrients are transported through the phloem. Removing a ring of bark (which includes the phloem) below the supplied leaf blocks the pathway for downward translocation, so no radioactivity can reach the roots.
- (3) is correct: Phloem translocation relies on active loading of sucrose into sieve tubes at the source, which requires metabolic energy (ATP) provided by companion cells.

Award 1 mark for the correct option D. No marks are awarded for incorrect options.

B is correct because:
1. Cutting down and burning or allowing forest vegetation to decay releases large amounts of carbon dioxide (\(\text{CO}_2\)) into the atmosphere, which increases the source of greenhouse gases.
2. Deforestation removes trees, which are major carbon sinks that absorb \(\text{CO}_2\) from the atmosphere via photosynthesis. Thus, it reduces the sink of greenhouse gases.

A, C, and D are incorrect because they increase the sources of greenhouse gases (carbon dioxide from fossil fuels, nitrous oxide from fertilizers, and methane from landfills, respectively) but do not directly reduce greenhouse gas sinks.

Award 1 mark for the correct option B. No marks are awarded for incorrect options.

(a) The layer of oil prevents the direct evaporation of water from the surface of the flask, ensuring that any water loss measured is solely due to absorption by the plant shoot.

(b) The rate of water uptake is higher in moving air than in still air. Moving air sweeps away the water vapor accumulating around the leaves. This maintains a steeper water vapor concentration gradient between the leaf interior and the atmosphere, increasing the rate of transpiration and thus water uptake.

(c) The potometer measures water uptake rather than transpiration. Some of the water taken up is used in photosynthesis or metabolic processes, and some is used to maintain cell turgidity.

(a) Prevents direct evaporation of water from flask surface (1); ensuring measured water loss is only due to plant absorption/transpiration (1).
(b) Higher water uptake in moving air (1); air movement sweeps away water vapor from leaf surface (1); maintains a steeper water vapor concentration gradient (1).
(c) Some water is used in photosynthesis/metabolism (1); some water is used to maintain cell turgidity/cell growth (1).

(a) The algal population increases rapidly, causing an algal bloom. This is because fertilizers contain high levels of nitrates and phosphates, which act as nutrients that promote rapid cell division and growth of algae.

(b) The thick layer of algae blocks sunlight from reaching submerged aquatic plants, preventing photosynthesis and causing them to die. When the algae and plants die, aerobic decomposers (bacteria) multiply rapidly to decompose the dead organic matter. These decomposers consume large amounts of dissolved oxygen during aerobic respiration.

(c) It can lead to a severe decline in biodiversity as sensitive species disappear, and it can disrupt food chains/webs as certain species lose their natural food sources or habitats.

(a) Rapid increase / algal bloom occurs (1); due to abundance of nitrates/phosphates as nutrients promoting growth (1).
(b) Algae block light, causing submerged plants to die (1); dead organic matter is decomposed by aerobic bacteria/decomposers (1); decomposers consume oxygen during aerobic respiration (1).
(c) Loss of biodiversity (1); disruption of food chains/webs (1); potential release of algal toxins (1) [Any two, max 2 marks].

(a) Before emulsification, lipids exist as large fat droplets, which have a small surface-area-to-volume ratio. After emulsification by bile salts, they are broken down into numerous tiny lipid droplets. This physical breakdown greatly increases the surface area of lipids available for lipase to act on, speeding up chemical digestion.

(b) Fatty acids and glycerol, being lipid-soluble, diffuse directly across the phospholipid bilayer of the microvilli into the epithelial cells. Inside these cells, they are recombined into triglycerides (lipids) and packaged into chylomicrons. They then leave the cells by exocytosis and enter the lacteals to be transported via the lymphatic system.

(c) The lacteal wall is highly permeable because its endothelial cells have larger intercellular gaps (pores) compared to blood capillaries. This allows the relatively large chylomicrons to enter the lymphatic vessel easily.

(a) Large fat droplets (before) vs tiny fat droplets (after) (1); increase in surface area (1); speeds up chemical digestion by lipase (1).
(b) Fatty acids and glycerol enter epithelial cells by diffusion (1); recombine into lipids/lipoproteins/chylomicrons inside cells (1); enter lacteals for lymphatic transport (1).
(c) Highly permeable wall / thin wall (1); presence of large pores/gaps between cells to allow large molecules to pass (1).

(a) Primers are short single-stranded DNA sequences that bind to the target DNA. They provide a free 3'-OH end, acting as starting points for DNA polymerase to bind and begin synthesizing the complementary DNA strands. Two different primers are required because the two separated DNA template strands are antiparallel (running in opposite directions) and have different, complementary base sequences at their respective 3' ends.

(b) The first step of each PCR cycle involves heating the mixture to around 95°C (denaturation) to separate the double-stranded DNA. Human DNA polymerase would denature and lose its activity at this temperature, whereas thermostable DNA polymerase (e.g., Taq polymerase) is stable and remains functional at high temperatures.

(c) In PCR, the number of DNA molecules doubles in each cycle. The formula is \(2^n\), where \(n\) is the number of cycles.
With 1 template molecule and 3 cycles:
Number of copies = \(2^3 = 8\) molecules.

(a) Provide starting points for DNA polymerase / allow binding of polymerase (1); two different primers are needed because template strands are antiparallel (1); and have different base sequences at their 3' ends (1).
(b) Denaturation step requires high temperatures (~95°C) (1); human polymerase would denature/lose function while Taq polymerase is heat-resistant and remains stable (1).
(c) Formula/working: \(2^3\) (1); Correct answer: 8 copies (1).

(a) Light energy absorbed by chlorophyll excites electrons to a higher energy level. These high-energy electrons are emitted and passed along a series of electron carriers. The energy released during this electron transport is used to synthesize ATP from ADP and inorganic phosphate. At the end of the pathway, the electrons and hydrogen ions (from photolysis of water) are accepted by NADP+ to form NADPH.

(b) In the carbon-fixing stage (Calvin cycle), carbon dioxide combines with RuBP to form unstable 6C compounds, which split into glycerate 3-phosphate (GP). ATP and NADPH produced in the light-dependent stage are then used: ATP provides energy and NADPH provides reducing power / hydrogen atoms to reduce GP into triose phosphate (3C sugar).

(c) In darkness, the light-dependent stage ceases, meaning no ATP and NADPH are produced. Without ATP and NADPH, GP cannot be reduced to triose phosphate, and RuBP cannot be regenerated. However, the existing RuBP continues to combine with carbon dioxide to form GP, leading to a rapid decline in RuBP concentration.

(a) Light energy excites electrons in chlorophyll which are passed along electron carriers (1); energy released is used to synthesize ATP (1); electrons and hydrogen ions are accepted by NADP+ to form NADPH (1).
(b) Carbon dioxide combines with RuBP to form GP (1); ATP provides energy and NADPH provides reducing power / hydrogen (1); to reduce GP to triose phosphate (1).
(c) No ATP and NADPH are produced in darkness (1); RuBP regeneration stops, but existing RuBP continues to combine with CO2, depleting its concentration (1).

(a) Bioaccumulation is the process by which substances (like microplastics or toxins) accumulate in the tissue of an organism over time. When lower-trophic-level organisms ingest microplastics, the particles cannot be digested or excreted, so they remain inside the body. When predators eat many of these prey, the microplastics are transferred and accumulated in much larger quantities at higher trophic levels.

(b) Bioaccumulation refers to the increase in concentration of a substance in a single organism's body over its lifespan. Biomagnification refers to the progressive increase in the concentration of a substance in organisms at successively higher trophic levels along a food chain.

(c) Humans can reduce plastic pollution by: 1. Implementing policies to ban or reduce single-use plastic items, and 2. Improving filtration technologies in municipal wastewater treatment plants to trap microplastics before they reach aquatic environments.

(a) Definition of bioaccumulation as accumulation of substance/toxin in an organism (1); microplastics are non-biodegradable/cannot be excreted (1); transferred up the food chain as predators consume many prey individuals (1).
(b) Bioaccumulation is within a single organism/over time (1); biomagnification is across different trophic levels/along a food chain (1).
(c) Reduce/ban single-use plastics (1); improve wastewater treatment filtration (1); promote plastic recycling (1) [Any two, max 2 marks].

(a) The structural adaptations of an alveolus include:
1. It has a wall that is only one-cell-thick, which greatly minimizes the diffusion distance for oxygen.
2. It is surrounded by a dense network of capillaries with flowing blood, which maintains a steep oxygen concentration gradient.
3. The inner surface is covered with a thin layer of moisture, allowing oxygen to dissolve before diffusing across the membrane.
4. There are millions of alveoli in the lungs, providing a very large surface area to increase the overall rate of diffusion.

(b) 1. Lack of a nucleus: This provides more internal space inside the cell to accommodate more hemoglobin molecules. As a result, each red blood cell can bind and carry a larger amount of oxygen.
2. Lack of mitochondria: This ensures that the red blood cell does not consume the oxygen it is transporting. Without mitochondria, the cell respires anaerobically, leaving all transported oxygen to be delivered to the body tissues.

(a) Wall is one-cell-thick / shortens diffusion distance (1); dense capillary network / maintains steep concentration gradient (1); moist inner surface / dissolves oxygen (1); numerous alveoli / large surface area (1).
(b) Lack of nucleus: provides more space (1); to pack more hemoglobin / increase oxygen-carrying capacity (1); Lack of mitochondria: prevents RBC from using oxygen (1); via aerobic respiration / leaves all oxygen for tissue delivery (1).

(a) This is a nonsense mutation (a type of substitution mutation). It introduces a premature stop codon, which causes the ribosome to stop translation earlier than normal. As a result, a truncated (shorter-than-normal) polypeptide chain is synthesized.

(b) The truncated polypeptide lacks many amino acids, including those needed for proper folding and active site formation. Therefore, it cannot fold into its correct three-dimensional conformation. This leads to a loss of the protein's biological function or activity.

(c) The genetic code is degenerate (redundant), which means that several different codons can code for the same amino acid. A mutation that changes a codon to another codon representing the same amino acid (silent mutation) will not affect the amino acid sequence.

(a) Nonsense mutation / substitution mutation (1); premature termination of translation (1); produces a truncated/shorter polypeptide (1).
(b) Lacks essential amino acids / functional domains (1); incorrect folding into 3D structure (1); loss of biological function/activity (1).
(c) Genetic code is degenerate / different codons can code for the same amino acid (1).

(a) In response to light, potassium ions (\(K^+\)) are actively transported from surrounding epidermal cells into the guard cells. This increases the solute concentration inside the guard cells, lowering their water potential. Consequently, water enters the guard cells by osmosis. The guard cells swell and become turgid; because their inner walls (facing the pore) are thicker and less elastic than their outer walls, the guard cells bend outward, opening the stomatal pore.

(b) During a severe drought, the plant experiences water stress and produces a plant hormone, abscisic acid (ABA). ABA triggers potassium ions to rapidly leave the guard cells, increasing their water potential and causing water to leave by osmosis. The guard cells become flaccid and the stomata close. The adaptive significance is to prevent excessive water loss via transpiration, conserving water and preventing the plant from wilting and dying under dry conditions.

(a) Active transport of potassium ions (\(K^+\)) into guard cells (1); increases solute concentration / lowers water potential inside guard cells (1); water enters guard cells by osmosis (1); guard cells become turgid and bend outwards because inner walls are thicker (1).
(b) Drought stress triggers ABA production causing stomatal closure (1); prevents excessive water loss/transpiration (1); avoids wilting / maintains water balance for survival (1).

(a) The highway acts as a physical barrier that prevents the small mammals on either side from crossing and mating with each other. This restricts gene flow between the two now-isolated populations. In each small population, inbreeding increases, and genetic drift is more pronounced. Over generations, this leads to the loss of alleles and an overall decrease in genetic diversity.

(b) 1. Constructing wildlife crossings, such as green bridges (ecoducts) or underpasses, to allow animals to safely cross over or under the highway.
2. Installing wildlife-proof fencing along the roadside to prevent animals from entering the highway, reducing roadkill and directing them toward the wildlife crossings.

(c) Conservationists could:
1. Install motion-activated camera traps at the wildlife crossings to document which species and how many individuals are using them.
2. Perform genetic sampling (e.g., extracting DNA from hair or feces) of the populations on both sides over several generations to test if gene flow has been re-established.
3. Attach GPS tracking collars to individual animals to study their movement patterns and confirm if they cross the highway using the provided crossings.

(a) Highway acts as barrier, preventing interbreeding/gene flow (1); leads to inbreeding within small isolated populations (1); increases genetic drift, causing loss of alleles/diversity (1).
(b) Green bridges/underpasses/ecoducts (1); wildlife fencing along the highway (1).
(c) Install camera traps to monitor animal passage (1); perform genetic analysis of populations over time to verify gene flow (1); track animal movements using GPS collars (1).

Water transport in xylem is a passive process driven by transpiration pull (created by evaporation of water from leaves) and maintained by cohesion between water molecules and adhesion to vessel walls, resulting in a continuous upward, unidirectional flow. In contrast, translocation of organic nutrients in phloem is active and bidirectional, explained by the pressure-flow hypothesis: sucrose is actively loaded into sieve tubes at the source (using ATP), generating high hydrostatic pressure as water enters by osmosis, which drives mass flow towards the lower-pressure sink. Under prolonged drought and high temperatures, plants close stomata to conserve water, which drastically reduces transpiration pull and xylem transport. Stomatal closure also limits carbon dioxide intake, decreasing photosynthesis rates, which leads to lower sugar production and consequently slows down phloem translocation.

Water transport in xylem (max 3 marks):
- Transpiration of water from leaves creates a tension / transpiration pull that pulls water upwards. (1 mark)
- Cohesion between water molecules and adhesion between water molecules and xylem walls maintain a continuous, unbroken column of water. (1 mark)
- It is a passive process driven by solar energy (evaporation) without expenditure of metabolic energy (ATP). (1 mark)

Translocation in phloem (max 3 marks):
- Sugars (sucrose) are actively loaded into sieve tubes at the source (leaves), requiring metabolic energy / ATP. (1 mark)
- This lowers the water potential, causing water to enter from adjacent xylem by osmosis, generating a high hydrostatic pressure. (1 mark)
- At the sink (e.g., roots), sugars are unloaded, raising water potential and causing water to leave, creating low hydrostatic pressure; the pressure gradient drives mass flow of materials from source to sink. (1 mark)

Comparison (max 1 mark):
- Xylem transport is strictly unidirectional (upward from roots to leaves), whereas phloem translocation is bidirectional (from source to sink). (1 mark)

Effects of drought and high temperature (max 2 marks):
- Under drought/high temperature, plants close stomata to prevent excessive water loss, which greatly reduces transpiration, thereby decreasing xylem water transport. (1 mark)
- Stomatal closure restricts carbon dioxide intake, reducing photosynthesis rate; this decreases sugar production at the source, thus slowing down phloem translocation. (1 mark)

Communication skills (2 marks):
- 1 mark for systematic and logical structure (distinct sections for xylem, phloem, comparison, and environmental effects).
- 1 mark for scientific clarity and appropriate terminology without major misconceptions.

(a) (i) During exercise, active skeletal muscles generate a large amount of metabolic heat. The high ambient temperature (33 degrees Celsius) reduces heat loss by radiation, conduction, and convection. Evaporation of sweat is the primary mechanism of heat loss in hot environments. However, high relative humidity (85%) significantly reduces the rate of sweat evaporation, meaning sweat drips off without cooling the body, leading to inefficient heat loss and risk of hyperthermia.\n(ii) Vasodilation of arterioles in the skin occurs, increasing blood flow to the skin to facilitate heat loss by radiation and convection. Since blood is diverted away from active skeletal muscles and internal organs toward the skin, this redirection of blood flow reduces the rate of oxygen and nutrient delivery to skeletal muscles, potentially speeding up the onset of fatigue.\n\n(b) (i) Water loss via sweat increases the solute concentration / osmotic pressure of the blood. This is detected by osmoreceptors in the hypothalamus. The hypothalamus stimulates the posterior pituitary gland to secrete / release more antidiuretic hormone (ADH) into the bloodstream. ADH travels via the blood to the kidneys, increasing the permeability of the cells in the collecting ducts (and distal convoluted tubules) to water by inserting more aquaporins into the cell membranes. Consequently, more water is reabsorbed from the filtrate back into the surrounding capillaries by osmosis, resulting in a smaller volume of highly concentrated urine.\n(ii) Trend: The solute concentration of the urine increases steadily over the 2-hour run, eventually reaching a high plateau. Explanation: As the run progresses, dehydration intensifies, leading to progressive increases in ADH secretion. Max water reabsorption occurs in the collecting ducts, making the urine hypertonic to blood plasma and highly concentrated.\n\n(c) Rapid intake of a large volume of pure water dilutes the extracellular fluid, significantly lowering its osmotic pressure. This creates an osmotic gradient where the water potential of the extracellular fluid is higher than that of the intracellular fluid, causing water to enter body cells (including brain cells) by osmosis, leading to cell swelling. In response, the low blood osmotic pressure is detected by hypothalamic osmoreceptors, which suppress the secretion of ADH. The kidney collecting ducts become less permeable to water, reducing water reabsorption, resulting in the excretion of a large volume of dilute urine to remove excess water and restore osmotic balance.

(a) (i) (Max 4 marks)\n- Active skeletal muscles generate a large amount of metabolic heat. (1)\n- High ambient temperature reduces heat loss by radiation, conduction, and convection. (1)\n- Evaporation of sweat is the primary mechanism of heat loss. (1)\n- High relative humidity significantly reduces the rate of sweat evaporation, leading to inefficient heat loss / risk of hyperthermia. (1)\n(a) (ii) (Max 3 marks)\n- Vasodilation of arterioles in the skin to increase skin blood flow / facilitate heat loss. (1)\n- Blood is diverted away from active skeletal muscles to the skin. (1)\n- This reduces the delivery of oxygen / nutrients to skeletal muscles, accelerating fatigue. (1)\n\n(b) (i) (Max 6 marks)\n- Sweating increases blood solute concentration / blood osmotic pressure. (1)\n- Detected by osmoreceptors in the hypothalamus. (1)\n- Stimulates posterior pituitary gland to secrete more antidiuretic hormone (ADH). (1)\n- ADH increases the permeability of collecting ducts / distal convoluted tubules to water. (1)\n- Insertion of aquaporins into the cell membrane. (1)\n- More water is reabsorbed into capillaries by osmosis, producing concentrated urine. (1)\n(b) (ii) (Max 3 marks)\n- Urine solute concentration increases steadily and plateaus. (1)\n- Dehydration intensifies, stimulating maximum ADH secretion. (1)\n- Max water reabsorption in collecting ducts makes urine highly hypertonic to plasma. (1)\n\n(c) (Max 4 marks)\n- Dilutes extracellular fluid / lowers its osmotic pressure. (1)\n- Water potential of extracellular fluid becomes higher than intracellular fluid, water enters cells by osmosis, causing swelling. (1)\n- Hypothalamic osmoreceptors detect low osmotic pressure, suppressing ADH secretion. (1)\n- Collecting ducts become less permeable to water, leading to excretion of large volume of dilute urine. (1)

(a) (i) Human genomic DNA contains introns (non-coding regions) and exons (coding regions). Bacteria (if used as hosts) lack the splicing machinery to remove introns from primary mRNA transcripts. Using cDNA synthesized from mRNA ensures that the synthetic gene consists only of coding sequences (exons) that can be correctly transcribed and translated into the complete, functional human protein without needing splicing.\n(ii) First, denaturation occurs by heating to high temperature (approx. 95 degrees Celsius) to separate the double-stranded cDNA template into single strands. Second, annealing occurs by cooling (approx. 50-60 degrees Celsius) to allow primers to bind specifically to the complementary sequences flanking the target gene on the single-stranded DNA templates. Third, extension occurs by heating (approx. 72 degrees Celsius) for thermostable DNA polymerase to synthesize the complementary strands by adding free nucleotides starting from the primers. Primers provide a starting point with a free 3'-OH group for the DNA polymerase; the thermostable DNA polymerase can withstand repeated exposure to high temperatures without denaturation.\n\n(b) (i) Using the same restriction enzyme ensures that both the target gene (cDNA) and the cut plasmid have complementary sticky ends (single-stranded overhangs). This allows the complementary bases of the sticky ends to pair up via hydrogen bonding (annealing). Then, DNA ligase can easily join them together by forming phosphodiester bonds, increasing the efficiency of ligation.\n(ii) Promoter: It is a specific DNA sequence that acts as a signal for RNA polymerase to bind and initiate transcription of the target gene (EPO). Without a promoter, the target gene cannot be expressed into mRNA. Selectable marker gene: It allows researchers to distinguish and select host cells that have successfully taken up the recombinant plasmid from those that have not. Host cells containing the plasmid can survive on a medium containing the corresponding antibiotic, while non-transformed cells die.\n\n(c) (i) E. coli is a prokaryote, which lacks membrane-bound organelles like the endoplasmic reticulum and Golgi apparatus. It cannot perform post-translational modifications, such as glycosylation (adding carbohydrate chains), which are crucial for the proper folding, stability, and biological activity of complex human glycoproteins like EPO.\n(ii) Alternative eukaryotic host system: Yeast (e.g., Saccharomyces cerevisiae) / Mammalian cell culture (e.g., Chinese Hamster Ovary cells) / Transgenic plants (e.g., tobacco). [Any one]\nBiosafety/ethical concerns (any two):\n1. Risk of gene flow / cross-pollination: The transgene from genetically modified plants could spread to wild relatives, potentially disrupting natural ecosystems.\n2. Potential allergenicity: The recombinant proteins produced in alternative hosts might trigger unexpected allergic reactions in humans due to altered glycosylation patterns.\n3. Animal welfare concerns: If transgenic animals are used, the insertion of foreign genes may cause physiological harm or suffering to the animals.\n4. 'Playing God' / Ethical boundary: Concerns regarding human interference with natural genomes and patenting of life forms.

(a) (i) (Max 3 marks)\n- Genomic DNA contains introns and exons. (1)\n- Eukaryotic splicing machinery is absent in bacterial host cells / in vitro systems cannot remove introns. (1)\n- cDNA (derived from mRNA) contains only exons, ensuring proper translation of the functional protein. (1)\n(a) (ii) (Max 4 marks)\n- Heating to 95 degrees Celsius denatures double-stranded cDNA into single strands. (1)\n- Cooling to 50-60 degrees Celsius allows primers to anneal to complementary sites flanking the target gene. (1)\n- Heating to 72 degrees Celsius allows Taq/DNA polymerase to synthesize complementary strands using free nucleotides. (1)\n- Primers provide a starting point (free 3'-OH); Taq polymerase is thermostable and resists denaturation at high temperatures. (1)\n\n(b) (i) (Max 3 marks)\n- Produces complementary sticky ends (single-stranded overhangs) on both cDNA and plasmid. (1)\n- Allows complementary base pairing (annealing) of sticky ends. (1)\n- DNA ligase can then join them efficiently by forming phosphodiester bonds. (1)\n(b) (ii) (Max 4 marks)\n- Promoter: Site where RNA polymerase binds (1) to initiate transcription/expression of the target gene. (1)\n- Selectable marker gene: Allows identification/selection of transformed cells (1) by survival on an antibiotic-containing medium. (1)\n\n(c) (i) (Max 2 marks)\n- E. coli (prokaryote) lacks endoplasmic reticulum and Golgi apparatus. (1)\n- Cannot perform glycosylation / post-translational modifications, leading to inactive / improperly folded protein. (1)\n(c) (ii) (Max 4 marks)\n- Eukaryotic host: Yeast / mammalian cell culture / plant cells. (1)\n- Biosafety/ethical concern 1: Gene flow / cross-pollination to wild relatives disrupting ecosystems. (1)\n- Biosafety/ethical concern 2: Altered glycosylation in non-human hosts may cause unexpected allergic reactions. (1)\n- (Accept other valid points like animal welfare, ethical concerns of 'playing God').