2025 Edexcel IGCSE Biology 模擬試題連答案詳解

Fungi and plants both possess cell walls, but the composition is different. All fungi have cell walls composed of chitin, while plants have cell walls made of cellulose. Some fungi (such as yeast) are unicellular, and not all plants reproduce using seeds (e.g., mosses and ferns reproduce via spores). Furthermore, plants store carbohydrate as starch, whereas fungi store carbohydrate as glycogen.

1 mark for selecting option A.

First, convert the image length to micrometres: 4 mm = 4,000 \(\mu\text{m}\). Use the formula: Actual size = Image size / Magnification. Actual size = 4,000 / 20,000 = 0.2 \(\mu\text{m}\).

1 mark for selecting option A.

Progesterone is secreted by the corpus luteum in the ovary following ovulation. Its main role is to maintain the thickness of the uterine lining (endometrium) in preparation for possible implantation of a fertilized egg. The pituitary gland secretes LH and FSH, while estrogen is secreted by the developing follicle to repair the lining.

1 mark for selecting option B.

During exhalation, the diaphragm relaxes and domes upwards, and the external intercostal muscles relax, moving the ribcage down and in. This decreases the volume of the thorax, which increases the air pressure inside the lungs relative to atmospheric pressure, forcing air out.

1 mark for selecting option B.

A solution with a very low water potential is highly concentrated. When a potato cylinder is placed in such a solution, water moves out of the potato cells by osmosis (from a region of higher water potential in the cells to a region of lower water potential in the solution). This loss of water causes a large decrease in mass. Solution X caused the greatest mass loss (-10%), indicating it has the lowest water potential.

1 mark for selecting option B.

The pyramid of numbers starts with a single oak tree (narrow base), goes to thousands of caterpillars (wide), hundreds of blue tits (narrower), and one sparrowhawk (very narrow). This is an inverted base pyramid of numbers. The pyramid of biomass is always upright because biomass decreases at each trophic level due to energy losses, meaning the single oak tree has a much greater biomass than all the caterpillars combined.

1 mark for selecting option D.

Arteries carry blood under high pressure away from the heart, so they have a thick muscular and elastic wall, a narrow lumen to maintain pressure, and do not have valves (except semilunar valves at the exit of the heart). Veins carry blood under low pressure back to the heart, so they have a thin muscular wall, a wide lumen to reduce resistance, and valves to prevent backflow.

1 mark for selecting option B.

Ultrafiltration is a size-selective filtration process under high pressure. Small molecules like glucose, water, ions, and urea are small enough to pass through the basement membrane into the Bowman's capsule, forming the glomerular filtrate. Large structures, such as blood cells and large plasma proteins (e.g., albumin, fibrinogen), are too large to pass through the membrane and remain in the capillary blood.

1 mark for selecting option D.

In a hypertonic solution (0.6 mol/dm³), the external solution has a lower water potential than the cell sap of the sweet potato cells. Therefore, water moves out of the vacuoles and cytoplasm of the plant cells down a water potential gradient by osmosis. This loss of water causes the volume of the vacuole and cytoplasm to shrink, and the cell membrane pulls away from the cell wall, a process known as plasmolysis.

1 mark for identifying the correct option (B).

Fungi are eukaryotic organisms that have cell walls made of chitin. While many fungi are multicellular (like mushrooms and moulds), some are unicellular (such as yeast). Since they lack chloroplasts, they cannot photosynthesise and are saprotrophic or parasitic. Protoctists have varied cell walls (or none) and do not typically contain chitin. Bacteria are prokaryotic. Plants are multicellular, eukaryotic, and have cellulose cell walls and chloroplasts.

1 mark for identifying the correct option (C).

During quiet exhalation, the external intercostal muscles and the diaphragm relax. As the diaphragm relaxes, it domes upwards into the thoracic cavity. Concurrently, the ribs move downwards and inwards due to gravity and relaxation of the external intercostals. These movements reduce the volume of the thoracic cavity, which increases the air pressure inside the lungs relative to the atmosphere, forcing air out.

1 mark for identifying the correct option (A).

First, convert the image length to the same unit as the actual length. Since \( 1\text{ cm} = 10,000\ \mu\text{m} \), the image length is \( 4.5 \times 10,000 = 45,000\ \mu\text{m} \). Next, use the formula for magnification: \( \text{Magnification} = \frac{\text{Image size}}{\text{Actual size}} \). Therefore, \( \text{Magnification} = \frac{45,000\ \mu\text{m}}{1.5\ \mu\text{m}} = 30,000 \). The magnification of the image is \( \times 30,000 \).

- 1 mark for showing correct unit conversion (e.g., converting \( 4.5\text{ cm} \) to \( 45,000\ \mu\text{m} \) or \( 1.5\ \mu\text{m} \) to \( 0.00015\text{ cm} \)) and dividing by actual size. - 0.5 mark for the correct numerical value of 30,000 (or \( \times 30,000 \)).

First, calculate the change in mass: \( 3.48\text{ g} - 4.00\text{ g} = -0.52\text{ g} \). Then, calculate the percentage change relative to the initial mass: \( \text{Percentage change} = \left( \frac{-0.52}{4.00} \right) \times 100 = -13\% \). This represents a \( 13\% \) loss in mass.

- 1 mark for calculating the correct percentage change value of \( 13\% \) or showing the formula \( \frac{0.52}{4.00} \times 100 \). - 0.5 mark for correctly identifying this as a \( 13\% \) loss (or indicating the negative change with \( -13\% \)).

First, calculate the change in volume: \( 3.6\text{ dm}^3 - 3.0\text{ dm}^3 = 0.6\text{ dm}^3 \). Next, calculate the percentage increase relative to the initial volume: \( \text{Percentage increase} = \left( \frac{0.6}{3.0} \right) \times 100 = 20\% \).

- 1 mark for calculating the correct volume change (\( 0.6\text{ dm}^3 \)) and setting up the fraction over the initial volume (\( 3.0\text{ dm}^3 \)). - 0.5 mark for the final correct answer of \( 20\% \).

First, calculate cardiac output in \( \text{cm}^3\text{ min}^{-1} \) by multiplying heart rate by stroke volume: \( 72\text{ beats min}^{-1} \times 75\text{ cm}^3\text{ beat}^{-1} = 5400\text{ cm}^3\text{ min}^{-1} \). Next, convert the volume from \( \text{cm}^3 \) to \( \text{dm}^3 \) by dividing by 1000: \( \frac{5400}{1000} = 5.4\text{ dm}^3\text{ min}^{-1} \).

- 1 mark for calculating the cardiac output in \( \text{cm}^3\text{ min}^{-1} \) (\( 5400 \)) or showing the multiplication of \( 72 \times 75 \). - 0.5 mark for dividing by 1000 to obtain the correct final value of \( 5.4\text{ dm}^3\text{ min}^{-1} \) with units.

First, calculate the total volume of carbon dioxide produced during the interval: \( 57\text{ cm}^3 - 12\text{ cm}^3 = 45\text{ cm}^3 \). Then, divide this volume by the duration of the experiment (\( 15\text{ minutes} \)) to find the rate: \( \text{Rate} = \frac{45\text{ cm}^3}{15\text{ min}} = 3\text{ cm}^3\text{ min}^{-1} \).

- 1 mark for calculating the total volume produced (\( 45\text{ cm}^3 \)) and dividing by \( 15 \) to get the numerical value \( 3 \). - 0.5 mark for providing the correct units, such as \( \text{cm}^3\text{ min}^{-1} \) or \( \text{cm}^3/\text{min} \).

First, find the area of a single quadrat: \( 0.5\text{ m} \times 0.5\text{ m} = 0.25\text{ m}^2 \). The total area sampled by 10 quadrats is \( 10 \times 0.25\text{ m}^2 = 2.5\text{ m}^2 \). The mean density of dandelions is \( \frac{15\text{ dandelions}}{2.5\text{ m}^2} = 6\text{ dandelions per m}^2 \). To estimate the total population, multiply the mean density by the total area of the field: \( 6\text{ dandelions per m}^2 \times 800\text{ m}^2 = 4800 \) dandelions.

- 1 mark for calculating the correct total sampled area (\( 2.5\text{ m}^2 \)) and the mean density (\( 6\text{ per m}^2 \)). - 0.5 mark for multiplying the density by the field area to get the final correct answer of \( 4800 \).

First, calculate the total energy absorbed by the water using the formula \( E = m \times c \times \Delta T \), where \( m = 20.0\text{ g} \), \( c = 4.2\text{ J g}^{-1}\ ^\circ\text{C}^{-1} \), and \( \Delta T = 15.0^\circ\text{C} \): \( E = 20.0 \times 4.2 \times 15.0 = 1260\text{ J} \). Next, divide this total energy by the mass of the food sample to find the energy released per gram: \( \text{Energy per gram} = \frac{1260\text{ J}}{0.5\text{ g}} = 2520\text{ J g}^{-1} \).

- 1 mark for calculating the total energy absorbed by the water (\( 1260\text{ J} \)) or showing the equation with correct values. - 0.5 mark for dividing by the mass of the food sample to obtain \( 2520\text{ J g}^{-1} \).

The parental plants are both heterozygous, meaning they have the genotype \(Tt\). When crossed (\(Tt \times Tt\)), the expected genetic combinations in the offspring can be shown using a Punnett square: \(1\ TT\) (tall), \(2\ Tt\) (tall), and \(1\ tt\) (dwarf). The probability of producing a dwarf offspring (genotype \(tt\)) is \(1\) out of \(4\), which is \(0.25\) or \(25\%\).

- 1 mark for determining the genotypes of the parents (\(Tt \times Tt\)) and the possible offspring genotypes, identifying \(tt\) as the dwarf phenotype. - 0.5 mark for expressing the probability correctly as \(25\%\).

When a plant cell is placed in a concentrated sucrose solution, the solution outside has a lower water potential than the cell cytoplasm and vacuole. Consequently, water moves out of the cell down its water potential gradient by osmosis across the selectively permeable cell membrane. This loss of water causes the cytoplasm to shrink and the cell membrane to pull away from the cell wall, resulting in a plasmolysed cell.

1.0 mark: Correctly describing that the cell membrane pulls away from the cell wall (or that the cell becomes plasmolysed). 0.5 marks: Explaining that this occurs because water leaves the cell by osmosis.

In the dark, the light-dependent stage of photosynthesis cannot occur, so the leaf no longer requires carbon dioxide as a reactant. To prevent water loss through transpiration when gas exchange is not needed, guard cells lose water by osmosis, become flaccid, and close the stomatal pores. This closure prevents the diffusion of carbon dioxide into the leaf.

1.0 mark: Explaining that stomata close in the dark as guard cells lose water and become flaccid. 0.5 marks: Stating that photosynthesis stops in the dark so there is no concentration gradient driving carbon dioxide diffusion into the leaf.

Untreated sewage contains high levels of organic matter. Decomposers (specifically aerobic bacteria) feed on this organic matter and multiply rapidly. As these bacteria break down the waste, they carry out aerobic respiration, which consumes large quantities of dissolved oxygen from the water, resulting in a severe drop in oxygen levels.

1.0 mark: Explaining that decomposers/bacteria multiply rapidly as they break down/feed on the organic waste in the sewage. 0.5 marks: Identifying that these microorganisms consume dissolved oxygen during aerobic respiration.

(a) Change in mass = \(3.61 - 4.20 = -0.59\text{ g}\). Percentage change = \((-0.59 / 4.20) \times 100 = -14.0476\%\). Rounded to one decimal place, this is \(-14.0\%\) (or a decrease of \(14.0\%\)). (b) The sucrose solution has a lower water potential than the cell sap of the potato cells. Water moves out of the potato cells down a water potential gradient, by osmosis, across the selectively permeable cell membrane. This net movement of water out of the cells causes them to lose mass and become flaccid.

Part (a): 2 marks total. 1 mark for correct working to calculate the percentage change: \((-0.59 / 4.20) \times 100\). 1 mark for correct final answer: \(-14.0\%\) (accept a decrease of \(14.0\%\); reject \(14\%\) or \(-14\%\) without the decimal place). Part (b): 2.5 marks total. 1 mark for stating that the sucrose solution has a lower water potential than the potato cells. 1 mark for stating that water moves out of the cells by osmosis. 0.5 mark for mentioning that this movement occurs across a selectively permeable membrane.

(a) First, convert units to be consistent: \(48\text{ mm} = 48 \times 1000 = 48000\text{ }\mu\text{m}\). Magnification = Image size / Actual size = \(48000 / 6 = 8000\). Thus, the magnification is \(\times 8000\). (b) The function of the nucleolus is to synthesize ribosomal RNA (rRNA) and assemble ribosomes.

Part (a): 3 marks total. 1 mark for converting \(48\text{ mm}\) to \(48000\text{ }\mu\text{m}\) (or \(6\text{ }\mu\text{m}\) to \(0.006\text{ mm}\)). 1 mark for correct formula or division: \(48000 / 6\). 1 mark for correct magnification of \(8000\) or \(\times 8000\). Part (b): 1.5 marks total. 1.5 marks for stating the synthesis of ribosomes / ribosomal RNA (rRNA) (accept production of ribosomes).

(a) Germination will only occur in Tube B. Germination requires three key conditions: water (moisture), oxygen, and warmth (a suitable temperature). Tube B is the only tube that provides all three conditions. Tube A lacks water, and Tube C lacks warmth (the temperature is too low for enzymes to work efficiently). (b) The layer of oil in Tube D forms a barrier that prevents oxygen from the air dissolving in the water and reaching the seeds. Without oxygen, the seeds cannot carry out aerobic respiration, which is necessary to release energy for growth and germination.

Part (a): 2.5 marks total. 1 mark for predicting Tube B (only). 1.5 marks for explaining that germination requires water, oxygen, and warmth, and showing how Tube A (lacks water) and Tube C (lacks warmth/too cold for enzymes) fail to meet these requirements. Part (b): 2 marks total. 1 mark for explaining that the oil layer blocks/prevents oxygen from reaching the seeds. 1 mark for stating that oxygen is needed for aerobic respiration to provide energy for germination.

(a) Ratio of lower to upper epidermis = \(98 : 12\). To write this in the form \(X : 1\), divide both sides by 12: \(98 / 12 = 8.1667\). Rounded to one decimal place, \(X = 8.2\). Thus, the ratio is \(8.2 : 1\). (b) The upper surface of the leaf is directly exposed to sunlight and higher temperatures, which increases evaporation of water. By having fewer stomata on the upper surface and more on the cooler, shaded lower surface, the plant reduces the rate of transpiration and minimizes water loss, preventing wilting.

Part (a): 2 marks total. 1 mark for showing correct working: \(98 / 12\) or \(98 : 12\). 1 mark for the correct ratio of \(8.2 : 1\) (accept \(8.2\) with working; reject other ratios like \(8.17 : 1\) or \(8 : 1\) unless working is shown). Part (b): 2.5 marks total. 1 mark for stating that the upper surface receives direct sunlight/heat. 1 mark for stating that having fewer stomata on the upper surface reduces water loss/transpiration rate. 0.5 mark for linking this to conserving water or preventing wilting.

(a) In bright light, nerve impulses stimulate the circular muscles of the iris to contract, while the radial muscles relax. This coordinated muscular activity reduces the diameter of the pupil (constriction), limiting the amount of light that enters the eye. (b) This reflex action protects the retina and its highly sensitive photoreceptor cells (rods and cones) from damage caused by high-intensity light, ensuring the individual's vision is preserved.

Part (a): 2.5 marks total. 1 mark for stating circular muscles contract. 1 mark for stating radial muscles relax. 0.5 mark for explaining this leads to pupil constriction / smaller pupil size. Part (b): 2 marks total. 1 mark for stating it protects the retina / photoreceptors from damage. 1 mark for stating that excessive light could cause blindness / permanent damage.

(a) Increase in red blood cells = \((6.2 \times 10^6) - (5.0 \times 10^6) = 1.2 \times 10^6\text{ cells per mm}^3\). Percentage increase = \((1.2 \times 10^6 / 5.0 \times 10^6) \times 100 = (1.2 / 5.0) \times 100 = 24\%\). (b) At high altitude, the atmospheric pressure is lower, and the oxygen concentration in the air is reduced. Increasing the number of red blood cells increases the hemoglobin content in the blood, enhancing its oxygen-carrying capacity. This ensures adequate delivery of oxygen to respiring tissues for aerobic respiration.

Part (a): 2 marks total. 1 mark for calculating the difference (\(1.2 \times 10^6\)) or showing the correct formula: \(((6.2 - 5.0) / 5.0) \times 100\). 1 mark for the correct percentage increase: \(24\%\). Part (b): 2.5 marks total. 1 mark for stating there is less oxygen available in the air at high altitudes. 1 mark for explaining that more red blood cells mean more hemoglobin to transport oxygen (increases oxygen-carrying capacity). 0.5 mark for linking this to sustaining aerobic respiration / preventing fatigue.

(a) Glucose molecules are small enough to pass through the walls of the glomerulus during ultrafiltration, so they enter the Bowman's capsule and are present in the filtrate. However, as the filtrate passes through the proximal convoluted tubule, all the glucose is selectively reabsorbed back into the blood capillaries by active transport, leaving none in the urine. (b) Factor of increase = Urine concentration / Filtrate concentration = \(2.00\text{ g/100 cm}^3 / 0.03\text{ g/100 cm}^3 = 66.6667\). Rounded to one decimal place, the factor is \(66.7\) times.

Part (a): 2.5 marks total. 1 mark for explaining that glucose is a small molecule so it is filtered during ultrafiltration into the Bowman's capsule. 1 mark for stating that glucose is selectively reabsorbed back into the blood in the proximal convoluted tubule. 0.5 mark for mentioning this reabsorption occurs via active transport. Part (b): 2 marks total. 1 mark for showing correct division: \(2.00 / 0.03\). 1 mark for correct final answer: \(66.7\) (accept \(67\) if working is shown; reject \(66\) without correct rounding).

(a) 1. Find the total area of the field: \(40\text{ m} \times 60\text{ m} = 2400\text{ m}^2\). 2. Find the area of one quadrat: \(0.5\text{ m} \times 0.5\text{ m} = 0.25\text{ m}^2\). 3. Find the total area sampled: \(15 \text{ quadrats} \times 0.25\text{ m}^2 = 3.75\text{ m}^2\). 4. Calculate the estimated total population: \((\text{Total dandelions found} / \text{Total area sampled}) \times \text{Total field area} = (30 / 3.75) \times 2400 = 8 \times 2400 = 19200\). (b) Random sampling is essential to avoid bias (e.g., consciously placing quadrats where there are many dandelions, which would overestimate the population). It ensures that the sample is representative of the entire field, making the estimate more valid.

Part (a): 3 marks total. 1 mark for calculating the total field area: \(2400\text{ m}^2\). 1 mark for showing correct working to find either the total sampled area (\(3.75\text{ m}^2\)) or the average density (\(8\text{ plants per m}^2\) or \(2\text{ plants per quadrat}\)). 1 mark for the correct final population estimate: \(19200\). Part (b): 1.5 marks total. 1 mark for stating that random placement avoids bias / ensures a representative sample. 0.5 mark for mentioning that this increases the validity of the population estimate.

1. Calculate the change in mass: \(4.3\ \text{g} - 5.0\ \text{g} = -0.7\ \text{g}\).
2. Calculate the percentage change in mass: \((-0.7 / 5.0) \times 100 = -14.0\%\) (a decrease of 14.0%).
3. Determine water potential relationship: The water potential of the potato cells was higher than that of the \(0.4\ \text{mol/dm}^{3}\) sucrose solution.
4. Explain: Water moved out of the potato cells into the sucrose solution by osmosis, down a water potential gradient, through the selectively permeable membrane, resulting in a loss of mass.

- Award 1 mark for calculating the correct change in mass (-0.7 g).
- Award 1 mark for the correct calculation of percentage change (-14.0% or 14% decrease).
- Award 1 mark for identifying that the potato cells had a higher water potential than the solution.
- Award 1.5 marks for explaining that water leaves the cells down a water potential gradient via osmosis through a selectively permeable cell membrane (0.5 marks per point).

1. Calculate the breathing rate during exercise: An increase of 150% means the new rate is \(12 + (1.50 \times 12) = 12 + 18 = 30\) breaths per minute.
2. Calculate the minute ventilation during exercise: \(\text{Breathing Rate} \times \text{Tidal Volume} = 30\ \text{breaths/min} \times 1.8\ \text{dm}^{3} = 54.0\ \text{dm}^{3}\text{/min}\).
3. Explain alveolar adaptations:
- Alveoli have walls that are only one cell thick, which minimizes the diffusion distance for oxygen and carbon dioxide.
- Alveoli are present in large numbers and have a folded structure, which maximizes the total surface area available for gas exchange.

- Award 1 mark for calculating the exercise breathing rate of 30 breaths per minute.
- Award 1 mark for the correct calculation of minute ventilation (54.0).
- Award 0.5 marks for stating the correct unit (dm³/min).
- Award 1 mark for explaining the adaptation of thin walls (one cell thick) to reduce diffusion distance.
- Award 1 mark for explaining the adaptation of large surface area to maximize diffusion rate.

1. Calculate the cross-sectional area of the afferent arteriole: \(A_{\text{afferent}} = \pi \times 15^{2} = 225\pi\ \mu\text{m}^{2}\).
2. Calculate the cross-sectional area of the efferent arteriole: \(A_{\text{efferent}} = \pi \times 10^{2} = 100\pi\ \mu\text{m}^{2}\).
3. Calculate the percentage reduction: \(((225\pi - 100\pi) / 225\pi) \times 100 = (125 / 225) \times 100 \approx 55.6\%\) (accept 56%).
4. Explain how this facilitates ultrafiltration: The narrower efferent arteriole creates a bottleneck effect, which generates a high hydrostatic pressure in the glomerulus, forcing water and small solute molecules out of the blood capillaries into the Bowman's capsule.
5. Name two substances: Glucose, urea, water, amino acids, or mineral ions.

- Award 1 mark for calculating the correct cross-sectional areas or their simplified ratio.
- Award 1 mark for calculating the correct percentage reduction of 55.6% (or 56%).
- Award 1.5 marks for explaining that the narrow efferent arteriole creates high hydrostatic pressure to force water and small molecules out of the glomerulus.
- Award 1 mark (0.5 marks each) for naming any two correct filtered substances (e.g., glucose, urea, water, amino acids, salts).

1. Find the radius of the capillary tube: \(r = 1.0\ \text{mm} / 2 = 0.5\ \text{mm}\).
2. Calculate the volume of water taken up (cylinder volume): \(V = 3.142 \times 0.5^{2} \times 45 = 3.142 \times 0.25 \times 45 = 35.3475\ \text{mm}^{3}\).
3. Calculate the rate of water uptake per minute: \(35.3475\ \text{mm}^{3} / 15\ \text{min} \approx 2.36\ \text{mm}^{3}\text{/min}\) (or 2.4).
4. List environmental factors: Increasing temperature and increasing wind speed (or increasing light intensity) will increase transpiration, thereby increasing water uptake.

- Award 1 mark for correctly determining the radius as 0.5 mm and calculating the volume of the cylinder as 35.35 mm³.
- Award 1 mark for dividing the volume by time (15 mins) to get approximately 2.36 mm³/min.
- Award 0.5 marks for the correct unit (mm³/min).
- Award 2 marks (1 mark each) for naming two correct environmental factors that increase water uptake when increased (accept temperature, wind speed/air flow, light intensity; reject humidity as increasing humidity decreases transpiration).

1. Calculate the number of blastocysts: \(180 \times 0.15 = 27\) blastocysts.
2. Calculate the number of live births: \(27 \times 0.222 = 5.994 \approx 6\) calves.
3. State genetic relationship with donor: The calves are genetically identical clones of the donor prized cow because they contain the diploid nucleus and genomic DNA derived entirely from the donor cow's somatic cell.
4. State genetic relationship with surrogate: The calves are genetically unrelated to the surrogate mothers because the surrogates only provide the uterine environment for gestation and do not contribute genetic material.

- Award 1 mark for calculating the correct number of blastocysts (27).
- Award 1.5 marks for calculating the correct number of live cloned calves (6).
- Award 1 mark for explaining that the calves are genetically identical to the donor cow due to receiving its diploid nucleus.
- Award 1 mark for explaining that the calves are genetically unrelated to the surrogate mothers because no nuclear genetic material is shared.

1. Use the magnification formula: \(\text{Actual size} = \text{Image size} / \text{Magnification}\).
2. Convert image size to micrometres: \(16\ \text{mm} = 16,000\ \mu\text{m}\).
3. Calculate actual size: \(16,000\ \mu\text{m} / 400 = 40\ \mu\text{m}\).
4. Explain adaptations:
- Palisade cells contain a high density of chloroplasts, which contain chlorophyll to absorb maximum light energy for photosynthesis.
- They have a tall, columnar shape, which allows many cells to be tightly packed together at the upper surface of the leaf, maximizing light capture.

- Award 1 mark for converting 16 mm into 16,000 micrometres.
- Award 1 mark for the correct calculation of actual size (40).
- Award 0.5 marks for stating the correct unit (Ιm).
- Award 1 mark for explaining that having many chloroplasts allows maximum light absorption.
- Award 1 mark for explaining that the tall/columnar shape allows dense packing to maximize light capture.

1. Calculate the absolute increase in progesterone: \(45.0\ \text{nmol/L} - 1.5\ \text{nmol/L} = 43.5\ \text{nmol/L}\).
2. Calculate the percentage increase: \((43.5 / 1.5) \times 100 = 2900\%\).
3. State the source: Progesterone is secreted by the corpus luteum, which develops from the ruptured follicle after ovulation on day 14.
4. Explain the biological role: Progesterone maintains the thickness of the uterus lining (endometrium) to prepare it for implantation of a fertilized egg.

- Award 1 mark for calculating the increase in concentration (43.5 nmol/L).
- Award 1 mark for calculating the correct percentage increase (2900%).
- Award 1 mark for identifying the source of progesterone as the corpus luteum.
- Award 1.5 marks for explaining that progesterone maintains the lining of the uterus to prepare for implantation of an embryo.

1. Calculate travel time along the neurone fibres: \(\text{Time} = \text{Distance} / \text{Speed} = 1.2\ \text{m} / 80\ \text{m/s} = 0.015\ \text{seconds}\).
2. Convert this travel time to milliseconds: \(0.015\ \text{s} \times 1000 = 15\ \text{ms}\).
3. Calculate the total synaptic delay: \(2 \times 2.5\ \text{ms} = 5\ \text{ms}\).
4. Calculate the total reflex time: \(15\ \text{ms} + 5\ \text{ms} = 20\ \text{ms}\).
5. Explain myelin sheath function: The myelin sheath acts as an electrical insulator around the axon. It allows the action potential / nerve impulse to jump from one node of Ranvier to the next (saltatory conduction), rather than traveling smoothly along the entire membrane, which dramatically increases conduction speed.

- Award 1 mark for calculating the transmission time along the neurones (15 ms or 0.015 s).
- Award 1 mark for calculating the total synaptic delay (5 ms).
- Award 0.5 marks for summing them to obtain the correct total time of 20 ms.
- Award 1 mark for stating that the myelin sheath acts as an electrical insulator.
- Award 1 mark for explaining that the impulse jumps between nodes of Ranvier (saltatory conduction), speeding up transmission.

For part (a): The surface area of Cube X is calculated as 6 * (1.5 cm * 1.5 cm) = 13.5 cm^2. The volume of Cube X is calculated as 1.5 cm * 1.5 cm * 1.5 cm = 3.375 cm^3. The surface area to volume ratio (SA:Vol) is therefore 13.5 / 3.375, which simplifies to 4:1. For part (b): Cube X has a larger surface-area-to-volume ratio than the larger 3.0 cm cube (which has a ratio of 2:1). Furthermore, the diffusion distance from the outer edge to the center of Cube X is only 0.75 cm, compared to 1.5 cm for the larger cube. This means the hydrochloric acid can diffuse to the center of Cube X much more rapidly, neutralizing the sodium hydroxide and turning the phenolphthalein indicator colorless in less time.

Part (a): 1 mark for correct calculation of surface area (13.5 cm^2) and volume (3.375 cm^3). 1 mark for the correct simplified ratio written as 4:1 (or 4). Part (b): 1 mark for identifying that Cube X has a larger surface-area-to-volume ratio than the 3.0 cm cube. 1 mark for explaining that Cube X has a shorter diffusion distance to its center. 0.5 marks for stating that the acid penetrates/diffuses to the center of Cube X faster, resulting in a quicker color change.

For part (a): Coleoptile P will bend towards the light source because its intact tip detects unilateral light and produces the plant hormone auxin. Auxin diffuses downwards and accumulates on the shaded side of the shoot, stimulating cells on the shaded side to elongate faster than those on the lit side, causing bending. Coleoptile Q does not bend (it grows straight or ceases growth) because the tip has been removed, so no auxin is produced to initiate a phototropic response. For part (b): Coleoptile R will grow straight upwards without bending. This is because the opaque metal cap prevents unilateral light from reaching the tip, so the light stimulus is not detected. Consequently, auxin remains evenly distributed down all sides of the shoot, causing equal elongation of cells.

Part (a): 1 mark for stating that Coleoptile P bends towards the light while Coleoptile Q grows straight (or does not bend). 1 mark for stating that the tip of Coleoptile P detects light and produces auxin (which is absent in Q). 0.5 marks for explaining that auxin accumulates on the shaded side of P and causes cell elongation. Part (b): 1 mark for stating that Coleoptile R grows straight up (or does not bend). 1 mark for explaining that the opaque cap blocks light detection by the tip, leading to an even distribution of auxin on all sides.

To investigate the effect of wind speed on transpiration, the student can use a potometer setup:

- **C (Change):** Vary the wind speed by placing an electric fan at different measured distances from the leafy shoot (e.g., 20 cm, 40 cm, 60 cm, 80 cm, and 100 cm) or by using different speed settings on the fan.
- **O (Organism):** Use leafy shoots from the same plant species, ensuring they have a similar number of leaves and total leaf surface area.
- **R (Repeat):** Repeat the experiment at least three times at each wind speed to calculate a mean and identify any anomalous results.
- **M1 (Measure 1):** Measure the distance moved by the air bubble in the capillary tube of the potometer.
- **M2 (Measure 2):** Record the distance moved over a fixed time period, such as 10 minutes.
- **S1 & S2 (Same):** Keep other environmental factors constant, such as temperature (by conducting the experiment in a temperature-controlled room) and light intensity (by keeping a lamp at a fixed distance from the shoot).

One mark for each point up to a maximum of 6 marks:
- **C**: Change the wind speed by using a fan at different distances (minimum of 5 distances) OR different fan speed settings.
- **O**: Use shoots from the same species / same age of plant OR control the leaf surface area / number of leaves.
- **R**: Repeat the measurement at least three times at each wind speed to calculate a mean / discard anomalies.
- **M1**: Measure the distance moved by the air bubble in a capillary tube / potometer (or measure the loss in mass of the potometer setup).
- **M2**: State an appropriate time period for the measurement (e.g., 5 to 30 minutes).
- **S1 & S2**: Control at least two environmental variables from: temperature (using a water bath/thermostat), light intensity (using a lamp at a constant distance), or humidity (using a clear plastic bag or humidifier).

To investigate the effect of wind speed on transpiration, the student can use a potometer setup:

- **C (Change):** Vary the wind speed by placing an electric fan at different measured distances from the leafy shoot (e.g., 20 cm, 40 cm, 60 cm, 80 cm, and 100 cm) or by using different speed settings on the fan.
- **O (Organism):** Use leafy shoots from the same plant species, ensuring they have a similar number of leaves and total leaf surface area.
- **R (Repeat):** Repeat the experiment at least three times at each wind speed to calculate a mean and identify any anomalous results.
- **M1 (Measure 1):** Measure the distance moved by the air bubble in the capillary tube of the potometer.
- **M2 (Measure 2):** Record the distance moved over a fixed time period, such as 10 minutes.
- **S1 & S2 (Same):** Keep other environmental factors constant, such as temperature (by conducting the experiment in a temperature-controlled room) and light intensity (by keeping a lamp at a fixed distance from the shoot).

One mark for each point up to a maximum of 6 marks:
- **C**: Change the wind speed by using a fan at different distances (minimum of 5 distances) OR different fan speed settings.
- **O**: Use shoots from the same species / same age of plant OR control the leaf surface area / number of leaves.
- **R**: Repeat the measurement at least three times at each wind speed to calculate a mean / discard anomalies.
- **M1**: Measure the distance moved by the air bubble in a capillary tube / potometer (or measure the loss in mass of the potometer setup).
- **M2**: State an appropriate time period for the measurement (e.g., 5 to 30 minutes).
- **S1 & S2**: Control at least two environmental variables from: temperature (using a water bath/thermostat), light intensity (using a lamp at a constant distance), or humidity (using a clear plastic bag or humidifier).

Luteinising hormone (LH) is released by the pituitary gland and triggers ovulation (the release of a mature egg from the ovary) around day 14 of the cycle. Progesterone is secreted by the corpus luteum and works to maintain the thickness of the uterus lining in preparation for a potential pregnancy.

Correct answer is A (1 mark). LH is responsible for triggering ovulation, whereas progesterone maintains the uterine lining. Options B, C, and D misidentify these primary roles.

Osmosis is the net movement of water molecules from an area of higher water potential (dilute solution) to an area of lower water potential (concentrated solution) across a partially permeable membrane. A solution of 0.8 mol dm\(^{-3}\) has the lowest water potential compared to the cytoplasm of the potato cells. Consequently, water will move out of the potato cells by osmosis down the water potential gradient, resulting in the largest loss of water and therefore the largest percentage decrease in mass.

Correct answer is D (1 mark). 0.8 mol dm\(^{-3}\) is the most hypertonic solution, driving the greatest net movement of water out of the potato cells. Option A causes water to enter the cells (mass increase), while options B and C cause less mass loss than D.

In bright light, the pupil constricts to reduce the amount of light entering the eye, protecting the retina from damage. This reflex is controlled by antagonistic muscles in the iris: the circular muscles contract and the radial muscles relax.

Correct answer is B (1 mark). Bright light triggers the pupil reflex, leading to circular muscle contraction and radial muscle relaxation to constrict the pupil. Other options describe the incorrect muscular coordination or the reaction to dim light.

During kidney function, ultrafiltration occurs at the Bowman's capsule where small molecules are squeezed out of the blood under high pressure. Useful substances, including all glucose, are then selectively reabsorbed back into the blood from the glomerular filtrate at the proximal convoluted tubule.

Correct answer is A (1 mark). Ultrafiltration occurs in the Bowman's capsule (glomerulus) and glucose selective reabsorption occurs specifically in the proximal convoluted tubule. Other options misidentify these specific locations.

The nutrient agar medium is autoclaved, which involves heating it to high temperatures under pressure to kill any microorganisms. Additionally, the instruments such as forceps are sterilised using ethanol and then flamed to eliminate any bacteria or fungi.

1 mark for stating that the agar medium is autoclaved or heated under pressure to kill microbes. 1 mark for mentioning that instruments are sterilised using alcohol, flame, or heat.

The scrotum keeps the testes at a temperature that is \(2\) to \(3^\circ\text{C}\) lower than the core body temperature of \(37^\circ\text{C}\). This lower temperature is optimal for sperm development and viability, as higher temperatures can inhibit spermatogenesis or damage sperm cells.

1 mark for mentioning that the scrotum maintains a lower temperature than the core body temperature. 1 mark for explaining that this lower temperature is required for successful sperm production or to prevent damage to sperm.

Domestic sewage contains organic matter which serves as a food source for aerobic bacteria. These decomposers multiply rapidly and break down the organic waste. During this process, they respire aerobically, consuming large amounts of dissolved oxygen from the water.

1 mark for explaining that decomposers or bacteria multiply to break down the organic matter in sewage. 1 mark for stating that these bacteria use up dissolved oxygen during aerobic respiration.

A single-celled organism has a very large surface area to volume ratio, meaning the rate of diffusion is fast enough to meet all its gas exchange needs. In contrast, large multicellular organisms have a much smaller surface area to volume ratio and a longer diffusion distance to internal cells, making simple diffusion too slow to sustain life.

1 mark for comparing surface area to volume ratio (stating single-celled organisms have a large SA:Vol ratio while multicellular organisms have a small SA:Vol ratio). 1 mark for stating that multicellular organisms have a larger diffusion distance, making simple diffusion too slow.

When the blood concentration is too high (dehydration), the pituitary gland releases more ADH into the bloodstream. ADH increases the permeability of the collecting ducts in the nephrons to water. This causes more water to be reabsorbed back into the blood by osmosis, resulting in a smaller volume of highly concentrated urine.

1 mark for stating that ADH increases the permeability of the collecting duct to water. 1 mark for stating that more water is reabsorbed into the blood by osmosis.

Auxin is produced at the tip of the shoot and diffuses downwards. When the light is unilateral, auxin accumulates on the shaded side of the shoot. The high concentration of auxin on the shaded side stimulates cell elongation, causing this side to grow faster and longer than the illuminated side, bending the shoot towards the light.

1 mark for stating that auxin accumulates or moves to the shaded side of the shoot. 1 mark for explaining that auxin causes cell elongation on the shaded side, causing the shoot to bend towards the light.

Quadrats must be placed randomly to avoid bias and ensure that the sample is representative of the entire area, allowing for a reliable estimate. This is achieved by setting up a grid system using tape measures at right angles and using a random number generator to select coordinates for placing the quadrats.

1 mark for stating that random placement avoids bias and ensures the sample is representative. 1 mark for describing the use of a grid with random number coordinates to place the quadrats.

(a) Actual size = Image size / Magnification.
Image size = \(36 \text{ mm} = 36 \times 1000 = 36000 \ \mu\text{m}\).
Actual size = \(36000 / 400 = 90 \ \mu\text{m}\).

(b) Palisade mesophyll cells have a cellulose cell wall, chloroplasts, and a large permanent vacuole, which are absent in animal cells.

(a) [3 marks]
- 1 mark for converting image size correctly to \(36000 \ \mu\text{m}\) OR showing rearrangement: \(\text{Actual} = \text{Image} / \text{Magnification}\).
- 1 mark for correct division: \(36000 / 400\) or \(36 / 400\) yielding \(0.09 \text{ mm}\).
- 1 mark for correct final answer: \(90 \ \mu\text{m}\).

(b) [2 marks]
- 1 mark for each correct structure (Accept: cell wall, chloroplasts, large permanent vacuole; Reject: vacuole alone, unless specified as permanent / cell sap vacuole).

Water is absorbed by the seed through the micropyle, which hydrates the seed tissues and activates digestive enzymes such as amylase. These enzymes break down starch to glucose, providing energy and building blocks for the growing embryo. Water also causes the seed coat (testa) to swell and rupture, allowing the radicle to emerge. Warm temperatures are required because germination depends on enzyme-catalysed reactions. Warmth increases the kinetic energy of enzymes and substrate molecules, leading to more frequent successful collisions and faster germination, up to an optimum temperature.

Any 5 marks from the following points:
- Water activates metabolic enzymes / hydrates tissues (1 mark)
- Water allows digestion / mobilisation of stored food reserves (e.g., starch to maltose/glucose) (1 mark)
- Dissolved nutrients can be transported to the growing embryo / plumule / radicle (1 mark)
- Water intake softens / swells / bursts the seed coat (testa) (1 mark)
- Temperature is needed to provide kinetic energy for enzymes (1 mark)
- Warmth increases collision rates between enzyme and substrate molecules (1 mark)
- Reject: Temperature 'kills' or 'creates' enzymes; Accept: high temperatures denature enzymes (1 mark)

Alveoli provide an incredibly large total surface area across which oxygen can diffuse. Both the alveolar wall and the capillary wall are extremely thin, being only one cell thick, which minimizes the diffusion pathway. Furthermore, the alveoli are lined with a thin layer of moisture in which oxygen dissolves before diffusing. The extensive capillary network ensures blood is constantly moving, bringing deoxygenated blood to the lungs and carrying oxygenated blood away. Combined with ventilation (breathing), this maintains a steep concentration gradient for oxygen between the alveolar air and the blood.

Max 5 marks from:
- Large surface area (due to millions of alveoli / folded structure) (1 mark)
- Short diffusion distance / thin walls (one cell thick / squamous epithelium) (1 mark)
- Moist lining allows oxygen to dissolve / diffuse more easily (1 mark)
- Rich capillary network / rapid blood flow maintains a steep concentration gradient (1 mark)
- Ventilation / breathing in oxygen-rich air maintains concentration gradient (1 mark)
- Accept: red blood cells being close to capillary walls reduces diffusion distance (1 mark)

In Sample A, the external solution has a lower water potential than the cytoplasm of the red blood cells. Water moves out of the cells by osmosis down a water potential gradient, causing the cells to shrink and shrivel (crenation).
In Sample B, distilled water has a higher water potential than the cell cytoplasm. Water moves into the red blood cells by osmosis. Because animal cells lack a rigid cellulose cell wall, the increasing internal turgor pressure causes the cell membrane to stretch and eventually burst (lysis).

Sample A (concentrated solution) [Max 3 marks]:
- Water moves out of the red blood cells (1 mark)
- By osmosis / from a high water potential to a lower water potential (1 mark)
- Cells shrink / shrivel / become crenated (1 mark)

Sample B (distilled water) [Max 3 marks]:
- Water moves into the red blood cells (1 mark)
- By osmosis (1 mark)
- Cells swell and burst / lyse (1 mark)
- Reason: no cell wall present to resist pressure (1 mark)

Maximum overall score of 5 marks.

The student should establish a coordinate grid system along two edges of the field using tape measures. A random number generator should be used to select coordinates to avoid bias. A quadrat (e.g., \(0.5 \text{ m} \times 0.5 \text{ m} = 0.25 \text{ m}^2\)) is placed at these random coordinates. This process is repeated to collect a large sample (at least 10–15 quadrats). The number of dandelions in each quadrat is counted and recorded. The mean number of dandelions per quadrat is calculated. Finally, the total population is estimated by multiplying the mean density per unit area by the total area of the field (\(80 \times 50 = 4000 \text{ m}^2\)).

Any 5 marks from the following points:
- Lay out tape measures to form a grid system / coordinate axes (1 mark)
- Use a random number generator to select coordinates (1 mark)
- This prevents bias / ensures representative sampling (1 mark)
- Use a large number of quadrats / repeat at least 10 times (1 mark)
- Count and record the number of dandelions in each quadrat (1 mark)
- Calculate the mean number of dandelions per quadrat (or per \(\text{m}^2\)) (1 mark)
- Multiply the mean by (total field area / quadrat area) to get the total estimated population (1 mark)

When the body is dehydrated, the water potential of the blood decreases. This change is detected by osmoreceptors in the hypothalamus. The hypothalamus signals the pituitary gland to release more ADH into the blood. ADH travels in the bloodstream to the kidneys, where it targets the collecting ducts of the nephrons. ADH increases the permeability of the collecting duct walls to water. As a result, more water is reabsorbed out of the filtrate by osmosis back into the blood capillaries. This restores blood water levels and produces a smaller volume of highly concentrated, dark urine.

Any 5 marks from the following points:
- Osmoreceptors in the hypothalamus detect a decrease in blood water potential / dehydration (1 mark)
- Pituitary gland secretes more ADH into the blood (1 mark)
- ADH travels to the kidneys / nephrons (1 mark)
- ADH increases the permeability of the collecting ducts to water (1 mark)
- More water is reabsorbed from the urine/filtrate into the blood via osmosis (1 mark)
- Blood water potential returns to normal / negative feedback loop (1 mark)
- Urine produced is more concentrated / smaller in volume (1 mark)

First, an unfertilised egg cell is harvested from a donor female, and its haploid nucleus is removed to create an enucleated egg cell. Next, a diploid nucleus is extracted from a mature body (somatic) cell, such as a skin cell, from the mammal that is to be cloned. This diploid nucleus is inserted into the enucleated egg cell, or the two cells are fused using an electric shock. Another mild electric shock is applied to stimulate the reconstructed egg cell to divide by mitosis. Once it has divided to form a ball of cells (an embryo), it is implanted into the uterus of a surrogate mother sheep, where it develops into a fetus.

Any 5 marks from the following points:
- Remove the nucleus from an unfertilised egg cell to produce an enucleated egg (1 mark)
- Obtain a diploid nucleus from a body (somatic) cell of the animal to be cloned (1 mark)
- Insert this diploid nucleus into the enucleated egg cell OR fuse the somatic cell and enucleated egg using electricity (1 mark)
- Apply an electric shock to stimulate cell division / mitosis (1 mark)
- Allow the cell to develop into an embryo in vitro (1 mark)
- Implant the embryo into the uterus/womb of a surrogate mother (1 mark)
- The offspring born is genetically identical to the somatic cell donor (1 mark)

An increase in wind speed increases the transpiration rate. Moving air blows away the humid water vapour accumulating directly outside the stomata. This maintains a steep concentration gradient of water vapour between the inside of the leaf and the surrounding atmosphere, leading to faster diffusion out of the stomata.
An increase in air temperature also increases transpiration. Heat energy increases the kinetic energy of water molecules inside the leaf, causing them to evaporate more rapidly from the surfaces of the spongy mesophyll cells into the air spaces, which speeds up diffusion.

Wind Speed (Max 3 marks):
- Wind speed increases the rate of transpiration (1 mark)
- Wind removes/blows away water vapour from around the leaf surface/stomata (1 mark)
- This maintains a steep concentration / water potential gradient (1 mark)
- Resulting in faster diffusion of water vapour out of the stomata (1 mark)

Temperature (Max 3 marks):
- Temperature increases the rate of transpiration (1 mark)
- Higher temperature increases the kinetic energy of water molecules (1 mark)
- Increases rate of evaporation from the spongy mesophyll cell walls (into intercellular air spaces) (1 mark)
- Faster diffusion out of the leaf (1 mark)

Maximum overall score of 5 marks.

Part (a):
Substance A is a large plasma protein (such as albumin). During ultrafiltration in the Bowman's capsule, high pressure forces water and small solutes through the capillary walls and the basement membrane. However, proteins are too large to pass through the pores of the basement membrane, so they remain in the blood plasma and do not enter the glomerular filtrate.

Part (b):
Substance B is glucose. Since glucose is a small soluble molecule, it easily passes through the basement membrane into the glomerular filtrate, matching the plasma concentration (1.0 g/dm³). However, in a healthy person, all glucose is selectively reabsorbed from the filtrate back into the blood capillaries surrounding the proximal convoluted tubule (PCT). This reabsorption occurs against a concentration gradient using active transport (utilising energy from ATP), resulting in a concentration of 0.0 g/dm³ in the final urine.

Part (a) [Maximum 2 marks]:
- Identifies Substance A as protein / polypeptide (1)
- Explains that protein molecules are too large to pass through the basement membrane / glomerulus wall (1)
- Reject: 'cell membrane' or 'cell wall'

Part (b) [Maximum 3 marks]:
- Identifies Substance B as glucose (1)
- Explains that glucose is selectively reabsorbed (1)
- Identifies the site of reabsorption as the proximal convoluted tubule (PCT) (1)
- Mentions that this process occurs via active transport / requires energy from respiration (1)

To clone plants using micropropagation, small pieces of plant tissue (explants) are taken from a healthy, disease-resistant parent plant (often from the shoot tip). Because microbes like bacteria and fungi thrive in growth media and can destroy the tissue, the explants must be thoroughly sterilised using a sterilising solution, such as diluted bleach or ethanol.

The sterilised explants are then transferred using sterile forceps (usually flamed in ethanol) onto a sterile agar medium in a laminar flow cabinet to prevent airborne contamination. The agar medium contains essential nutrients like glucose (for respiration and energy), amino acids, mineral ions (e.g., nitrates for proteins), and plant hormones (auxins and cytokinins) to stimulate cell division and differentiation.

The cultures are incubated under controlled conditions of light and temperature to allow photosynthesis and growth. After the explants develop roots and shoots (forming small plantlets), they are carefully transferred into sterile compost or greenhouse soil to grow into mature, genetically identical plants.

Award 1 mark for each correct point up to a maximum of 5 marks:
- MP1: Explants / small pieces of tissue are cut from the parent plant (1)
- MP2: Explants are sterilised / washed in bleach / sodium hypochlorite / ethanol to kill bacteria / fungi / pathogens (1)
- MP3: Explants are placed on a sterile agar / nutrient medium (1)
- MP4: Medium contains glucose / mineral ions / amino acids AND plant growth hormones / auxins (1)
- MP5: Use of aseptic techniques, such as working in a laminar flow hood / sterilising tools in a flame (1)
- MP6: Once plantlets have grown roots/shoots, they are transferred to compost/soil (1)
- Accept: sterilising agent instead of bleach/ethanol.
- Reject: autoclaving the plant tissue itself.