2025 Cambridge IGCSE Biology (0610) 模擬試題連答案詳解

At high temperatures, the thermal energy causes the weak bonds holding the enzyme's tertiary structure to break. This denatures the enzyme, altering the shape of the active site so that the substrate is no longer complementary and cannot bind.

1 mark for the correct option (A).

In the dark, plants close their stomata to conserve water, which dramatically cuts down transpiration. Additionally, high humidity reduces the water vapor concentration gradient between the inside of the leaf and the surrounding air, further decreasing the rate of diffusion.

1 mark for the correct option (C).

The left ventricle pumps blood through the systemic circulation to the entire body, which presents a much higher resistance than the pulmonary circulation to the lungs. Consequently, the left ventricle requires a thicker muscle wall to generate the higher force and pressure needed.

1 mark for the correct option (B).

In yeast, anaerobic respiration (fermentation) produces ethanol (alcohol) and carbon dioxide. In human muscle cells, anaerobic respiration produces lactic acid only, with no carbon dioxide produced.

1 mark for the correct option (B).

The prostate gland secretes seminal fluid, which contains nutrients (like sugars) and enzymes that nourish, protect, and assist the sperm in swimming.

1 mark for the correct option (A).

Diffusion is the net movement of particles from a region of their higher concentration to a region of their lower concentration down a concentration gradient. Gas exchange in the lungs occurs via diffusion.

1 mark for the correct option (B).

Adrenaline is the fight-or-flight hormone. It increases heart rate to deliver more oxygen and glucose to active muscles, and dilates pupils to allow more light in, improving visual awareness.

1 mark for the correct option (D).

Energy transfer between trophic levels is inefficient (usually about 10%). Energy is lost at each step through metabolic processes (respiration as heat), excretion, and because not all parts of the organism are eaten or digested.

1 mark for the correct option (C).

High temperatures break the weak bonds holding the enzyme's three-dimensional structure together, denaturing the enzyme. This permanently changes the shape of the active site so that the substrate can no longer fit.

Award 1 mark for choosing B: explanation of denaturation and change in the active site shape.

Lipase breaks down fats (lipids) into fatty acids and glycerol. Amylase breaks down starch into maltose (not amino acids). Protease breaks down proteins into amino acids (not simple sugars). Maltase breaks down maltose into glucose.

Award 1 mark for choosing B: correct identification of lipase, fats, and fatty acids and glycerol.

High humidity decreases the concentration gradient of water vapour between the inside and outside of the leaf. Low wind speed allows water vapour to accumulate around the stomata, and low temperature reduces the kinetic energy of water molecules, reducing evaporation.

Award 1 mark for choosing A: high humidity, low wind speed, and low temperature.

Water evaporates from the damp cell walls of the spongy mesophyll into the intercellular air spaces of the leaf, and then diffuses out of the leaf through the stomata down a concentration gradient.

Award 1 mark for choosing A: correct pathway from mesophyll cell walls, to air spaces, and out through stomata.

The sperm duct (vas deferens) is responsible for carrying sperm from the testes to the urethra. The prostate gland secretes seminal fluid (not storing sperm). The scrotum holds the testes outside the body cavity to keep them cool. The testes produce sperm and testosterone.

Award 1 mark for choosing C: correct pairing of sperm duct and its transport function.

Glucose and oxygen diffuse down their concentration gradients from the maternal blood into the fetal capillaries. Carbon dioxide and urea move in the opposite direction. Red blood cells and large proteins cannot pass through the placental barrier.

Award 1 mark for choosing B: correct diffusion of glucose and oxygen into fetal capillaries.

The left ventricle must pump blood through the systemic circulation to the entire body, which has high resistance, and therefore requires higher pressure. The right ventricle only pumps blood to the lungs, which are nearby and at much lower pressure.

Award 1 mark for choosing C: left ventricle wall generates the high pressure needed to pump blood to all parts of the body except the lungs.

In yeast, anaerobic respiration (fermentation) produces ethanol (alcohol) and carbon dioxide. In human muscle cells, anaerobic respiration produces lactic acid only, with no carbon dioxide produced.

Award 1 mark for choosing B: Yeast produces alcohol and carbon dioxide; Muscle produces lactic acid only.

As temperature increases towards the optimum, the molecules gain kinetic energy and move faster. This increases the frequency of successful collisions between the active sites of enzymes and substrate molecules, thereby increasing the rate of reaction.

Award 1 mark for the correct option (C). Reject all other options.

The rate of transpiration is lowest under conditions of high humidity (which reduces the water vapour concentration gradient between the leaf interior and the atmosphere), still air (which allows water vapour to accumulate around the stomata), and low temperature (which reduces the rate of evaporation from mesophyll cells).

Award 1 mark for the correct option (A). Reject all other options.

Sperm cells are much smaller, motile (using a flagellum to swim), and can carry either an X or a Y sex chromosome. Egg cells are larger, non-motile (stationary), and always carry an X chromosome.

Award 1 mark for the correct option (C). Reject all other options.

Aerobic respiration completely breaks down glucose in the presence of oxygen, releasing a large amount of energy, and producing carbon dioxide and water as products.

Award 1 mark for the correct option (C). Reject all other options.

Oxygenated blood returns from the lungs via the pulmonary vein into the left atrium. It then passes into the left ventricle before being pumped out to the body through the aorta.

Award 1 mark for the correct option (B). Reject all other options.

The 'lock and key' hypothesis states that the active site of a specific enzyme has a precise, complementary shape that fits only one specific substrate molecule, allowing them to bind.

Award 1 mark for the correct option (B). Reject all other options.

The evaporation of water vapour from leaves during transpiration creates a tension (pull) that is transmitted down the continuous column of water molecules in the xylem, drawing water up from the roots.

Award 1 mark for the correct option (C). Reject all other options.

Anaerobic respiration in yeast (alcoholic fermentation) produces alcohol (ethanol) and carbon dioxide. Anaerobic respiration in human muscle cells (lactic acid fermentation) produces lactic acid only.

Award 1 mark for the correct option (B). Reject all other options.

High temperatures cause the weak bonds maintaining the three-dimensional structure of the enzyme protein to break, altering the shape of the active site irreversibly. This process is called denaturation. Consequently, the substrate (starch) can no longer fit into the active site.

1 mark for correct option C.

Pepsin (the protease enzyme found in the mammalian stomach) has an optimum pH of around 2 (highly acidic) and operates best at body temperature (around \( 37^\circ\text{C} \)). Boiling the enzyme denatures it, rendering it completely inactive.

1 mark for correct option A.

High humidity decreases the water vapour concentration gradient between the air spaces inside the leaf and the external atmosphere, reducing diffusion. Low temperature decreases the kinetic energy of water molecules, reducing evaporation. Still air allows a boundary layer of humid air to accumulate around the stomata, which further lowers the rate of transpiration.

1 mark for correct option B.

Water is first absorbed from the soil by root hair cells via osmosis. It then moves across the root cortex cells before entering the xylem vessels in the center of the root, which transport it up the stem.

1 mark for correct option B.

In human males, the urethra is a single tube connecting to both the bladder (carrying urine) and the sperm ducts (carrying semen) to transport these fluids out of the body at different times.

1 mark for correct option C.

Glucose is a nutrient required by the fetus for respiration and growth, so it diffuses across the placenta from the maternal blood into the fetal blood. In contrast, carbon dioxide and urea are metabolic waste products that diffuse from the fetal blood into the maternal blood. Glycogen is too large to cross the placenta.

1 mark for correct option B.

Anaerobic respiration in human muscle cells produces lactic acid only. In yeast cells, anaerobic respiration (fermentation) produces ethanol and carbon dioxide.

1 mark for correct option C.

Deoxygenated blood entering the right atrium passes through the atrioventricular valve to enter the right ventricle. When the ventricle contracts, the blood is pumped through the semi-lunar valve into the pulmonary artery, which carries it to the lungs.

1 mark for correct option A.

Beyond the optimum temperature, high thermal energy breaks the bonds holding the enzyme's tertiary structure together. This causes the active site to lose its complementary shape to the substrate (denaturation), preventing substrate binding.

1 mark for selecting option B.

Gastric juice in the stomach contains hydrochloric acid, making it highly acidic (typically pH 1.5 to 2.0). Enzymes found here, like pepsin (a protease), are adapted to function optimally in highly acidic environments.

1 mark for selecting option A.

Transpiration is slowest when the concentration gradient of water vapour between the inside of the leaf and the outside air is lowest. High humidity decreases this gradient; low temperature decreases the kinetic energy of water molecules; still air allows water vapour to accumulate around stomata, further reducing the diffusion gradient.

1 mark for selecting option A.

Xylem vessels are reinforced with a tough, woody substance called lignin. This lignification provides mechanical strength, preventing the vessel walls from collapsing inward when water is pulled upward under high tension.

1 mark for selecting option B.

Glucose is a nutrient required by the fetus for respiration and growth, so it moves from maternal blood to fetal blood across the placenta. Carbon dioxide and urea are fetal waste products that move in the opposite direction. Glycogen is a large macromolecule stored in cells and does not diffuse across the placenta.

1 mark for selecting option B.

In human males, the urethra is a dual-purpose tube. It connects to both the urinary bladder (to excrete urine) and the sperm ducts (to ejaculate semen), carrying both fluids out of the body at different times.

1 mark for selecting option C.

Anaerobic respiration in yeast produces ethanol and carbon dioxide, whereas anaerobic respiration in human muscle cells produces only lactic acid. Thus, yeast produces carbon dioxide as a product, but human muscle cells do not.

1 mark for selecting option B.

The right ventricle only pumps blood a short distance to the lungs (pulmonary circulation), requiring less pressure. The left ventricle must pump blood throughout the entire body (systemic circulation), requiring much higher pressure to overcome vascular resistance over a longer distance. Thus, the left ventricle has a thicker muscular wall to contract with more force.

1 mark for selecting option B.

An increase in temperature from 20 °C to 35 °C increases the kinetic energy of both the enzyme and substrate molecules. This causes them to move faster, leading to more frequent and successful collisions between the substrates and the active sites of the enzymes, which increases the rate of reaction.

Award 1 mark for the correct answer B. Option A is incorrect because kinetic energy increases, not decreases. Option C is incorrect because human enzymes typically denature at temperatures above 40-45 °C. Option D is incorrect because enzymes lower the activation energy, and temperature does not alter this requirement.

The movement of the air bubble represents water uptake, which is closely related to the rate of transpiration. Transpiration is slowest in conditions of low temperature (less kinetic energy for water molecules), high humidity (decreased water vapour concentration gradient between the inside of the leaf and the outside air), and still air (allowing water vapour to accumulate on the leaf surface, further reducing the concentration gradient).

Award 1 mark for the correct answer B. Option A is incorrect because high temperature and wind increase transpiration. Option C is incorrect because low humidity and wind increase transpiration. Option D is incorrect because high temperature and low humidity increase transpiration.

The scrotum is the outer sac of skin containing the testes. Its function is to keep the testes at a temperature slightly below normal body temperature, which is optimal for the production of healthy sperm cells.

Award 1 mark for the correct answer B. Option A is incorrect because the testes produce sperm cells, whereas the prostate gland secretes seminal fluid. Option C is incorrect because the sperm duct transports sperm from the testes to the urethra, but does not produce seminal fluid. Option D is incorrect because the bladder stores urine, and the testes produce sperm.

When the ventricles contract (systole), the high pressure of the blood forces the atrioventricular valves to close to prevent blood flowing back into the atria. At the same time, this high pressure forces the semilunar valves to open, allowing blood to flow out of the heart into the aorta and pulmonary artery.

Award 1 mark for the correct answer C. Options A and B are incorrect because atrioventricular valves must close during ventricular contraction to prevent backflow into the atria. Option D is incorrect because semilunar valves must open to allow blood to exit the heart into the arteries.

In human muscle cells during vigorous exercise, anaerobic respiration breaks down glucose in the absence of oxygen to produce lactic acid, releasing a small amount of energy compared to aerobic respiration.

Award 1 mark for the correct answer B. Option A is incorrect because carbon dioxide is not produced in human muscles during anaerobic respiration. Option C is incorrect because ethanol and carbon dioxide are products of anaerobic respiration in yeast and plants, not human muscles. Option D is incorrect because anaerobic respiration occurs in the absence of oxygen.

The lock-and-key hypothesis states that the active site of an enzyme has a highly specific shape that is complementary to the shape of a specific substrate molecule. Only this specific substrate can fit into the active site to form an enzyme-substrate complex, much like a key fits into a specific lock.

Award 1 mark for the correct answer C. Option A is incorrect because the hypothesis describes a rigid complementary fit, not a flexible fit for any substrate. Option B is incorrect because enzymes are biological catalysts and are not destroyed in the reactions they catalyse. Option D is incorrect because enzymes are highly specific to their substrates.

Water is absorbed from the soil by root hair cells. It then travels across the root cortex into the xylem vessels. From the xylem vessels, it is transported upwards to the leaves, where it enters the mesophyll cells before evaporating and diffusing out of the leaf through the stomata.

Award 1 mark for the correct answer B. Option A is incorrect because mesophyll cells are located in the leaves, so water must reach the xylem before reaching these cells. Options C and D are incorrect because xylem is located after root hair cells in the pathway of water uptake.

Sperm cells are highly adapted for movement (motile) because they possess a flagellum (tail) that enables them to swim towards the egg cell. Egg cells, on the other hand, are non-motile and are swept along the oviduct by ciliated cells.

Award 1 mark for the correct answer C. Option A is incorrect because sperm cells are much smaller than egg cells. Option B is incorrect because egg cells do not possess a flagellum. Option D is incorrect because egg cells contain a large food store (yolk) to support the early embryo, whereas sperm cells do not.

At 65 °C, which is well above the optimum temperature for amylase, the shape of the enzyme's active site changes permanently. This denaturation prevents starch from binding to the active site, so no reaction occurs and starch remains present.

Award 1 mark for the correct option A. Reject any other options.

According to the lock and key hypothesis, the active site of the enzyme has a complementary shape to the specific substrate molecule, allowing them to fit together precisely.

Award 1 mark for the correct option A. Reject any other options.

Transpiration and water uptake are lowest when the concentration gradient of water vapour between the inside of the leaf and the external air is reduced. This occurs in high humidity, low temperature, and calm air.

Award 1 mark for the correct option A. Reject any other options.

Water leaves the xylem vessels in the leaf, moves into the mesophyll cells, evaporates from their wet walls into the air spaces of the spongy mesophyll, and then diffuses out of the leaf through the stomata.

Award 1 mark for the correct option A. Reject any other options.

Sperm are produced in the testes, travel along the sperm duct where they mix with fluids from the seminal vesicles/prostate gland, and are then ejaculated through the urethra.

Award 1 mark for the correct option A. Reject any other options.

Fertilisation (the fusion of the nuclei of the male and female gametes) occurs in the oviduct. The resulting zygote divides to form an embryo, which then travels to the uterus and implants into the lining.

Award 1 mark for the correct option A. Reject any other options.

Aerobic respiration fully oxidises glucose to carbon dioxide and water, releasing a relatively large amount of energy. Anaerobic respiration in human muscle cells only partially breaks down glucose to lactic acid, releasing much less energy per molecule.

Award 1 mark for the correct option B. Reject any other options.

The left ventricle must contract with more force to generate the high blood pressure needed to pump blood to all the organs of the body, whereas the right ventricle only pumps blood at a lower pressure to the nearby lungs.

Award 1 mark for the correct option A. Reject any other options.

Protease enzymes in the stomach, such as pepsin, are adapted to function in highly acidic conditions. Their optimum pH is around 1.5 to 2.0. Therefore, the highest rate of reaction will be observed at pH 2.0.

Award 1 mark for the correct option A.

High air humidity decreases the concentration gradient of water vapour between the inside of the leaf and the outside air. Low temperature decreases the kinetic energy of water molecules, reducing evaporation. Still air allows water vapour to accumulate around the stomata, further reducing the concentration gradient. This combination minimizes transpiration.

Award 1 mark for the correct option B.

The left ventricle has the thickest muscular wall because it must generate high pressure to pump blood to the rest of the body (systemic circulation). The right ventricle only pumps blood to the lungs, which are closer to the heart and require less pressure.

Award 1 mark for the correct option B.

Anaerobic respiration in human muscle cells produces only lactic acid. Because glucose is not completely broken down, it releases a much smaller amount of energy compared to aerobic respiration, and it does not produce carbon dioxide.

Award 1 mark for the correct option C.

Progesterone is the hormone responsible for maintaining the vascularized uterine lining during the luteal phase (second half of the cycle). It is primarily produced by the corpus luteum in the ovary after ovulation.

Award 1 mark for the correct option B.

The lock-and-key hypothesis states that the substrate molecule (the key) has a complementary shape that fits precisely into the active site (the lock) of the enzyme molecule.

Award 1 mark for the correct option C.

Water is absorbed from the soil into the root hair cell by osmosis. It then moves through the root cortex cells, enters the xylem vessel in the centre of the root, and is transported upwards towards the mesophyll cells of the leaves.

Award 1 mark for the correct option B.

The umbilical artery carries deoxygenated blood and metabolic waste products (like carbon dioxide and urea) away from the fetus to the placenta. The umbilical vein carries oxygenated blood rich in nutrients (like glucose and amino acids) from the placenta to the fetus.

Award 1 mark for the correct option A.

According to the lock-and-key hypothesis, the enzyme's active site has a specific, complementary 3D shape (the lock) into which only a specific substrate (the key) fits perfectly to form an enzyme-substrate complex.

1 mark for correct option A.

The rate of transpiration is lowest when humidity is high (decreasing the water potential gradient between the inside of the leaf and the atmosphere), wind speed is low (allowing a boundary layer of humid air to remain near the stomata), and light intensity is low (causing stomata to close or narrow).

1 mark for correct option B.

The left ventricle pumps oxygenated blood out of the heart through the aorta to the rest of the body. The semi-lunar valve at the base of the aorta prevents the backflow of blood into the left ventricle when it relaxes.

1 mark for correct option A.

Anaerobic respiration in human muscle cells uses glucose as a substrate, breaking it down incompletely to produce lactic acid. Because glucose is only partially broken down, the energy released per glucose molecule is much less than in aerobic respiration.

1 mark for correct option B.

Fertilisation (the fusion of the nuclei of the male and female gametes) normally occurs in the oviduct (fallopian tube). The resulting embryo then travels down to the uterus, where it implants into the endometrium (uterus lining).

1 mark for correct option B.

Water leaves the xylem vessels and passes into the wet cell walls of the mesophyll cells. It then evaporates from these cell walls into the intercellular air spaces of the spongy mesophyll as water vapour, before diffusing out of the leaf through open stomata.

1 mark for correct option A.

Progesterone is secreted by the corpus luteum after ovulation. Its main function is to maintain and further prepare the uterus lining (endometrium) for the implantation of a fertilised egg. If fertilisation does not occur, progesterone levels drop, triggering menstruation.

1 mark for correct option B.

At high temperatures, the increased thermal energy causes the enzyme molecules to vibrate violently. This breaks the weak chemical bonds (such as hydrogen bonds) holding the specific tertiary structure of the enzyme together. The active site permanently loses its shape (denaturation), preventing the substrate from binding.

1 mark for correct option C.

At high temperatures, human enzymes undergo denaturation because the thermal energy disrupts their shape, altering the active site so that the substrate can no longer bind. Below the optimum temperature, enzymes are inactive but not destroyed (ruling out B). Increasing temperature increases the kinetic energy of the molecules (ruling out C). Human enzymes have an optimum temperature of around \( 37\ ^\circ\text{C} \), not \( 100\ ^\circ\text{C} \) (ruling out D).

Award 1 mark for the correct option A.

High humidity decreases the concentration gradient of water vapour between the inside and outside of the leaf. Low wind speed allows water vapour to accumulate near the stomatal openings, further lowering the gradient. Low temperature decreases the kinetic energy of water molecules, reducing evaporation. Together, these conditions result in the lowest rate of transpiration.

Award 1 mark for the correct option A.

The testes are the site of sperm production in the human male reproductive system. The urethra is the common tube that transports both urine from the bladder and semen from the sperm duct out of the body through the penis.

Award 1 mark for the correct option C.

Anaerobic respiration in yeast (also known as fermentation) breaks down glucose to produce ethanol and carbon dioxide, releasing a relatively small amount of energy compared to aerobic respiration.

Award 1 mark for the correct option B.

The left ventricle must pump blood throughout the systemic circulation to the rest of the body, which requires high pressure to overcome high resistance over a long distance. In contrast, the right ventricle only pumps blood a short distance to the lungs under much lower pressure.

Award 1 mark for the correct option B.

The lock-and-key hypothesis states that the active site of the enzyme has a highly specific, complementary shape to the substrate molecule, allowing them to fit together precisely to form an enzyme-substrate complex.

Award 1 mark for the correct option C.

Water enters the plant via the root hair cells by osmosis, travels across the root cortex cells to the xylem vessels, rises up the stem into the mesophyll cells of the leaf, evaporates into the leaf's intercellular air spaces, and finally diffuses out through the stomata into the atmosphere.

Award 1 mark for the correct option A.

The acrosome is a specialized vesicle at the tip of the sperm's head that contains digestive enzymes. These enzymes are released to break down the jelly layer surrounding the egg cell, allowing the sperm nucleus to enter.

Award 1 mark for the correct option A.

1 (a) Two marks for defining enzyme as a protein and biological catalyst/speeds up reactions.
(b) Two marks for explaining the breakdown of pectin in cell walls and the release of juice/clarification.
(c)(i) Three marks for describing the overall trend: increases up to 40 °C (maximum) and decreases above 40 °C, citing specific data from Table 1.1 with units.
(ii) Two marks for correct calculation: showing the working (31.5 - 12.5) / 12.5 * 100 and the final answer of 152%.
(iii) Two marks for explaining denaturation: high temperature causes the active site to change shape/denature, preventing substrate binding.

1 (a)
- protein; [1]
- biological catalyst / speeds up reactions; [1]
(b)
- breaks down pectin (in cell walls); [1]
- releases more juice / makes juice clearer; [1]
(c)(i)
- volume of juice increases as temperature increases from 20 °C to 40 °C; [1]
- maximum volume extracted at 40 °C (31.5 cm³); [1]
- volume of juice decreases as temperature increases above 40 °C; [1]
(c)(ii)
- correct working shown: (31.5 - 12.5) / 12.5 * 100 or 19.0 / 12.5 * 100; [1]
- 152 (%); [1]
(c)(iii)
- enzyme / pectinase is denatured (at 60 °C); [1]
- active site changed shape so substrate / pectin can no longer bind; [1]

2 (a) Two marks for correct substrate and products of catalase.
(b) Three marks for identifying key controlled variables like temperature, concentration of reactant, and source/amount of enzyme.
(c)(i) One mark for multiplying rate by time to get 21 cm³.
(ii) Two marks for explaining optimum pH as the pH of maximum activity/highest rate where the enzyme is not denatured.
(iii) Three marks for explaining denaturation at extreme pH (pH 2): change in shape of active site, inability of substrate to bind, resulting in lower activity.

2 (a)
- substrate: hydrogen peroxide; [1]
- products: water AND oxygen; [1]
(b) Any three from:
- temperature; [1]
- concentration of hydrogen peroxide; [1]
- volume of hydrogen peroxide; [1]
- surface area / mass / number of potato discs; [1]
- age / variety / source of potato; [1]
(c)(i)
- 21 (cm³); [1]
(c)(ii)
- the pH at which the enzyme is most active / has maximum rate of reaction; [1]
- the active site has the correct shape / is not denatured; [1]
(c)(iii)
- pH 2 is far from the optimum pH / too acidic; [1]
- enzyme is denatured / active site changes shape; [1]
- substrate can no longer fit / bind to the active site; [1]

3 (a) Two marks for identifying the elements in proteins: carbon, hydrogen, oxygen, and nitrogen (any two for 1 mark, all four for 2 marks).
(b) Three marks for explaining digestion of proteins to soluble amino acids and their removal by dissolution.
(c) Four marks for describing the complementary shapes, the active site, the lock-and-key fit, the formation of the enzyme-substrate complex, and product release.
(d) Two marks for explaining denaturation of enzymes at 80 °C and their stability/optimal activity at 40 °C.

3 (a)
- carbon, hydrogen, oxygen; [1]
- nitrogen (allow sulfur); [1]
(b)
- protease digests / breaks down large, insoluble proteins into small, soluble molecules / amino acids; [1]
- soluble molecules dissolve in water; [1]
- easily washed away / removed from fabric; [1]
(c)
- enzyme active site has a specific shape; [1]
- complementary to the shape of the substrate; [1]
- substrate fits into the active site / forms enzyme-substrate complex; [1]
- products are released and enzyme remains unchanged; [1]
(d)
- enzymes are denatured at 80 °C; [1]
- active site changes shape so substrate can no longer bind / no digestion of stains occurs; [1]

4 (a) Four marks for correctly filling the blanks: xylem, spongy, stomata, diffusion.
(b) Two marks for correct directions of transpiration rate change: (i) increases, (ii) decreases.
(c) Three marks for explaining humidity's effect on water vapour concentration gradient and diffusion rate.
(d) Two marks for explaining that the waxy cuticle is waterproof/impermeable and reduces evaporation from the leaf surface.

4 (a)
- xylem; [1]
- spongy; [1]
- stomata; [1]
- diffusion; [1]
(b)(i)
- increases; [1]
(b)(ii)
- decreases; [1]
(c)
- higher humidity means more water vapour in the air outside the leaf; [1]
- decreases the water vapour concentration gradient (between inside and outside); [1]
- reduces the rate of diffusion of water vapour out of the stomata; [1]
(d)
- waxy cuticle is waterproof / barrier to water; [1]
- reduces evaporation / water loss from the epidermal surface; [1]

5 (a) Two marks for any two valid functions of water (e.g., reactant in photosynthesis, turgidity/support, solvent for mineral transport).
(b) Three marks for explaining osmosis in guard cells, turgidity vs flaccidity, and the mechanical opening/closing of the stoma.
(c) Four marks for describing wilting/drooping and explaining it through loss of water, flaccidity, and loss of turgor pressure which provides structural support.
(d) One mark for identifying phloem.
(e) One mark for defining translocation as movement of sucrose and amino acids from source to sink.

5 (a) Any two from:
- photosynthesis / raw material; [1]
- turgidity / structural support; [1]
- transport of dissolved mineral ions / sucrose; [1]
- solvent for metabolic reactions / medium; [1]
(b)
- guard cells take in water by osmosis / become turgid; [1]
- thicker inner wall causes them to bend / curve, opening stoma; [1]
- guard cells lose water / become flaccid, closing stoma; [1]
(c)
- plant wilts / leaves and stem droop; [1]
- cells lose water (by osmosis); [1]
- cells become flaccid / lose turgor pressure; [1]
- no longer support the plant stem / leaves; [1]
(d)
- phloem; [1]
(e)
- movement of sucrose AND amino acids in phloem from source to sink; [1]

6 (a) Two marks for defining fertilisation: fusion of nuclei, of gametes / sperm and egg.
(b) Two marks: (i) oviduct/fallopian tube, (ii) uterus/womb.
(c) Five marks for explaining the placenta (diffusion barrier, thin membrane, large surface area) and umbilical cord (blood transport), and identifying the direction of exchange of nutrients/oxygen and waste products (urea, carbon dioxide).
(d) Two marks for listing two maternal-to-fetal substances: oxygen, glucose, amino acids, antibodies (any two).

6 (a)
- fusion of nuclei; [1]
- of gametes / sperm and egg (to form a zygote); [1]
(b)(i)
- oviduct / fallopian tube; [1]
(b)(ii)
- uterus / womb; [1]
(c)
- placenta has large surface area / thin membrane separating maternal and fetal blood; [1]
- exchange occurs by diffusion; [1]
- umbilical cord contains blood vessels to transport substances between fetus and placenta; [1]
- oxygen / glucose / nutrients diffuse from maternal blood to fetal blood; [1]
- carbon dioxide / urea / wastes diffuse from fetal blood to maternal blood; [1]
(d) Any two from:
- oxygen; [1]
- glucose; [1]
- amino acids; [1]
- antibodies; [1]
- water / mineral ions; [1]

7 (a) Two marks for identifying secreting organs: (i) pituitary gland, (ii) ovary.
(b) Four marks for oestrogen (uterus lining repair, feedback on FSH/LH) and progesterone (maintenance of uterus lining, inhibition of FSH/LH).
(c) Three marks: (i) two marks for menstruation details (shedding of lining, blood loss, caused by low hormone levels); (ii) one mark for ovulation details (release of egg from ovary into oviduct).
(d) Two marks for female secondary sexual characteristics (breast development, hip widening, pubic hair, etc.).

7 (a)(i)
- pituitary gland; [1]
(a)(ii)
- ovary / ovaries; [1]
(b)
- oestrogen stimulates repair / thickening of uterus lining; [1]
- oestrogen inhibits FSH / stimulates LH; [1]
- progesterone maintains thickness of uterus lining; [1]
- progesterone inhibits FSH AND LH secretion; [1]
(c)(i)
- uterus lining breaks down / is shed; [1]
- loss of blood / tissue through vagina; [1]
(c)(ii)
- release of a mature egg from the ovary (into the oviduct); [1]
(d) Any two from:
- breast development; [1]
- widening of hips; [1]
- growth of pubic / axillary hair; [1]
- fat deposition on hips / thighs; [1]

8 (a) Three marks for correctly identifying: (i) vena cava, (ii) aorta, (iii) coronary artery.
(b) Three marks for explaining the difference in ventricle wall thickness: left ventricle pumps to body (further distance), requires higher pressure, more force, hence more muscle.
(c) Two marks for describing how pressure differences open/close valves to prevent backflow.
(d) Three marks for listing three CHD risk factors: high-fat diet, lack of exercise, smoking, stress, genetic predisposition, obesity.

8 (a)(i)
- vena cava; [1]
(a)(ii)
- aorta; [1]
(a)(iii)
- coronary artery; [1]
(b)
- left ventricle pumps blood to the whole body / right ventricle only pumps to the lungs; [1]
- left ventricle needs to generate much higher pressure; [1]
- requires more muscle to contract with greater force; [1]
(c)
- valves open / close in response to pressure differences; [1]
- prevent backflow of blood (from ventricles to atria / arteries to ventricles); [1]
(d) Any three from:
- diet high in saturated fat / cholesterol / salt; [1]
- lack of exercise / sedentary lifestyle; [1]
- smoking (tobacco); [1]
- stress; [1]
- obesity; [1]
- genetic predisposition / family history; [1]

(a) An enzyme is defined as a biological catalyst (1) made of protein (1) that increases the rate of chemical reactions without being changed.

(b) At low temperatures, enzymes and substrates have low kinetic energy, meaning fewer collisions and a low rate of reaction (1). As temperature increases, kinetic energy increases, causing more frequent and successful collisions (1). At optimum temperature, rate of reaction is at its highest (1). Beyond the optimum, high temperatures break the chemical bonds within the enzyme, denaturing it so the active site changes shape and the substrate can no longer bind (1).

(c) Any two from: pH, concentration of hydrogen peroxide (substrate), volume/concentration of catalase extract (enzyme), same sweet potato variety/age (2).

(d) Formula: Rate = Volume / Time (1). Calculation: 15 / 3.0 = 5.0 (1). Units: cm³/minute (1).

1 (a) Max 2 marks:
- Protein that acts as a biological catalyst / speeds up reactions (1)
- Remains unchanged at the end of the reaction (1)

1 (b) Max 4 marks:
- Low temperature = low kinetic energy / fewer successful collisions (1)
- Increasing temperature increases kinetic energy / more frequent successful collisions (1)
- Peak rate at optimum temperature (1)
- Excessively high temperature denatures enzyme / changes active site shape / substrate cannot fit (1)

1 (c) Max 2 marks:
- Any two valid control variables, e.g. volume of substrate, concentration of substrate, pH, volume of enzyme extract (2)

1 (d) Max 3 marks:
- Correct working: 15 / 3.0 (1)
- Correct calculation: 5.0 (accept 5) (1)
- Correct units: cm³/minute / cm³ min⁻¹ (1)

(a) Guard cells (1).

(b) Increased wind speed sweeps away water vapour from the boundary layer of the leaf (1), which maintains a steep concentration gradient of water vapour (1). This increases the diffusion of water vapour out of the stomata (1), thus increasing the transpiration rate (1).

(c) Water on the wet surfaces of spongy mesophyll cell walls evaporates into the air spaces (1). Water vapour accumulates in these intercellular spaces and diffuses down a water vapour concentration gradient (1) out of the leaf through open stomata into the atmosphere (1).

(d) Any three from: thick waxy cuticle, sunken stomata, rolled/curled leaves, hairs on the leaf surface, leaves modified into spines/reduced surface area (3).

2 (a) 1 mark:
- Guard cells (1)

2 (b) Max 4 marks:
- Wind speed increases transpiration rate (1)
- Sweeps away the water vapour / boundary layer of air on the leaf surface (1)
- Maintains a steep water vapour concentration gradient (1)
- Increases rate of diffusion of water vapour out of stomata (1)

2 (c) Max 3 marks:
- Evaporation of water from spongy mesophyll cell walls / surfaces into air spaces (1)
- Diffusion / movement down a water vapour concentration gradient (1)
- Passes out through stomata (1)

2 (d) Max 3 marks:
- Any three valid adaptations: thick cuticle, sunken stomata, rolled leaves, leaf hairs, reduced leaf area / needles (3)

(a) Fertilization involves the sperm cell swimming to and penetrating the egg cell (1), resulting in the fusion of the haploid nuclei of the sperm and egg to form a diploid zygote (1).

(b) The placenta acts as an exchange site, supplying the fetus with dissolved nutrients and oxygen (1), removing metabolic waste products like urea and carbon dioxide (1), and producing progesterone to maintain the pregnancy (1). It also acts as a barrier to some harmful pathogens (1).

(c) In the umbilical artery, oxygen concentration is low and urea concentration is high (2). In the umbilical vein, oxygen concentration is high and urea concentration is low (2).

(d) Progesterone (1). Accept estrogen (1).

3 (a) Max 2 marks:
- Fusion of nuclei (1)
- Of sperm (male gamete) and egg (female gamete) (1)

3 (b) Max 3 marks:
- Transfer of oxygen / glucose / amino acids / water to fetus (1)
- Removal of urea / carbon dioxide / wastes from fetus (1)
- Secretion of hormones / progesterone (1)
- Barrier to some toxins / pathogens / mother's high blood pressure (1)

3 (c) Max 4 marks:
- Umbilical artery: low oxygen concentration AND high urea concentration (2)
- Umbilical vein: high oxygen concentration AND low urea concentration (2)

3 (d) Max 2 marks:
- Progesterone (1) / Estrogen (1)

(a) C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O (2). 1 mark for correct formulas, 1 mark for correct balancing.

(b) Yeast: produces ethanol and carbon dioxide (1); Muscle cells: produces lactic acid only, no carbon dioxide (1). Yeast does not have an oxygen debt, whereas muscle cells accumulate an oxygen debt (1).

(c) During a sprint, anaerobic respiration occurs, building up lactic acid in muscles (1). The runner accumulates an oxygen debt (1). Post-exercise, heart rate remains high to pump blood carrying oxygen to the muscles and liver (1) to break down lactic acid aerobically into carbon dioxide and water (1).

(d) Any three from: muscle contraction, protein synthesis, active transport, cell division/growth, passage of nerve impulses, maintenance of constant body temperature (3).

4 (a) Max 2 marks:
- C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O
- 1 mark for correct reactants and products (1)
- 1 mark for correct balancing (1)

4 (b) Max 2 marks:
- Yeast produces ethanol AND carbon dioxide (1)
- Muscle cells produce lactic acid only (no CO₂) (1)
- Reference to oxygen debt in muscle cells vs no debt in yeast (1)

4 (c) Max 4 marks:
- Sprinting causes anaerobic respiration / lactic acid accumulation (1)
- Creates an oxygen debt (1)
- High heart rate continues to supply oxygen to liver / muscles (1)
- To aerobically break down / oxidize lactic acid into carbon dioxide and water (1)

4 (d) Max 3 marks:
- Any three valid uses: muscle contraction, active transport, cell division, protein synthesis, temperature maintenance, nerve impulse transmission (3)

(a) The left ventricle pumps blood to the systemic circulation (the whole body) (1), which is a much longer distance than the pulmonary circulation to the lungs (1). Therefore, it must contract with more force to create a higher blood pressure (1).

(b) The septum separates the left and right sides of the heart (1) to prevent oxygenated and deoxygenated blood from mixing, keeping the concentration gradient of oxygen high for body tissues (1).

(c) Coronary arteries supply the heart muscle (myocardium) (1) with oxygen and glucose (1) for aerobic respiration to release energy for contraction (1).

(d) Saturated fat intake increases blood cholesterol levels (1), causing cholesterol deposits (plaques/atheroma) to build up in the inner walls of the coronary arteries (1). This narrows the lumen of the coronary artery (1), reducing blood and oxygen supply to the cardiac muscle, causing ischemia / angina / heart attack (1).

5 (a) Max 3 marks:
- Left ventricle pumps blood to the whole body / systemic circulation (1)
- Right ventricle only pumps blood to the lungs (1)
- Left ventricle must generate higher pressure / force to overcome greater resistance over a longer distance (1)

5 (b) Max 2 marks:
- Separates the right side from the left side of the heart (1)
- Prevents mixing of oxygenated and deoxygenated blood (1)

5 (c) Max 2 marks:
- Deliver blood directly to the cardiac / heart muscle (1)
- Supply oxygen / glucose / nutrients for respiration (1)

5 (d) Max 4 marks:
- High saturated fat intake increases blood cholesterol (1)
- Cholesterol deposits / plaques build up in coronary artery walls (1)
- Narrows the artery lumen / restricts blood flow (1)
- Reduces oxygen supply to heart muscle, leading to angina / myocardial infarction / heart attack (1)

(a) Phototropism is a growth response (1) where the direction of growth is determined by the direction of the light stimulus (1).

(b) Auxin is produced in the shoot tip (1) and moves downwards by diffusion (1). Under unilateral light, auxin accumulates on the shaded side of the shoot (1). Auxin stimulates cell elongation in shoots (1), so cells on the shaded side elongate more than cells on the light side (1), causing the shoot to bend towards the light source (1).

(c) Shoots grow towards light to maximize light absorption by leaves (1) for a higher rate of photosynthesis (1).

(d) Roots grow downwards / show positive gravitropism (1). This anchors the plant firmly in the soil / helps roots absorb water and mineral ions from deep within the soil (1).

6 (a) Max 2 marks:
- Growth response of a plant (1)
- Stimulated by / directed towards light (1)

6 (b) Max 5 marks:
- Auxin is made in the shoot tip (1)
- Diffuses downwards (1)
- Accumulates / concentrates on the shaded side (1)
- High auxin concentration stimulates cell elongation in shoots (1)
- Shaded side cells elongate / grow faster than light side cells (1)
- Causes shoot to bend / grow towards the light (1)

6 (c) Max 2 marks:
- Leaves absorb more light / sunlight (1)
- Increases rate of photosynthesis / glucose production (1)

6 (d) Max 2 marks:
- Roots grow downwards / show positive gravitropism (1)
- Advantage: anchors plant / absorbs water / mineral ions from soil (1)

(a) Independent variable: Trypsin concentration (\%). Dependent variable: Time taken for casein to clear (s) / Rate of digestion (\text{s}^{-1}).
(b) Rate of digestion = 1 / time = 1 / 50 = 0.0200 \text{ s}^{-1}.
(c) Plot trypsin concentration (0.2, 0.4, 0.6, 0.8, 1.0) on x-axis with linear scale. Plot rate of digestion on y-axis with linear scale (from 0 to 0.025). Draw a smooth curve or line of best fit.
(d) As trypsin concentration increases, the rate of digestion increases. This is because there are more active sites available to bind with casein molecules, leading to more frequent successful collisions.
(e) Source of error: Judging 'completely clear' is subjective and varies between trials. Improvement: Use a colorimeter to measure light absorbance/transmission at fixed intervals, or place a black cross behind the tube and record the time when it becomes clearly visible.

(a) Max 2 marks:
- Trypsin concentration [1]
- Time taken to clear / rate of digestion [1]
(b) Max 3 marks:
- Correct formula use (1/time) [1]
- Calculation: 1/50 = 0.02 [1]
- Four decimal places: 0.0200 (allow ecf if working is correct) [1]
(c) Max 4 marks:
- Axes labelled correctly with units [1]
- Suitable linear scale occupying more than half the grid [1]
- Points plotted accurately to within 0.5 small square [1]
- Smooth line of best fit or points joined with straight ruled lines [1]
(d) Max 2 marks:
- Rate of reaction increases as trypsin concentration increases [1]
- More active sites / enzyme-substrate complex formations / successful collisions [1]
(e) Max 2 marks:
- Subjectivity of deciding when the tube is clear [1]
- Use a colorimeter / place a cross behind the tube [1]

(a) To prevent moisture from the surrounding air reacting with the cobalt chloride paper, ensuring that the colour change is only due to water loss (transpiration) from the leaf surface.
(b) Sum = 780 + 820 + 750 + 800 + 850 = 4000. Mean = 4000 / 5 = 800 s.
(c) The lower surface loses water much faster than the upper surface because the mean time for the cobalt chloride paper to turn pink on the lower surface is 130 s, which is significantly shorter than the 800 s for the upper surface.
(d) Most stomata are located on the lower epidermis to reduce water loss from direct sunlight, while the upper surface has very few stomata and is covered by a thick waxy cuticle which is impermeable to water.
(e) Temperature, light intensity, humidity, leaf area covered by the tape/paper, species of plant used.

(a) Max 2 marks:
- To prevent moisture in the air from turning the paper pink [1]
- Ensure only water from the leaf surface is measured [1]
(b) Max 3 marks:
- Summing the values (4000) [1]
- Dividing by 5 [1]
- Correct mean: 800 s [1]
(c) Max 2 marks:
- Water loss is faster from the lower surface / slower from the upper surface [1]
- Supported by reference to data (e.g., lower surface mean of 130 s is much faster than upper surface mean of 800 s) [1]
(d) Max 3 marks:
- More stomata on the lower surface [1]
- Waxy cuticle on the upper surface prevents water loss [1]
- Upper surface is exposed to more direct light, so stomata are on the bottom to limit transpiration [1]
(e) Max 3 marks:
- Species of plant / same plant [1]
- Temperature [1]
- Light intensity / wind speed / humidity [1]

(a) 74 bpm (at 5 minutes of recovery, the heart rate reaches its stable baseline).
(b) X-axis: Time / min (0 to 5). Y-axis: Heart rate / bpm (70 to 170). Both curves plotted clearly with a key to distinguish light and vigorous exercise.
(c) Heart rate at 0 min = 165 bpm. Heart rate at 3 min = 98 bpm.
Decrease = 165 - 98 = 67 bpm.
Percentage decrease = (67 / 165) * 100 = 40.606% = 40.6%.
(d) Vigorous exercise causes anaerobic respiration, leading to the build-up of lactic acid in muscles. The heart rate remains high to deliver oxygen to the muscles and liver to break down lactic acid, which is known as repaying the oxygen debt.
(e) Repeat the experiment multiple times with the same student (and calculate a mean), or use a larger sample of students of the same age and fitness level.

(a) Max 1 mark:
- 74 bpm [1]
(b) Max 5 marks:
- Axes correctly labelled with units (Time / min and Heart rate / bpm) [1]
- Appropriate linear scale occupying more than half the grid [1]
- All points for light exercise plotted correctly [1]
- All points for vigorous exercise plotted correctly [1]
- Clean lines connecting points and key/labels clearly showing which line is which [1]
(c) Max 3 marks:
- Difference calculated correctly (67) [1]
- Division by initial value (67/165 * 100) [1]
- Correct answer to 1 decimal place: 40.6% [1]
(d) Max 3 marks:
- Muscles respired anaerobically / built up lactic acid [1]
- Oxygen debt created [1]
- Oxygen needed to transport to liver / break down lactic acid / return body to homeostasis [1]
(e) Max 1 mark:
- Repeat the test and calculate a mean / use more participants [1]

(a) Gas: Carbon dioxide. Test: Bubble the gas through limewater. Expected result: Limewater turns cloudy/milky.
(b) Rate = 2.0 / 20 = 0.10 \text{ cm}^3\text{ min}^{-1}.
(c) Yeast cells do not secrete amylase in sufficient quantities to quickly digest starch into glucose. Starch is a large polymer (polysaccharide) and is too large to pass through the yeast cell membrane directly. Therefore, it cannot be respired.
(d) Plan:
- Independent variable: Temperature of the water-bath (e.g., 20, 30, 40, 50, 60 ^\circ\text{C}).
- Dependent variable: Volume of carbon dioxide gas collected in a gas syringe in a set time (e.g., 10 minutes), or counting the number of bubbles per minute.
- Constants: Yeast concentration and volume, glucose concentration and volume, pH (using a buffer).
- Method: Set up yeast-glucose mixture in tubes. Place in different temperature-controlled water-baths. Allow 5 minutes to reach temperature (equilibration). Connect to gas syringe, start timer, and record volume of gas after 10 minutes.
- Replicates: Repeat the experiment at least 3 times at each temperature and calculate a mean.
- Safety: Wear safety goggles, take care with hot water-baths.

(a) Max 3 marks:
- Carbon dioxide [1]
- Test: Limewater [1]
- Result: Turns cloudy / milky / white precipitate [1]
(b) Max 1 mark:
- 0.1 or 0.10 [1]
(c) Max 3 marks:
- Starch is a polysaccharide / large / complex molecule [1]
- Cannot be absorbed across yeast cell membrane / yeast lacks amylase [1]
- Must be broken down into simple sugars (like glucose) first before respiration [1]
(d) Max 6 marks:
- 1: Independent variable: at least 3 different temperatures specified [1]
- 2: Dependent variable: measuring volume of gas or counting bubbles in a set time [1]
- 3: Method: use of water-baths to control temperature [1]
- 4: Control/constant variables: volume/concentration of yeast/sugar, or pH [1]
- 5: Replication: repeat at each temperature and calculate mean [1]
- 6: Safety precaution: e.g., hand protection with hot water [1]

(a) Starch test: Iodine solution. Expected result: remains orange-brown (no starch present). Glucose test: Benedict's reagent (heated in water bath). Expected result: turns green/yellow/orange/red (glucose is present).
(b) Glucose molecules are small and soluble, so they can pass through the microscopic pores of the selectively permeable dialysis tubing. Starch molecules are large and insoluble polymers, so they cannot fit through the pores and remain trapped inside the tubing.
(c) X-axis: Time / min (0 to 30). Y-axis: Glucose concentration / mg dm^-3 (0 to 7). Points plotted accurately, connected with a smooth line or ruled straight lines.
(d) Pattern: Glucose concentration increases rapidly at first, then the rate of increase slows down. Explanation: As glucose diffuses out, the concentration inside decreases and the concentration outside increases. This reduces the concentration gradient, which decreases the rate of diffusion.

(a) Max 4 marks:
- Starch reagent: Iodine solution AND result: orange / brown / yellow [1]
- Glucose reagent: Benedict's solution AND heated / hot water bath [1]
- Glucose result: green / yellow / orange / red [1]
- Shows starch does not pass through but glucose does [1]
(b) Max 3 marks:
- Dialysis tubing is selectively permeable / has tiny pores [1]
- Glucose is small / simple sugar / monomer and can pass through [1]
- Starch is a large / polymer and cannot pass through [1]
(c) Max 4 marks:
- Axes correctly labelled with units [1]
- Suitable linear scale on both axes [1]
- All points plotted correctly [1]
- Points joined with a smooth curve or straight lines [1]
(d) Max 2 marks:
- Concentration of glucose increases [1]
- Rate slows down as concentration gradient decreases [1]

(a) To provide a constant supply of dissolved carbon dioxide (so it is not a limiting factor).
(b) Rate = 6 / 3 = 2.0 bubbles \text{ min}^{-1}.
(c) As the distance of the lamp increases, the rate of photosynthesis decreases. This is because light intensity is inversely proportional to the square of the distance, so less light energy is available for photosynthesis at greater distances.
(d) Errors and improvements:
- Error 1: Bubbles are of different sizes, so counting bubbles is not a precise measure of gas volume. Improvement: Collect the oxygen in a capillary tube/graduated syringe and measure the actual volume.
- Error 2: The lamp emits heat, which increases the temperature of the water at closer distances. Improvement: Place a clear glass block or beaker of water between the lamp and the test-tube to act as a heat shield.
(e) Plan:
- Independent variable: Colour of light (e.g., red, blue, green, white). Varied by using different coloured filters wrapped around the test-tube or placed over the lamp.
- Dependent variable: Number of oxygen bubbles counted per minute, or volume of gas collected in 5 minutes.
- Constants: Distance of lamp (intensity), temperature of water-bath, concentration of sodium hydrogencarbonate, same piece of pondweed.
- Method: Place pondweed tube in water-bath. Wrap red filter around tube. Acclimate for 5 minutes. Count bubbles for 3 minutes. Repeat using blue, green, and white filters.
- Replicates: Repeat at least 3 times for each colour and calculate a mean.

(a) Max 1 mark:
- To supply carbon dioxide / ensure carbon dioxide is not limiting [1]
(b) Max 1 mark:
- 2.0 or 2 [1]
(c) Max 2 marks:
- As distance increases, the rate decreases [1]
- Distance increases from 10 to 50 cm, rate drops from 45.0 to 2.0 bubbles/min [1]
(d) Max 4 marks (2 marks for error, 2 marks for corresponding improvement):
- Error 1: Bubbles may be different sizes [1] -> Improvement: Measure volume using gas syringe / capillary tube [1]
- Error 2: Temperature increases closer to the lamp [1] -> Improvement: Use a heat shield / water screen / LED lamp [1]
(e) Max 5 marks:
- 1: Independent variable: different colours of light / filters [1]
- 2: Dependent variable: bubbles counted per minute / volume of gas in set time [1]
- 3: Key constant: distance from light source / temperature / CO2 concentration [1]
- 4: Acclimation period (e.g., 5 mins) before measuring [1]
- 5: Repeats and calculation of a mean [1]