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

Photosynthesis is an enzyme-controlled process. As temperature increases from 15 °C to 35 °C, the kinetic energy of molecules increases, leading to more frequent successful collisions between enzymes and substrates, which raises the rate of photosynthesis. However, at 45 °C, the temperature exceeds the optimum range, causing the active sites of these enzymes to denature (change shape permanently). Consequently, the substrate can no longer bind, leading to a drastic decline in the rate of photosynthesis.

1 mark for identifying that enzymes involved in photosynthesis are denatured at 45 °C.

A limiting factor is something present in the environment in such short supply that it restricts life processes. Since increasing the concentration of carbon dioxide from 0.03% to 0.15% resulted in a significant increase in the rate of photosynthesis, carbon dioxide concentration was the factor limiting the rate under the initial conditions.

1 mark for selecting carbon dioxide concentration as the limiting factor.

A steep concentration gradient is maintained by rapidly removing oxygen from the alveolar surface via an excellent, continuous blood supply (capillaries) and by constantly replacing air within the alveoli with fresh, oxygen-rich air through ventilation (breathing).

1 mark for identifying a good blood supply and continuous ventilation as the features that maintain the steep concentration gradient.

During exhalation at rest, the diaphragm muscle relaxes (causing it to dome upwards), and the external intercostal muscles relax (letting the ribs move downwards and inwards). These actions decrease the volume of the thoracic cavity, which increases the air pressure inside the thorax relative to atmospheric pressure, forcing air out of the lungs.

1 mark for the correct combination: diaphragm relaxes, external intercostals relax, and pressure in the thorax increases.

Wind-pollinated flowers do not need to attract insects, so they typically have small, dull green petals. Their anthers hang outside the flower to let the wind carry away pollen grains easily. Their stigmas are feathery with a large surface area to maximize the chance of catching airborne pollen grains.

1 mark for identifying Flower C as the wind-pollinated flower.

Using the formula: \(\text{actual size} = \text{image size} \div \text{magnification}\).
1. Convert the image size from centimetres to millimetres: \(4.5\text{ cm} = 45\text{ mm}\).
2. Calculate actual size in millimetres: \(45\text{ mm} \div 1500 = 0.03\text{ mm}\).
3. Convert millimetres to micrometres: \(0.03\text{ mm} \times 1000 = 30\text{ }\mu\text{m}\).

1 mark for correct calculation and unit conversion to obtain 30 micrometres.

In cold conditions, body heat must be conserved. Arterioles supplying the skin capillaries constrict (vasoconstriction) to divert blood away from the skin surface, reducing heat loss by radiation. Concurrently, hair erector muscles contract, causing the hairs on the skin to stand upright, which traps a layer of warm insulating air next to the skin.

1 mark for selecting the correct row: arterioles constrict and hair erector muscles contract.

Since the black lamb displays the recessive phenotype, its genotype must be homozygous recessive (\(ww\)). This offspring must have inherited one recessive allele (\(w\)) from each parent. Since both parents have white wool (the dominant phenotype), they must also contain at least one dominant allele (\(W\)). Therefore, both parents must be heterozygous (\(Ww\)).

1 mark for identifying both parents as heterozygous (\(Ww\)).

To find the magnification, use the formula:

$$\text{Magnification} = \frac{\text{image size}}{\text{actual size}}$$

First, convert the image size from centimetres to micrometres:

$$4.8\text{ cm} = 48\text{ mm} = 48\,000\ \mu\text{m}$$

Now, calculate the magnification:

$$\text{Magnification} = \frac{48\,000\ \mu\text{m}}{120\ \mu\text{m}} = 400$$

Therefore, the magnification is \(\times 400\).

1 mark for the correct answer C. Award 0 marks for incorrect options.

At low light intensities, light is the limiting factor. However, once the light intensity is high enough, increasing it further does not increase the rate of photosynthesis because the rate is now limited by another factor, such as the concentration of carbon dioxide in the water or the temperature. Since temperature is stated as constant and warm, carbon dioxide concentration is the limiting factor.

1 mark for the correct answer A. Award 0 marks for incorrect options.

The diffusion distance is minimized by structural features that make the barrier between air and blood as thin as possible. These features are: the single-cell thickness of both the alveolar wall (1) and the capillary wall (2). The concentration gradient (3) affects the rate of diffusion but not the distance. The surface area (4) affects the total capacity of diffusion but not the pathway distance.

1 mark for the correct answer A. Award 0 marks for incorrect options.

In insect-pollinated flowers:
- The stigma receives pollen grains during pollination (Correct).
- The anther produces pollen grains (A is incorrect).
- The ovary contains ovules and produces female gametes (B is incorrect).
- The sepal protects the flower when in bud (C is incorrect).

1 mark for the correct answer D. Award 0 marks for incorrect options.

Drinking water increases blood water potential. To restore normal osmotic balance, the pituitary gland secretes less ADH (decreases). This decreases the permeability of the collecting ducts in the kidney, resulting in less water reabsorption and a larger volume of more dilute (decreased concentration) urine being produced.

1 mark for the correct answer A. Award 0 marks for incorrect options.

The genotype of a heterozygous tall plant is \(Tt\), and a short plant is homozygous recessive, with genotype \(tt\). Crossing these genotypes:

- Parental genotypes: \(Tt \times tt\)
- Offspring gamete combinations: \(Tt, Tt, tt, tt\)
- Offspring phenotypic ratio: 50% Tall (\(Tt\)) and 50% Short (\(tt\)).

Thus, the probability of obtaining tall offspring is 50%.

1 mark for the correct answer C. Award 0 marks for incorrect options.

Active transport is the process that requires metabolic energy (ATP) from respiration. Both the absorption of mineral ions by root hair cells (1) and the reabsorption of glucose in kidney tubules (2) occur via active transport. The movement of carbon dioxide (3) occurs by passive diffusion, and water uptake (4) occurs by osmosis (passive).

1 mark for the correct answer A. Award 0 marks for incorrect options.

The rate of transpiration is highest when conditions encourage rapid evaporation and diffusion of water vapour:
1. Higher temperatures (\(30\ ^\circ\text{C}\) vs. \(15\ ^\circ\text{C}\)) increase the kinetic energy of water molecules, increasing evaporation.
2. Lower relative humidity (30% vs. 80%) creates a steeper water potential gradient between the leaf interior and the outside air.
3. Higher wind speed (\(5\text{ m/s}\) vs. \(1\text{ m/s}\)) continually removes humid air from near the stomata, maintaining a steep concentration gradient.

Thus, option D represents the combination that will lead to the highest transpiration rate.

1 mark for the correct answer D. Award 0 marks for incorrect options.

At high light intensities, light is no longer the limiting factor. Since increasing the temperature from \(15\ ^\circ\text{C}\) to \(25\ ^\circ\text{C}\) increases the rate of photosynthesis, temperature must have been the limiting factor at \(15\ ^\circ\text{C}\).

Award 1 mark for the correct option (C).
Reject other options:
- A is incorrect because at high light intensities, light is not limiting.
- B is incorrect because at low light intensities, temperature is not the main limiting factor.
- D is incorrect because water is not a limiting factor for an aquatic plant.

During inspiration, the diaphragm contracts and flattens (moves downwards), and the external intercostal muscles contract (pulling the ribcage upwards and outwards). This increases the volume of the thorax, causing the pressure inside the thorax to decrease below atmospheric pressure, drawing air into the lungs.

Award 1 mark for the correct option (A).

After a pollen grain lands on a compatible stigma, a pollen tube grows down through the style and enters the ovule through a tiny opening called the micropyle. The male gamete nucleus then fuses with the female gamete nucleus inside the ovule to complete fertilisation.

Award 1 mark for the correct option (A).
Reject other options because the whole pollen grain does not travel down the style, the pollen tube does not grow down the filament, and meiosis does not occur within the pollen tube to produce diploid nuclei.

Using the formula: \(\text{Actual size } (A) = \frac{\text{Image size } (I)}{\text{Magnification } (M)}\).
First, convert the image length from millimetres to micrometres: \(45\text{ mm} \times 1000 = 45\,000\ \mu\text{m}\).
Now, calculate the actual size: \(A = \frac{45\,000\ \mu\text{m}}{5000} = 9.0\ \mu\text{m}\).

Award 1 mark for the correct option (C).
Option A is incorrect due to a lack of unit conversion.
Option B is incorrect due to an incorrect conversion factor of 100 instead of 1000.
Option D is incorrect due to an incorrect conversion factor of 10,000.

In a cold environment, shunt vessels dilate to divert blood flow away from the capillaries near the skin surface, reducing heat loss by radiation. Sweat glands secrete less (or no) sweat to prevent evaporative cooling. Hair erector muscles contract to pull hairs upright, trapping an insulating layer of warm air next to the skin.

Award 1 mark for the correct option (B).

Since the woman has dry earwax (the recessive phenotype), her genotype must be \(\text{ww}\). The man has wet earwax, so his genotype is either \(\text{WW}\) or \(\text{Ww}\). Because his father had dry earwax (\(\text{ww}\)), the man must have inherited a recessive \(\text{w}\) allele from him, making the man's genotype heterozygous \(\text{Ww}\). Crossing \(\text{Ww} \times \text{ww}\) yields a \(1:1\) phenotypic ratio, so there is a \(0.50\) (or \(50\%\)) chance of having a child with dry earwax (\(\text{ww}\)).

Award 1 mark for the correct option (B).

At high temperatures, the heat energy disrupts the hydrogen bonds holding the enzyme's specific three-dimensional shape. This results in denaturation, causing the active site to permanently change shape so that the substrate can no longer fit.

Award 1 mark for the correct option (B).
Reject A (kinetic energy increases at high temperatures, though denaturation stops the reaction).
Reject C (activation energy is not affected in this way).
Reject D (enzymes are proteins, not carbohydrates, and do not break down into fatty acids).

The rate of diffusion is directly proportional to the concentration gradient, temperature (which increases the kinetic energy of particles), and the surface area of the membrane available for diffusion. Therefore, a higher concentration gradient, higher temperature, and larger surface area maximize the diffusion rate.

Award 1 mark for the correct option (A).

First, convert the image length to the same unit as the actual length:
\(45\text{ mm} = 45000\text{ }\mu\text{m}\).

Next, use the formula for magnification:
\(\text{Magnification} = \frac{\text{Image size}}{\text{Actual size}} = \frac{45000\text{ }\mu\text{m}}{150\text{ }\mu\text{m}} = 300\).

Therefore, the magnification is \(\times 300\).

1 mark for the correct calculation of magnification and selecting option B.

Photosynthesis requires both light energy and carbon dioxide. Sodium hydrogencarbonate solution acts as a source of carbon dioxide, whereas distilled water contains very little of it. Bright light provides the energy needed for the reaction. Therefore, the shoot in 1% sodium hydrogencarbonate solution in bright light will photosynthesise at the highest rate.

1 mark for identifying the correct combination of a high carbon dioxide source and bright light.

During expiration, the diaphragm and external intercostal muscles relax, causing the ribcage to move down and in, and the diaphragm to dome upwards. This decreases the volume of the thorax. The decrease in volume increases the air pressure inside the lungs relative to atmospheric pressure, pushing air out.

1 mark for selecting the option that correctly identifies thorax volume decrease and lung pressure increase.

When body temperature rises, the hypothalamus coordinates responses to increase heat loss. Arterioles supplying skin capillaries dilate (vasodilation) to allow more blood to flow near the skin surface, losing heat by radiation. Sweat glands also secrete more sweat, cooling the skin as it evaporates.

1 mark for identifying vasodilation of skin arterioles and increased sweat secretion as the correct combination of cooling responses.

The heterozygous white sheep has the genotype \(Ww\). The black sheep must be homozygous recessive (\(ww\)) because black wool is recessive. Crossing \(Ww \times ww\) results in offspring genotypes: \(50\%\) heterozygous white (\(Ww\)) and \(50\%\) homozygous black (\(ww\)). Thus, the probability of having a black offspring is \(0.50\).

1 mark for calculating the probability as 0.50 and selecting option C.

The stigma is the female structure of the flower that receives pollen during pollination. The anther is the male structure that produces and releases pollen grains.

1 mark for identifying that the stigma receives pollen and the anther produces pollen.

Photosynthesis is controlled by enzymes. As temperature increases, the kinetic energy of the molecules increases, leading to a faster rate of reaction up to an optimum temperature. Beyond this optimum, enzymes denature, causing the rate of photosynthesis to decrease rapidly.

1 mark for identifying the correct relationship involving enzyme activity, optimum temperature, and denaturation.

The walls of the alveoli and the walls of the surrounding capillaries are each only one cell thick. This extremely thin boundary minimizes the distance over which oxygen and carbon dioxide must diffuse, maximizing the rate of diffusion.

1 mark for identifying the thinness (one cell thick) of the walls of alveoli and capillaries as the feature reducing diffusion distance.

As light intensity increases, the rate of photosynthesis increases proportionally up to a certain point. Beyond this point, the rate plateaus (levels off) because another factor—in this case, carbon dioxide concentration—is in short supply and acts as the limiting factor. Graph B represents this relationship correctly.

1 mark for the correct option B.

During expiration (breathing out), the diaphragm muscle relaxes and dome-shapes upwards, while the external intercostal muscles relax, causing the ribcage to move downwards and inwards. This action decreases the volume of the thoracic cavity, thereby increasing the air pressure inside the lungs, forcing air out to the atmosphere.

1 mark for the correct option D.

In insect-pollinated flowers, petals are brightly colored to attract insect vectors, and the stigma is typically sticky or positioned internally to successfully receive pollen grains transferred from the insect's body.

1 mark for the correct option B.

Use the formula: \(\text{Magnification} = \frac{\text{Image size}}{\text{Actual size}}\).
First, convert the image size from millimeters to micrometers:
\(15\text{ mm} = 15 \times 1000\text{ }\mu\text{m} = 15000\text{ }\mu\text{m}\).
Now, calculate the magnification:
\(\text{Magnification} = \frac{15000\text{ }\mu\text{m}}{6\text{ }\mu\text{m}} = 2500\).
Therefore, the magnification is \(\times 2500\).

1 mark for the correct calculation leading to option C.

To lose excess heat, the arterioles supplying skin capillaries dilate (vasodilation) to increase blood flow near the skin surface, maximizing heat loss via radiation. Simultaneously, sweat glands secrete more sweat, which absorbs thermal energy from the body to evaporate, cooling the skin.

1 mark for the correct option D.

The parental genotypes are \(Tt\) (heterozygous tall) and \(tt\) (homozygous recessive dwarf). The cross \(Tt \times tt\) yields offspring with genotypes \(Tt\) and \(tt\) in equal proportion. Thus, the phenotypic ratio is \(1\text{ tall} : 1\text{ dwarf}\) (or \(1 : 1\)).

1 mark for the correct Punnett square analysis leading to option A.

Stomata, located mainly in the lower epidermis, form pores that open to allow the diffusion of carbon dioxide gas into the air spaces of the spongy mesophyll, from where it diffuses directly into photosynthesizing cells.

1 mark for identifying the stomata and mesophyll air spaces (option B).

Self-pollination involves the fusion of gametes produced by the same individual. Although meiosis still introduces some variation, self-pollination significantly reduces overall genetic diversity within the population and increases homozygosity over generations.

1 mark for the correct option D.

(a) The balanced equation for photosynthesis is:
$$6\text{CO}_2 + 6\text{H}_2\text{O} \xrightarrow{\text{light, chlorophyll}} \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2$$

(b) (i) Based on Table 1.1, as light intensity increases from 10 to 40 arbitrary units, the rate of oxygen production increases rapidly and proportionally from 1.2 to 4.0 \(\text{cm}^3\,\text{min}^{-1}\). From 40 to 60 arbitrary units, the rate increases very slightly from 4.0 to 4.1 \(\text{cm}^3\,\text{min}^{-1}\). Above 60 arbitrary units, the rate remains constant at 4.1 \(\text{cm}^3\,\text{min}^{-1}\), showing a plateued rate of photosynthesis.

(b) (ii) At 0.02% \(\text{CO}_2\), the rate levels off because carbon dioxide concentration becomes the limiting factor. Even if light intensity is increased further, the enzymes involved in photosynthesis cannot work any faster because there are not enough carbon dioxide molecules. At 0.10% \(\text{CO}_2\), carbon dioxide is no longer as limiting, so the rate continues to rise with light intensity until light or another factor (like temperature) becomes limiting at 6.5 \(\text{cm}^3\,\text{min}^{-1}\).

(c) Leaf adaptations:
1. Large surface area to maximize light absorption.
2. Leaves are thin, ensuring light can penetrate to all cells.
3. Upper epidermis is transparent to allow light to reach the palisade layer.
4. Palisade mesophyll cells are tightly packed, vertically aligned, and contain numerous chloroplasts positioned near the upper surface of the leaf.

(a) [Total: 3 marks]
- 1 mark for correct reactants (CO2 and H2O) and products (C6H12O6 and O2)
- 1 mark for correct chemical formulas
- 1 mark for correct balancing (6, 6, 1, 6)

(b) (i) [Total: 3 marks]
- 1 mark for stating that as light intensity increases, rate of photosynthesis increases up to 40/50 arbitrary units
- 1 mark for stating that above 50/60 arbitrary units, the rate levels off / remains constant
- 1 mark for quoting correct data with units (e.g. constant at 4.1 cm3/min or increases from 1.2 to 4.0 cm3/min)

(b) (ii) [Total: 4 marks]
- 1 mark for identifying carbon dioxide as the limiting factor at 0.02% CO2
- 1 mark for explaining that increasing light intensity above 40 a.u. has no effect because CO2 is in short supply
- 1 mark for stating that at 0.10% CO2, the concentration of the limiting factor is increased
- 1 mark for explaining that this allows the rate of photosynthesis to increase further until a new plateau is reached at 6.5 cm3/min

(c) [Total: 4 marks]
- Max 2 marks for naming adaptations (e.g. large surface area, thinness, transparent epidermis, palisade mesophyll packed with chloroplasts)
- Max 2 marks for explanation of how each adaptation maximizes light absorption (e.g. thinness allows light penetration, transparent cuticle allows light passage, palisade cells at the top absorb maximum light)

(a) Gas exchange surface adaptations:
1. **Thin wall**: Alveolar wall and capillary wall are each only one cell thick, providing a very short diffusion pathway for oxygen and carbon dioxide.
2. **Large surface area**: The spherical shape of millions of alveoli creates a massive total surface area, allowing a large volume of gas to be exchanged simultaneously.
3. **Excellent blood supply**: A dense network of capillaries constantly moves deoxygenated blood to the alveoli and oxygenated blood away, maintaining a steep concentration gradient.
4. **Moist surface**: A thin layer of water lines the alveolus, allowing gases to dissolve before diffusing across the membrane.

(b) Pathway of oxygen:
1. Oxygen flows through the trachea, into the bronchi, and through the bronchioles into the alveoli.
2. The oxygen molecule dissolves in the film of moisture lining the alveolus.
3. It diffuses down its concentration gradient across the thin alveolar epithelial wall.
4. It diffuses through the thin capillary endothelial wall into the blood plasma.
5. It passes through the cell membrane of a red blood cell and binds chemically to haemoglobin to form oxyhaemoglobin.

(c) Differences between inspired and expired air:
- **Oxygen content**: Inspired air has approximately 21% oxygen, whereas expired air has around 16% oxygen.
- **Carbon dioxide content**: Inspired air has approximately 0.04% carbon dioxide, whereas expired air has around 4% carbon dioxide.
- **Water vapour**: Inspired air has variable water vapour depending on humidity, whereas expired air is saturated with water vapour.
- **Temperature**: Expired air is usually warmer (close to body temperature) compared to inspired air.

(a) [Total: 6 marks]
- 1 mark for each correct adaptation identified (max 3)
- 1 mark for each corresponding explanation of how it increases diffusion efficiency (max 3)
- Features: thin walls / one cell thick (short diffusion distance); large surface area (more diffusion can occur); moist surface (gases dissolve); high capillary density / blood flow (maintains concentration gradient)

(b) [Total: 4 marks]
- 1 mark for correct anatomical order: trachea -> bronchus -> bronchiole -> alveolus
- 1 mark for oxygen dissolving in moisture on the alveolar surface
- 1 mark for diffusion across alveolar and capillary walls
- 1 mark for entering the red blood cell and binding to haemoglobin

(c) [Total: 3 marks]
- 1 mark for comparing oxygen content (higher in inspired / lower in expired)
- 1 mark for comparing carbon dioxide content (lower in inspired / higher in expired)
- 1 mark for comparing water vapour content (lower/variable in inspired / saturated in expired)
- Accept comparisons of temperature (warmer in expired)

(a) Completed Table 3.1:
- **Petals**:
- Insect-pollinated flower: Large, brightly coloured, often scented, with nectar guides to attract insect pollinators.
- Wind-pollinated flower: Small, dull, green or brown, without scent or nectar.
- **Position of anthers and stigmas**:
- Insect-pollinated flower: Located deep inside the flower so that insects must brush past them to reach nectar.
- Wind-pollinated flower: Hang loosely outside the flower on long filaments (anthers) to release pollen easily, and stigmas are long, feathery, and exposed outside to catch drifting pollen.

(b) Events after pollination:
1. A pollen grain lands on the receptive surface of the stigma.
2. The pollen grain germinates, absorbing nutrients and moisture from the stigma.
3. It grows a long pollen tube down through the tissues of the style, directed by chemical signals toward the ovary.
4. The pollen tube enters the ovary and reaches an ovule, entering through a small opening called the micropyle.
5. The male gamete nucleus travels down the pollen tube.
6. The tip of the pollen tube ruptures inside the ovule, releasing the male nucleus.
7. Fertilisation occurs when the male nucleus fuses with the haploid nucleus of the female egg cell, forming a diploid zygote.

(c) Conditions for seed germination:
1. **Water**: Needed to activate enzymes, swell the seed coat (testa), and transport digested nutrients.
2. **Oxygen**: Required for aerobic respiration to release energy for cell division and growth.
3. **Suitable temperature**: Warmth is needed for optimal enzyme activity; extremely high temperatures denature enzymes, and low temperatures slow down reactions.

(a) [Total: 4 marks]
- 1 mark for correct description of petals in insect-pollinated (brightly coloured/large) and wind-pollinated (dull/small/green)
- 1 mark for correct comparison of petals
- 1 mark for describing position of anthers/stigmas in insect-pollinated (enclosed/inside flower)
- 1 mark for describing position of anthers/stigmas in wind-pollinated (exposed/hanging outside)

(b) [Total: 6 marks]
- 1 mark for pollen grain landing on the stigma
- 1 mark for growth of the pollen tube down the style
- 1 mark for pollen tube entering the ovary/ovule through the micropyle
- 1 mark for male nucleus/gamete traveling down the pollen tube
- 1 mark for fusion of male and female nuclei
- 1 mark for naming fertilisation as the fusion of nuclei to form a zygote

(c) [Total: 3 marks]
- 1 mark for water / moisture
- 1 mark for oxygen
- 1 mark for suitable temperature / warmth
- Reject: sunlight / soil / carbon dioxide

(a) (i) Calculation of magnification:
- **Formula**:
$$\text{Magnification} = \frac{\text{Image size (size of drawing)}}{\text{Actual size}}$$
- **Working**:
$$\text{Magnification} = \frac{64\text{ mm}}{0.08\text{ mm}} = 800$$
- **Answer**: 800 (or 800)

(ii) Conversion of units:
- Since \(1\text{ mm} = 1000\,\mu\text{m}\):
- **Working**:
$$0.08\text{ mm} \times 1000 = 80\,\mu\text{m}$$
- **Answer**: \(80\,\mu\text{m}\)

(b) (i) Plant-only cell structures in Fig. 4.1:
1. Cell wall
2. Chloroplast
3. Large permanent vacuole

(b) (ii) Functions:
- **Cell wall**: Made of cellulose, provides structural strength and rigidity, prevents the cell from bursting when it absorbs water by osmosis (turgidity).
- **Chloroplast**: Contains chlorophyll which absorbs light energy to carry out photosynthesis, converting carbon dioxide and water into glucose and oxygen.
- **Large permanent vacuole**: Contains cell sap; helps support the cell shape by keeping it turgid, and stores pigments, waste products, and water.

(c) Root hair cell specialization:
- Has a long, thin, finger-like projection (root hair) which significantly increases the surface area-to-volume ratio, allowing faster absorption of water by osmosis.
- Cell membrane has a high density of carrier proteins for active transport of mineral ions, which lowers the water potential inside the cell to promote osmosis.

(a) (i) [Total: 3 marks]
- 1 mark for correct formula (Magnification = Image / Actual)
- 1 mark for correct working (64 / 0.08)
- 1 mark for correct answer: x800 / 800 (accept any correct representation)

(a) (ii) [Total: 2 marks]
- 1 mark for multiplying by 1000
- 1 mark for correct answer: 80 micrometers

(b) (i) [Total: 2 marks]
- 1 mark for each correct structure identified (max 2): Cell wall / Chloroplast / Large permanent vacuole

(b) (ii) [Total: 4 marks]
- 2 marks for description of the function of the first identified structure
- 2 marks for description of the function of the second identified structure
- Cell wall: prevents bursting / provides support / maintains shape / cellulose structure
- Chloroplast: site of photosynthesis / contains chlorophyll / absorbs light to make glucose
- Vacuole: stores cell sap / maintains turgor pressure / stores water and mineral ions

(c) [Total: 2 marks]
- 1 mark for referencing the long extension which increases the surface area
- 1 mark for thin cell wall / thin membrane / cell sap lowering water potential to draw in water by osmosis

(a) Definition and principle:
- **Homeostasis**: The maintenance of a constant internal environment (such as body temperature, blood glucose levels, and water potential) within narrow limits.
- **Negative feedback**: A control system where any deviation of a factor from its normal set point is detected by receptors. This triggers a response from effectors that counteracts the change, bringing the factor back to its normal level, thus stabilizing the system.

(b) Responses to cold:
1. **Vasoconstriction**: Arterioles supplying the skin capillaries constrict (narrow). This reduces the volume of warm blood flowing close to the skin surface, minimizing heat loss by radiation, convection, and conduction.
2. **Shivering**: Rapid, involuntary contractions of skeletal muscles are stimulated. This increase in muscle activity raises the rate of respiration, releasing thermal energy as a metabolic byproduct to warm up the blood.
3. **Hair erection (piloerection)**: Hair erector muscles in the skin contract, causing hairs to stand upright. This traps a thick layer of still air close to the skin surface, which acts as an effective thermal insulator.
4. **Reduction in sweating**: Sweat glands stop secreting sweat, preventing cooling by evaporation.

(c) Regulation of high blood glucose:
1. Receptors in the pancreas detect the increase in blood glucose concentration above normal levels (e.g., after a meal).
2. The islet cells of the pancreas secrete the hormone **insulin** directly into the blood.
3. Insulin travels in the blood to target organs, primarily the liver and skeletal muscles.
4. It stimulates liver and muscle cells to increase glucose uptake from the blood.
5. It accelerates the conversion of excess glucose into the insoluble storage carbohydrate **glycogen** (glycogenesis), lowering the concentration of glucose in the blood back to normal.

(a) [Total: 4 marks]
- 1 mark for defining homeostasis as maintenance of constant internal environment
- 1 mark for reference to maintaining factors within narrow limits / set points
- 1 mark for negative feedback detecting change / deviation from set point
- 1 mark for corrective mechanism being triggered to reverse the change

(b) [Total: 6 marks]
- Max 3 marks for identifying corrective mechanisms: vasoconstriction / shivering / hair erection / reduced sweating
- Max 3 marks for correct explanations:
- Vasoconstriction: arterioles narrow, less blood to capillaries, reduces heat loss by radiation
- Shivering: muscle contraction, increases respiration, releases heat energy
- Hair erection: erector muscles contract, traps air, acts as insulator
- Reject: capillaries constrict / move up and down

(c) [Total: 4 marks]
- 1 mark for pancreas detecting high glucose levels
- 1 mark for pancreas secreting insulin
- 1 mark for liver / muscle cells taking up more glucose
- 1 mark for conversion of glucose to glycogen

(a) Completed Genetic Diagram:
- **Gametes**: Parents produce gametes containing either allele \(F\) or allele \(f\).
- **Punnett Square**:
| | \(F\) | \(f\) |
| :---: | :---: | :---: |
| \(F\) | \(FF\) | \(Ff\) |
| \(f\) | \(Ff\) | \(ff\) |
- **Offspring Genotypes**: \(FF\), \(Ff\), \(Ff\), \(ff\)
- **Offspring Phenotypes**:
- \(FF\) and \(Ff\): Normal lung and digestive function (3 out of 4)
- \(ff\): Cystic fibrosis (1 out of 4)
- **Probability of child with cystic fibrosis**: \(0.25\) / \(25\%\) / \(1\text{ in } 4\).

(b) Definitions:
- (i) **heterozygous**: Having two different alleles of a particular gene (e.g., \(Ff\)).
- (ii) **phenotype**: The observable physical or physiological features of an organism, determined by its genotype and environment.

(c) Codominance:
- (i) The alleles for red pigment (\(C^R\)) and white pigment (\(C^W\)) are codominant. Neither allele is completely dominant over the other, so both alleles are expressed in the heterozygous state (\(C^R C^W\)), resulting in a pink intermediate color.
- (ii) Cross between two pink-flowered plants (\(C^R C^W \times C^R C^W\)):
- Gametes: \(C^R\) and \(C^W\) from both parents.
- Offspring genotypes: \(1\, C^R C^R\) : \(2\, C^R C^W\) : \(1\, C^W C^W\)
- **Phenotypic ratio**: 1 Red : 2 Pink : 1 White.

(a) [Total: 6 marks]
- 1 mark for correct identification of gametes (F and f for both parents)
- 2 marks for correct offspring genotypes in the Punnett square (FF, Ff, Ff, ff)
- 1 mark for linking genotypes to phenotypes correctly (FF and Ff are Normal, ff is Cystic fibrosis)
- 1 mark for correct phenotypic ratio (3 Normal : 1 Cystic fibrosis)
- 1 mark for stating correct probability (25% / 0.25 / 1 in 4)

(b) (i) [Total: 1 mark]
- 1 mark for 'having two different alleles of a gene'

(b) (ii) [Total: 1 mark]
- 1 mark for 'physical / observable features / characteristics of an organism'

(c) (i) [Total: 2 marks]
- 1 mark for stating that both alleles (CR and CW) are codominant / both are expressed
- 1 mark for explaining that the combination results in an intermediate / blended phenotype (pink)

(c) (ii) [Total: 3 marks]
- 1 mark for showing correct offspring genotypes (CRCR, CRCW, CWCW)
- 1 mark for linking genotypes to correct phenotypes (CRCR is Red, CRCW is Pink, CWCW is White)
- 1 mark for the final correct phenotypic ratio: 1 Red : 2 Pink : 1 White

(a)(i) Table:
| Sucrose concentration / mol/dm³ | Initial mass / g | Final mass / g | Change in mass / g | Percentage change in mass / % |
| :---: | :---: | :---: | :---: | :---: |
| 0.0 | 4.00 | 4.32 | +0.32 | +8.0 |
| 0.2 | 4.00 | 4.12 | +0.12 | +3.0 |
| 0.4 | 4.00 | 3.96 | -0.04 | -1.0 |
| 0.6 | 4.00 | 3.76 | -0.24 | -6.0 |
| 0.8 | 4.00 | 3.60 | -0.40 | -10.0 |

(a)(ii) Working:
Change in mass = \(3.60\text{ g} - 4.00\text{ g} = -0.40\text{ g}\)
Percentage change = \(\frac{-0.40}{4.00} \times 100 = -10.0\%\)

(a)(iii) Graph details:
- x-axis: Sucrose concentration (mol/dm³)
- y-axis: Percentage change in mass (%)
- Linear scales starting from negative values to positive values on the y-axis, covering at least half the grid
- Points plotted accurately and joined with a straight line or smooth curve.

(a)(iv) The isotonic concentration is around 0.38 mol/dm³ (accept 0.36 to 0.39 mol/dm³), which is the concentration where there is 0% change in mass. At this concentration, there is no net movement of water by osmosis into or out of the cells because the water potential inside the sweet potato cells is equal to the water potential of the sucrose solution.

(a)(v) Source of error: Blotting with paper towels was not standardised, leading to varying amounts of water remaining on the surface of each cylinder.
Improvement: Standardise the blotting method by rolling each cylinder exactly 3 times on dry paper towel with a constant force.

(b)(i) Drawing of a plasmolysed cell should show:
- Large size, filling more than half of the space.
- Continuous clear lines (no sketching).
- Double line for cell wall.
- Shrunken vacuole/cytoplasm showing cell membrane clearly pulled away from the cell wall in several places.
- Correct labels pointing to the outer cell wall and the inner, detached cell membrane.

(b)(ii) Calculation:
Magnification = \(\frac{\text{size of image}}{\text{actual size}} = \frac{48\text{ mm}}{0.08\text{ mm}} = 600\)
Magnification is \(\times 600\).

**(a)(i) Table of results [4 marks total]**
- Column headings with appropriate units (sucrose concentration / mol/dm³, mass / g, percentage change / %) [1]
- All concentrations listed and correct initial/final mass values recorded [1]
- Correct calculation of change in mass (with positive/negative signs) [1]
- Correct calculation of percentage change in mass (with positive/negative signs) [1]

**(a)(ii) Calculation [2 marks total]**
- Correct substitution: \((-0.40 / 4.00) \times 100\) [1]
- Correct final answer: \(-10\%\) or \(-10.0\%\) (must include the negative sign) [1]

**(a)(iii) Graph plotting [4 marks total]**
- Axes labelled correctly with units [1]
- Appropriate linear scales where plotted data occupies at least half the grid in both directions (including negative region on y-axis) [1]
- All points plotted accurately [1]
- Correct line of best fit drawn through points or points connected with straight ruled lines [1]

**(a)(iv) Estimate and explanation [2 marks total]**
- Correct reading from their graph where line crosses y = 0 (accept 0.36 to 0.39 mol/dm³) [1]
- Explanation: No net movement of water by osmosis / water potential is equal [1]

**(a)(v) Error and Improvement [2 marks total]**
- Identifies error: Blotting variation / size variation / time variation [1]
- Suggests matching improvement: Standardised blotting technique / precise cutting with a guide / timed intervals [1]

**(b)(i) Cell drawing [4 marks total]**
- Outline: single clear continuous lines with double line representing cell wall [1]
- Proportions: shrunken cytoplasm detached from the cell wall [1]
- Size: drawing occupies at least half the space available [1]
- Labels: correct labels pointing to cell wall and cell membrane [1]

**(b)(ii) Magnification calculation [2 marks total]**
- Correct division: \(48 / 0.08\) [1]
- Correct final magnification: \(\times 600\) [1]

(a)(i) Table:
| Sugar substrate | Volume of gas produced in 10 minutes / cm³ | Rate of gas production / cm³/minute |
| :--- | :---: | :---: |
| Glucose | 18.0 | 1.8 |
| Fructose | 15.0 | 1.5 |
| Maltose | 6.0 | 0.6 |
| Sucrose | 12.0 | 1.2 |
| Distilled water | 0.0 | 0.0 |

(a)(ii) Working:
Rate = \(\frac{\text{Volume of gas}}{\text{Time}} = \frac{18.0\text{ cm}^3}{10\text{ minutes}} = 1.8\text{ cm}^3/\text{minute}\)

(a)(iii) Independent variable: Type of sugar substrate.
Dependent variable: Volume of gas produced / rate of gas production.

(a)(iv) The distilled water acts as a control treatment to prove that gas is produced due to the metabolism of sugar by yeast cells, and not from factors like temperature change or residual nutrients in the yeast suspension.

(a)(v) Constant variables (any two):
- Temperature of the water bath / environment
- Concentration of the yeast suspension (5%)
- Volume of the yeast suspension (10 cm³)
- Volume of the sugar solution (10 cm³)
- Concentration of the sugar solution (5%)

(a)(vi) Test: Bubble the gas into limewater (calcium hydroxide solution).
Expected result: The limewater changes from clear/colourless to cloudy/milky.

(b) Plan for investigation:
- **Independent variable**: Temperature (choose at least five different temperatures, e.g., 20°C, 30°C, 40°C, 50°C, 60°C).
- **Control of temperature**: Use thermostatically controlled water baths at each temperature to keep conditions stable.
- **Dependent variable**: Measure the volume of carbon dioxide gas produced in a set time (e.g., 10 minutes) using a gas syringe.
- **Controlled variables**: Keep the volume and concentration of the yeast suspension constant, use the same sugar substrate (e.g., 5% glucose) and keep its volume constant, maintain the same pH.
- **Standardised step**: Equilibrate the yeast and glucose solutions separately in the water baths for 5 minutes before mixing to ensure they reach the target temperature.
- **Reliability**: Repeat each temperature treatment at least three times, discard anomalies, and calculate a mean value for each temperature.
- **Safety**: Wear eye protection / gloves; be cautious around hot water baths.

**(a)(i) Table of results [4 marks total]**
- Suitable table drawn with neat borders [1]
- Column headings with correct units: Volume / cm³ and Rate / cm³/min (or cm³ min⁻¹) [1]
- All 5 substrates listed with correct volumes recorded [1]
- All rate values calculated correctly [1]

**(a)(ii) Calculation [2 marks total]**
- Shows dividing volume by time (\(18.0 / 10\)) [1]
- Correct answer: \(1.8\) with units (\(\text{cm}^3/\text{minute}\)) [1]

**(a)(iii) Variables [2 marks total]**
- Independent: type of sugar substrate [1]
- Dependent: volume of gas produced (or rate of gas production) [1]

**(a)(iv) Distilled water explanation [1 mark total]**
- Distilled water serves as a control / to confirm gas is produced from respiring sugar and not other factors [1]

**(a)(v) Constant variables [2 marks total]**
- Identifies any two constant variables (e.g., temperature, concentration of yeast, volume of yeast, volume of sugar solution) [2]

**(a)(vi) Carbon dioxide test [3 marks total]**
- Use limewater [1]
- Bubble gas through/into limewater [1]
- Limewater turns cloudy / milky [1]

**(b) Planning [6 marks total]**
- Range of independent variable: at least 5 different temperatures [1]
- Method of controlling temperature: water baths [1]
- Dependent variable: volume of gas collected in gas syringe over a fixed time [1]
- Keep other key variables constant: volume/concentration of yeast and glucose [1]
- Standardised step: equilibrate yeast/glucose at target temperature before mixing [1]
- Reliability: repeat at least 3 times at each temperature and calculate a mean [1]