2024 Cambridge IGCSE Biology (0610) 模拟试题及答案详解

An increase in concentration gradient increases the rate of diffusion as the difference in concentration is greater. An increase in surface area increases the rate of diffusion because there is more area across which molecules can pass. An increase in membrane thickness decreases the rate of diffusion because the diffusion path is longer. A decrease in temperature decreases the kinetic energy of the molecules, thus decreasing the rate of diffusion. Therefore, only changes 1 and 3 will cause an increase in the rate of diffusion.

B is correct (1 mark).

The ovary produces female gametes (eggs) and the female hormones estrogen and progesterone. The oviduct is the site of fertilization, the uterus is the site of implantation and development of the embryo, and the cervix is the ring of muscle at the neck of the uterus.

A is correct (1 mark).

In bright light, the pupil constricts to protect the retina from damage. This is achieved by the contraction of the circular muscles and the relaxation of the radial muscles in the iris.

A is correct (1 mark).

Boiling the leaf in ethanol removes chlorophyll, decolorizing the leaf so that any subsequent color change when iodine solution is added (to test for starch) can be clearly seen.

B is correct (1 mark).

Auxin is produced in the shoot tip and moves away from light, accumulating on the shaded side. In shoots, a higher concentration of auxin stimulates cell elongation. The cells on the shaded side elongate more than those on the illuminated side, causing the shoot to bend towards the light source.

B is correct (1 mark).

Anaerobic respiration in yeast (alcoholic fermentation) produces ethanol and carbon dioxide. Anaerobic respiration in human muscle cells (lactic acid fermentation) produces only lactic acid. Both release a relatively small amount of energy compared to aerobic respiration.

C is correct (1 mark).

The blood capillaries in a villus absorb water-soluble nutrients such as amino acids, glucose, water, and inorganic ions. The lacteal (lymphatic capillary) absorbs fat-soluble nutrients, namely fatty acids and glycerol (re-assembled into fats).

A is correct (1 mark).

The movement of carbon dioxide (a gas) down its concentration gradient into a mesophyll cell occurs by simple diffusion. The movement of water through a partially permeable membrane is osmosis. The active uptake of mineral ions and glucose against a concentration gradient are active transport.

B is correct (1 mark).

Diffusion rate increases when the concentration gradient is steeper (increased), the surface area is larger (increased), and the diffusion distance is shorter (decreased membrane thickness).

A is correct because all three changes increase the rate of diffusion according to Fick's Law / general diffusion principles.

Fertilisation (the fusion of the nuclei of the male and female gametes) occurs in the oviduct. The zygote then travels to the uterus, where implantation occurs and the fetus develops.

A is the correct answer because fertilisation occurs in the oviduct, and the fetus develops in the uterus.

In bright light, the pupil constricts (decreases in diameter) to protect the retina from damage. This is achieved by the contraction of the circular muscles and the relaxation of the radial muscles in the iris.

A is the correct answer. In bright light, circular muscles contract and radial muscles relax, causing the pupil to decrease in size.

Since the plant already has high light intensity and plenty of water, these are not limiting factors. Carbon dioxide is a reactant in photosynthesis and is often the limiting factor under these conditions. Increasing carbon dioxide concentration will increase the rate of photosynthesis.

A is the correct answer because carbon dioxide is a limiting factor for photosynthesis under high light intensity.

Plant shoots grow towards light, which is a positive phototropic response. They grow away from gravity (upwards), which is a negative gravitropic response.

A is correct because shoots are positively phototropic and negatively gravitropic.

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

A is correct as yeast produces ethanol + CO2, and human muscles produce lactic acid.

The lacteal (lymphatic capillary) in the center of the villus is responsible for absorbing fatty acids and glycerol (products of fat digestion). The blood capillaries absorb water-soluble nutrients, including glucose and amino acids.

A is correct because lacteals absorb lipids/fats (fatty acids and glycerol) while capillaries absorb glucose and amino acids.

Because the sugar solution is highly concentrated, it has a lower water potential than the plant cell's cytoplasm and vacuole. Water moves down the water potential gradient out of the cell by osmosis. This causes the cell membrane to pull away from the cell wall, making the cell plasmolysed.

A is correct because water leaves the cell when placed in a hypertonic (concentrated) solution, causing plasmolysis.

Diffusion is faster at higher temperatures because particles have more kinetic energy. It is also faster when there is a steeper (higher) concentration gradient and a larger surface area available for movement. Therefore, Setup D (high temperature, high concentration gradient, and large surface area) will have the fastest rate of diffusion.

D is correct. 1 mark for identifying the combination of high temperature, high concentration gradient, and large surface area.

The umbilical vein carries oxygenated and nutrient-rich blood from the placenta to the fetus. Therefore, glucose concentration is higher in the umbilical vein than in the umbilical artery. The umbilical artery carries deoxygenated blood containing metabolic waste products, such as urea, from the fetus to the placenta. Thus, urea concentration is lower in the umbilical vein compared to the umbilical artery.

B is correct. 1 mark for recognizing that glucose is higher in the umbilical vein and urea is lower in the umbilical vein.

In bright light, the pupil constricts to protect the retina from damage. This pupil reflex is controlled by antagonistic muscles in the iris: circular muscles contract and radial muscles relax, resulting in a decrease in pupil diameter (constriction).

A is correct. 1 mark for correct actions of circular muscles, radial muscles, and pupil size change.

A limiting factor is something present in the environment in such short supply that it restricts life processes. Since increasing the carbon dioxide concentration causes the rate of photosynthesis to increase, carbon dioxide was the factor limiting the rate of photosynthesis at the lower concentration.

A is correct. 1 mark for identifying carbon dioxide as the limiting factor when its increase raises the photosynthetic rate.

In phototropism, auxin is produced at the shoot tip and diffuses to the shaded side of the shoot, resulting in a higher concentration of auxin on the shaded side. In shoots, a high concentration of auxin stimulates cell elongation, causing the shaded side to grow faster and bend the shoot towards the light source.

D is correct. 1 mark for correctly matching the shaded side with stimulation of cell elongation in shoots.

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

B is correct. 1 mark for identifying ethanol and carbon dioxide in yeast, and lactic acid in humans.

Glucose and amino acids are water-soluble molecules that are absorbed directly into the blood capillary network inside each villus. Fatty acids, glycerol, and fats enter the lymphatic system via the lacteals. Water is absorbed into the blood capillary network by osmosis.

A is correct. 1 mark for correctly matching glucose with the capillary network.

A 10% salt solution has a lower water potential than the cytoplasm of the red blood cells. Therefore, water leaves the red blood cells by osmosis (moving from a higher water potential to a lower water potential). Because red blood cells do not have a cell wall, the loss of water causes them to shrink and shrivel.

B is correct. 1 mark for identifying that water leaves by osmosis and causes the cells to shrivel.

The rate of diffusion is increased by a higher temperature (which increases the kinetic energy of the molecules), a steeper concentration gradient, and a larger surface area of the membrane across which diffusion occurs.

1 mark for the correct choice B.

Ovulation occurs in the ovary when an egg is released. It travels into the oviduct (where fertilisation typically happens), and then implants into the lining of the uterus.

1 mark for selecting A.

In bright light, the pupil constricts to reduce the amount of light entering the eye. This is achieved by the contraction of circular muscles and the relaxation of radial muscles in the iris.

1 mark for selecting A.

Both temperature (15 °C) and carbon dioxide concentration (0.04%) are below optimal levels and acting as limiting factors. Increasing them will increase photosynthesis. Since light intensity is already high, increasing it further will not significantly affect the rate.

1 mark for selecting A.

Auxin is redistributed to the shaded side of the shoot. In shoots, higher auxin concentrations promote cell elongation, causing the shaded side to grow faster and bend the shoot toward the light.

1 mark for selecting A.

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

1 mark for selecting A.

Microvilli are folds on the surface of epithelial cells of the villi that increase surface area for faster diffusion and active transport. Capillaries absorb glucose and amino acids. Lacteals absorb fatty acids and glycerol.

1 mark for selecting C.

Smaller cubes have a larger surface-area-to-volume ratio (X is 6:1, Y is 3:1, Z is 2:1). Because Cube X is the smallest, the acid has the shortest distance to diffuse to reach the center, and it has the greatest relative surface area, allowing it to be penetrated first.

1 mark for selecting A.

The time taken for a substance to diffuse to the center of a cube depends on the diffusion distance (determined by the cube's size), the temperature, and the concentration gradient. Cube X takes the shortest time because it has the smallest size (1 cm × 1 cm × 1 cm, meaning a shorter distance to the center of 0.5 cm), the highest temperature (40 °C, which increases the kinetic energy and speed of the molecules), and the steepest concentration gradient (0.5 mol/dm³).

1 mark for the correct option B.

Fertilisation typically occurs in the oviduct (fallopian tube). After ovulation, the remains of the follicle form the corpus luteum in the ovary, which secretes progesterone to maintain the uterus lining in early pregnancy. Implantation of the embryo occurs in the lining of the uterus.

1 mark for the correct option A.

For distant vision, the ciliary muscles relax. This relaxation increases the tension on the suspensory ligaments, pulling them tight. The tightened suspensory ligaments pull on the lens, causing it to become less rounded (thinner and flatter) so that light is refracted less.

1 mark for the correct option C.

Oxygen is a product of photosynthesis. Photosynthesis requires light energy (provided by a bright lamp), carbon dioxide (supplied by dissolved sodium hydrogencarbonate in the water), and a suitable temperature for enzyme action (20 °C is much closer to the optimum than 4 °C). Boiled and cooled water lacks dissolved carbon dioxide, which severely limits the rate of photosynthesis.

1 mark for the correct option B.

This is a gravitropic response. Gravity causes auxin to accumulate on the lower side of the horizontally placed shoot. In shoots, a higher concentration of auxin stimulates cell elongation. Consequently, the cells on the lower side elongate more than those on the upper side, causing the shoot to bend and grow upwards.

1 mark for the correct option B.

Anaerobic respiration in yeast cells (fermentation) produces ethanol and carbon dioxide as waste products. In contrast, anaerobic respiration in human muscle cells produces lactic acid (lactate) only, without releasing any carbon dioxide.

1 mark for the correct option B.

Fatty acids and glycerol are absorbed into the epithelial cells of the villi, where they are reassembled into fats and enter the lacteals. The lacteals are part of the lymphatic system, which eventually drains into the blood circulatory system.

1 mark for the correct option B.

Gas exchange in the lungs is a classic example of simple diffusion. Oxygen moves from a region of higher concentration in the alveoli to a region of lower concentration in the blood of the surrounding capillaries down its concentration gradient.

1 mark for the correct option C.

(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, as a result of their random movement.

(b) The mass of glucose diffused increases linearly as temperature increases. This is because higher temperatures provide glucose molecules with more kinetic energy, causing them to move faster and diffuse at a quicker rate.

(c) Percentage increase: \(\frac{9.6 - 2.4}{2.4} \times 100 = 300\%\).

(d) 1. Concentration gradient.
2. Surface area of the membrane.
3. Diffusion distance (membrane thickness).

(e) Active transport requires energy (ATP) and moves substances against a concentration gradient (from a region of lower concentration to a region of higher concentration) using protein carrier pumps in the cell membrane, whereas diffusion is passive and moves substances down a concentration gradient.

(a) Max [2 marks]
- Net movement of particles from a region of higher concentration to lower concentration; [1]
- Down a concentration gradient; [1]
- Result of random movement of particles; [1]

(b) Max [4 marks]
- Description: rate of diffusion / mass of glucose increases as temperature increases; [1]
- Linear relationship / quantitative ref to Table 1.1; [1]
- Explanation: higher temperature increases kinetic energy of molecules; [1]
- Molecules move faster; [1]
- Resulting in more frequent passages through pores / faster rate of diffusion; [1]

(c) Max [2 marks]
- Correct substitution / working shown: \(\frac{9.6 - 2.4}{2.4} \times 100\); [1]
- Correct answer: 300; [1]

(d) Max [3 marks]
- Surface area / membrane surface area; [1]
- Concentration gradient; [1]
- Diffusion distance / membrane thickness; [1]

(e) Max [3 marks]
- Active transport requires energy / ATP (whereas diffusion is passive); [1]
- Active transport moves substances against a concentration gradient / from low to high concentration; [1]
- Active transport requires protein carriers / pumps in the cell membrane; [1]

(a) The placenta is adapted for exchange by having:
1. Many chorionic villi which provide a very large surface area.
2. A thin barrier (only one or two cells thick) ensuring a short diffusion distance.
3. A rich blood supply on both maternal and fetal sides to maintain steep concentration gradients.

(b) (i) Progesterone maintains the lining of the uterus (endometrium) to prevent miscarriage and inhibits uterine contractions.
(ii) Early pregnancy: Corpus luteum. Later pregnancy: Placenta.

(c) Table 2.1 completed:
- Oxygen: From mother to fetus (✓)
- Carbon dioxide: From fetus to mother (✓)
- Urea: From fetus to mother (✓)
- Glucose: From mother to fetus (✓)
- Antibodies: From mother to fetus (✓)

(a) Max [4 marks]
- Chorionic villi provide a large surface area; [1]
- Thin exchange membrane / thin walls (only one or two cells thick) provides a short diffusion path; [1]
- Maternal and fetal blood flow in opposite directions / Counter-current flow maintains steep concentration gradient; [1]
- Extensive network of capillaries / rich blood supply ensures rapid transport; [1]

(b)(i) Max [2 marks]
- Maintains uterine lining / endometrium; [1]
- Prevents menstruation / prevents uterine contractions; [1]

(b)(ii) Max [2 marks]
- First few weeks: Corpus luteum; [1]
- Later: Placenta; [1]

(c) [5 marks]
- Award 1 mark for each correctly placed tick:
- Oxygen: Mother to fetus; [1]
- Carbon dioxide: Fetus to mother; [1]
- Urea: Fetus to mother; [1]
- Glucose: Mother to fetus; [1]
- Antibodies: Mother to fetus; [1]

(a) Walking into bright sunlight triggers pupil constriction. The circular muscles in the iris contract while the radial muscles relax, reducing the size of the pupil. This limits the amount of light entering the eye, preventing damage to the retina.

(b) (i) Structure [X] is the relay neurone.
(ii) The effector is the iris (or iris muscles). Its response is that circular muscles contract and radial muscles relax, causing the pupil to constrict.

(c) (i) Rods are highly sensitive to low light intensities (working well in dim light), while cones are only sensitive to high light intensities (bright light).
(ii) Rods cannot distinguish colors (providing monochrome vision), whereas cones allow color perception (three types: red, green, blue cones).
(iii) Cones are highly concentrated at the fovea, while rods are found at the periphery of the retina and are completely absent from the fovea.

(a) Max [4 marks]
- Pupil diameter decreases / constricts; [1]
- Circular muscles contract; [1]
- Radial muscles relax; [1]
- Prevents too much light entering the eye / protects the retina from damage; [1]

(b)(i) [1 mark]
- Relay neurone / coordinator; [1]

(b)(ii) Max [2 marks]
- Effector: Iris / iris muscles; [1]
- Response: Circular muscles contract OR radial muscles relax; [1]

(c) Max [6 marks] - 2 marks for each section:
- (i) Sensitivity:
- Rods sensitive to low light intensity / work in dim light; [1]
- Cones sensitive to high light intensity / work in bright light; [1]
- (ii) Color:
- Cones detect color / three different types (red, green, blue); [1]
- Rods only detect grayscale / black and white / do not detect color; [1]
- (iii) Distribution:
- Cones concentrated in the fovea / yellow spot; [1]
- Rods absent from fovea / distributed around periphery; [1]

(a) \(6\text{CO}_2 + 6\text{H}_2\text{O} \xrightarrow{\text{light, chlorophyll}} \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2\)

(b) At high light intensities, light is no longer the factor restricting the rate. If carbon dioxide is only present at 0.04%, the rate plateaus because there are not enough carbon dioxide molecules to react with the hydrogen from water. Increasing the concentration of carbon dioxide to 0.2% increases the rate of photosynthesis, demonstrating that CO2 was previously the limiting factor.

(c) Photosynthesis is controlled by enzymes. Temperature changes enzyme kinetic energy; keeping it constant ensures that changes in the rate of photosynthesis are due only to changes in light intensity or carbon dioxide concentration, and not due to temperature affecting enzyme activity / denaturation.

(d) Carbon dioxide in the air moves through stomata via diffusion, travels into the air spaces in the spongy mesophyll layer, dissolves in the moist lining of the mesophyll cell walls, and diffuses through the cell membrane and cytoplasm into the chloroplasts.

(a) Max [2 marks]
- Correct reactants and products: \(\text{CO}_2 + \text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + \text{O}_2\); [1]
- Correct balancing: \(6\text{CO}_2 + 6\text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2\); [1]

(b) Max [3 marks]
- At high light intensities, light is no longer limiting; [1]
- Low carbon dioxide concentration restricts the carbon-fixing reactions / dark reactions; [1]
- Increasing CO2 concentration increases the rate (until another factor becomes limiting); [1]

(c) Max [4 marks]
- Enzymes control the chemical reactions of photosynthesis; [1]
- Temperature affects the kinetic energy of enzymes and substrates; [1]
- High temperatures can denature enzymes; [1]
- Low temperatures reduce molecular collisions / rate; [1]
- Keeping it constant ensures valid results / only one independent variable is altered; [1]

(d) Max [4 marks]
- Diffusion from high concentration in atmosphere to low in leaf; [1]
- Passes through stomata; [1]
- Diffuses through intercellular air spaces / spongy mesophyll; [1]
- Dissolves in the layer of water/moisture on cell walls; [1]
- Diffuses across cell wall, cell membrane, and cytoplasm into chloroplasts; [1]

(a) Phototropism is a growth response to a directional light stimulus, where shoots grow towards the light (positive phototropism). Gravitropism is a growth response to gravity, where shoots grow away from gravity (negative gravitropism) and roots grow towards gravity (positive gravitropism).

(b) Auxin is produced at the shoot tip. When unilateral light shines on the shoot, auxin diffuses to the shaded side of the shoot. This high concentration of auxin on the shaded side stimulates rapid cell elongation, causing the shaded side to grow faster than the lit side, which makes the shoot bend towards the light.

(c) Place several germinating bean seeds horizontally on a damp paper towel inside a petridish. Place the petridish vertically inside a dark box (to prevent phototropic responses). For the control, place identical seeds on a rotating clinostat in the same dark box. After 24-48 hours, observe that the roots of stationary seeds grow downwards, while the roots of the rotating seeds grow straight outwards because they experience gravitational pull equally on all sides.

(d) Positive gravitropism ensures roots grow downwards into the soil, anchoring the plant securely and allowing it to absorb water and dissolved mineral ions. Negative gravitropism ensures shoots grow upwards, reaching the surface to get sunlight for photosynthesis.

(a) Max [3 marks]
- Phototropism is a response to light direction, whereas gravitropism is a response to gravity; [1]
- Shoots show positive phototropism / negative gravitropism; [1]
- Roots show negative phototropism / positive gravitropism; [1]

(b) Max [4 marks]
- Auxin is synthesized / produced in the tip of the shoot; [1]
- Light causes auxin to move / diffuse to the shaded side; [1]
- Auxin stimulates cell elongation on the shaded side; [1]
- Unequal growth / faster growth on shaded side causes bending towards light; [1]

(c) Max [4 marks]
- Place germinating seeds horizontally; [1]
- Keep in darkness / eliminate light factor; [1]
- Use of clinostat as a control (rotating seeds); [1]
- Observe root bending downwards in stationary seeds vs straight in control; [1]

(d) Max [2 marks]
- Roots grow downwards for anchorage; [1]
- Roots reach water / minerals deep in soil; [1]
- Shoots grow upwards to find light / for photosynthesis; [1]

(a) \(\text{C}_6\text{H}_{12}\text{O}_6 \rightarrow 2\text{C}_2\text{H}_5\text{OH} + 2\text{CO}_2\)

(b) Anaerobic respiration is less efficient because glucose is only partially broken down. Consequently, much less energy is released per glucose molecule, and most of the energy remains stored in the chemical bonds of the organic end-products (lactic acid or ethanol).

(c) Lactic acid produced in muscles enters the bloodstream and is transported to the liver. Aerobic respiration resumes fully after exercise, and the oxygen debt must be paid. Oxygen is used to break down / oxidize the lactic acid in the liver to carbon dioxide and water.

(d) The ileum is adapted for maximum absorption by having:
1. Villi and microvilli, which enormously increase the surface area available for absorption.
2. A single-cell thick epithelium which minimizes the diffusion distance.
3. A rich network of blood capillaries inside each villus to rapidly carry away glucose and amino acids, maintaining a concentration gradient.
4. Lacteals inside each villus to absorb fatty acids and glycerol.
5. Numerous mitochondria in epithelial cells to provide ATP for active transport.

(a) Max [2 marks]
- Correct reactant and products: \(\text{C}_6\text{H}_{12}\text{O}_6 \rightarrow \text{C}_2\text{H}_5\text{OH} + \text{CO}_2\); [1]
- Correct balancing: \(\text{C}_6\text{H}_{12}\text{O}_6 \rightarrow 2\text{C}_2\text{H}_5\text{OH} + 2\text{CO}_2\); [1]

(b) Max [2 marks]
- Incomplete breakdown of glucose molecule; [1]
- Less energy released per molecule / energy remains locked in products (lactic acid/ethanol); [1]

(c) Max [4 marks]
- Lactic acid builds up in muscles / enters blood; [1]
- Transported to the liver; [1]
- Aerobic respiration/oxygen used to break down lactic acid; [1]
- Reference to 'oxygen debt' / breathing rate remains high to provide oxygen; [1]

(d) Max [6 marks]
- Large surface area due to villi / microvilli / folding of the wall; [1]
- One-cell thick wall / epithelium gives short diffusion pathway; [1]
- Capillaries inside villi absorb glucose / amino acids; [1]
- Lacteals inside villi absorb fatty acids / glycerol; [1]
- Constant blood flow maintains steep concentration gradient; [1]
- Epithelial cells have many mitochondria to provide energy for active transport; [1]

(a) (i)
- For Cylinder X: Change in mass = 2.30 g - 2.48 g = -0.18 g. Percentage change = (-0.18 / 2.48) * 100 = -7.258% -> -7.3%
- For Cylinder Y: Change in mass = 2.20 g - 2.55 g = -0.35 g. Percentage change = (-0.35 / 2.55) * 100 = -13.725% -> -13.7%

(ii) The initial masses of the zucchini cylinders were not identical. Calculating the percentage change in mass allows for a fair comparison of the changes across different cylinders.

(b) Graph plotting requirements:
- Axes correctly labelled with units: 'Concentration of sucrose solution / mol/dm³' on the x-axis and 'Percentage change in mass / %' on the y-axis.
- Suitable linear scale chosen so that the plotted points occupy more than half of the grid in both directions.
- All 5 points plotted accurately using small crosses or dots inside circles.
- A smooth line of best fit drawn through the points, showing intercept at 0.4 mol/dm³.

(c) (i) Estimate of concentration of zucchini cell sap = 0.40 mol/dm³ (where the line of best fit crosses the x-axis, i.e., 0% change in mass).

(ii) The zucchini cells are plasmolysed / flaccid. This is because water moved out of the cells by osmosis down a water potential gradient (from a region of higher water potential inside the cells to a lower water potential in the external 0.8 mol/dm³ solution).

(d) (i) Biological drawing:
- Clear, continuous outline with no sketching or shading.
- Size larger than half the space provided.
- Relative proportions correct (e.g., large vacuole, double line for cell wall).
- Correct polygonal/rectangular shape of the plant cell.

(ii) Length of line AB = 45 mm (accept 44-46 mm).
Magnification = Image Size / Actual Size = 45 mm / 0.15 mm = x300.

(e) Experimental design:
- Independent variable: At least 5 different temperatures (e.g., 10°C, 20°C, 30°C, 40°C, 50°C using thermostatically-controlled water baths).
- Dependent variable: Measure the percentage change in mass of the zucchini cylinders over a set period of time (e.g., 30 minutes).
- Control variables: Zucchini cylinders must have the same initial dimensions/surface area, be cut from the same zucchini, and be immersed in the same concentration of sucrose solution (e.g., 0.4 mol/dm³ or distilled water).
- Replicates: Repeat the experiment at each temperature at least 3 times and calculate average values to ensure reliability.

Part (a)
- (i) 2 marks: 1 mark for each correct value with shown working (-7.3% and -13.7%).
- (ii) 1 mark: Identifies that initial masses of the tissue cylinders are different/unequal, so percentage change standardises the comparison.

Part (b)
- 4 marks: 1 mark for correct axes labelling with units; 1 mark for linear, appropriate scale occupying >50% of grid; 1 mark for accurate plotting of all 5 points; 1 mark for smooth best-fit line.

Part (c)
- (i) 1 mark: Correct value from candidate's graph (expected: 0.40 mol/dm³).
- (ii) 2 marks: State 'plasmolysed / flaccid' (1 mark); explain that water moved out of the cells via osmosis down a water potential gradient (1 mark).

Part (d)
- (i) 4 marks: Clear line quality with no shading (1 mark); large size (1 mark); cell wall shown as double line (1 mark); correct overall proportions and polygonal shape (1 mark).
- (ii) 2 marks: 1 mark for correct measurement of line AB (45 mm ± 1 mm); 1 mark for correct calculation of magnification with working (x300).

Part (e)
- 4 marks (Max 4 from following points):
- Use of at least 5 different temperatures (1 mark).
- Method for measuring dependent variable (e.g., percentage change in mass over a fixed time) (1 mark).
- Control of at least two key variables (e.g., dimensions of zucchini, volume/concentration of solution) (1 mark).
- Replication of trials (at least 3 times at each temperature) to calculate averages (1 mark).

(a) (i) Volumes of gas collected:
- Glucose: 36.0 cm³
- Sucrose: 24.0 cm³
- Lactose: 3.0 cm³

(ii) Rate of gas production:
- Glucose: 36.0 cm³ / 15 minutes = 2.4 cm³/minute
- Sucrose: 24.0 cm³ / 15 minutes = 1.6 cm³/minute

(b) (i) Gas: Carbon dioxide.
Test: Bubble the gas through limewater.
Positive result: Limewater turns cloudy / milky.

(ii) The paraffin oil layer acts as a barrier to oxygen. It prevents oxygen from entering the mixture, ensuring anaerobic conditions are maintained.

(c) Bar chart plotting requirements:
- Axes labelled with units: 'Type of sugar' on the x-axis and 'Rate of gas production / cm³/minute' on the y-axis.
- Suitable linear scale on the y-axis starting from 0.0.
- Bars drawn with equal width, distinct gaps between them (non-touching bars), and plotted accurately (Glucose = 2.4, Sucrose = 1.6, Lactose = 0.2).
- Neat construction using a ruler.

(d) Importance: Temperature affects enzyme activity and the kinetic energy of yeast cells, which alters the rate of respiration. Therefore, it must be kept constant to ensure the validity of the results.
Method: Use a thermostatically-controlled water bath, or monitor the temperature of a water bath with a thermometer and add hot or cold water as needed.

(e) Experimental design:
- Independent variable: At least 5 different pH levels (e.g., pH 4, 5, 6, 7, and 8) established using buffer solutions.
- Dependent variable: Measure the volume of carbon dioxide gas produced in a gas syringe over a fixed time period (e.g., 10 minutes).
- Control variables: Keep the concentration and volume of both the yeast suspension and the sugar solution constant, and maintain a constant temperature (e.g., 35°C in a water bath).
- Replicates: Repeat the experiment at least 3 times at each pH level and calculate the average rate of respiration.

Part (a)
- (i) 2 marks: All 3 values correct (Glucose = 36.0, Sucrose = 24.0, Lactose = 3.0) = 2 marks. 1 or 2 correct = 1 mark.
- (ii) 2 marks: Correct calculations of rate (Glucose = 2.4 cm³/min, Sucrose = 1.6 cm³/min) with units shown (1 mark each).

Part (b)
- (i) 2 marks: Identify 'carbon dioxide' (1 mark); test using limewater and state positive result as cloudy/milky (1 mark).
- (ii) 1 mark: Explain that it prevents oxygen from entering/ensures anaerobic conditions.

Part (c)
- 4 marks: 1 mark for correct axes labels with units; 1 mark for linear scale on y-axis; 1 mark for accurate bar heights; 1 mark for bars of equal width with gaps between them.

Part (d)
- 3 marks: 1 mark for explanation that temperature affects enzyme/kinetic energy/respiration rate; 2 marks for outlining a practical method to maintain constant temperature (e.g., thermostatically-controlled water bath or manual adjustment using hot/cold water with thermometer monitoring).

Part (e)
- 6 marks (Max 6 from following points):
- Use of buffer solutions to establish at least 5 different pH values (1 mark).
- Method to measure dependent variable (volume of gas in syringe over a set time) (1 mark).
- Control of sugar concentration and volume (1 mark).
- Control of yeast suspension concentration and volume (1 mark).
- Maintaining a constant temperature using a water bath (1 mark).
- Repeating trials at each pH at least 3 times to calculate average (1 mark).
- Relevant safety precaution (e.g., safety goggles) (1 mark).