2024 HKDSE Biology Practice Paper with Answers

Statement (1) is correct: Two hours after a meal, products of carbohydrate digestion (glucose) are actively absorbed in the small intestine and transported to the liver via the hepatic portal vein. In the liver, insulin stimulates the conversion of excess glucose into glycogen (glycogenesis), so the concentration of glucose leaving via the hepatic vein is lower.

Statement (2) is correct: Excess amino acids absorbed from the gut undergo deamination in the liver to form urea. Consequently, blood leaving the liver via the hepatic vein contains a higher concentration of urea than the incoming blood in the hepatic portal vein.

Statement (3) is incorrect: Amino acids are absorbed from the small intestine into the hepatic portal vein. In the liver, some amino acids are used for protein synthesis or deaminated, so their concentration in the hepatic vein is lower than in the hepatic portal vein.

Therefore, only (1) and (2) are correct.

C (1 mark) - Deduce that glucose and amino acids are higher in the hepatic portal vein due to absorption, while urea is higher in the hepatic vein due to deamination in the liver.

Statement (1) is correct: Both aerobic respiration and lactic acid fermentation share the initial stage of glycolysis, which takes place in the cytoplasm and breaks down glucose into pyruvate.

Statement (2) is correct: Glycolysis produces a net yield of 2 molecules of ATP per glucose molecule in both pathways (aerobic respiration produces additional ATP downstream via oxidative phosphorylation).

Statement (3) is correct: In aerobic respiration, NADH is oxidised back to \(\text{NAD}^+\) via the electron transport chain on the inner mitochondrial membrane. In lactic acid fermentation, NADH is oxidised back to \(\text{NAD}^+\) when pyruvate is reduced to lactate in the cytoplasm, allowing glycolysis to continue.

Therefore, (1), (2), and (3) are all correct.

D (1 mark) - Identify that glycolysis, net ATP generation, and the regeneration of NAD+ from NADH occur in both aerobic and anaerobic pathways in muscle cells.

First, the trait must be recessive because unaffected parents (P and Q, as well as R and S) have affected offspring (their son and daughter T respectively).

To distinguish between autosomal recessive and X-linked recessive:
If the condition were X-linked recessive, an affected female (daughter T) must possess two recessive alleles (\(X^a X^a\)). She must inherit one \(X^a\) allele from her father (S). However, father S is unaffected, meaning his genotype must be \(X^A Y\) and he cannot transmit an \(X^a\) allele. The presence of an affected daughter (T) born to an unaffected father (S) conclusively disproves X-linked recessive inheritance and confirms that the allele is located on an autosome.

Option A and D explain why the allele is recessive, but not why it is autosomal. Option B is compatible with both autosomal and X-linked recessive patterns.

C (1 mark) - Recognise that an affected female must inherit a recessive allele from her father, so an unaffected father producing an affected daughter rules out X-linked recessive inheritance.

Water-soluble nutrients such as glucose and amino acids are absorbed directly into the blood capillaries of the villi and are transported via the hepatic portal vein to the liver before entering the general circulation. In contrast, lipid-soluble nutrients such as fatty acids, glycerol, and fat-soluble vitamins (e.g., Vitamin D) are packaged into chylomicrons and enter the lacteals (lymphatic system), passing through lymphatic vessels and the thoracic duct before draining into the subclavian vein.

C (1 mark)

Statement (1) is correct because oxygen release (a product of photolysis of water) occurred despite the complete absence of carbon dioxide. Statement (2) is correct because illumination led to the decolorisation (reduction) of DCPIP, demonstrating that the transfer of electrons requires light energy. Statement (3) is incorrect because ATP synthesis during photophosphorylation is part of the light-dependent reactions and does not require carbon dioxide; carbon dioxide is consumed during the Calvin cycle.

A (1 mark)

A cross between two heterozygous pink-flowered plants (\(Rr \times Rr\)) yields offspring in the genotypic ratio of \(1\,RR : 2\,Rr : 1\,rr\). Because allele expression shows incomplete dominance, the phenotypic ratio is \(1\text{ red} : 2\text{ pink} : 1\text{ white}\). The expected proportion of pink-flowered offspring is \(\frac{2}{4} = \frac{1}{2}\). Thus, out of 240 offspring, the expected number with pink flowers is \(240 \times \frac{1}{2} = 120\).

B (1 mark)

Statement (1) is correct: At the arterial end of capillaries, the hydrostatic pressure of the blood generated by the contraction of the heart is greater than the opposing osmotic pressure (water potential gradient), forcing water and small dissolved solutes out of the capillary walls into the surrounding tissue spaces to form tissue fluid.

Statement (2) is correct: At the venous end of capillaries, the hydrostatic pressure drops significantly due to resistance and fluid loss, whereas the presence of plasma proteins creates an osmotic pressure pulling water back. As a result, the majority (around 85–90%) of tissue fluid is reabsorbed directly into the blood capillaries by osmosis.

Statement (3) is correct: The remaining tissue fluid (around 10–15%) drains into blind-ended lymphatic capillaries as lymph. Lymph is transported along lymphatic vessels and eventually emptied back into the blood system at the subclavian veins near the heart.

Therefore, statements (1), (2), and (3) are all correct.

D (1 mark)
- (1), (2) and (3) are all physiologically accurate descriptions of capillary microcirculation and lymphatic drainage.

Statement (1) is correct: Respiration is an enzyme-controlled process. Lowering the temperature from \(25\ ^\circ\text{C}\) to \(15\ ^\circ\text{C}\) reduces the kinetic energy of respiratory enzymes and substrates, resulting in a lower rate of aerobic respiration, lower rate of \(\text{O}_2\) consumption, and thus a smaller volume change and smaller distance moved per unit time.

Statement (2) is correct: The volume of oxygen consumed relates to distance moved by \(\Delta V = \pi r^2 \times d\). For the same rate of oxygen consumption (same \(\Delta V\)), a capillary tube with a larger internal diameter (larger \(r\)) results in a smaller linear distance (\(d\)) moved by the liquid drop per unit time.

Statement (3) is correct: Dormant seeds have a much lower metabolic rate than actively germinating seeds, so they consume oxygen at a much slower rate, resulting in a decreased distance moved by the droplet per unit time.

Therefore, (1), (2), and (3) will all decrease the distance moved by the liquid drop per unit time.

D (1 mark)
- All three modifications lead to a reduced rate of movement / smaller linear displacement of the meniscus per unit time.

Let the pink-flowered, tall parent plant have genotype \(C^R C^W T\_\).
The white-flowered, dwarf parent plant has genotype \(C^W C^W tt\).

1. Petal colour inheritance:
\(C^R C^W \times C^W C^W \rightarrow \frac{1}{2} C^R C^W\) (pink) : \(\frac{1}{2} C^W C^W\) (white).
This matches the observed 1 pink : 1 white ratio.

2. Plant height inheritance:
All offspring are tall. The dwarf parent can only donate a recessive \(t\) allele. For all offspring to be tall (\(Tt\)), the tall parent must donate a dominant \(T\) allele to every offspring, which means the tall parent must be homozygous dominant (\(TT\)). If the tall parent were heterozygous (\(Tt\)), approximately half of the offspring would be dwarf (\(tt\)).

Combining both traits, the genotype of the pink-flowered, tall parent plant is \(C^R C^W TT\).

A (1 mark)
- Correct deduction that pink requires \(C^R C^W\) and 100% tall offspring requires the tall parent to be homozygous dominant \(TT\).

Statement (1) is correct: Boiling denatures enzymes and photosynthetic electron transport proteins in the thylakoid membranes, preventing the reduction of DCPIP.
Statement (2) is correct: Tube 3 contains intact chloroplasts and DCPIP but lacks light; the absence of decolorisation confirms that light energy is indispensable for driving the reaction.
Statement (3) is correct: During the light-dependent reactions, light energy absorbed by chlorophyll initiates the photolysis of water, releasing electrons and protons. These electrons travel through the electron transport chain and reduce DCPIP (acting as an artificial electron acceptor in place of \(\text{NADP}^+\)), turning it colourless.

Therefore, all three statements (1), (2), and (3) are correct.

D (1 mark) — All three statements (1), (2), and (3) are scientifically accurate.

Two hours after a meal rich in carbohydrates and lipids:
1. Glucose is actively absorbed from the small intestine into the capillary network and transported to the liver via the hepatic portal vein (Vessel P). In the liver, excess glucose is converted into glycogen under the stimulation of insulin. Therefore, glucose concentration in Vessel P is higher than that in the hepatic vein (Vessel Q) leaving the liver (P > Q).
2. Digested lipids (fatty acids and glycerol) are resynthesised into triglycerides within epithelial cells, packaged into chylomicrons, and absorbed predominantly into the lacteals (lymphatic system) rather than directly into blood capillaries. Hence, the concentration of lipids/triglycerides is much higher in the lymphatic vessel (Vessel R) than in the hepatic portal vein (Vessel P) (R > P).

Thus, Option A correctly describes both comparisons.

A (1 mark) — Correctly identifies that glucose in the hepatic portal vein exceeds that in the hepatic vein after absorption, and lipid concentration in lymphatic vessels draining lacteals exceeds that in the blood of the hepatic portal vein.

1. Determination of recessive vs. dominant: Unaffected parents (1 and 2) produce affected children (3 and 5). This proves that the allele responsible for the condition must be recessive, as unaffected parents can only be heterozygous carriers.
2. Determination of autosomal vs. X-linked: If the condition were X-linked recessive, an affected female (Individual 5, \(X^a X^a\)) must inherit one recessive allele (\(X^a\)) from her father (Individual 1). This would mean her father must have genotype \(X^a Y\) and be affected. However, Individual 1 is unaffected (genotype \(X^A Y\)). Therefore, the condition cannot be X-linked recessive and must be autosomal recessive.

Thus, Option A is the correct combination.

A (1 mark) — Correctly identifies autosomal recessive inheritance and the deduction that an affected daughter having an unaffected father excludes X-linked recessive inheritance.

Statement (1) is correct: Two hours after a carbohydrate-rich meal, glucose is actively absorbed in the ileum and transported directly to the liver via the hepatic portal vein. In the liver, excess glucose is converted into glycogen under the action of insulin, resulting in a lower concentration of glucose in the hepatic vein.

Statement (2) is correct: The liver is the primary site of deamination of excess amino acids, forming urea. As urea is released into the bloodstream from hepatocytes, the hepatic vein has a higher concentration of urea than the hepatic portal vein.

Statement (3) is incorrect: Plasma proteins (such as albumin and fibrinogen) are synthesised by the liver and secreted into the blood, meaning blood leaving the liver does not have a lower concentration of plasma proteins than blood entering it.

Therefore, (1) and (2) only are correct.

B (1 mark)
- (1) is correct: Glucose absorption from the gut elevates glucose in the hepatic portal vein relative to the hepatic vein post-meal.
- (2) is correct: Urea is produced by deamination in the liver, so hepatic vein urea is higher.
- (3) is incorrect: Plasma proteins are synthesised by the liver.

At a light intensity of 12 a.u., the plant is at its light compensation point, where the net rate of carbon dioxide exchange is zero. This occurs when the rate of photosynthesis (which consumes \(\text{CO}_2\) and produces \(\text{O}_2\)) is exactly equal to the rate of respiration (which produces \(\text{CO}_2\) and consumes \(\text{O}_2\)).

- Option A is incorrect because photosynthesis is actively occurring, meaning photolysis of water takes place.
- Option C is incorrect because light-independent reactions actively fix the carbon dioxide produced by respiration.
- Option D is incorrect because the rate of oxygen production by chloroplasts equals the rate of oxygen consumption by mitochondria.

B (1 mark)
- Light compensation point is reached when the rate of photosynthetic carbon dioxide uptake equals the rate of respiratory carbon dioxide release (rate of photosynthesis = rate of respiration).

1. Determine parental genotypes:
The child with blood group O has genotype \(ii\). Each parent must contribute an \(i\) allele. Therefore, the father (blood group A) has genotype \(I^A i\) and the mother (blood group B) has genotype \(I^B i\).

2. Determine probability of blood group AB:
Cross: \(I^A i \times I^B i\)
Offspring genotypes and probabilities:
- \(I^A I^B\) (group AB): \(\frac{1}{4}\)
- \(I^A i\) (group A): \(\frac{1}{4}\)
- \(I^B i\) (group B): \(\frac{1}{4}\)
- \(ii\) (group O): \(\frac{1}{4}\)
Probability of blood group AB = \(\frac{1}{4}\).

3. Determine probability of being a boy:
Probability of having a male child = \(\frac{1}{2}\).

4. Combined probability:
\(\text{Probability} = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}\) (or 12.5%).

B (1 mark)
- Deduction of parental genotypes as \(I^A i\) and \(I^B i\).
- Probability of genotype \(I^A I^B\) is 1/4.
- Probability of male sex is 1/2.
- Overall probability = \(1/4 \times 1/2 = 1/8\).

Statement (1) is correct: In aerobic respiration, oxygen serves as the final electron acceptor in the electron transport chain located on the inner mitochondrial membrane.
Statement (2) is correct: As substrate (glucose) concentration increases, the rate of respiration increases until all active sites of the relevant metabolic enzymes are saturated, resulting in a plateau.
Statement (3) is incorrect: At 60 °C, respiratory enzymes undergo denaturation due to disruption of their tertiary structure, causing a sharp decrease or cessation in oxygen uptake rather than an increase.
Therefore, (1) and (2) only are correct.

Award 1 mark for option B.

The liver is the primary organ where deamination of excess amino acids occurs, converting amino groups into urea via the ornithine cycle. As a result, blood leaving the liver via the hepatic vein consistently has a higher concentration of urea than the blood entering through the hepatic portal vein.
Regarding other options:
- B is incorrect: Liver cells undergo aerobic respiration and produce carbon dioxide, so hepatic vein blood has a higher concentration of \(\text{CO}_2\).
- C is incorrect: During fasting or between meals, liver glycogen is broken down into glucose and released into the blood, making glucose concentration in the hepatic vein higher than in the hepatic portal vein.
- D is incorrect: Most absorbed lipids are transported via the lymphatic system (lacteals) to the subclavian vein, bypassing the hepatic portal vein.

Award 1 mark for option A.

When heterozygous \(F_1\) plants (\(Rr\)) self-fertilise (\(Rr \times Rr\)), the expected genotypic ratio in the \(F_2\) generation is:
\(1\ RR : 2\ Rr : 1\ rr\)
Because the alleles show incomplete dominance, the phenotypic ratio corresponds directly to the genotypic ratio:
- Red (\(RR\)): \(\frac{1}{4} \times 400 = 100\)
- Pink (\(Rr\)): \(\frac{1}{2} \times 400 = 200\)
- White (\(rr\)): \(\frac{1}{4} \times 400 = 100\)
Thus, the correct option is B.

Award 1 mark for option B.

Statement (1) is correct: Glucose from digested carbohydrates is absorbed into the blood capillaries in the villi of the small intestine, so blood leaving the small intestine has a higher glucose concentration than blood entering it.
Statement (2) is incorrect: The intestinal wall cells actively respire to provide energy (e.g. for active transport of nutrients), consuming oxygen. Thus, blood leaving the small intestine has a lower concentration of oxygen.
Statement (3) is incorrect: Blood leaving the small intestine travels via the hepatic portal vein to the liver, where it passes through a second capillary network (hepatic sinusoids) before entering the hepatic vein and eventually the inferior vena cava.

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

In the light-dependent reactions of photosynthesis, \(\text{NADP}^+\) acts as the terminal electron acceptor. When the reduction of \(\text{NADP}^+\) is blocked, electrons cannot be transferred from photosystem I to \(\text{NADP}^+\). As a result, the electron transport chain becomes fully reduced and electron flow from photosystem II halts. Because photolysis of water relies on electron replenishment into photosystem II, the rate of photolysis of water decreases rapidly. NADPH production ceases rather than accumulates, and the Calvin cycle reactions (carbon dioxide fixation and triose phosphate synthesis) will eventually decrease rather than increase.

A (1 mark): The rate of photolysis of water decreases due to the blockage of the terminal electron acceptor.

Let \(A\) be the dominant normal allele and \(a\) be the recessive mutant allele.
Since both parents are phenotypically normal but have an affected daughter (genotype \(aa\)), both parents must be heterozygous (genotype \(Aa\)).

For any single child produced by this couple:
- Probability of affected phenotype (\(aa\)) = \(\frac{1}{4}\)
- Probability of normal phenotype (\(AA\) or \(Aa\)) = \(\frac{3}{4}\)

Since the trait is autosomal, sex does not alter the phenotypic probability. Furthermore, fraternal (dizygotic) twins arise from two independent fertilisation events.

Therefore, the probability that both the boy and the girl have the normal phenotype is:
\(P = \frac{3}{4} \times \frac{3}{4} = \frac{9}{16}\).

C (1 mark): \(\frac{3}{4} \times \frac{3}{4} = \frac{9}{16}\).

Condition 3 provides both moving air (which blows away humid air near the leaf surface, steepening the water vapour concentration gradient) and bright light (which stimulates the opening of stomata to their maximum aperture), resulting in the highest transpiration and water absorption rate. Condition 2 has moving air but dimmer light, so stomata are less widely open than in Condition 3. Condition 1 has still air and dim light, leading to the accumulation of a humid boundary layer and lower transpiration. Condition 4 has petroleum jelly blocking all stomata on both surfaces, virtually preventing transpiration; thus water uptake and bubble movement are minimal. Therefore, the sequence from fastest to slowest is Condition 3 > Condition 2 > Condition 1 > Condition 4.

A (1 mark)
- Condition 3 > Condition 2 > Condition 1 > Condition 4 correctly deduced based on stomatal aperture and water vapour concentration gradient.

Statement (1) is incorrect because at light intensities below the compensation point (as long as light is present), photosynthesis still takes place, but its rate is lower than the rate of respiration. Statement (2) is correct because at the compensation point, the rate of photosynthesis equals the rate of respiration; hence, the volume of oxygen produced by photolysis equals the volume consumed by cellular respiration. Statement (3) is correct because above the light saturation point, the photosynthetic machinery is saturated and another factor (e.g., carbon dioxide concentration or temperature) becomes limiting, so increasing light intensity does not raise the rate of photosynthesis further.

C (1 mark)
- (2) and (3) only.

During early ventricular systole (isovolumetric contraction), the left ventricle contracts and ventricular pressure rises. As soon as the ventricular pressure exceeds the pressure in the left atrium, the bicuspid (mitral) valve is forced shut to prevent backflow into the atrium. However, because ventricular pressure has not yet exceeded the higher pressure in the aorta, the aortic semi-lunar valve remains closed. Therefore, both the bicuspid valve and the aortic semi-lunar valve are closed during this phase.

B (1 mark)
- Correct deduction that high ventricular pressure closes atrioventricular valves while remaining below arterial pressure keeps semi-lunar valves closed.

At the light compensation point, the rate of gross photosynthesis is equal to the rate of cellular respiration, resulting in zero net gas exchange.

- (1) is correct: Higher temperature (25°C vs 15°C) increases respiratory enzyme activity, so the rate of cellular respiration is higher (indicated by the higher \(\text{O}_2\) uptake in darkness, \(2.2\text{ }\mu\text{mol cm}^{-2}\text{ h}^{-1}\) vs \(1.0\text{ }\mu\text{mol cm}^{-2}\text{ h}^{-1}\)).
- (2) is correct: Because cellular respiration proceeds at a higher rate at 25°C, a greater rate of gross photosynthesis (and hence a higher light intensity) is required to produce sufficient \(\text{O}_2\) to offset respiratory consumption.
- (3) is incorrect: At 15°C, enzymes are less active due to lower kinetic energy, but they are not denatured (denaturation occurs at abnormally high temperatures).

Therefore, (1) and (2) only are correct.

B (1 mark)
- (1) and (2) are correct biological deductions.
- (3) is scientifically incorrect (low temperatures reduce kinetic energy, not cause denaturation).

- (1) is correct: Bile salts act as emulsifiers, breaking large lipid droplets into smaller droplets. This increases the total surface area accessible to lipase, increasing the rate of digestion and thus causing a faster drop in pH.
- (2) is correct: Lipase catalyses the hydrolysis of lipids (triglycerides) into fatty acids and glycerol. Fatty acids release hydrogen ions in solution, lowering the pH.
- (3) is incorrect: The pH in Tube 3 remains unchanged because lipase (a protein enzyme) was denatured by boiling, rendering it inactive. Bile salts are steroid-derived salts, not enzymes, and do not catalyze digestion on their own.

Therefore, (1) and (2) only are correct.

A (1 mark)
- (1) correctly explains the physical effect of emulsification.
- (2) correctly identifies the acidic product of lipid digestion.
- (3) incorrectly attributes inactivity to bile salts rather than the denatured enzyme (lipase).

- (1) is correct: Both parents (1 and 2) are unaffected but have affected children (3 and 5), which indicates that the allele for the condition must be recessive.
- (2) is incorrect: If the condition were X-linked recessive, the affected daughter (Individual 3, \(X^a X^a\)) must inherit an \(X^a\) allele from her father. Her father (Individual 1) would then be \(X^a Y\) and show the condition. Since the father is unaffected, the gene must be located on an autosome (autosomal recessive).
- (3) is correct:
- Parents 1 and 2 are both heterozygous carriers (\(Aa\)).
- Individual 4 is an unaffected son from this cross, so his possible genotypes are \(AA\) (\(\frac{1}{3}\)) or \(Aa\) (\(\frac{2}{3}\)). Thus, \(P(\text{Individual 4 is } Aa) = \frac{2}{3}\).
- Individual 6 is unaffected, but her father was affected (\(aa\)), so she must carry one recessive allele (\(Aa\)). Thus, \(P(\text{Individual 6 is } Aa) = 1\).
- The probability of two \(Aa\) parents having an affected (\(aa\)) child is \(\frac{1}{4}\).
- Overall probability = \(\frac{2}{3} \times 1 \times \frac{1}{4} = \frac{1}{6}\).

Therefore, (1) and (3) only are correct.

C (1 mark)
- (1) is correct: unaffected parents have affected offspring -> recessive.
- (2) is incorrect: unaffected father with affected daughter excludes X-linked recessive inheritance.
- (3) is correct: \(\frac{2}{3} \times \frac{1}{4} = \frac{1}{6}\).

(1) is correct: Large plasma proteins (such as albumin and fibrinogen) are too large to pass through the pores of the capillary wall, so tissue fluid contains very little plasma protein compared to blood plasma.
(2) is correct: Surrounding body cells continuously consume glucose for cellular respiration, maintaining a lower glucose concentration in tissue fluid than in the incoming arterial blood plasma, which facilitates diffusion of glucose into tissue fluid.
(3) is incorrect: At the arterial end of a capillary bed, the hydrostatic pressure of blood plasma is high (around 35 mmHg due to ventricular contraction) and is significantly higher than that of the surrounding tissue fluid (close to 0 mmHg), forcing fluid out into the intercellular space.

Award 1 mark for option B.
- Statement (1) is correct: capillary endothelium is impermeable to large proteins.
- Statement (2) is correct: cellular uptake of glucose maintains a concentration gradient.
- Statement (3) is incorrect: blood hydrostatic pressure exceeds tissue fluid hydrostatic pressure.

- If the condition were X-linked dominant, an affected father would carry the dominant allele on his only X chromosome (\(X^A Y\)). He must pass his X chromosome to all of his daughters, meaning all daughters must inherit the condition. The presence of an unaffected daughter in Family 1 rules out X-linked dominant inheritance (Statement 2).
- If the condition were X-linked recessive, an affected mother must be homozygous recessive (\(X^a X^a\)). She must pass an \(X^a\) chromosome to all her sons, who receive a Y chromosome from the father (\(X^a Y\)), meaning all sons must be affected. The presence of an unaffected son in Family 2 rules out X-linked recessive inheritance (Statement 3).
- Autosomal dominant and autosomal recessive inheritance are both possible for both families (e.g., heterozygous affected parents or carriers).
Therefore, both (2) and (3) can be ruled out.

Award 1 mark for option D.
- Deduce that an unaffected daughter from an affected father rules out X-linked dominant inheritance.
- Deduce that an unaffected son from an affected mother rules out X-linked recessive inheritance.

(1) is correct: Potassium hydroxide (\(\text{KOH}\)) reacts with and absorbs carbon dioxide gas released during aerobic respiration.
(2) is correct: Because carbon dioxide produced is removed by the \(\text{KOH}\), any net reduction in gas volume inside the tube is solely due to oxygen uptake, pulling the coloured droplet inwards towards the chamber.
(3) is correct: Physical changes in temperature and barometric pressure cause expansion or contraction of gas inside the sealed apparatus; using boiled, sterilised seeds of equal volume/mass accounts for non-biological volume fluctuations.

Award 1 mark for option D.
- Statement (1): function of KOH in CO2 absorption.
- Statement (2): principle of measuring rate of oxygen uptake.
- Statement (3): purpose of an inert control for environmental pressure/temperature fluctuations.

Physical (mechanical) digestion involves breaking down large food pieces or droplets into smaller ones without breaking chemical bonds, thereby increasing the surface area for enzyme action.
- (1) is correct: Bile salts emulsify large lipid droplets into smaller droplets mechanically via amphipathic properties without hydrolysing ester bonds.
- (2) is correct: Churning by the muscular wall of the stomach physically breaks down food and mixes it with gastric juice.
- (3) is incorrect: Pancreatic lipase catalyses the chemical hydrolysis of ester bonds in triglycerides to form fatty acids and glycerol, which is chemical digestion.

Therefore, (1) and (2) only are physical in nature.

B (1 mark)
- (1) and (2) are physical digestion processes.
- (3) is chemical digestion (enzymatic hydrolysis).

Photosynthesis requires light, chlorophyll, and carbon dioxide to produce starch:
- Part X possesses chlorophyll and is supplied with \(\text{CO}_2\) and light. Photosynthesis occurs, producing starch, which stains blue-black with iodine solution.
- Part Y lacks chlorophyll (white area) and therefore cannot carry out photosynthesis. No starch is produced, so it remains brown (the colour of iodine solution).
- Part Z is enclosed with potassium hydroxide solution, which absorbs all carbon dioxide. Without \(\text{CO}_2\), photosynthesis cannot occur, so no starch is produced and it remains brown.

Thus, the result is Part X: Blue-black, Part Y: Brown, Part Z: Brown.

B (1 mark)
- Part X: Photosynthesis occurs \(\rightarrow\) starch present \(\rightarrow\) Blue-black
- Part Y: No chlorophyll \(\rightarrow\) no starch \(\rightarrow\) Brown
- Part Z: No \(\text{CO}_2\) (absorbed by KOH) \(\rightarrow\) no starch \(\rightarrow\) Brown

- (1) is correct: Both unaffected parents have an affected daughter. This demonstrates that the allele causing the disorder is masked in the parents, proving it is recessive.
- (2) is correct: If the condition were X-linked recessive, the affected daughter would have the genotype \(X^a X^a\), meaning she must receive one \(X^a\) allele from her father. The father would then have the genotype \(X^a Y\) and would express the disorder. Since the father is unaffected, the gene cannot be on the X chromosome and must be autosomal.
- (3) is incorrect: Both parents are heterozygous (\(Aa\)). The unaffected son can have either the genotype \(AA\) (probability 1/3) or \(Aa\) (probability 2/3). He is not guaranteed to be heterozygous.

Therefore, only statements (1) and (2) are definitely correct.

A (1 mark)
- (1) is correct: Unaffected parents having affected offspring confirms recessive inheritance.
- (2) is correct: Normal father having an affected daughter rules out X-linked recessive inheritance; it must be autosomal.
- (3) is incorrect: Normal son has a 1/3 chance of being homozygous dominant and a 2/3 chance of being heterozygous.

Lipase catalyses the chemical breakdown of lipids (triglycerides in milk) into glycerol and free fatty acids. The accumulation of fatty acids lowers the pH of the mixture, neutralising the alkaline sodium carbonate and causing the phenolphthalein indicator to become colourless. Bile salts do not contain enzymes but act as an emulsifier, breaking large lipid droplets into smaller droplets. This substantially increases the surface area available for lipase action, speeding up digestion. Tube 3 contains both active lipase and bile salts, so lipid hydrolysis occurs at the highest rate, leading to the most rapid decolourisation.

Award 1 mark for option C. Options A, B, and D are incorrect because boiled lipase is denatured (Tube 1), lipase without bile salts acts more slowly due to lower surface area (Tube 2), and bile salts alone cannot hydrolyse fats without lipase (Tube 4).

In the light-dependent reactions of photosynthesis, absorbed light energy drives the photolysis of water into protons, electrons, and oxygen gas (statement 1 is correct). The flow of electrons through the electron transport chain drives photophosphorylation to produce ATP from ADP and \(\text{P}_i\), and reduces \(\text{NADP}^+\) to \(\text{NADPH}\) (statement 2 is correct). However, the synthesis of triose phosphate occurs during the light-independent reactions (Calvin cycle), which requires carbon dioxide as the substrate for carbon fixation by RuBisCO. In the absence of \(\text{CO}_2\), no carbon fixation can occur and triose phosphate cannot be produced (statement 3 is incorrect). Therefore, only statements (1) and (2) are correct.

Award 1 mark for option A: (1) and (2) only. Deduct 0 marks for wrong alternatives. Statements (1) and (2) correctly identify light-dependent processes that do not directly require carbon dioxide, whereas statement (3) requires carbon dioxide fixation.

1. ABO Blood group inheritance:
- Father has blood group AB: genotype \(I^A I^B\).
- Mother has blood group B and her father had blood group O (\(ii\)), so she must carry the recessive allele \(i\). Her genotype is \(I^B i\).
- Crossing \(I^A I^B \times I^B i\) yields possible genotypes: \(I^A I^B\) (group AB), \(I^A i\) (group A), \(I^B I^B\) (group B), and \(I^B i\) (group B) in equal proportions.
- Probability of child having blood group A (genotype \(I^A i\)) = \(\frac{1}{4}\).

2. Colour blindness inheritance:
- Let \(X^N\) be the normal vision allele and \(X^n\) be the colour blindness allele.
- Father: \(X^N Y\); Mother (carrier): \(X^N X^n\).
- Offspring combinations: \(X^N X^N\) (normal female, \(\frac{1}{4}\)), \(X^N X^n\) (carrier female, \(\frac{1}{4}\)), \(X^N Y\) (normal male, \(\frac{1}{4}\)), \(X^n Y\) (colour-blind male, \(\frac{1}{4}\)).
- Probability of having a colour-blind son (\(X^n Y\)) = \(\frac{1}{4}\).

3. Combined probability:
Since the ABO gene and the X-linked colour vision gene assort independently:
\(\text{Probability} = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}\).

Award 1 mark for option A (1/16). Options B (1/8), C (1/4), and D (1/2) represent common calculation errors such as failing to account for sex determination or parental heterozygosity.

(a) A potometer measures the rate of water uptake by the shoot, which is slightly higher than the transpiration rate. A small portion of absorbed water is retained by plant cells for metabolic reactions (e.g., photosynthesis) and for maintaining turgidity/growth.

(b) Description: Increasing wind speed from 0.0 to 3.0 m/s causes a steady and significant increase in the rate of water uptake from 1.8 mm/min to 5.1 mm/min.
Explanation: In still air (0.0 m/s), water vapour accumulates outside the stomatal pores, forming a humid boundary layer. Increased wind blows away this boundary layer, maintaining a steep concentration gradient of water vapour between the intercellular air spaces of the leaf and the surrounding air, which increases the rate of transpiration and thus the rate of water uptake.

(c) Reasons for plateau: At very high wind speeds, water loss is so rapid that the rate of water loss exceeds water uptake, causing guard cells to lose turgor and stomata to close to prevent dehydration. Alternatively, the boundary layer of water vapour has already been completely blown away at 4.5 m/s, so increasing wind speed further does not increase the diffusion gradient.

(a)
- Not all absorbed water is transpired / some water is retained by the plant (1 mark)
- Absorbed water is used for photosynthesis / metabolic reactions / maintaining turgidity / growth (1 mark)

(b)
- Description: Rate of water uptake increases from 1.8 mm/min to 5.1 mm/min / increases proportionally with wind speed (1 mark)
- Wind removes the layer of humid air / water vapour accumulating around the stomata / leaf surface (1 mark)
- This maintains / steepens the concentration gradient of water vapour between the leaf interior and the atmosphere (1 mark)

(c) Any two of the following (1 mark each, max 2 marks):
- Guard cells lose turgor / stomata close (partially or fully) to prevent excessive water loss / desiccation (1 mark)
- The boundary layer is already completely dissipated / minimum boundary layer thickness is reached, so the diffusion gradient cannot be further increased (1 mark)
- Internal resistance to water movement in the xylem vessels limits further increases in flow rate (1 mark)

(a) Hydrolysis of triglycerides (olive oil) by lipase produces fatty acids and glycerol. Fatty acids ionise in solution to release \(\text{H}^+\) ions, which increases the acidity and lowers the pH.

(b) Tube Q contains bile salts, whereas Tube P contains only distilled water. Bile salts emulsify large lipid droplets into numerous tiny droplets. This physical process significantly increases the surface area-to-volume ratio of the lipid substrate, providing more binding sites for lipase and thus increasing the rate of enzymatic hydrolysis into fatty acids.

(c) In Tube R, the lipase was boiled prior to the experiment. High temperature breaks hydrogen and ionic bonds, permanently disrupting the tertiary structure (denaturation) and altering the specific shape of the active site so it is no longer complementary to the substrate. Therefore, no digestion occurs and the pH remains unchanged.

(d) Resynthesised triglycerides are packaged into chylomicrons and enter the lacteal located in the centre of each intestinal villus.

(a)
- Hydrolysis of triglycerides produces fatty acids (which release \(\text{H}^+\) / increase acidity) (1 mark)

(b)
- Bile salts emulsify oil / break large oil droplets into smaller droplets (1 mark)
- This increases the surface area (to volume ratio) of lipids available for lipase action (1 mark)
- Higher frequency of effective collisions / faster rate of enzymatic hydrolysis of triglycerides into fatty acids (1 mark)

(c)
- pH remains unchanged / stays constant (1 mark)
- Lipase is denatured / active site is deformed by high temperature, so no substrate can bind / no reaction occurs (1 mark)

(d)
- Lacteal / central lymphatic vessel (1 mark) [Reject: capillary / blood vessel]

(a) Step 1: Identify that parents I-1 and I-2 are phenotypically normal.
Step 2: Note that they produce an affected child, II-2.
Step 3: Conclude that the allele for the disorder is recessive because it is masked in the heterozygous parents and expressed only when homozygous.

(b) Step 1: Assume the condition is X-linked recessive.
Step 2: Affected female III-1 must have the genotype \(X^a X^a\), requiring one \(X^a\) from her mother (II-1) and one \(X^a\) from her father (II-4).
Step 3: This would mean father II-4 has genotype \(X^a Y\) and should show the disorder. However, II-4 is unaffected (phenotypically normal with genotype \(X^A Y\)). Hence, the condition must be autosomal.

(c) Step 1: Determine the genotype probability of II-3.
Parents I-1 and I-2 are both carriers (\(Aa\)). A cross of \(Aa \times Aa\) gives offspring in the ratio \(1 AA : 2 Aa : 1 aa\).
Since II-3 is phenotypically normal, he cannot be \(aa\). Thus, \(P(\text{II-3 is } Aa) = 2/3\).
Step 2: Cross between II-3 (\(Aa\) with probability \(2/3\)) and carrier wife (\(Aa\)):
\(P(\text{child is } aa) = 1/4\).
Step 3: Probability that the child is male = \(1/2\).
Step 4: Combined probability = \(2/3 \times 1/4 \times 1/2 = 2/24 = 1/12\) (or \(0.0833\) / \(8.33\%\)).

(a)
- State that the allele is recessive (1 mark)
- Parents I-1 and I-2 are unaffected / do not have the disorder (1 mark)
- But they produce an affected offspring II-2 / the disease allele is masked in the parents (1 mark)

(b)
- State that the gene is autosomal / located on an autosome (1 mark)
- If it were X-linked recessive, affected female III-1 must receive an affected allele from her father II-4 (1 mark)
- Father II-4 would have to be affected, but he is unaffected / normal (1 mark)

(c)
- Probability of II-3 being a carrier (\(Aa\)) is \(2/3\) (1 mark)
- Calculation showing \(2/3 \times 1/4 \times 1/2 = 1/12\) / \(0.083\) / \(8.33\%\) (1 mark)

(a) Bile salts emulsify fats (break large lipid droplets into smaller droplets), which greatly increases the surface area for pancreatic lipase to act on.

(b) (i) Tube P will show a faster and greater decrease in pH compared to tube Q.
(ii) The emulsification by bile salts in Tube P speeds up the enzymatic hydrolysis of triglycerides into fatty acids and glycerol. As fatty acids are acidic, their faster release causes a faster drop in pH.

(c) (i) Lacteal (or lymphatic vessel).
(ii) Under high hydrostatic pressure at the arterial end of blood capillaries, fluid is filtered out to form tissue fluid. About 10–15% of excess tissue fluid drains into the lymphatic system to return to the blood circulation. When lymphatic vessels are blocked, this excess fluid cannot be drained and builds up in intercellular / tissue spaces, resulting in swelling (edema).

(a) Emulsify lipids / fats into smaller droplets / droplets with larger surface area for lipase action (1)

(b) (i) pH in tube P drops faster / drops to a lower value than in tube Q (1)
(ii) Emulsification increases surface area for lipase action (1); faster production / release of fatty acids which lowers pH (1)

(c) (i) Lacteal / lymphatic vessel (1) [Reject: blood capillary]
(ii) Tissue fluid is filtered out of blood capillaries under hydrostatic pressure (1); blockage prevents excess tissue fluid from draining into lymph vessels / returning to the blood circulation, causing fluid accumulation in interstitial spaces (1)

(a) Photolysis of water occurs during the light-dependent stage (in the thylakoid membrane), releasing oxygen gas as a by-product.

(b) (i) Light intensity is the limiting factor.
(ii) When light intensity is low (< 200 a.u.), changing the CO2 concentration from 0.03% to 0.15% has no effect on the photosynthetic rate. This proves that CO2 is not the factor limiting the rate; rather, the low availability of light energy limits ATP and NADPH generation.

(c) (i) Carbon dioxide is fixed by RuBisCO in the Calvin cycle (light-independent reaction). At high light intensity, light is abundant, so carbon dioxide availability becomes the limiting factor. Higher CO2 concentration increases the rate of carbon fixation, utilising more ATP and NADPH, thus accelerating the entire photosynthetic pathway.
(ii) Temperature (or light wavelength/spectrum) must be kept constant because temperature affects the kinetic energy and catalytic rate of photosynthetic enzymes (e.g., RuBisCO).

(a) Light-dependent stage / photolysis of water (1)

(b) (i) Light intensity (1)
(ii) Increasing CO2 concentration does not change / increase the rate of photosynthesis (1); showing that light energy is insufficient / restricting the rate of ATP and NADPH synthesis (1)

(c) (i) CO2 is the substrate for carbon fixation in the light-independent stage / Calvin cycle (1); higher CO2 increases the turnover of ATP / NADPH / enables faster enzyme action, allowing a higher overall photosynthetic rate (1)
(ii) Temperature / light wavelength (1) [Reject: light intensity, CO2 concentration]

(a) The condition is dominant. Individual 3 (affected) and Individual 7 (unaffected) produce an unaffected child, Individual 9. Individual 7 can only contribute normal recessive alleles. Individual 9 must have received a normal recessive allele from Individual 3. Since Individual 3 displays the condition while carrying a normal allele, the allele for the condition must be dominant over the normal allele.

(b) The gene is X-linked (located on the X chromosome). Individual 1 is an affected male. In X-linked dominant inheritance, an affected father passes his X chromosome to all his daughters and his Y chromosome to all his sons. Here, all daughters (3 and 4) are affected and all sons (5 and 6) are unaffected, which is characteristic of X-linked dominant inheritance.

(c) The probability is 1.0 (or 100%). Individual 8 has the genotype X^D Y (where X^D is the dominant mutant allele). He must pass the X^D chromosome to every daughter. Since the allele is dominant, possessing one copy is sufficient for the daughter to express the condition.

(a) Dominant allele (1);
Individual 3 is affected and Individual 7 is unaffected, but they have an unaffected daughter / child (Individual 9) (1);
Individual 9 must have inherited a normal allele from Individual 3, showing Individual 3 is heterozygous and expresses the trait (1)

(b) X-linked / located on the X chromosome (1);
Individual 1 (affected male) passed the condition to all his daughters (3 and 4) but none of his sons (5 and 6) (1);
Because father passes X chromosome to daughters and Y chromosome to sons (1)

(c) Probability = 1 / 100% (1);
Individual 8 has genotype X^D Y and passes his X^D chromosome to all daughters (1)

(a) 1. Cut the stem of the leafy shoot underwater (to prevent air bubbles entering and blocking xylem vessels). 2. Ensure all connections/joints of the apparatus are airtight (using petroleum jelly/Vaseline).

(b) Increased wind speed blows away water vapour accumulating around the stomata on the leaf surface, which steepens the concentration gradient of water vapour between the intercellular air spaces and the surrounding air. This increases the rate of diffusion/evaporation of water vapour through stomata (transpiration). The resulting greater transpiration pull draws more water up the xylem, leading to a higher rate of water uptake.

(c) The claim is incorrect because not all water absorbed by the plant is lost through transpiration; a small proportion (around 1–2%) is retained and used for metabolic activities (e.g., photosynthesis) and for maintaining cell turgidity/growth.

(a) Any two of the following (1 mark each, max 2 marks):
- Cut the stem underwater / trim the cut end underwater (1) [to prevent air entry into xylem vessels]
- Seal all joints with Vaseline / petroleum jelly to ensure airtightness (1)
- Dry the leaves before starting the experiment (1)

(b)
- Wind sweeps away accumulated water vapour from the leaf surface / boundary layer (1)
- This maintains / steepens the concentration gradient of water vapour between the sub-stomatal air space and the outside atmosphere (1)
- Increases rate of transpiration / evaporation, creating a stronger transpiration pull to draw up more water (1)

(c)
- Rate of water uptake is slightly higher than / not equal to rate of transpiration (1)
- Some water absorbed is used for photosynthesis / metabolic reactions / maintaining cell turgidity / cell expansion (1)

(a) When a nerve impulse arrives at the synaptic terminal of the sensory neurone, synaptic vesicles fuse with the presynaptic membrane and release neurotransmitters into the synaptic cleft by exocytosis. The neurotransmitters diffuse across the synaptic cleft and bind to specific receptor proteins on the postsynaptic membrane of the interneurone, causing depolarization and generating a new nerve impulse.

(b) Neurotransmitters are only synthesized and stored in synaptic vesicles within the presynaptic terminal, and specific neurotransmitter receptors are only present on the postsynaptic membrane.

(c) (i) No nerve impulses will be generated in the motor neurone because neurotransmitters cannot bind to the blocked receptors to cause depolarization.
(ii) The biceps muscle will not contract, so the withdrawal reflex response fails to occur.

(a)
- Arrival of nerve impulse causes synaptic vesicles to release neurotransmitter into synaptic cleft via exocytosis (1)
- Neurotransmitter diffuses across the synaptic cleft (1)
- Neurotransmitter binds to specific receptors on postsynaptic membrane, triggering a new nerve impulse (1)

(b)
- Synaptic vesicles / neurotransmitters are only present in the presynaptic neurone (1)
- Receptor sites are only located on the postsynaptic membrane (1)

(c)
(i) No nerve impulse is generated / triggered in the motor neurone (1)
(ii) No contraction of the biceps muscle / withdrawal reflex does not occur (1)

(a) The allele causing alkaptonuria is recessive. Both Individuals 1 and 2 are unaffected, but they produced an affected son (Individual 3). This indicates that both parents must be heterozygous carriers who pass the hidden/masked recessive allele to Individual 3.

(b) The gene is autosomal. If it were X-linked recessive, an affected male (Individual 3, X^a Y) must inherit the recessive allele from his mother (Individual 2, X^A X^a), which is possible; however, if an affected female (Individual 7, X^a X^a) is born to an affected father (Individual 6, X^a Y) and an unaffected mother (Individual 4), Individual 4 must be heterozygous (X^A X^a). Furthermore, if Individual 3 were affected and had an unaffected daughter, or in a reciprocal cross where an unaffected father has an affected daughter, it rules out X-linked dominance. Since unaffected parents 1 and 2 produced an affected child 3 without sex bias, and an affected female 7 has an affected father 6, it is consistent with autosomal recessive inheritance.

(c) (i) Individual 1: Aa
(ii) Individual 4: Aa (since she is unaffected but gave birth to an affected child, Individual 7 who is aa).

(d) Cross between Individual 4 (Aa) and Individual 6 (aa):
Gametes from Individual 4: A, a
Gametes from Individual 6: a
Offspring genotypes: 1/2 Aa (unaffected), 1/2 aa (affected).
Probability of being unaffected = 1/2 (50%).
Probability of being a boy = 1/2 (50%).
Overall probability = (1/2) × (1/2) = 1/4 (or 25% or 0.25).

(a)
- Recessive (1)
- Unaffected parents (Individuals 1 and 2) produced an affected offspring (Individual 3), showing that the allele for alkaptonuria was masked in the parents (1)

(b)
- Autosomal (1)
- If the condition were X-linked recessive, an affected female (Individual 7) must have inherited one recessive allele from each parent; her mother (Individual 4) is unaffected, meaning Individual 4 is a carrier (Aa). Moreover, both unaffected parents 1 and 2 produce affected male 3, showing the trait is autosomal recessive / equal transmission across sexes (1)

(c)
(i) Aa (1)
(ii) Aa (1)

(d)
- Probability of unaffected child (Aa) = 1/2 (1)
- Probability of being a boy = 1/2, so total probability = 1/2 × 1/2 = 1/4 / 0.25 / 25% (1)

(a) Cutting the stem under water prevents air bubbles from entering the lumen of xylem vessels, thereby maintaining the continuous water column essential for transpiration pull and accurate potometer readings.

(b) (i) As light intensity increases from 0 to 400 a.u., the average stomatal aperture increases significantly (from 1.2 \(\mu\text{m}\) to 9.4 \(\mu\text{m}\)).

(ii) In the presence of light, guard cells carry out photosynthesis and accumulate solutes (e.g., potassium ions/sugars), which decreases the water potential inside the guard cells. Water enters the guard cells from adjacent epidermal cells by osmosis. As the guard cells become turgid, their thicker inner cell walls resist stretching while the thinner outer walls expand outwards, causing the guard cells to curve and open the stomatal pore.

(c) In darkness, water loss still occurs because stomata are not completely closed (aperture is 1.2 \(\mu\text{m}\)) and water vapor can also evaporate directly through the leaf cuticle (cuticular transpiration). In addition, a small amount of water is retained to maintain cell turgidity.

(d) Deduction: The density of stomata is much higher on the lower epidermis than on the upper epidermis.
Explanation: Applying petroleum jelly to the lower epidermis (Shoot B) led to a substantial decrease in water uptake (a drop of \(56\text{ mm}^3\text{ min}^{-1}\), from 82 to 26), whereas applying it to the upper epidermis (Shoot A) only caused a small drop (a drop of \(8\text{ mm}^3\text{ min}^{-1}\), from 82 to 74). This indicates that the majority of transpiration occurs through the lower leaf surface.

+---------------------------------------------------------------------------------------------------------------+--------------------------------------------------------------------+-------+
| Concept for mark award | Example | Marks |
+---------------------------------------------------------------------------------------------------------------+--------------------------------------------------------------------+-------+
| (a) Prevent air entry into xylem | To prevent air / air bubbles from entering the xylem vessels / | 1 |
| | to maintain a continuous water column. | |
| | | |
| (b)(i) Correct description of trend | Stomatal aperture increases (from 1.2 to 9.4 μm) with light. | 1 |
| | | |
| (b)(ii) Osmotic entry of water into guard cells | Light causes accumulation of solutes / photosynthesis in guard | 1 |
| | cells, lowering their water potential so water enters by osmosis. | |
| Differential wall expansion / turgidity leading to opening | Guard cells become turgid; because the inner wall is thicker / | 1 |
| | less elastic than the outer wall, the cells curve outward to open. | |
| | | |
| (c) Reason for water uptake in darkness | Transpiration still occurs via cuticle (cuticular transpiration) / | 1 |
| | stomata are partially open (1.2 μm) / water used for metabolism. | |
| | | |
| (d) Correct deduction regarding stomatal distribution | More stomata / higher density of stomata on the lower epidermis | 1 |
| | than on the upper epidermis. | |
| Supporting explanation with comparative data | Blocking the lower epidermis resulted in a much greater reduction | 1 |
| | in water uptake (from 82 to 26 vs. 82 to 74 mm³ min⁻¹). | |
+---------------------------------------------------------------------------------------------------------------+--------------------------------------------------------------------+-------+

Structural adaptations of leaves to minimise water loss:
- A thick, waxy cuticle covers the epidermis, which acts as an impermeable barrier to reduce non-stomatal cuticular evaporation.
- Stomata are predominantly located on the lower epidermis (or are sunken in pits / protected by epidermal hairs), which reduces the exposure to direct sunlight and wind, trapping a layer of humid air and reducing the water vapour concentration gradient.

Physiological response of guard cells:
- Under severe water deficit, guard cells lose water by osmosis to adjacent epidermal cells, causing a loss of turgor.
- As guard cells become flaccid, the stomatal aperture closes, significantly reducing transpiration.

Effect on carbon fixation in chloroplasts:
- Stomatal closure restricts the entry of carbon dioxide into the intercellular air spaces of the leaf by diffusion.
- The decreased concentration of carbon dioxide in the stroma slows down the light-independent reactions (Calvin cycle / carbon fixation), thereby reducing the overall rate of photosynthesis.

Effect on phloem translocation:
- Translocation of sugars occurs by mass flow driven by a hydrostatic pressure gradient between source and sink.
- Active loading of sugars into sieve tubes at the source lowers the water potential, which usually draws water from the xylem by osmosis to generate high turgor pressure.
- Under severe water deficit, the low water potential in surrounding tissues reduces water movement into the sieve tubes, lowering the hydrostatic pressure gradient and slowing the rate of mass flow of organic nutrients to sink tissues.

Biological Content (max. 8 marks):

Leaf Structural Adaptations (max 2 marks):
- Thick / waxy cuticle on epidermis provides an impermeable layer to reduce cuticular evaporation. (1 mark)
- Higher density of stomata on lower epidermis / sunken stomata / trichomes (hairs) reduce air movement / trap moist air around stomata to lower water vapour concentration gradient. (1 mark)

Physiological Response of Guard Cells (max 2 marks):
- Guard cells lose turgor / become flaccid as water exits by osmosis under water deficit. (1 mark)
- Loss of turgidity causes guard cells to close the stomatal pore / aperture, reducing transpiration. (1 mark)

Impact on Carbon Fixation (max 2 marks):
- Stomatal closure restricts the diffusion / entry of carbon dioxide into leaf mesophyll / intercellular spaces. (1 mark)
- Decreased carbon dioxide availability in the stroma limits the light-independent reaction / Calvin cycle / carbon fixation. (1 mark)

Impact on Phloem Translocation (max 2 marks):
- Phloem translocation / mass flow relies on osmotic entry of water into sieve tubes at the source (leaf) to generate high hydrostatic / turgor pressure. (1 mark)
- Severe water deficit / lower water potential reduces water entry into sieve tubes, thereby decreasing the hydrostatic pressure gradient between source and sink and slowing mass flow. (1 mark)

Effective Communication (3 marks):
- 3 marks: Candidate presents answers logically and systematically, using accurate biological terminology throughout, with no irrelevant details.
- 2 marks: Candidate presents answers clearly with minor disorganisation or minor inaccuracies in terminology.
- 1 mark: Candidate presents answers with poor organisation, making comprehension difficult, or includes substantial irrelevant material.
- 0 marks: Incomprehensible response or completely irrelevant content.

(a)(i)
• Location of receptors: Hypothalamus (osmoreceptors).
• Endocrine gland: Posterior lobe of pituitary gland (posterior pituitary).

(a)(ii)
• The increase in plasma osmolarity (from 288 to 316 mOsm L^-1) stimulates the secretion of ADH into the bloodstream, increasing its concentration from 1.8 to 8.6 pg mL^-1.
• ADH increases the permeability of the collecting duct and distal convoluted tubule cells to water (by promoting insertion of aquaporins).
• More water is reabsorbed by osmosis into the hypertonic medullary interstitial fluid and capillaries.
• Consequently, less water remains in the filtrate, leading to a reduced urine production rate (from 1.2 to 0.25 mL min^-1) and a higher urine osmolarity / more concentrated urine (from 350 to 1180 mOsm L^-1).

(a)(iii)
• To continuously excrete toxic metabolic waste products (such as urea, uric acid, and creatinine) and prevent their toxic accumulation in body fluids.

(a)(iv)
• Activation of the thirst mechanism in the hypothalamus to stimulate water intake / drinking behavior.
• Activation of the renin-angiotensin-aldosterone system (RAAS) to promote vasoconstriction (arteriolar constriction) and sodium reabsorption, helping to elevate/maintain blood pressure.

(b)(i)
• Cold thermoreceptors in the skin detect the drop in temperature and send nerve impulses along sensory neurones to the thermoregulatory centre in the hypothalamus.
• The hypothalamus sends nerve impulses along sympathetic motor neurones to superficial blood vessels in the skin.
• Arterioles in the skin constrict (vasoconstriction) and precapillary sphincters close, reducing blood flow to superficial capillary networks / diverting blood to deeper core tissues, thereby reducing conductive and convective heat loss to the cold water.

(b)(ii)
• Shivering involves involuntary, rapid, rhythmic contractions of skeletal muscles.
• This increases the rate of cellular respiration, which generates heat as a byproduct to warm the blood and maintain core body temperature.

(b)(iii)
• During intense exercise, increased cellular respiration produces large amounts of carbon dioxide, which dissolves in plasma and lowers blood pH / increases H⁺ concentration.
• Central chemoreceptors in the medulla oblongata and peripheral chemoreceptors in the carotid and aortic bodies detect the elevated arterial CO₂ level / decreased pH.
• Nerve impulses are sent to the respiratory control centre (medulla oblongata), which sends more frequent impulses via motor nerves to the intercostal muscles and diaphragm.
• This increases the rate and depth of ventilation to remove CO₂ and restore normal blood pH.

(a)(i)
• Hypothalamus [1]
• Posterior pituitary gland [1]

(a)(ii)
• Increased plasma osmolarity triggers greater secretion of ADH [1]
• ADH increases water permeability of collecting ducts / distal convoluted tubules [1]
• More water is reabsorbed by osmosis into the medullary capillaries / bloodstream [1]
• Results in smaller urine volume / lower production rate AND higher concentration / osmolarity [1]

(a)(iii)
• Elimination of toxic metabolic wastes (e.g. urea, creatinine) [1]
• Prevents toxic build-up in blood / maintains homeostatic balance of solutes [1]

(a)(iv)
• Any two (1 mark each):
- Stimulation of thirst centre / thirst sensation leading to fluid ingestion [1]
- Secretion of aldosterone / activation of RAAS to enhance Na⁺ reabsorption and arterial vasoconstriction [1]
- Generalized peripheral vasoconstriction to maintain central blood pressure [1]

(b)(i)
• Detection of cold by skin thermoreceptors and transmission of impulses to hypothalamus [1]
• Sympathetic stimulation leading to constriction of superficial arterioles / vasoconstriction [1]
• Diverting blood flow away from skin surface capillaries to deep shunt vessels, reducing heat dissipation [1]

(b)(ii)
• Involuntary skeletal muscle contraction [1]
• Increases metabolic respiration, producing heat to warm core body [1]

(b)(iii)
• High CO₂ / H⁺ / low pH produced by muscle respiration [1]
• Detected by central chemoreceptors (medulla) and peripheral chemoreceptors (carotid/aortic bodies) [1]
• Impulses sent to respiratory centre in medulla oblongata [1]
• Increased impulse frequency via phrenic/intercostal nerves to diaphragm and intercostal muscles to increase rate and depth of ventilation [1]

(a)(i)
• BOD measures the amount of dissolved oxygen consumed by aerobic microorganisms (bacteria) to decompose organic matter in a unit volume of water over a specific period (typically 5 days at 20°C).
• At Point X, the untreated livestock sewage introduces a huge influx of organic waste (feces and unconsumed feed).
• This provides an abundant food source, causing a population explosion of aerobic decomposers, which drastically increases the demand for oxygen.

(a)(ii)
• The dissolved oxygen level is a balance between the rate of oxygen consumption by decomposers and the rate of oxygen dissolution/reaeration from the atmosphere and photosynthesis.
• Immediately at Point X, bacterial decomposition is initiating, but it takes time for the bacterial population to multiply to peak numbers and consume oxygen at the maximum rate.
• At 2 km downstream, cumulative bacterial aerobic respiration exceeds atmospheric reaeration to the greatest extent, resulting in the dissolved oxygen minimum (oxygen sag).

(a)(iii)
• Severely polluted zone (0–3 km): Low DO levels support only pollution-tolerant organisms with specialized adaptations for low oxygen, such as tubifex worms (sludge worms) or rat-tailed maggots / bloodworms (chironomid larvae).
• Semi-polluted / recovery zone (3–6 km): As DO increases and organic load drops, moderately tolerant species appear, such as freshwater shrimps / water louse (Asellus).
• Clean recovery zone (6–8 km): Fully recovered DO supports pollution-sensitive species that require high oxygen levels, such as mayfly nymphs, stonefly nymphs, or caddisfly larvae.

(a)(iv)
• Anaerobic digestion / biogas digester to break down concentrated organic solids into methane and stable sludge.
• Aerobic biological treatment (e.g. activated sludge system or trickling filter) to oxidise remaining organic pollutants and nitrify ammonium to nitrate before discharge.

(b)(i)
• Provides complex three-dimensional hard substrates that offer shelter, breeding sites, and nurseries for fish and benthic organisms, increasing biodiversity.
• Physical barrier function: The rugged structures prevent destructive bottom trawling gears from operating in the area, protecting seabed habitats.

(b)(ii)
• Within the no-take MPA, fish populations are protected from fishing mortality, allowing individuals to survive longer, grow larger, and reproduce with higher fecundity.
• High population density and competition inside the reserve lead to the outward migration of adult and juvenile fish into adjacent unprotected fishing grounds (the "spillover effect").
• Additionally, planktonic fish eggs and larvae produced in the reserve drift into surrounding waters, enhancing recruitment and boosting catch outside the reserve.

(b)(iii)
• 1. Regulating mesh size of fishing nets (minimum mesh size regulations): Allows juvenile and small immature fish to escape and grow to reproductive maturity before being caught, ensuring continuous recruitment.
• 2. Imposing closed fishing seasons (moratorium during breeding/spawning seasons): Prevents disturbance and harvest of breeding adults and spawning aggregations, maximizing spawning success and stock replenishment.

(a)(i)
• Definition of BOD: Amount of oxygen consumed by aerobic microorganisms to break down organic matter [1]
• Influx of large amounts of organic waste from livestock sewage [1]
• Rapid proliferation / high metabolic activity of aerobic decomposers increases oxygen demand [1]

(a)(ii)
• DO depends on dynamic balance between reaeration rate and rate of deoxygenation/consumption [1]
• Microorganisms require time/distance to proliferate to peak population biomass [1]
• At 2 km downstream, cumulative oxygen consumption by maximal bacterial density exceeds atmospheric reaeration rate [1]

(a)(iii)
• Characteristic tolerant organism in polluted zone: Tubifex worms / bloodworms / rat-tailed maggots [1]
• Characteristic sensitive organism in clean zone: Mayfly nymphs / stonefly nymphs / caddisfly larvae [1]
• Description of progressive recovery: Shift from pollution-tolerant detritivores to moderate species to diverse oxygen-sensitive species as DO rises [1]

(a)(iv)
• Any two (1 mark each):
- Anaerobic digestion / biogas tank for bulk organic sludge removal [1]
- Aerobic activated sludge / biofilter / aeration pond for nitrification and BOD reduction [1]
- Constructed wetland / reed bed for phytoremediation and nutrient uptake [1]

(b)(i)
• Any two (1 mark each):
- Creates physical niches / attachment surfaces / shelter for marine life to enhance habitat complexity [1]
- Physically obstructs bottom trawling nets, preventing seabed habitat destruction [1]

(b)(ii)
• Protection from fishing enables fish to grow older/larger with higher reproductive capacity [1]
• Spillover effect: Net movement/migration of adult and juvenile fish from crowded MPA to adjacent open areas [1]
• Larval export: High production of eggs/larvae inside MPA drift out to replenish outside fishing grounds [1]

(b)(iii)
• Measure 1 + biological explanation (2 marks):
- Imposing minimum mesh size / gear restrictions [1]; allows juvenile fish to escape and reach sexual maturity before capture [1]
• Measure 2 + biological explanation (2 marks):
- Closed fishing season (moratorium during spawning period) [1]; protects breeding adults during critical reproductive phase to maximize recruitment [1]
(OR Catch quotas / Total Allowable Catch (TAC) [1]; limits total mortality to below maximum sustainable yield [1])