2025 Cambridge IGCSE Biology (0610) Practice Paper with Answers

At pH 7.0, the time taken for starch to be completely digested is 1.5 minutes, which is the shortest time of all the pH values tested. Since rate is inversely proportional to time, the rate of starch breakdown is highest at pH 7.0, indicating it is the optimum pH. Amylase is active at pH 5.0 (it takes 6 minutes), so it is not completely denatured. Amylase works slower at pH 8.0 (4 minutes) than at pH 6.0 (3 minutes).

1 mark: Correct identification of option A as the only scientifically valid conclusion based on the experimental data.

Statement 1 is correct because enzyme specificity relies on complementary shapes of the active site and substrate. Statement 3 is correct because successful kinetic collisions result in the formation of enzyme-substrate complexes. Statement 2 is incorrect because enzymes are biological catalysts and remain unchanged chemically. Statement 4 is incorrect because temperatures above the optimum cause denaturation, altering the active site shape and stopping the reaction.

1 mark: Correct selection of option A (statements 1 and 3).

High humidity increases the water vapour concentration in the air surrounding the leaf. This reduces the concentration gradient between the internal air spaces of the leaf and the external atmosphere, decreasing the rate of diffusion of water vapour out of the stomata, which reduces water uptake. Wind increases transpiration rate by removing water vapour from around the leaf, increasing the gradient.

1 mark: Correct selection of option B.

Water first moves up the stem inside the xylem vessels (3). From the xylem in the leaf, water is drawn into the mesophyll cells by osmosis (4). Water then evaporates from the wet cell walls of these mesophyll cells into the air spaces inside the leaf (2). Finally, water vapour diffuses out of the leaf through the stomata into the atmosphere (1). Thus, the correct pathway sequence is 3 → 4 → 2 → 1.

1 mark: Correct sequence identification of option A.

In the female reproductive system, fertilization of the egg by a sperm normally occurs in the oviduct. The prostate gland produces seminal fluid, but sperm is produced and stored in the testes. Progesterone and estrogen are produced by the ovaries, not the uterus. The male urethra carries both urine and semen.

1 mark: Correct match of structure and function (option A).

The fetal blood contains metabolic waste products such as carbon dioxide and urea, which must diffuse across the placenta into the maternal blood to be excreted. Conversely, the fetus requires oxygen and nutrients (like glucose and amino acids) for aerobic respiration and growth, which diffuse from the mother's blood into the fetal blood.

1 mark: Correctly identifying the direction of diffusion for wastes and nutrients (option A).

In humans, anaerobic respiration occurs in muscle cells during vigorous exercise and produces lactic acid only. In yeast, anaerobic respiration (fermentation) produces ethanol (alcohol) and carbon dioxide. Anaerobic respiration in both organisms does not require oxygen and releases significantly less energy per glucose molecule than aerobic respiration.

1 mark: Correctly distinguishing the products of anaerobic respiration in humans and yeast (option A).

The atria (right and left) have very thin walls because they only pump blood to the ventricles. The left ventricle has the thickest muscle wall because it must generate high pressure to pump blood to the entire body (systemic circulation). The right ventricle only pumps blood a short distance to the lungs (pulmonary circulation) under lower pressure.

1 mark: Correctly identifying the thinnest/thickest walls and the correct physiological reason (option A).

At low temperatures (\(10^\circ\text{C}\) and \(20^\circ\text{C}\)), the kinetic energy of the molecules is low, leading to fewer effective collisions and a slower rate of reaction. The enzyme is NOT denatured at low temperatures (ruling out A). At \(30^\circ\text{C}\), the molecules have more kinetic energy than at lower temperatures, resulting in more frequent collisions and a higher rate of product formation (ruling out B and D). At \(60^\circ\text{C}\), the high thermal energy breaks the bonds maintaining the tertiary structure of the enzyme, permanently changing the shape of the active site (denaturation). This prevents the substrate from binding, so no product is formed.

1 mark for the correct option.
- Accept C: identify that high temperature (\(60^\circ\text{C}\)) denatures the enzyme by altering the active site's shape.
- Reject A: low temperature inactivates but does not denature enzymes.
- Reject B: higher temperatures increase kinetic energy, not decrease it.
- Reject D: rate of collisions is higher at the higher temperature (\(30^\circ\text{C}\)).

Pepsin is a protease enzyme produced in the stomach, meaning its optimum pH is acidic (around pH 2.0). Like most human enzymes, its optimum temperature is body temperature (\(37^\circ\text{C}\)). When pepsin active sites are fully functional, they digest the insoluble egg white proteins into small, soluble peptides, turning the cloudy mixture clear. At \(85^\circ\text{C}\), the enzyme is denatured regardless of pH. At pH 8, pepsin is denatured or inactive because the pH is too alkaline.

1 mark for the correct option.
- Accept A: pH 2 is the optimum pH for stomach pepsin, and \(37^\circ\text{C}\) is the optimum temperature.
- Reject B and D: \(85^\circ\text{C}\) will denature the enzyme.
- Reject C: pH 8 is too alkaline for pepsin to function efficiently.

A warm, dry (low humidity), and windy (moving air) day increases the water potential gradient between the inside of the leaf and the atmosphere, which significantly increases the rate of transpiration. Since a higher rate of transpiration means water is being taken up much faster, the potometer bubble will travel the fixed distance of \(20\text{ mm}\) in the shortest amount of time. Therefore, low humidity, moving air, and the shortest time (15 seconds) is the correct combination.

1 mark for the correct option.
- Accept D: relates low humidity and moving air to the highest transpiration rate, hence the shortest time (15 s).
- Reject A and B: high humidity slows down transpiration, resulting in longer times.
- Reject C: still air decreases the rate of transpiration compared to moving air.

During transpiration, water is delivered to the leaf via the xylem vessels in the veins. It then travels into the mesophyll cells, evaporating from the wet cell walls into the intracellular air spaces as water vapour. Finally, this water vapour diffuses out of the leaf down a water potential gradient through the stomatal pores into the atmosphere.

1 mark for the correct option.
- Accept A: correctly sequences the pathway from xylem to mesophyll cell walls, into air spaces, and out through stomata.
- Reject B, C, and D: describe incorrect physiological pathways of water movement.

The scrotum is a sac of skin holding the testes outside the abdominal cavity, which keeps them at a temperature roughly \(2^\circ\text{C}\) to \(3^\circ\text{C}\) below core body temperature. This lower temperature is required for healthy sperm development. The prostate gland secretes prostate fluid (it does not store sperm, which is stored in the epididymis/testes). The testes produce sperm and testosterone. In males, the urethra carries both urine and semen at different times.

1 mark for the correct option.
- Accept B: scrotum's function is temperature regulation for spermatogenesis.
- Reject A: the prostate gland does not store sperm.
- Reject C: the testis produces sperm, whereas accessory glands secrete seminal fluid.
- Reject D: the male urethra is a common pathway for both urine and semen.

Fertilisation (the fusion of the nuclei of the male gamete and female gamete to form a zygote) normally occurs inside the oviduct (fallopian tube). The fertilised egg undergoes cell division as it travels down the oviduct to become a blastocyst. It then implants itself into the endometrium, which is the lining of the uterus wall.

1 mark for the correct option.
- Accept A: fertilisation in the oviduct and implantation in the endometrium.
- Reject B: implantation does not occur in the ovary.
- Reject C: fertilisation does not typically occur inside the main uterine cavity.
- Reject D: fertilisation does not occur in the vagina, and implantation is specific to the endometrium layer.

Aerobic respiration releases energy in cells by breaking down nutrient molecules in the presence of oxygen. This released energy is used to power energy-requiring metabolic processes inside the organism, such as active transport, protein synthesis, cell division, growth, and muscle contraction.

1 mark for the correct option.
- Accept B: correctly identifies active transport and cell division as processes powered by respiratory energy.
- Reject A: glucose is a substrate of respiration, not a product.
- Reject C: respiration does not store all energy as thermal energy in the lungs, nor is it entirely lost.
- Reject D: animal tissues synthesize glycogen, not starch, and respiration is not used to split water into gases in this manner.

Deoxygenated blood returns from the body via the vena cava into the right atrium. It passes through the tricuspid valve into the right ventricle, which pumps it via the pulmonary artery to the lungs. Oxygenated blood returns from the lungs via the pulmonary vein into the left atrium, moves into the left ventricle, and is pumped into the aorta to the rest of the body.

1 mark for the correct option.
- Accept A: correct sequential pathway of blood through chambers, lungs, and major vessels.
- Reject B: blood cannot flow directly from right atrium to left atrium.
- Reject C: the vena cava enters the right atrium first, not the left.
- Reject D: blood flows from atrium to ventricle, not ventricle to atrium.

Enzymes are proteins (not carbohydrates), so statement 1 is incorrect. They are biological catalysts that speed up biochemical reactions and have an active site that is complementary in shape to a specific substrate, so statements 2 and 3 are correct. Extremely high temperatures alter the shape of the active site (denaturation) by breaking weak hydrogen bonds, but they do not break all covalent peptide bonds or cause the enzyme to dissolve, making statement 4 incorrect.

1 mark for the correct option B. Options A, C, and D are incorrect as they include incorrect statements (1 or 4).

The rate of reaction is inversely proportional to the time taken for the substrate to be broken down. Since the shortest time (1.5 minutes) was recorded at pH 7, the rate of reaction is highest at this pH. We cannot conclude that the optimum pH is exactly 7.0 because intermediate values (like pH 6.8 or 7.2) were not tested. The enzyme is not completely denatured at pH 3 because starch was still broken down. Increasing pH from 7 to 9 increases the time taken, meaning it decreases the rate of reaction.

1 mark for the correct option C. Option A is incorrect because we cannot be certain the exact optimum is 7.0. Option B is incorrect because there is still reaction activity. Option D is incorrect because rate decreases from pH 7 to 9.

The bubble moves the fastest when the rate of water uptake (and therefore transpiration) is highest. High temperatures increase the kinetic energy of water molecules, increasing evaporation. Low humidity increases the water potential gradient between the inside of the leaf and the external air. High wind speed removes saturated air from the leaf surface, maintaining a steep diffusion gradient. Therefore, high temperature, low humidity, and high wind speed produce the fastest rate.

1 mark for the correct option B. Options A, C, and D represent conditions that would reduce the rate of transpiration compared to B.

Water enters root hair cells and travels across the root cortex into the xylem by osmosis down a water potential gradient. It is pulled up the xylem vessels in the stem as a continuous column due to cohesion and tension (transpiration pull). Finally, water vapor diffuses out of the stomata from the air spaces inside the leaf to the atmosphere.

1 mark for the correct option A. Options B, C, and D describe incorrect mechanisms for one or more steps of the pathway.

The placenta prevents the mixing of maternal and fetal blood to protect the fetus from high maternal blood pressure and maternal immune responses (1). It also facilitates the exchange of waste materials (urea, carbon dioxide) from fetal blood to maternal blood (3). The amniotic fluid (inside the amniotic sac), not the placenta, protects the fetus from mechanical shock (2).

1 mark for the correct option B. Options A, C, and D are incorrect because they include function 2, which is performed by the amniotic fluid.

Aerobic respiration completely oxidizes glucose to carbon dioxide and water, releasing a high yield of energy (about 36-38 ATP per glucose molecule). Anaerobic respiration in humans only partially breaks down glucose to lactic acid, yielding a much smaller amount of energy (2 ATP per glucose molecule). Therefore, statement B is correct.

1 mark for the correct option B. Option A is incorrect because human anaerobic respiration produces lactic acid, not alcohol. Option C is incorrect because respiration does not require carbon dioxide. Option D is incorrect because anaerobic respiration occurs entirely in the cytoplasm.

Germinating seeds are alive and respire aerobically, releasing carbon dioxide gas. Carbon dioxide turns limewater cloudy. Boiled seeds are dead and cannot respire; they do not produce carbon dioxide, so the limewater in Flask Y remains clear.

1 mark for the correct option A. Options B, C, and D are incorrect because they fail to correctly predict carbon dioxide production only in the living setup (Flask X).

Oxygenated blood from the lungs returns to the left side of the heart via the pulmonary vein (P). It is received by the left atrium and passes into the left ventricle (Q). From the left ventricle, it is pumped under high pressure through the aorta (R) to be distributed to the body tissues. After delivering oxygen, deoxygenated blood returns from the body tissues to the right atrium via the vena cava (S).

1 mark for the correct option A. Options B, C, and D represent incorrect pathways or structures mismatched with the circulatory route.

At high temperatures such as $55\text{ }^{\circ}\text{C}$, enzyme molecules gain too much kinetic energy, causing the weak bonds holding their three-dimensional shape together to break. This denatures the enzyme, permanently changing the shape of its active site so that the substrate (starch) can no longer fit. Starch is a polysaccharide and does not denature. At $10\text{ }^{\circ}\text{C}$, low kinetic energy explains the slow rate, not at $55\text{ }^{\circ}\text{C}$.

B is correct (1 mark). A is incorrect as kinetic energy is high at 55 °C. C is incorrect as starch does not denature. D is incorrect as enzymes lowering activation energy increases reaction rate, and denaturation stops the reaction because the active site is lost.

Transpiration is the loss of water vapour from plant leaves by evaporation and diffusion. High temperature increases the kinetic energy of water molecules, increasing evaporation from cell walls. High wind speed sweeps away water vapour from the leaf surface, and low humidity ensures a steep concentration gradient between the air spaces inside the leaf and the external air. This combination (low humidity, high temperature, high wind speed) maximizes the diffusion rate of water vapour out of the stomata.

C is correct (1 mark). A, B and D contain factors (such as high humidity, low temperature or low wind speed) that decrease the rate of transpiration.

The scrotum is a sac of skin that holds the testes outside the main body cavity, ensuring sperm are kept at a temperature slightly below $37\text{ }^{\circ}\text{C}$ which is ideal for healthy sperm development. The testes (not prostate gland) produce sperm and testosterone. The sperm duct carries sperm from the testes to the urethra. The bladder stores urine.

B is correct (1 mark). A is incorrect as testes produce sperm, not the prostate gland. C is incorrect as sperm duct/urethra path is misdescribed. D is incorrect as the bladder stores urine.

Anaerobic respiration in yeast produces ethanol and carbon dioxide:
$$\text{glucose} \rightarrow \text{ethanol} + \text{carbon dioxide}$$
In human muscle cells, anaerobic respiration produces only lactic acid:
$$\text{glucose} \rightarrow \text{lactic acid}$$
Therefore, yeast produces carbon dioxide while muscle cells do not. Both pathways release the same small amount of energy per glucose molecule because glucose is only partially broken down. Neither process requires any oxygen.

B is correct (1 mark). A is reversed. C is incorrect as both release similar small amounts of energy per glucose. D is incorrect as both are completely anaerobic.

The left ventricle has a much thicker muscular wall than the right ventricle because it must pump blood at high pressure throughout the entire body (systemic circulation), whereas the right ventricle only pumps blood at lower pressure to the lungs (pulmonary circulation). The left side of the heart receives oxygenated blood from the lungs via the pulmonary vein and pumps it out, so the blood in the left ventricle is oxygenated.

A is correct (1 mark). B, C, and D are incorrect because the left ventricle is thicker than the right, and the left side contains oxygenated blood while the right side contains deoxygenated blood.

A significant change in pH (either an increase or a decrease) away from the optimum pH alters the charges on the amino acids making up the enzyme. This disrupts the ionic and hydrogen bonds that maintain the specific three-dimensional shape of the active site, causing the enzyme to denature. As a result, the substrate can no longer fit into the active site, and the reaction stops. pH changes do not affect kinetic energy, nor do they hydrolyse peptide bonds into individual amino acids under normal physiological conditions.

B is correct (1 mark). A is incorrect because temperature (not pH) determines kinetic energy. C is incorrect because denaturation does not break primary peptide bonds into individual amino acids. D is incorrect as denaturation increases activation energy by making the catalyst non-functional.

Water is transported to the leaf through xylem vessels. It then moves into the mesophyll cells, evaporating from the wet surface of the mesophyll cell walls into the intercellular air spaces of the spongy mesophyll. Finally, the water vapour diffuses out of the leaf through the stomata into the surrounding atmosphere.

A is correct (1 mark). B, C, and D list the steps in incorrect or reversed physiological order.

Sperm cells are highly motile (using their flagellum/tail to swim), produced in millions daily by the testes, and are extremely small. In contrast, egg cells (ova) are non-motile (moved along the oviduct by cilia and peristalsis), typically released only once a month from the ovaries, and are much larger because they contain a yolk/food store in their cytoplasm to sustain the early embryo.

A is correct (1 mark). B, C, and D contain incorrect statements regarding mobility, relative numbers produced, or relative size.

At lower temperatures, the kinetic energy of both the enzyme and substrate molecules is low, which reduces the rate of successful collisions and thus decreases the rate of reaction. However, low temperature does not denature the enzyme, so it remains functional and its active site is undamaged.

1 mark for the correct option (B).

Enclosing the shoot in a clear plastic bag increases the humidity around the leaves. This reduces the concentration gradient of water vapor between the inside of the leaf and the external air, thereby reducing the rate of transpiration and subsequent water uptake.

1 mark for the correct option (A).

Glucose, oxygen, and antibodies are useful substances that are transferred from the maternal blood to the fetal blood across the placenta to support the growth, respiration, and passive immunity of the fetus. Urea and carbon dioxide are excretory products that move in the opposite direction.

1 mark for the correct option (B).

During vigorous exercise, human muscle cells respire anaerobically. This pathway produces lactic acid as the only product (no carbon dioxide is produced) and releases a relatively low amount of energy per glucose molecule compared to aerobic respiration because glucose is only partially broken down.

1 mark for the correct option (B).

Deoxygenated blood in the right ventricle is pumped into the pulmonary artery to be carried to the lungs. After gas exchange occurs in the capillaries of the lungs, the oxygenated blood returns to the left atrium of the heart via the pulmonary vein.

1 mark for the correct option (B).

Enzymes are highly specific because the shape of their active site is complementary to the shape of their specific substrate molecule, allowing them to bind and form an enzyme-substrate complex.

1 mark for the correct option (A).

Water moves from the xylem vessels into the surrounding spongy mesophyll cells, evaporates from their wet cell walls into the intercellular air spaces, and then diffuses as water vapor through the stomata into the atmosphere.

1 mark for the correct option (A).

FSH is secreted by the pituitary gland. Its primary role in the menstrual cycle is to stimulate the growth and development of follicles in the ovaries.

1 mark for the correct option (A).

As the temperature rises, the kinetic energy of both the enzyme and substrate molecules increases. This leads to faster molecular movement, causing more frequent and energetic collisions. Consequently, the rate of successful enzyme-substrate complex formation increases, which raises the overall rate of reaction.

1 mark for selecting A. Reject other choices because temperature does not change enzyme specificity (B), does not raise activation energy (C), and denaturation (D) reduces activity rather than increasing it.

Pepsin is an acidic protease found in the stomach. It functions optimally at an acidic pH (provided by hydrochloric acid) and at normal body temperature (37 degrees C). Boiling denatures enzymes, and pepsin does not digest starch.

1 mark for choosing A. Reject B because sodium hydroxide is alkaline and denatures/inactivates pepsin. Reject C because boiling denatures the enzyme. Reject D because pepsin acts on protein, not starch.

All enzymes are proteins (statement 1) and act as biological catalysts (statement 3). They are highly sensitive to pH changes (statement 2 is false) and are not used up in the reactions they catalyse, allowing them to be reused (statement 4 is false).

1 mark for A. Reject options containing statements 2 or 4 as they are incorrect statements regarding the general properties of enzymes.

Transpiration pull is the primary driver of water uptake. High humidity decreases the concentration gradient of water vapour between the leaf interior and the air. Low wind speed allows water vapour to accumulate near the stomata, further reducing diffusion. Low light causes stomata to close, restricting water vapour exit. Thus, this combination leads to the slowest rate of transpiration and water uptake.

1 mark for A. Correctly identifying that high humidity, low wind speed, and low light intensity each independently lower the rate of transpiration.

During transpiration, water evaporates from the surfaces of the mesophyll cells into the air spaces of the leaf, and then diffuses out of the leaf through the stomatal pores into the surrounding atmosphere.

1 mark for A. Reject B because active transport does not move water vapour, reject C because phloem transports sucrose and amino acids, reject D because root hairs absorb water rather than releasing it during transpiration.

Progesterone is secreted by the corpus luteum (the remains of the follicle after ovulation). Its primary role is to maintain and further vascularise the lining of the uterus in preparation for potential implantation of an embryo.

1 mark for A. Reject B because estrogen is responsible for initially rebuilding the endometrium before ovulation. Reject C (FSH) and D (LH) as they regulate follicle development and trigger ovulation, respectively, but do not maintain the uterus lining.

In yeast, anaerobic respiration (fermentation) produces ethanol and carbon dioxide. In human muscle cells, anaerobic respiration produces only lactic acid (lactate) and does not release carbon dioxide.

1 mark for A. Reject options B, C, and D because they associate incorrect metabolic products with anaerobic pathways in yeast or humans (e.g., suggesting muscle cells produce carbon dioxide or water during anaerobic respiration).

The vena cava is the main vein that returns deoxygenated blood from the upper and lower body tissues directly into the right atrium of the heart.

1 mark for A. Reject B because the pulmonary artery carries deoxygenated blood away from the right ventricle to the lungs. Reject C because the pulmonary vein carries oxygenated blood to the left atrium. Reject D because the aorta carries oxygenated blood away from the left ventricle.

At \(40^\circ\text{C}\), the starch disappears in the shortest time (2 minutes), meaning the rate of reaction is highest. Therefore, the optimum temperature lies in the range surrounding \(40^\circ\text{C}\), which is between \(30^\circ\text{C}\) and \(50^\circ\text{C}\). At \(50^\circ\text{C}\), the enzyme is active but working slower (15 minutes), so it is not completely denatured. At \(60^\circ\text{C}\), starch is not broken down, likely due to complete denaturation.

1 mark for identifying the correct statement. C is correct because the shortest time for starch breakdown (2 minutes) is at \(40^\circ\text{C}\), which falls within the range of \(30^\circ\text{C}\) to \(50^\circ\text{C}\).

According to the lock-and-key hypothesis, the active site of the enzyme has a specific shape that is complementary to its substrate, allowing them to fit together. The enzyme itself remains chemically unchanged after the reaction, and high temperatures will denature the active site, disrupting the complementary fit.

1 mark for identifying that the active site is complementary to the substrate. B is correct.

Transpiration rate is highest when the concentration gradient of water vapour between the inside of the leaf and the external air is steepest. Warm temperature increases the rate of evaporation; dry air decreases external humidity; and wind moves water vapour away from the leaf surface, preventing it from accumulating. Therefore, Shoot 1 under warm, dry, and windy conditions will transpire fastest, resulting in the fastest water uptake and bubble movement.

1 mark for identifying Shoot 1 as the correct option. A is correct.

The continuous column of water in the xylem is maintained by cohesion (water molecules sticking to one another due to hydrogen bonding) and adhesion (water molecules sticking to the cellulose walls of the xylem vessels). This allows the transpiration pull to draw water upwards without the column breaking.

1 mark for selecting B. Cohesion and adhesion are the key physical forces maintaining the transpiration stream.

Useful materials and protective substances like oxygen, glucose, amino acids, mineral ions, and antibodies diffuse from the mother's blood to the fetus's blood. Waste products such as carbon dioxide and urea diffuse in the opposite direction, from the fetus to the mother.

1 mark for identifying the group containing only substances diffusing maternal-to-fetal. B is correct.

Sperm are highly motile due to their flagellum (tail). Eggs are much larger, non-motile (sessile), and rely on the cilia and muscular peristalsis of the oviduct to move towards the uterus. Sperm are small and produced in millions; eggs are large, contain nutrient stores, and are typically released one per month.

1 mark for identifying row B as the correct comparison of gametes.

The left ventricle pumps blood into the aorta to supply the systemic circulation (the whole body), which has high resistance and requires high blood pressure. The right ventricle only pumps blood to the lungs (pulmonary circulation), which is much closer and has lower resistance, requiring less pressure.

1 mark for selecting B. The thickness of the left ventricle wall is directly related to generating high systemic pressure.

In humans, anaerobic respiration produces lactic acid only (no carbon dioxide is released). In yeast, anaerobic respiration (fermentation) produces ethanol (alcohol) and carbon dioxide.

1 mark for identifying the correct products. A is correct.

Raising the temperature of the reaction mixture increases the kinetic energy of both the enzyme and substrate molecules. Consequently, they move faster, resulting in more frequent collisions per unit time with sufficient energy to overcome the activation energy barrier (successful collisions).

1 mark for selecting the correct option A. Incorrect options are B, C, and D.

An extreme change in pH alters the ionic charges on the amino acids making up the enzyme. This disrupts the chemical bonds holding the tertiary structure of the protein together, particularly at the active site. This changes the shape of the active site (denaturation), meaning the substrate can no longer fit.

1 mark for selecting the correct option B. Incorrect options are A, C, and D.

Transpiration rate is lowest under conditions of high humidity (which decreases the water vapor concentration gradient between the leaf interior and the atmosphere), low temperature (which decreases the rate of evaporation of water), and still air (which allows water vapor to accumulate around the stomata, further reducing the concentration gradient).

1 mark for selecting the correct option B. Incorrect options are A, C, and D.

The transpiration stream begins with the evaporation of water from the damp cell walls of mesophyll cells into the air spaces of the leaf, followed by diffusion through the stomata. This creates a water potential gradient that draws water from the leaf xylem. Due to cohesion (the attraction between water molecules), this pulling force (tension) is transmitted all the way down the water column in the xylem vessels.

1 mark for selecting the correct option A. Incorrect options are B, C, and D.

The placenta allows passive and active exchange of substances. Useful substances such as oxygen, glucose, amino acids, vitamins, and maternal antibodies pass from the maternal blood to the fetal blood. Metabolic waste products such as carbon dioxide and urea pass from the fetal blood to the maternal blood for excretion.

1 mark for selecting the correct option A. Incorrect options are B, C, and D.

Germinating seeds carry out aerobic respiration, in which they consume oxygen and produce carbon dioxide. Since the carbon dioxide produced is immediately absorbed by the chemical in the tube (such as potassium hydroxide), the net volume of gas inside the test tube decreases because the oxygen used is not replaced by free gaseous carbon dioxide. This decrease in gas volume decreases the internal pressure, causing the colored liquid drop to move toward the seeds.

1 mark for selecting the correct option C. Incorrect options are A, B, and D.

During anaerobic respiration in yeast (alcoholic fermentation), glucose is broken down to produce ethanol and carbon dioxide: \(\text{glucose} \rightarrow \text{ethanol} + \text{carbon dioxide}\). In contrast, anaerobic respiration in human muscle cells produces only lactic acid: \(\text{glucose} \rightarrow \text{lactic acid}\). Neither process requires oxygen, and both yield a very small amount of energy compared to aerobic respiration.

1 mark for selecting the correct option A. Incorrect options are B, C, and D.

Deoxygenated blood returns from the body tissues to the heart through the vena cava. It enters the right atrium, passes through the tricuspid valve into the right ventricle, and is then pumped out through the semi-lunar valve into the pulmonary artery, which carries it to the lungs for oxygenation.

1 mark for selecting the correct option B. Incorrect options are A, C, and D.

The rate of enzyme activity is inversely proportional to the time taken for the substrate (starch) to be completely broken down. A shorter time indicates a faster rate of reaction. Since pH 7 took the shortest time (30 seconds), the rate of enzyme activity was highest at pH 7.

Award 1 mark for identifying C (pH 7) as the correct option because it represents the shortest reaction time, corresponding to the highest rate of enzyme activity.

When an enzyme is heated significantly above its optimum temperature, the high thermal energy breaks the bonds maintaining its tertiary structure. This alters the shape of the active site permanently (denaturation), meaning the substrate can no longer fit.

Award 1 mark for selecting B because denaturation changes the active site shape, preventing substrate binding.

Warm, sunny, and well-ventilated (windy) conditions increase the rate of transpiration. Wind moves water vapor away from the leaf surface, keeping the humidity of the surrounding air low. This maintains a steep concentration gradient of water vapor between the inside of the leaf and the outside air, increasing diffusion.

Award 1 mark for option A, correctly linking moving air (wind/ventilation) with the removal of water vapor to maintain a steep diffusion gradient.

A potometer measures the rate of water uptake by the leafy shoot. While water uptake is closely correlated with transpiration (water loss), a small fraction of the absorbed water is used in photosynthesis and to maintain turgor pressure within plant cells. Therefore, the direct measurement is water absorption/uptake.

Award 1 mark for option C. Distinguish between water uptake (measured directly) and actual water loss via transpiration (estimated).

The prostate gland produces seminal fluid, which contains nutrients to nourish and help transport sperm. Option B is incorrect because sperm are stored in the epididymis/sperm duct, not the testis. Option C is incorrect because the urethra, not the sperm duct, carries both urine and semen. Option D is incorrect because the testis, not the urethra, produces testosterone and sperm.

Award 1 mark for option A, which accurately describes the function of the prostate gland according to the syllabus.

Aerobic respiration completely oxidizes glucose to carbon dioxide and water, releasing a relatively large amount of energy (around 30-32 ATP equivalents). Anaerobic respiration in humans only partially breaks down glucose to lactic acid, releasing a much smaller amount of energy (2 ATP equivalents).

Award 1 mark for option B, reflecting the comparative energy yields of aerobic and anaerobic respiration in human muscle tissue.

The left ventricle has the thickest muscular wall of all four chambers because it must generate enough high pressure to pump blood to the entire body. It pumps oxygenated blood directly into the aorta.

Award 1 mark for A. The left ventricle has a thicker wall than the right ventricle and pumps blood into the aorta.

Boiling denatures the enzyme by destroying the specific shape of its active site. This denaturation is irreversible, so even after cooling back down, the enzyme cannot catalyze the reaction. This demonstrates that enzymes are proteins denatured by high temperatures.

Award 1 mark for option B. Boiling causing irreversible loss of activity demonstrates thermal denaturation of protein enzymes.

At temperatures above the optimum, the increased kinetic energy causes bonds holding the enzyme's specific 3D structure to break. This changes the shape of the active site (denaturation), meaning the substrate can no longer fit into it, and the rate of reaction drops rapidly.

Award 1 mark for identifying that the active site denatures (changes shape) at high temperatures, preventing substrate binding.

Because only substance Y was digested, the experiment demonstrates enzyme specificity. This is due to the complementary shapes of the substrate and the enzyme's active site (the lock-and-key hypothesis).

Award 1 mark for recognizing that selectivity among different substrates demonstrates enzyme specificity.

Transpiration is fastest when temperature is high (increasing the kinetic energy and rate of evaporation of water molecules), humidity is low (maintaining a steep water potential gradient between the inside of the leaf and the atmosphere), and wind speed is high (moving water vapor away from the leaf surface to maintain the gradient).

Award 1 mark for selecting the combination of high temperature, low humidity, and high wind speed.

Water evaporates from the wet cell walls of mesophyll cells into the air spaces of the leaf, and then diffuses out through the stomata. This loss of water creates a tension (pull) that is transmitted down the continuous column of water in the xylem, drawing water upwards (transpiration pull).

Award 1 mark for identifying evaporation of water from mesophyll cell surfaces as the direct driver of the transpiration pull.

Fertilisation normally takes place in the oviduct. Implantation of the embryo occurs in the lining of the uterus (endometrium). Progesterone is secreted by the corpus luteum in the ovary during the menstrual cycle and early pregnancy. Therefore, row A is correct.

Award 1 mark for correctly matching fertilisation to the oviduct, implantation to the uterus lining, and progesterone secretion to the ovary.

Sperm cells are produced in the testes (testis). They then travel along the sperm duct (vas deferens), where fluids are added by glands, and finally pass out of the body through the urethra.

Award 1 mark for identifying the pathway testis to sperm duct to urethra.

During vigorous exercise, human muscle cells can undergo anaerobic respiration when oxygen supply is insufficient. This process breaks down glucose partially to produce lactic acid, releasing only a small fraction of the energy (2 ATP per glucose) compared to aerobic respiration (which fully oxidizes glucose to CO2 and H2O, releasing a large amount of energy).

Award 1 mark for identifying that anaerobic respiration in human muscle cells produces lactic acid and a small amount of energy.

When the left ventricle relaxes (diastole), the blood pressure in the aorta is higher than in the ventricle. To prevent blood from flowing backward into the left ventricle, the semi-lunar valve at the base of the aorta closes.

Award 1 mark for identifying the semi-lunar valve as the valve preventing backflow from the aorta into the left ventricle.

(a) The volume of oxygen produced increases as temperature rises from 10 degrees C to 40 degrees C, where it reaches a maximum of 38.0 cm3. Above 40 degrees C, the volume of oxygen produced decreases rapidly, reaching 0.0 cm3 at 60 degrees C. (b) At 40 degrees C, the molecules have higher kinetic energy and move faster, leading to more frequent successful collisions between catalase and hydrogen peroxide, forming more enzyme-substrate complexes. At 60 degrees C, the enzyme is denatured because high temperatures break the bonds maintaining its shape, causing the active site to change shape so that the substrate can no longer fit. (c) Concentration of hydrogen peroxide substrate, and pH of the reaction mixture.

(a) Max 4 marks: activity/volume increases as temperature increases up to 40 degrees C (1); maximum activity / optimum temperature is at 40 degrees C / 38.0 cm3 (1); activity decreases above 40 degrees C (1); no activity / 0 cm3 at 60 degrees C (1). (b) Max 5 marks: at 40 degrees C, higher kinetic energy of molecules (1); faster molecular movement (1); more frequent successful collisions / more enzyme-substrate complexes formed (1); at 60 degrees C, high temperature denatures the enzyme (1); active site changes shape (1); substrate can no longer fit/bind (1). (c) Max 2.4 marks: concentration of hydrogen peroxide (1.2); volume of hydrogen peroxide (1.2); volume of potato extract/catalase (1.2); pH of the mixture (1.2); surface area of potato tissue if raw tissue was used (1.2).

(a) Starch is completely broken down into maltose by amylase, so there is no starch left to react with iodine. (b) At pH 2, the acidic conditions denature amylase. The active site changes shape, meaning starch can no longer bind. Starch remains undigested, so it reacts with iodine to turn blue-black. (c) Test at smaller pH intervals around the optimum, such as every 0.2 units between pH 6.0 and pH 7.0. (d) Amylase has a specific active site shape that is complementary only to the starch substrate. Proteins have a different shape that does not fit into amylase's active site.

(a) Max 2 marks: starch is fully broken down/hydrolysed (1); maltose does not react with iodine (1). (b) Max 4 marks: extreme pH denatures amylase (1); active site changes shape (1); substrate/starch cannot fit into the active site (1); starch remains undigested/present to react with iodine (1). (c) Max 2 marks: test at smaller/narrower intervals of pH (1); e.g., pH 6.2, 6.4, 6.6, 6.8 (1). (d) Max 3.4 marks: enzyme has a specific 3D shape/active site (1.2); active site is complementary to starch substrate (1.1); protein has a different shape and cannot fit into the active site/cannot form enzyme-substrate complexes (1.1).

(a) Water is absorbed from the soil into root hair cells by osmosis. It moves across the root cortex into the xylem vessels. Water is pulled upwards through the xylem in the stem by a transpiration pull to the mesophyll cells of the leaf. (b) Percentage increase = ((90 - 45) / 45) * 100 = 100%. (c) Wind moves water vapour away from the leaf surface. This maintains a steep water vapour concentration gradient between the inside of the leaf and the outside air, increasing the rate of diffusion of water vapour through the stomata.

(a) Max 4 marks: water enters root hair cells by osmosis (1); moves across cortex cells (1); enters xylem vessels (1); pulled up xylem due to transpiration pull/cohesion (1). (b) Max 3 marks: working showing increase in distance (90 - 45 = 45) (1); formula (increase / original) * 100 (1); correct answer: 100% (1). (c) Max 4.4 marks: wind blows away water vapour/moist air from the leaf surface (1.1); maintains a steep concentration gradient (1.1); between inside of leaf and outside air (1.1); increases rate of diffusion through stomata (1.1).

(a) Transpiration is the loss of water vapour from plant leaves by evaporation of water at the surfaces of the mesophyll cells followed by diffusion of water vapour through the stomata. (b) Guard cells take up water by osmosis and become turgid, which causes them to curve outward and open the pore because their inner cell walls are thicker than their outer walls. When water is lost, guard cells become flaccid, causing them to straighten and close the pore. (c) 1. Thick waxy cuticle, which acts as a barrier to reduce evaporation from the leaf surface. 2. Rolled leaves, which trap moist air to reduce the water vapour concentration gradient. 3. Hairs on leaves, which trap moisture near the stomata to reduce the rate of diffusion.

(a) Max 2 marks: loss of water vapour from leaves (1); evaporation at mesophyll surface followed by diffusion through stomata (1). (b) Max 4 marks: guard cells take up water by osmosis (1); guard cells become turgid (1); inner walls are thicker/less flexible than outer walls causing bending (1); when guard cells lose water they become flaccid and close (1). (c) Max 5.4 marks: award 1.8 marks for each correct adaptation paired with explanation: thick waxy cuticle reduces evaporation (1.8); rolled leaves trap humid air / reduce concentration gradient (1.8); hairy leaves trap moisture / reduce concentration gradient (1.8); sunken stomata trap water vapour (1.8); needles/spines reduce surface area (1.8).

(a) (i) Ovary, (ii) Oviduct (or Fallopian tube), (iii) Uterus (lining). (b) Sperm cells (male gametes) are much smaller than egg cells (female gametes), which contain a large food store. Sperm cells are motile because they have a flagellum to swim, whereas egg cells are non-motile. Millions of sperm are produced continuously, whereas usually only one egg is released per month. (c) The amniotic sac encloses the amniotic fluid. The amniotic fluid acts as a shock absorber to protect the developing fetus from physical damage, maintains a constant temperature, and allows the fetus to move freely to develop muscles and joints.

(a) Max 3 marks: (i) ovary (1); (ii) oviduct / Fallopian tube (1); (iii) uterus (lining) / endometrium (1). (b) Max 5 marks: sperm is smaller / egg is larger (1); sperm has flagellum/tail / is motile (1); egg is non-motile (1); sperm produced in millions / egg released singly/monthly (1); egg contains nutrient/yolk store (1). (c) Max 3.4 marks: amniotic sac encloses/protects the fluid (1.1); amniotic fluid acts as a shock absorber/protects against mechanical shock (1.1); maintains constant temperature / prevents dehydration / allows movement (1.2).

(a) Estrogen stimulates the repair and rebuilding of the uterus lining after menstruation. Progesterone maintains the thickness of the uterus lining, making it vascularised and ready for implantation of a fertilized egg. (b) FSH stimulates the maturation of a follicle in the ovary and stimulates the ovary to secrete estrogen. LH triggers ovulation (the release of the mature egg) and stimulates the development of the remaining follicle tissue into the corpus luteum. (c) If fertilization does not occur, the corpus luteum breaks down, causing a sharp drop in progesterone and estrogen levels. Without these hormones, the uterus lining cannot be maintained, so it breaks down and is shed during menstruation.

(a) Max 4 marks: estrogen repairs/rebuilds uterus lining (1); makes lining thick (1); progesterone maintains lining (1); prepares lining for implantation / increases blood vessels (1). (b) Max 4 marks: FSH stimulates follicle development/egg maturation (1); FSH stimulates estrogen release (1); LH triggers ovulation (1); LH stimulates corpus luteum formation (1). (c) Max 3.4 marks: uterus lining breaks down/sheds/menstruation occurs (1.1); caused by degeneration of corpus luteum (1.1); leading to decrease in progesterone/estrogen levels (1.2).

(a) C6H12O6 + 6O2 -> 6CO2 + 6H2O. (b) Aerobic respiration requires oxygen, whereas anaerobic respiration occurs in the absence of oxygen. Both processes use glucose as a reactant. Aerobic respiration produces carbon dioxide and water, while anaerobic respiration in yeast produces ethanol and carbon dioxide. Aerobic respiration releases a large amount of energy, whereas anaerobic respiration releases a much smaller amount. (c) Bubble the gas produced by the yeast through limewater. If carbon dioxide is present, the limewater will turn from clear to cloudy/milky.

(a) Max 3 marks: reactants: C6H12O6 and O2 (1); products: CO2 and H2O (1); correct balancing: C6H12O6 + 6O2 -> 6CO2 + 6H2O (1). (b) Max 5 marks: aerobic requires oxygen / anaerobic does not (1); both use glucose (1); aerobic products are carbon dioxide and water (1); anaerobic products in yeast are ethanol and carbon dioxide (1); aerobic releases much more energy / anaerobic releases less energy (1). (c) Max 3.4 marks: use limewater / calcium hydroxide solution (1.2); pass gas through / bubble gas into it (1.1); positive result: turns cloudy / milky / forms precipitate (1.1).

(a) The left ventricle must pump blood under high pressure to the entire body, which is a much longer distance with higher resistance. The right ventricle only pumps blood to the lungs, which is close to the heart and requires much lower pressure. (b) Valves ensure blood flows in one direction only. When the ventricles contract, the atrioventricular valves close to prevent blood flowing back into the atria. When the ventricles relax, the semilunar valves close to prevent blood flowing back into the ventricles from the aorta and pulmonary artery. (c) Double circulation means that for every complete circuit of the body, blood passes through the heart twice. The advantage is that it maintains high blood pressure to the body tissues, which ensures faster delivery of oxygen and nutrients and rapid removal of waste products.

(a) Max 3 marks: left ventricle pumps blood to the rest of the body / systemic circulation (1); needs to generate higher pressure / overcome more resistance (1); right ventricle only pumps to lungs / pulmonary circulation which is low pressure (1). (b) Max 4 marks: valves prevent backflow / ensure one-way flow of blood (1); atrioventricular valves close during ventricular contraction (1); semilunar valves close during ventricular relaxation (1); pressure of blood pushes valves open/shut (1). (c) Max 4.4 marks: blood passes through heart twice for one full circuit (1.1); pulmonary and systemic systems (1.1); advantage: higher pressure maintained in systemic circulation (1.1); faster delivery of oxygen/nutrients to tissues / higher metabolic rate supported (1.1).

(a) An enzyme is a protein molecule that serves as a biological catalyst. It increases the rate of specific metabolic reactions without being permanently altered or consumed in the process.

(b)(i) According to Table 1.1, as temperature increases from 20 °C to 40 °C, the volume of apple juice collected increases from 12.0 \(\text{cm}^3\) to 31.0 \(\text{cm}^3\). The optimum temperature is 40 °C, where the maximum volume of 31.0 \(\text{cm}^3\) is obtained. Above 40 °C (from 50 °C to 60 °C), the volume of juice decreases sharply, falling to 5.0 \(\text{cm}^3\) at 60 °C.

(b)(ii) Between 40 °C and 60 °C, the thermal energy is high enough to break the chemical bonds (e.g., hydrogen bonds) that maintain the specific three-dimensional shape of the pectinase enzyme. This leads to denaturation of the enzyme. The active site permanently changes its shape, which means the substrate (pectin) is no longer complementary and cannot bind. This prevents the formation of enzyme-substrate complexes, slowing down or halting the digestion of cell walls, resulting in less juice being released.

(c) Control variables in this experiment include the pH of the reaction mixture, the concentration of the pectinase enzyme, the volume of the pectinase solution added, and the incubation time (which must be exactly 15 minutes for all trials).

(a)
- 1 mark: protein;
- 1 mark: biological catalyst / speeds up reactions. [Max 2]

(b)(i)
- 1 mark: volume of juice increases as temperature increases from 20 °C to 40 °C / peak at 40 °C;
- 1 mark: volume of juice decreases from 40 °C to 60 °C;
- 1 mark: correct comparative data point quoted from Table 1.1 with units (e.g., peak of 31.0 \(\text{cm}^3\) at 40 °C or minimum of 5.0 \(\text{cm}^3\) at 60 °C). [Max 3]

(b)(ii)
- 1 mark: (at high temperatures) enzyme is denatured;
- 1 mark: thermal energy breaks internal chemical bonds;
- 1 mark: active site changes shape;
- 1 mark: substrate (pectin) can no longer fit into the active site / no enzyme-substrate complexes form. [Max 4]

(c)
- 1.2 marks for each correct variable (up to 2): pH, concentration of enzyme, volume of enzyme, incubation time, source/type of apple. [Max 2.4]

(a) Amylase is a digestive enzyme that catalyses the hydrolysis of starch (the substrate) into the reducing sugar maltose (the product).

(b) The lock and key hypothesis states that the active site of an enzyme represents the 'lock' and the substrate represents the 'key'. The active site possesses a highly specific, three-dimensional shape that is complementary only to the shape of its specific substrate molecule. When they collide, the substrate fits precisely into the active site to form an enzyme-substrate complex. Since other substances have different chemical structures and shapes, they cannot fit into the active site, meaning the enzyme cannot catalyse other reactions.

(c) Enzymes have an optimum pH at which they function most efficiently. Extreme changes in pH (either highly acidic or highly alkaline) disrupt the delicate ionic and hydrogen bonds that hold the enzyme protein in its specific tertiary conformation. This causes the enzyme to denature, permanently altering the shape of the active site. Because the shape is altered, the substrate is no longer complementary, preventing the formation of enzyme-substrate complexes and reducing the reaction rate to zero.

(d) Amylase is secreted in the mouth by the salivary glands, where the pH is neutral (approx. pH 7), or by the pancreas into the duodenum (small intestine), where the pH is slightly alkaline (approx. pH 8) due to the secretion of sodium hydrogencarbonate.

(a)
- 1 mark: starch (as substrate);
- 1 mark: maltose (as product; accept maltose / glucose / reducing sugars). [Max 2]

(b)
- 1 mark: active site has a specific 3D shape;
- 1 mark: shape of active site is complementary to substrate;
- 1 mark: reference to lock and key analogy (substrate fits into active site like a key in a lock);
- 1 mark: only the correct substrate can bind to form an enzyme-substrate complex. [Max 4]

(c)
- 1 mark: enzymes have an optimum pH;
- 1 mark: extreme pH alters charges / disrupts hydrogen and ionic bonds;
- 1 mark: enzyme is denatured / active site changes shape;
- 0.4 marks: substrate can no longer bind / fit into active site. [Max 3.4]

(d)
- 1 mark: mouth / salivary glands OR pancreas / small intestine / duodenum;
- 1 mark: matching pH (neutral / pH 7 for mouth OR alkaline / pH 8 for pancreas/duodenum). [Max 2]

(a) Water travels up the xylem and exits into the surrounding mesophyll cells of the leaf via osmosis. On reaching the spongy mesophyll, water evaporates from the wet cell walls into the internal intercellular air spaces, turning into water vapour. This water vapour then diffuses down a concentration gradient from the air spaces out through open stomatal pores into the drier atmosphere.

(b) Increased wind speed significantly increases the rate of transpiration. In still air, water vapour diffusing out of the stomata accumulates on the leaf surface, forming a localized region of high humidity (boundary layer). This reduces the concentration gradient. Wind blows away this humid air layer, replacing it with drier air. This maintains a steep water vapour concentration gradient between the internal air spaces and the external atmosphere, allowing water vapour to diffuse out of the stomata at a faster rate.

(c) Plants in dry habitats (xerophytes) have several leaf adaptations to conserve water:
1. A thick, waxy cuticle on the epidermis provides a waterproof barrier, preventing non-stomatal cuticular evaporation.
2. Sunken stomata located in pits create microclimates that trap moist, humid air next to the stomata, reducing the concentration gradient.
3. Rolled or curled leaves enclose the stomata in a humid inner chamber, shielding them from wind.
4. Spines or needles instead of broad leaves dramatically reduce the total surface area available for transpiration.

(a)
- 1 mark: water moves from xylem into mesophyll cells (by osmosis);
- 1 mark: water evaporates from mesophyll cell walls into intercellular air spaces;
- 1 mark: water vapour diffuses out through stomata. [Max 3]

(b)
- 1 mark: increases rate of transpiration;
- 1 mark: wind blows away accumulated water vapour / removes the boundary layer;
- 1 mark: maintains a steep water vapour concentration gradient;
- 1.4 marks: between the air spaces inside the leaf and the external air, resulting in a faster rate of diffusion out of the stomata. [Max 4.4]

(c)
- 1 mark each for any 4 described adaptations (up to 4):
- thick, waxy cuticle to act as a waterproof barrier;
- sunken stomata / stomata in pits to trap moist air / reduce concentration gradient;
- hairs on leaves to trap moisture / block wind;
- rolled/curled leaves to trap humid air / shield stomata;
- leaves reduced to spines/needles to reduce surface area for transpiration;
- stomata closed during the day (CAM-like physiology). [Max 4]

(a) Transpiration is defined as the loss of water vapour from plant leaves, caused by the evaporation of water at the wet surfaces of the mesophyll cells, followed by the diffusion of this water vapour through the stomatal pores into the atmosphere.

(b)(i) Species C is most likely to be a xerophyte. The data shows that Species C has zero stomata (0 per \(\text{mm}^2\)) on its upper epidermis, which is directly exposed to sunlight and heat, and concentrates all of its 145 stomata per \(\text{mm}^2\) on the lower epidermis. Having stomata restricted to the shaded, cooler lower epidermis reduces water loss from transpiration. Species D has substantial numbers of stomata on both surfaces (90 and 95), which is typical of plants in wetter habitats.

(b)(ii) Guard cells flank each stoma and regulate its opening. When water is plentiful, guard cells actively pump in ions, causing water to enter them by osmosis. This makes the guard cells swell and become turgid. Because the inner cell walls of the guard cells (bordering the pore) are thick and less elastic than the outer walls, the cells bow outward, pulling the stomatal pore open. Under dry conditions or in darkness, guard cells lose water, become flaccid, and the outer walls relax back, closing the pore to prevent water loss.

(c) Environmental factors affecting transpiration include:
1. Temperature: An increase in temperature increases the kinetic energy of water molecules, leading to faster evaporation and diffusion, increasing transpiration.
2. Light intensity: An increase in light intensity stimulates guard cells to open stomata wider for photosynthesis, increasing transpiration. (Alternatively, humidity: an increase in humidity decreases the concentration gradient, decreasing transpiration).

(a)
- 1 mark: loss of water vapour from leaves / plant parts;
- 1 mark: evaporation of water at the surfaces of mesophyll cells;
- 1 mark: diffusion of water vapour through the stomata. [Max 3]

(b)(i)
- 1 mark: Species C;
- 1.4 marks: correct justification referencing data (0 stomata on upper epidermis vs 145 on lower epidermis) explaining that placing stomata on the lower, shaded surface minimizes water loss by solar heating. [Max 2.4]

(b)(ii)
- 1 mark: guard cells take up water by osmosis / become turgid;
- 1 mark: inner wall is thicker / less elastic than outer wall;
- 1 mark: turgid guard cells bend / curve outwards to open the stomatal pore;
- 1 mark: loss of water makes guard cells flaccid, closing the pore. [Max 4]

(c)
- 1 mark for each factor named with its correct effect (up to 2):
- Light intensity: increase increases transpiration rate;
- Temperature: increase increases transpiration rate;
- Humidity: increase decreases transpiration rate. [Max 2]

(a) Fertilisation is the fusion of the haploid nucleus of a male gamete (sperm) with the haploid nucleus of a female gamete (ovum/egg) to produce a diploid zygote.

(b)(i) The amniotic sac is a tough, flexible membrane that grows to enclose the developing embryo and fetus. Its primary function is to contain the amniotic fluid and act as a physical barrier that prevents pathogens from entering from the vagina.

(ii) The amniotic fluid acts as a critical support system:
- It absorbs physical shocks and impact, protecting the fetus from mechanical damage.
- It maintains a constant, warm temperature surrounding the fetus.
- It prevents the fetal skin from adhering to the amniotic sac and allows free fetal movement, which is essential for muscular and skeletal development.

(c) The placenta is highly adapted to facilitate fast and efficient exchange:
1. Large Surface Area: The placenta contains thousands of highly folded, finger-like projections called chorionic villi, which maximize the surface area available for diffusion.
2. Short Diffusion Distance: The membrane separating maternal and fetal blood is extremely thin (only one or two layers of cells thick), meaning gases and nutrients can diffuse rapidly.
3. Concentration Gradient Maintenance: Fetal capillaries inside the villi bring blood high in carbon dioxide and low in oxygen close to maternal blood pools (lacunae) that are high in oxygen and low in carbon dioxide. This close proximity ensures rapid exchange down steep gradients.
4. Active Transport: Cells of the placental membrane are rich in transport proteins and contain many mitochondria to generate the ATP necessary for the active transport of essential substances like amino acids and glucose.

(a)
- 1 mark: fusion of nuclei of male gamete / sperm and female gamete / egg (ovum);
- 1 mark: produces a zygote. [Max 2]

(b)(i)
- 1.4 marks: membrane that encloses/contains the amniotic fluid and embryo / protects embryo from mechanical damage / prevents entry of pathogens. [Max 1.4]

(b)(ii)
- 1 mark each for any two of (up to 2):
- absorbs physical shock / cushions fetus from external impacts;
- maintains constant temperature;
- allows movement (vital for muscle/joint development);
- prevents adhesion of fetus to sac. [Max 2]

(c)
- 1 mark for each adaptation explained (up to 6):
- (chorionic) villi provide a very large surface area;
- thin membrane / barrier (only 1-2 cells thick) provides short diffusion distance;
- rich blood supply on both sides / fetal capillaries and maternal blood pools close together;
- blood flows in opposite directions / counter-current mechanism maintains steep concentration gradients;
- presence of transport/carrier proteins in membranes for active transport;
- abundant mitochondria to release energy/ATP for active transport (of glucose/amino acids). [Max 6]

(a) Coronary arteries branch off the aorta to supply the cardiac muscle cells (myocardium) with oxygen, glucose, and other nutrients needed for continuous aerobic respiration, while also removing waste products like carbon dioxide.

(b) The left ventricle has a much thicker muscular wall because it must generate high pressure to pump blood throughout the systemic circulation (to the entire body, excluding the lungs). This requires overcoming greater resistance over a longer distance. Conversely, the right ventricle has a thinner wall because it only pumps blood to the lungs (pulmonary circulation), which is a short distance. High pressure in the pulmonary system would damage the delicate capillaries in the lungs and cause fluid buildup in alveoli.

(c) Coronary heart disease (CHD) occurs when fatty substances, such as cholesterol, build up inside the walls of the coronary arteries, forming plaques (atherosclerosis). This narrows the lumen of the arteries, restricting blood flow and reducing the oxygen supply to the heart muscle cells. This can lead to chest pain (angina) or a heart attack (myocardial infarction). Lifestyle changes to reduce risk include: eating a diet low in saturated fats/cholesterol, engaging in regular cardiovascular exercise, quitting smoking, and reducing stress.

(d) During physical exercise, muscle cells contract more frequently and strongly, requiring more energy. This increases the rate of aerobic respiration in the muscles. To meet this demand, the body must supply more oxygen and glucose, and remove carbon dioxide and lactic acid more rapidly. The brain detects the elevated carbon dioxide concentration in the blood and signals the pacemaker of the heart to increase heart rate, pumping blood faster.

(a)
- 1 mark: supply blood/oxygen/glucose to the cardiac muscle;
- 1 mark: for aerobic respiration / muscle contraction. [Max 2]

(b)
- 1 mark: left ventricle pumps blood to the whole body / systemic circulation;
- 1 mark: right ventricle pumps blood only to the lungs / pulmonary circulation;
- 1 mark: left ventricle needs to generate much higher pressure to overcome resistance;
- 0.4 marks: right ventricle requires lower pressure to prevent damage to lung capillaries. [Max 3.4]

(c)
- 1 mark: narrowing/blockage of coronary arteries due to cholesterol / fatty plaque buildup;
- 1 mark: results in reduced blood flow / oxygen delivery to heart muscle cells;
- 1 mark for any two correct lifestyle changes (0.5 marks each): stop smoking, low-cholesterol diet, regular exercise, lose weight, reduce stress. [Max 3]

(d)
- 1 mark: heart rate increases;
- 1 mark: muscles require more energy / respire faster during exercise;
- 1 mark: faster delivery of oxygen / glucose OR faster removal of carbon dioxide. [Max 3]

To carry out this investigation:
1. Prepare five test-tubes, each containing 5 \(\text{cm}^3\) of a different buffer solution (pH 3.0, 5.0, 7.0, 9.0, and 11.0).
2. Use a cork borer to cut potato cylinders, then use a ruler and scalpel on a cutting tile to measure and cut them to exactly 2.0 cm in length. This standardizes the surface area and enzyme concentration.
3. Add one potato cylinder to the first buffer tube.
4. Add 5 \(\text{cm}^3\) of \(1.0\%\) hydrogen peroxide solution to the tube and quickly seal it with a rubber bung connected by a delivery tube to a gas syringe.
5. Start the stopwatch immediately and record the volume of oxygen gas collected in the syringe after exactly 3 minutes.
6. Repeat steps 3-5 for all pH buffer solutions.
7. Keep the temperature constant by placing all tubes in a thermostatically controlled water bath set at \(25^\circ\text{C}\).
8. Set up a control experiment using boiled and cooled potato cylinders in pH 7.0 buffer to show that no gas is produced when the catalase is denatured.
9. Perform three replicates for each pH to check for reliability and calculate mean volumes of gas.
10. Safety: Wear safety goggles as hydrogen peroxide is a tissue irritant, and use the scalpel with care, cutting away from fingers.

Award marks for:
- **Independent Variable (2 marks):** Using at least 5 different pH buffers (pH 3, 5, 7, 9, 11) to vary pH.
- **Dependent Variable (2 marks):** Measuring the volume of gas produced in a set time (e.g., 3 minutes) using a gas syringe or measuring cylinder over water.
- **Standardisation of Source (2 marks):** Use of a cork borer to maintain constant cylinder diameter and a ruler/scalpel to cut to equal length (e.g., 2.0 cm).
- **Controlled Variables (2 marks):** Keeping temperature constant (using a water bath) and maintaining constant concentration/volume of hydrogen peroxide.
- **Control Experiment (2 marks):** Using boiled potato cylinders to show that gas production ceases when the enzyme is denatured.
- **Reliability (2 marks):** Specifying that the experiment must be repeated at least three times at each pH to calculate a mean and identify anomalies.
- **Safety (1.3 marks):** Wearing safety goggles to protect against corrosive/irritant peroxide and using a cutting tile/cutting away from the body with the scalpel.

a) Table of results:
| Temperature / \(^\circ\text{C}\) | Time to complete starch digestion / \(\text{s}\) | Rate of digestion / \(\text{s}^{-1}\) |
|---|---|---|
| 10 | 480 | 2.08 |
| 20 | 240 | 4.17 |
| 30 | 120 | 8.33 |
| 40 | 60 | 16.67 |
| 50 | 180 | 5.56 |
| 60 | >600 | 0.00 |

b) Trend: The rate of starch digestion increases with temperature up to a maximum rate of 16.67 \(\text{s}^{-1}\) at \(40^\circ\text{C}\). Above this temperature, the rate decreases sharply to 5.56 \(\text{s}^{-1}\) at \(50^\circ\text{C}\) and falls to 0 at \(60^\circ\text{C}\).

c) Explanation:
- From \(10^\circ\text{C}\) to \(40^\circ\text{C}\), higher temperatures increase the kinetic energy of both amylase and starch molecules, resulting in faster movement and more frequent successful collisions.
- Above \(40^\circ\text{C}\), the high temperature causes the active site of amylase to lose its specific shape (denaturation) due to the rupture of weak hydrogen bonds. The substrate (starch) can no longer fit into the active site, stopping the reaction.

d) Errors and Improvements:
1. Error: Time interval between adding mixture to iodine is too large, leading to inaccurate measurements. Improvement: Decrease the testing interval from every 30 or 60 seconds to every 10 seconds.
2. Error: Subjective determination of the end-point color of iodine. Improvement: Use a colorimeter to provide quantitative measurements of color change.

Award marks for:
- **Table Construction (2 marks):** Clear table with columns, including correct units in headers: Temperature / \(^\circ\text{C}\), Time / \(\text{s}\), and Rate / \(\text{s}^{-1}\) (or arbitrary units).
- **Calculations (2 marks):** Correctly calculated rate values: 2.08, 4.17, 8.33, 16.67, 5.56, 0.00.
- **Trend Description (2 marks):** Identifying optimum rate at \(40^\circ\text{C}\), increasing rate below optimum, and decreasing rate above optimum.
- **Kinetic Explanation (2.3 marks):** Linking low-to-medium temperatures with kinetic energy, molecular movement, and collisions.
- **Structural Explanation (2 marks):** Explaining denaturation (loss of active site shape) above the optimum temperature.
- **Errors and Improvements (3 marks):** Identifying two distinct valid errors (such as subjective endpoint color or fluctuating temperature) with corresponding improvements (use of colorimeter, use of controlled water bath, or narrower testing intervals).

To conduct this investigation:
1. Cut the leafy shoot under water at an angle to prevent air bubbles from entering the xylem vessel, which would block water movement.
2. Assemble the bubble potometer under water and insert the shoot's stem into the tube. Apply vaseline/petroleum jelly to the joins to make the system completely airtight.
3. Dry the leaves with paper towels to ensure stomata are not blocked by liquid water.
4. Introduce an air bubble into the capillary tube by briefly lifting the tube out of the water, then replacing it.
5. Position a fan at a fixed distance (e.g., 50 cm) directly facing the shoot. Use the fan speed settings: 0 (No Fan), Low, Medium, and High to vary the wind speed.
6. For each setting, allow the system to adapt for 5 minutes, then record the distance the bubble travels along the capillary tube over a 5-minute period using a ruler and stopwatch.
7. Use the reservoir to reset the bubble to the start of the scale.
8. Repeat the measurement three times for each wind speed setting to calculate a mean value.
9. Control variables: Ensure room temperature is kept constant (monitored via a thermometer), keep light intensity constant by maintaining the same distance to any bench lamps, and ensure the same leafy shoot is used throughout.

Award marks for:
- **Airtight Setup (3 marks):** Cutting shoot under water at an angle, sealing connections with petroleum jelly, and drying leaves to prevent blocked stomata.
- **Independent Variable (2 marks):** Varying wind speed using different fan settings (low, medium, high) and including a control with no fan, keeping distance constant.
- **Dependent Variable (2 marks):** Measuring the distance moved by the bubble in a set time interval (e.g., 5 minutes) using a ruler and stopwatch.
- **Controlled Variables (3 marks):** Controlling temperature, light intensity, and using the same shoot throughout to keep surface area constant.
- **Reliability (1.3 marks):** Repeating measurements at each setting three times using the reservoir to reset the bubble, and calculating mean rates.
- **Safety (2 marks):** Standard laboratory safety precautions, such as keeping water away from the electrical fan and handling scalpel cutting implements with care.

a) Calculating percentage mass loss is necessary because leaves naturally vary in their initial mass and size. Comparing absolute mass differences directly would be misleading, as a heavier leaf might lose more water simply because it has a larger overall surface area. Percentage mass loss normalizes the data.

b) Calculations:
- Leaf A: \(\frac{4.20 - 4.16}{4.20} \times 100 = \frac{0.04}{4.20} \times 100 = 0.95\%\)
- Leaf B: \(\frac{4.50 - 3.82}{4.50} \times 100 = \frac{0.68}{4.50} \times 100 = 15.11\%\)
- Leaf C: \(\frac{4.10 - 3.98}{4.10} \times 100 = \frac{0.12}{4.10} \times 100 = 2.93\%\)
- Leaf D: \(\frac{4.40 - 3.61}{4.40} \times 100 = \frac{0.79}{4.40} \times 100 = 17.95\%\)

c) Conclusions:
- The vast majority of stomata are located on the lower surface of the leaf.
- This is supported by the fact that sealing the upper surface while keeping the lower surface open (Leaf B) resulted in a massive percentage mass loss of \(15.11\%\), which is almost as high as when both sides were open (Leaf D, \(17.95\%\)).
- Conversely, when the lower surface was sealed and the upper surface was left open (Leaf C), water loss was extremely low (\(2.93\%\)), indicating very few functional stomata exist on the upper surface.
- The small water loss in Leaf A (\(0.95\%\)) represents residual cuticular transpiration since all stomata were fully sealed.

Award marks for:
- **Why use Percentage (2 marks):** Leaves have different starting masses/sizes. Percentage mass loss standardizes the comparison to ensure a fair test.
- **Calculations (4 marks):** 1 mark for each correct percentage value (A: 0.95%, B: 15.11%, C: 2.93%, D: 17.95%) accompanied by clear intermediate working (mass loss calculation).
- **Stomata Distribution Conclusion (3 marks):** Concluding that there are significantly more stomata on the lower surface than the upper surface.
- **Data Evidence (3 marks):** Comparing the high rate of mass loss in Leaf B (lower surface open) with the very low rate of mass loss in Leaf C (upper surface open).
- **Control validation (1.3 marks):** Using Leaf A to show that petroleum jelly is highly effective at preventing transpirational water loss.

a) Purpose of paraffin oil: Paraffin oil is highly hydrophobic and forms a physical barrier on top of the aqueous solution. This prevents atmospheric oxygen from diffusing into the yeast suspension, forcing the yeast to respire anaerobically rather than aerobically.

b) Gas test: Collect the gas produced using a syringe or delivery tube and bubble it through clear, colorless calcium hydroxide solution (limewater). If the gas is carbon dioxide, a white precipitate of calcium carbonate forms, making the limewater turn cloudy/milky.

c) Relationship: From \(0\%\) to \(6\%\) glucose concentration, there is a direct, linear increase in the volume of carbon dioxide gas produced (from \(0.1\text{ cm}^3\) to \(8.0\text{ cm}^3\)). Between \(6\%\) and \(8\%\) glucose concentration, the volume of gas levels off, plateauing at approximately \(8.1\text{ cm}^3\).

d) Explanation: Between \(6\%\) and \(8\%\) glucose, the curve plateaus because glucose is in excess and is no longer the limiting factor for the reaction. The rate is restricted by other factors, such as the fixed concentration of yeast enzymes, meaning all enzyme active sites are fully saturated with substrate. Alternatively, other factors like temperature or the toxic accumulation of ethanol might be limiting.

Award marks for:
- **Paraffin Oil Purpose (2 marks):** Excludes oxygen / prevents oxygen dissolution; ensures anaerobic conditions are maintained.
- **CO2 Test Method (2 marks):** Bubbling the gas into limewater (calcium hydroxide).
- **CO2 Test Observation (1.3 marks):** Limewater turning from clear to cloudy, milky, or chalky.
- **Trend Description - Linear phase (2 marks):** Explaining that increasing glucose concentration up to \(6\%\) leads to a proportional increase in gas production.
- **Trend Description - Plateau (2 marks):** Noting that gas production levels off / remains constant between \(6\%\) and \(8\%\) glucose.
- **Plateau Explanation (4 marks):** Glucose is no longer the limiting factor; enzyme active sites are fully saturated; enzyme concentration or yeast density has become the limiting factor.

a) Calculation of magnification:
1. Convert the measured image size to micrometers: \(65\text{ mm} \times 1000 = 65000\ \mu\text{m}\).
2. Apply the formula: \(\text{Magnification} = \frac{\text{Image Size}}{\text{Actual Size}}\).
3. Calculate: \(\text{Magnification} = \frac{65000}{50} = \times 1300\).

b) Constant variables:
- Temperature (must be maintained at physiological temperature, \(37^\circ\text{C}\), using an incubator or heated stage).
- Concentration of nutrients / glucose in the buffer (sperm require an energy source to swim).
- Initial concentration of sperm cells and exposure time (exactly 2 hours).

c) Relationship to the reproductive tract:
- The optimum pH is 7.0 (neutral), showing the highest percentage of motile sperm (\(80\%\)).
- The vagina is naturally acidic (pH 4.0–5.0) to inhibit pathogen growth, which is hostile to sperm cells (as shown by only \(5\%\) survival at pH 5.0).
- Semen is alkaline (pH 7.2–8.0), neutralizing the vaginal acid to create a neutral/alkaline microenvironment (near pH 7.0–8.0) that ensures sperm survival and high motility for fertilization.

d) Control experiment: Incubate sperm in a standard, physiological saline solution (or nutrient medium without chemical buffers) adjusted to pH 7.0 using very dilute hydrochloric acid or sodium hydroxide. If the sperm display similarly high motility, it proves the buffer salts were not toxic.

Award marks for:
- **Magnification Conversion (1 mark):** Correctly converting 65 mm to \(65000\ \mu\text{m}\).
- **Magnification Formula (1 mark):** Stating or using \(\text{Magnification} = \text{Image Size} / \text{Actual Size}\).
- **Magnification Value (2 marks):** Correct calculation of \(\times 1300\) (allow error carried forward from incorrect measurements).
- **Controlled Variables (2 marks):** Identifying temperature (kept at \(37^\circ\text{C}\)) and concentration of glucose/nutrients as crucial controls.
- **Optimum pH Identification (1 mark):** Stating that pH 7.0 is the optimum.
- **Vaginal Environment Connection (2 marks):** Explaining that the vagina is acidic to prevent infections, which kills sperm.
- **Semen Neutralisation Connection (1.3 marks):** Explaining that alkaline semen neutralizes this acid, raising the local pH to the sperm's optimal neutral range.
- **Control Setup (3 marks):** Describing a test using an unbuffered saline/medium at pH 7.0 to confirm the buffers were non-toxic.