2024 Cambridge IGCSE Biology (0610) Practice Paper with Answers

In Condition 2, despite the light intensity being high, the rate of photosynthesis did not increase compared to Condition 1 (low light) because the carbon dioxide concentration (0.02%) was limiting. When carbon dioxide concentration was increased to 0.15% in Condition 3, the rate increased significantly. This shows that in Condition 2, carbon dioxide concentration was the limiting factor.

Award 1 mark for selecting B (carbon dioxide is the limiting factor in Condition 2 because increasing its concentration while keeping light intensity high increases the photosynthetic rate).

The correct sequence of steps to test a leaf for starch is:
1. **Boil in water**: This kills the cells and breaks down cell membranes, making them permeable.
2. **Boil in ethanol** (using a water bath for safety): This dissolves and removes the green chlorophyll so that the color change of the iodine can be clearly seen.
3. **Dip in hot water**: This softens the leaf, which becomes stiff and brittle after being boiled in ethanol.
4. **Add iodine solution**: To test for starch (color change from brown to blue-black if starch is present).

Award 1 mark for the correct sequence: boiling in water to kill the cell membranes, boiling in ethanol to decolorize, and dipping in hot water to soften.

During expiration (breathing out):
- The external intercostal muscles relax, causing the ribcage to move downwards and inwards.
- The diaphragm relaxes, dome-shaping upwards.
- This decreases the volume of the thorax, which increases the pressure inside the thorax relative to atmospheric pressure, forcing air out of the lungs.

Award 1 mark for C, because expiration involves the relaxation of the external intercostal muscles and the diaphragm, resulting in a decrease in thoracic volume and a subsequent increase in thoracic air pressure.

Expired air contains approximately 78% nitrogen, 16% oxygen, 4% carbon dioxide, and is saturated with water vapour due to evaporation from the warm, moist surfaces of the alveoli. Sample A represents inspired (inhaled) atmospheric air.

Award 1 mark for B (expired air contains roughly 16% oxygen, 4% carbon dioxide, and is saturated with water vapour).

Wind-pollinated flowers do not need to attract insects, so their petals are typically small and dull (often green). To facilitate wind pollination, their anthers are suspended outside the flower on long, flexible filaments so that the pollen can be easily blown away. Their stigmas are feathery and hang outside the flower to provide a large surface area to trap airborne pollen grains.

Award 1 mark for C, recognizing that wind-pollinated flowers have reduced petals, dangling anthers on flexible filaments, and protruding feathery stigmas.

First, convert both values to the same unit (e.g., micrometres, \(\mu\text{m}\)):
- Image size = \(15\text{ mm} = 15 \times 1000 = 15\,000\ \mu\text{m}\).
- Actual size = \(5\ \mu\text{m}\).

Now, apply the magnification formula:
$$\text{Magnification} = \frac{\text{Image size}}{\text{Actual size}} = \frac{15\,000\ \mu\text{m}}{5\ \mu\text{m}} = \times 3000$$

Award 1 mark for calculating the correct magnification of \(\times 3000\) (converting 15 mm to 15,000 \(\mu\text{m}\) and dividing by 5 \(\mu\text{m}\)).

In a cold environment, the body needs to reduce heat loss by radiation from the skin surface:
- Arterioles supplying skin capillaries undergo **vasoconstriction** (they constrict) to reduce blood flow near the surface.
- **Shunt vessels dilate** (widen) to allow blood to bypass these surface capillaries.
- **Hair erector muscles contract**, causing the skin hairs to stand upright, trapping an insulating layer of warm air next to the skin.

Award 1 mark for A, representing the correct physiological responses of the skin to cold conditions to conserve core body heat.

A cross between two heterozygotes (\(Bb \times Bb\)) produces a classic \(3:1\) phenotypic ratio:
- Black phenotype (\(BB\) and \(Bb\)): \(\frac{3}{4}\) (75%)
- Brown phenotype (\(bb\)): \(\frac{1}{4}\) (25%)

For a litter size of 8:
- Expected number of black offspring = \(\frac{3}{4} \times 8 = 6\)
- Expected number of brown offspring = \(\frac{1}{4} \times 8 = 2\)

Award 1 mark for selecting B, demonstrating understanding of the \(3:1\) phenotypic ratio in a heterozygous cross and applying it to a litter size of 8 (6 black, 2 brown).

Using the formula: Image size = Actual size \( \times \) Magnification. Here, actual size = 0.06 mm and magnification = 800. Therefore, Image size = 0.06 mm \( \times \) 800 = 48 mm.

1 mark for the correct calculation yielding 48 mm.

Because increasing the carbon dioxide concentration causes an increase in the rate of photosynthesis while light intensity and temperature are kept constant, carbon dioxide concentration was the limiting factor at 0.03%.

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

The short diffusion distance is maintained because the pathway consists only of the alveolar wall (one cell thick) and the capillary wall (one cell thick).

1 mark for identifying the one-cell-thick walls of the alveoli and capillaries.

To lose heat, arterioles supplying skin capillaries dilate (vasodilation) and shunt vessels constrict, forcing more blood to flow close to the skin surface where heat can be radiated away.

1 mark for selecting arterioles dilate and shunt vessels constrict.

The cross is Tt (heterozygous tall) \( \times \) tt (short). The offspring genotypes will be 50% Tt (tall) and 50% tt (short), which gives a phenotypic ratio of 1 tall : 1 short.

1 mark for the correct 1:1 phenotypic ratio.

After landing on the stigma, the pollen tube grows down through the style into the ovary, and finally enters the ovule where fertilization takes place.

1 mark for the correct anatomical sequence: stigma, style, ovary, ovule.

Glucose is converted to starch for storage because starch is insoluble. It is transported in the phloem as sucrose, and is converted into cellulose to build plant cell walls.

1 mark for identifying starch (storage), sucrose (transport), and cellulose (cell walls).

During inspiration, the diaphragm contracts (moving downwards) and the external intercostal muscles contract (pulling the ribcage upwards and outwards), increasing thoracic volume.

1 mark for identifying that both the diaphragm and external intercostal muscles contract.

The flask contains sodium hydroxide, which absorbs carbon dioxide. Since carbon dioxide is necessary for photosynthesis, the leaf inside the flask cannot photosynthesize and produce starch, even in the green parts containing chlorophyll. The white, variegated parts of the leaf also cannot photosynthesize because they lack chlorophyll. Therefore, neither part will contain starch, and both parts will test negative (yellow-brown) with iodine solution.

Award 1 mark for the correct option (D).
- Reject other options because both CO2 and chlorophyll are required for starch production via photosynthesis.

At point P (low light intensity), the rate increases as light intensity increases, indicating that light intensity is the limiting factor. At point Q, the light intensity is high, but the rate is restricted at the lower carbon dioxide concentration. Increasing the carbon dioxide concentration from \(0.04\%\) to \(0.15\%\) increases the rate of photosynthesis, demonstrating that carbon dioxide concentration is the limiting factor at point Q.

Award 1 mark for the correct option (C).
- Reject options where point P is carbon dioxide or point Q is light intensity/temperature.

Features 2 (alveolar walls are one cell thick) and 3 (capillary walls are one cell thick) minimize the distance over which gases must diffuse, making gas exchange highly efficient. Feature 1 (large surface area) increases the rate of diffusion but does not decrease the physical diffusion distance. Feature 4 (capillary network) helps maintain a steep concentration gradient but does not reduce the diffusion distance itself.

Award 1 mark for the correct option (B).
- Reject options containing statement 1 and 4 as they do not affect the diffusion distance.

During expiration, the diaphragm relaxes and moves upwards to its natural dome shape, while the external intercostal muscles relax, causing the ribcage to move downwards and inwards. This decreases the volume of the thoracic cavity, causing the air pressure in the thorax to increase above atmospheric pressure, which forces air out of the lungs.

Award 1 mark for the correct option (B).
- Reject options where the diaphragm contracts during expiration, or where the pressure in the thorax decreases during expiration.

Feathery stigmas in wind-pollinated flowers provide a large surface area to catch pollen grains blowing in the wind. Large, brightly colored petals are a feature of insect-pollinated flowers. Short, stiff filaments would prevent anthers from hanging outside to catch the wind. Wind-pollinated flowers have light, smooth pollen grains, not sticky or heavy ones.

Award 1 mark for the correct option (A).
- Reject B, C, and D as they confuse wind-pollinated and insect-pollinated characteristics or provide incorrect structural details.

First, convert both values to the same unit (millimeters):
Image size = \(4.8\text{ cm} = 48\text{ mm}\)
Actual size = \(0.08\text{ mm}\)

Now, calculate the magnification using the formula:
\(\text{Magnification} = \frac{\text{Image size}}{\text{Actual size}} = \frac{48\text{ mm}}{0.08\text{ mm}} = 600\)

Therefore, the magnification is \(\times 600\).

Award 1 mark for the correct option (C).
- Reject other options due to errors in unit conversion (e.g., using 4.8 mm or failing to convert cm to mm correctly).

When body temperature rises, the arterioles supplying the capillaries near the skin surface dilate (vasodilation) to allow more blood to flow close to the skin surface, where heat can be lost by radiation. Simultaneously, shunt vessels constrict, directing more blood into the surface capillaries rather than bypassing them. This increases total blood flow to the surface capillaries.

Award 1 mark for the correct option (A).
- Reject option B as this describes vasoconstriction (the response to cold temperatures).
- Reject options C and D because they describe incorrect combinations of arteriole and shunt vessel actions.

The cross is between two heterozygotes (\(Bb \times Bb\)). The Punnett square predicts offspring genotypes in the ratio of \(1\,BB : 2\,Bb : 1\,bb\). This results in a theoretical phenotypic ratio of \(3\) black-furred (dominant phenotype) to \(1\) brown-furred (recessive phenotype).
For a litter size of 8, the expected numbers are:
Black-furred offspring: \(\frac{3}{4} \times 8 = 6\)
Brown-furred offspring: \(\frac{1}{4} \times 8 = 2\)

Award 1 mark for the correct option (B).
- Reject A (corresponds to crossing homozygous dominant with any genotype, or pure chance variation not aligning with theoretical ratios).
- Reject C (corresponds to a test cross, \(Bb \times bb\)).
- Reject D (corresponds to a ratio favoring the recessive phenotype).

In bright light, the rate of photosynthesis in the water plant is higher than its rate of respiration. This results in a net uptake of carbon dioxide from the surrounding indicator solution, causing the pH to rise and turning the red hydrogencarbonate indicator purple.

Award 1 mark for identifying that Tube 1 (bright light) turns purple because the rate of photosynthesis exceeds respiration, decreasing the concentration of carbon dioxide.

Boiling the leaf in ethanol extracts and dissolves the chlorophyll, which decolourises the leaf. This is necessary so that the final colour change (from orange-brown to blue-black with iodine) can be easily seen.

Award 1 mark for identifying that boiling in ethanol dissolves and removes chlorophyll to make the colour change visible.

Goblet cells produce mucus to trap pathogens and dust, while ciliated cells have cilia that beat rhythmically to sweep the mucus upwards away from the lungs towards the throat.

Award 1 mark for selecting the option with the correct functions for both cells.

Flower X exhibits classic adaptations for wind pollination: small, dull petals (energy is not wasted on attracting insects), stamens hanging outside on long filaments to release pollen into the wind, and a feathery stigma hanging outside to catch drifting pollen grains.

Award 1 mark for identifying Flower X as the wind-pollinated flower based on its structural adaptations.

The correct sequence begins with pollination, where the pollen grain lands on the stigma. Then, a pollen tube grows down through the style to reach the ovary. The male gamete nucleus travels down this tube and enters the ovule, where it fuses with the female gamete nucleus.

Award 1 mark for the correct sequence showing pollen landing on stigma, tube growth down the style, entry into ovule, and fusion of nuclei.

Convert the image size to the same units as the actual size: \(4.0\text{ cm} = 40\text{ mm}\). Using the magnification formula: \(M = \frac{\text{Image size}}{\text{Actual size}} = \frac{40\text{ mm}}{0.08\text{ mm}} = 500\). Therefore, the magnification is \(\times 500\).

Award 1 mark for showing the correct conversion of units and applying the formula to obtain \(\times 500\).

To conserve heat in a cold environment, vasoconstriction occurs. Arterioles supplying the skin capillaries constrict (narrow), while the shunt vessels dilate (widen). This combination diverts blood flow away from the capillaries close to the skin surface, minimizing heat loss via radiation.

Award 1 mark for selecting the response where arterioles constrict and shunt vessels dilate, leading to reduced capillary blood flow.

The heterozygous red parent has the genotype \(Rr\). The white parent must be homozygous recessive, \(rr\). Crossing \(Rr \times rr\) yields offspring genotypes in a 1 \(Rr\) : 1 \(rr\) ratio. This corresponds to a phenotypic ratio of 1 red-flowered plant : 1 white-flowered plant.

Award 1 mark for the correct cross analysis resulting in a 1:1 ratio.

When light intensity is increased but the rate of photosynthesis no longer increases, light is no longer the limiting factor. At this point, some other factor required for photosynthesis, such as carbon dioxide concentration or temperature, is in short supply and limits the rate of the reaction.

1 mark for the correct option D.
- Reject A: High light intensity in a classroom setup does not typically damage chlorophyll to cause a flat plateau.
- Reject B: Chloroplasts do not become 'saturated' in a way that stops further light absorption without a limiting factor being involved.
- Reject C: Respiration rates do not increase to match photosynthesis just because light intensity increases.

The cartilage rings are rigid structures that keep the trachea open. During inspiration, the pressure inside the trachea drops below atmospheric pressure, which would cause the airway to collapse if it were not supported by these rings.

1 mark for the correct option B.
- Reject A: Cartilage is a structural tissue and does not actively contract to push air.
- Reject C: Mucus is secreted by goblet cells, not cartilage.
- Reject D: While the rings are C-shaped to allow the esophagus to expand, the primary function of the cartilage itself is supporting the trachea from collapsing.

Wind-pollinated flowers do not need to attract insects, so they have small, inconspicuous (often green) petals and no scent. To maximize the chance of pollen capture, they produce large amounts of light, smooth pollen that can drift easily, and have feathery stigmas that hang outside the flower.

1 mark for the correct option B.
- Reject A, C, and D: These feature combinations of large petals, sticky/heavy pollen, or internal stigmas are adaptations for insect pollination.

Using the magnification formula: \(\text{Magnification} = \frac{\text{Image size}}{\text{Actual size}}\). Here, Image size = 32 mm and Actual size = 0.04 mm. \(\text{Magnification} = \frac{32}{0.04} = 800\). Thus, the magnification is \(\times 800\).

1 mark for the correct option D.
- Reject A: This is the inverse calculation (0.04 / 32).
- Reject B: This is the product of the two values (32 * 0.04).
- Reject C: This is a calculation error where 32 was divided by 0.4 instead of 0.04.

To minimize heat loss by radiation from the skin surface during cold conditions, arterioles supplying the skin surface capillaries constrict (vasoconstriction), decreasing blood flow to the surface. Simultaneously, shunt vessels dilate, allowing blood to bypass the surface capillaries and remain deeper within the tissues.

1 mark for the correct option A.
- Reject B: Dilating skin arterioles and constricting shunt vessels is vasodilation, which occurs when body temperature is high to promote heat loss.
- Reject C and D: Arterioles and shunt vessels must coordinate in opposite ways (one constricts while the other dilates) to effectively redirect blood flow.

The genotype of the roan bull is \(C^R C^W\). The genotype of the white cow is \(C^W C^W\). Crossing these two individuals: \(C^R C^W \times C^W C^W\) yields offspring genotypes of \(50\%\) \(C^R C^W\) (roan) and \(50\%\) \(C^W C^W\) (white). The phenotypic ratio is therefore 1 roan : 1 white.

1 mark for the correct option C.
- Reject A: An 'all roan' phenotype is only expected from homozygous red (\(C^R C^R\)) crossed with homozygous white (\(C^W C^W\)).
- Reject B: No red offspring (\(C^R C^R\)) can be produced because the white cow only passes on the \(C^W\) allele.
- Reject D: This is the ratio expected from crossing two roan individuals (\(C^R C^W \times C^R C^W\)).

- In Tube 1 (bright light), the rate of photosynthesis is much greater than respiration, consuming \(\text{CO}_2\). The decrease in \(\text{CO}_2\) concentration increases the pH, turning the indicator purple.
- In Tube 2 (dark), only respiration occurs, producing \(\text{CO}_2\). The increase in \(\text{CO}_2\) concentration lowers the pH, turning the indicator yellow.
- In Tube 3 (dim light), photosynthesis and respiration occur at equal rates (compensation point), resulting in no net change in \(\text{CO}_2\) concentration. The indicator remains orange-red.

1 mark for the correct option A.
- Reject B, C, and D: These options confuse the color changes corresponding to increases and decreases in carbon dioxide concentration.

During aerobic respiration, cells consume oxygen and produce carbon dioxide and water. Consequently, expired air has less oxygen (used by tissues), more carbon dioxide (released from tissues into the blood and expired), and more water vapor (evaporated from the moist surfaces of the alveoli).

1 mark for the correct option C.
- Reject A: Expired air contains more water vapor, not less, due to evaporation from the respiratory tract.
- Reject B and D: Expired air contains less oxygen, not more, because oxygen is absorbed into the blood.

(a) A limiting factor is an environmental factor that is in the shortest supply and directly restricts or limits the rate of a physiological process such as photosynthesis. (b)(i) Counting bubbles is inaccurate because bubbles can vary in volume, some of the produced oxygen dissolves in the surrounding water, and the gas inside the bubbles may contain other gases like nitrogen. An improvement is to collect the gas using a gas syringe or a graduated capillary tube to measure the actual volume of oxygen gas produced over a set time. (b)(ii) As the distance from the light source decreases, the light intensity increases, which increases the rate of photosynthesis. Initially, light intensity is the limiting factor because there is insufficient light energy to drive the light-dependent stage. Above a certain light intensity, the rate of bubbling plateaus and remains constant. In this phase, light intensity is no longer the limiting factor; instead, another factor, such as temperature or carbon dioxide concentration, is in short supply and limits the overall rate. (c) Glucose is converted to starch for energy storage because starch is insoluble and does not affect the water potential or osmotic balance of the cell. Glucose is converted to cellulose, which provides high tensile strength to construct plant cell walls. To synthesize proteins, glucose is combined with nitrate ions (absorbed from the soil) to form amino acids, which are then polymerised into proteins. Active transport is required to absorb the nitrate ions against their concentration gradient.

(a) 1 mark for stating that it is a factor in shortest supply; 1 mark for stating that it restricts/limits the rate of a process. (b)(i) 1 mark for noting bubble size variation or gas dissolving; 1 mark for mentioning gas syringe/graduated capillary tube; 1 mark for measuring volume over time. (b)(ii) 1 mark for stating rate increases as distance decreases; 1 mark for stating rate plateaus; 1 mark for identifying light as the limiting factor initially; 1 mark for explaining that increasing light provides more energy; 1 mark for identifying another factor (such as temperature or CO2) as limiting in the plateau. (c) 1 mark for glucose to starch (insoluble storage); 1 mark for glucose to cellulose (cell walls); 1 mark for glucose + nitrates to amino acids/proteins; 0.33 marks for identifying the active transport of nitrate ions from the soil.

(a) Structural features include: 1. Large surface area: millions of alveoli and folding allow a larger area for gas diffusion; 2. Thin wall: alveolar wall is only one cell thick, which minimizes the diffusion distance; 3. Dense capillary network: continuous blood flow maintains a steep concentration gradient; 4. Moist lining: oxygen dissolves in the moisture layer before diffusing across the alveolar membrane. (b) Goblet cells in the epithelium of the trachea and bronchi secrete sticky mucus which traps inhaled dust particles, bacteria, and pathogens. Ciliated epithelial cells possess hair-like projections called cilia that beat rhythmically to move the dust-laden mucus upwards away from the lungs towards the pharynx, where it can be swallowed or coughed out. (c)(i) Tobacco smoke triggers an inflammatory response in the lungs, causing phagocytes to release elastase enzymes. These enzymes break down the elastic fibres in the alveolar walls, causing them to rupture and merge into larger, irregular air spaces. This significantly reduces the surface area available for gas exchange, causing shortness of breath. (ii) Carbon monoxide binds irreversibly to haemoglobin to form carboxyhaemoglobin, reducing the oxygen-carrying capacity of the blood. Nicotine is an addictive stimulant that constricts blood vessels, increasing heart rate and blood pressure, which increases the risk of coronary heart disease. Tar contains carcinogens that cause cellular mutations leading to lung cancer.

(a) Award 1 mark for each correct pair of feature and explanation, up to 4 marks. (b) 1 mark for goblet cells secreting mucus; 1 mark for mucus trapping particles/pathogens; 1 mark for cilia being hair-like structures; 1 mark for cilia beating to sweep mucus away from lungs (max 4 marks). (c)(i) 1 mark for phagocytes/elastase breaking down elastic fibres; 1 mark for alveolar walls rupturing/merging; 1 mark for reduced surface area causing less gas exchange/shortness of breath. (c)(ii) 1 mark each for correct component with its corresponding effect (up to 2 marks); 0.33 marks for naming haemoglobin as the specific target of carbon monoxide or highlighting the addictive stimulant nature of nicotine.

(a) Insect-pollinated flowers have large, brightly coloured, scented petals, stiff stamens enclosed inside the flower, sticky or spiky pollen grains, and a sticky stigma situated inside the flower. In contrast, wind-pollinated flowers have small, dull, green petals with no scent, stamens with versatile anthers hanging on long filaments outside the flower, smooth, light pollen grains produced in huge quantities, and a feathery stigma hanging outside the flower to catch drifting pollen. (b) Self-pollination is the transfer of pollen grains from the anther of a flower to the stigma of the same flower or another flower on the same plant. Advantages include preservation of successful genotypes that are highly adapted to a stable environment and reproduction without relying on external pollinators. Disadvantages include reduced genetic variation in offspring, which limits adaptation to environmental change, and a higher probability of expressing harmful homozygous recessive genetic disorders. (c) A pollen grain absorbs moisture and germinates on the stigma. A pollen tube grows down through the style under the control of the tube nucleus, guided by chemical signals. The pollen tube enters the ovary and reaches the ovule, entering through a small pore called the micropyle. The male gamete nucleus travels down the pollen tube and fuses with the female gamete nucleus inside the ovule, resulting in the formation of a diploid zygote.

(a) 1 mark for contrasting petals (bright/large vs dull/small); 1 mark for contrasting stamens (enclosed/stiff vs hanging/versatile); 1 mark for contrasting pollen (sticky/large vs smooth/light/abundant); 1 mark for contrasting stigma (sticky/enclosed vs feathery/exposed). (b) 1 mark for correct definition of self-pollination; 1 mark for advantage (preserves adapted genes or no pollinator dependency); 1 mark for disadvantage of lower genetic variation; 1 mark for higher risk of homozygous recessive diseases (max 4 marks). (c) 1 mark for germination of pollen on stigma; 1 mark for pollen tube growth down style; 1 mark for entering ovule via micropyle; 1 mark for male nucleus traveling down tube; 1 mark for fusion of male and female nuclei; 0.33 marks for identifying the zygote as diploid.

(a)(i) Magnification = Image size / Actual size. Image size = 48 mm, Actual size = 0.04 mm. Magnification = 48 / 0.04 = x1200. (a)(ii) Palisade mesophyll cells are located near the upper surface of the leaf to maximize light absorption. They contain a high density of chloroplasts to maintain a high rate of photosynthesis. (b)(i) Magnification = x15000. Image size = 3.6 cm = 36 mm = 36000 micrometres. Actual size = Image size / Magnification. Actual size = 36000 / 15000 = 2.4 micrometres. (b)(ii) Mitochondria are the sites of aerobic respiration. They release energy in the form of ATP for cellular metabolic activities. (c)(i) Actual size = 7 micrometres. Drawing size = 3.5 cm = 35 mm = 35000 micrometres. Magnification = Drawing size / Actual size = 35000 / 7 = x5000. (c)(ii) A red blood cell has a biconcave disc shape to increase its surface area to volume ratio, which speeds up the rate of oxygen diffusion into and out of the cell. It also lacks a nucleus to provide more space for haemoglobin.

(a)(i) 1 mark for correct formula or conversion; 1 mark for correct calculation showing x1200. (a)(ii) 1 mark for stating maximum light absorption; 1 mark for linking to chloroplasts/photosynthesis. (b)(i) 1 mark for converting cm to micrometres (36000); 1 mark for correct formula rearranged; 1 mark for correct answer of 2.4 micrometres. (b)(ii) 1 mark for aerobic respiration; 1 mark for releasing energy. (c)(i) 1 mark for converting cm to micrometres (35000); 1 mark for correct formula; 1 mark for correct answer of x5000. (c)(ii) 1 mark for identifying a red blood cell adaptation and explaining its function; 0.33 marks for naming haemoglobin as the oxygen-carrying protein.

(a) Homeostasis is the maintenance of a constant internal environment within narrow limits around a physiological set point. (b)(i) When temperature rises, sweat glands secrete sweat onto the surface of the skin. The water in sweat evaporates, absorbing latent heat energy from the skin and blood vessels, thereby cooling the body. (b)(ii) Arterioles in the skin undergo vasodilation, where their smooth muscle walls relax and dilate, while shunt vessels constrict. This directs a larger volume of blood flow to the superficial capillaries near the skin surface. Heat is lost from the blood to the cooler surroundings via radiation and convection. (c) When blood glucose levels fall below normal, the islets of Langerhans in the pancreas detect this change and secrete the hormone glucagon into the bloodstream. Glucagon travels to the liver, where it binds to specific receptors and stimulates liver cells to break down stored glycogen into glucose. This glucose is then released into the blood, raising blood glucose concentration back to the normal range through a negative feedback mechanism.

(a) 1 mark for maintenance of a constant internal environment; 1 mark for within narrow limits or around a set point. (b)(i) 1 mark for sweat glands secreting sweat; 1 mark for sweat evaporating; 1 mark for evaporation absorbing heat energy. (b)(ii) 1 mark for skin arterioles dilating (vasodilation); 1 mark for shunt vessels constricting; 1 mark for increased blood flow in capillaries near the skin surface; 1 mark for heat loss by radiation or convection. (c) 1 mark for pancreas/islets of Langerhans detecting low glucose; 1 mark for secreting glucagon; 1 mark for glucagon acting on the liver; 1 mark for glycogen breaking down to glucose; 0.33 marks for identifying this as an example of negative feedback.

(a)(i) An allele is an alternative or different form of a gene. (a)(ii) Heterozygous means having two different alleles of a particular gene (e.g. Ff). (a)(iii) Phenotype refers to the physical or observable characteristics of an organism, which is determined by its genotype and environmental influences. (b)(i) Parent phenotypes: Carrier x Carrier; Parent genotypes: Ff x Ff; Gametes: F and f from both parents; Punnett square showing offspring genotypes: FF, Ff, Ff, ff; Offspring phenotypes and ratio: 3 healthy (or carrier) : 1 cystic fibrosis (or 75% healthy, 25% with cystic fibrosis). (b)(ii) The probability is 2 in 4, which is 0.5 or 50% or 1 in 2. (c) If two unaffected parents have an affected child, the disease allele must be recessive, because the parents must carry the allele without showing symptoms (carriers). If a disease is dominant, every affected individual must have at least one affected parent, meaning the disease does not skip generations. Genetic counselors use this to calculate the exact probability of prospective parents passing on the disease to their future children, which is useful for family planning and making informed reproductive decisions.

(a)(i) 1 mark for different form of a gene. (a)(ii) 1 mark for having two different alleles of a gene. (a)(iii) 1 mark for physical/observable features of an organism. (b)(i) 1 mark for correct parent phenotypes; 1 mark for parent genotypes; 1 mark for gametes; 1 mark for offspring genotypes; 1 mark for correct phenotypes matched to genotypes with 3:1 ratio (max 5 marks). (b)(ii) 1 mark for 0.5 / 50% / 1 in 2. (c) 1 mark for explaining that unaffected parents having an affected child indicates a recessive allele; 1 mark for stating that carriers carry but do not express the recessive phenotype; 1 mark for noting dominant traits do not skip generations; 1 mark for explaining that this calculates probability of offspring inheritance; 0.33 marks for linking to family planning decisions.

**(a)(i) Table of Results:**

| Concentration of \(\text{NaHCO}_3\) solution / % | Number of bubbles counted in 3 minutes | Mean rate of photosynthesis / bubbles per minute |
| :---: | :---: | :---: | :---: | :---: |
| | **Trial 1** | **Trial 2** | **Trial 3** | |
| 0.1 | 12 | 11 | 13 | 4 |
| 0.2 | 22 | 24 | 23 | 8 |
| 0.3 | 35 | 33 | 37 | 12 |
| 0.4 | 45 | 47 | 46 | 15 |
| 0.5 | 46 | 48 | 47 | 16 |

*Working for mean rates per minute (rounded to nearest whole number):*
- **0.1%:** \((12+11+13) / 3 = 12\) bubbles in 3 min \(\rightarrow 12 / 3 = 4\) bubbles/min.
- **0.2%:** \((22+24+23) / 3 = 23\) bubbles in 3 min \(\rightarrow 23 / 3 = 7.67 \approx 8\) bubbles/min.
- **0.3%:** \((35+33+37) / 3 = 35\) bubbles in 3 min \(\rightarrow 35 / 3 = 11.67 \approx 12\) bubbles/min.
- **0.4%:** \((45+47+46) / 3 = 46\) bubbles in 3 min \(\rightarrow 46 / 3 = 15.33 \approx 15\) bubbles/min.
- **0.5%:** \((46+48+47) / 3 = 47\) bubbles in 3 min \(\rightarrow 47 / 3 = 15.67 \approx 16\) bubbles/min.

**(a)(ii)**
- **Independent Variable:** Concentration of sodium hydrogencarbonate (\(\text{NaHCO}_3\)) solution (or carbon dioxide concentration).
- **Dependent Variable:** Rate of photosynthesis (or number of bubbles / oxygen gas released per minute).

**(a)(iii)** Any two of:
- Light intensity (distance of the light source from the plant).
- Temperature of the solution.
- Species of the pondweed plant used.
- Length / mass / surface area of the pondweed.
- Volume of the solution used.

**(b)**
- **Formula:** \(\text{Magnification} = \frac{\text{Image size}}{\text{Actual size}}\)
- **Substitution:** \(\text{Magnification} = \frac{42\text{ mm}}{0.07\text{ mm}}\)
- **Calculation:** Magnification = \(\times 600\) (or 600)

**(c) Plan Investigation:**
- **Independent Variable:** Set up the investigation at five different temperatures (e.g., 10 °C, 20 °C, 30 °C, 40 °C, and 50 °C).
- **Temperature Control:** Use water baths (thermostatically controlled or containing ice/hot water as needed) to maintain the temperatures, monitoring with a thermometer.
- **Dependent Variable:** Count the number of oxygen bubbles released from the cut stem of *Cabomba* in a set time period (e.g., 2 minutes) or measure the volume of gas collected in a gas syringe.
- **Controlled Variables:**
- Keep light intensity constant by placing an LED lamp at a fixed distance (e.g., 20 cm) from each tube.
- Place a transparent heat shield (beaker of water) between the lamp and the plant to prevent the lamp's heat from affecting the water temperature.
- Keep the carbon dioxide concentration constant by using the same concentration of sodium hydrogencarbonate solution (e.g., 0.2% \(\text{NaHCO}_3\)) in all tubes.
- Use a piece of *Cabomba* of the same length (e.g., 5 cm) from the same source.
- **Reliability:** Leave the plant to equilibrate at each temperature for 5 minutes before counting bubbles. Repeat the measurement at each temperature three times and calculate a mean.
- **Safety:** Wear safety goggles and heatproof gloves when dealing with hot water baths.

**(a)(i)**
- **M1 Table Design:** Clear column and row headings, independent variable with unit (Concentration of \(\text{NaHCO}_3\) solution / %) and dependent variable with unit (Number of bubbles counted in 3 minutes [Trial 1, 2, 3] AND Mean rate / bubbles per minute). [1]
- **M2 Raw Data:** Correct transcription of all 15 raw bubble counts. [1]
- **M3 Mean (0.1%):** Correct calculation of 4 bubbles per minute. [1]
- **M4 Mean (0.2% & 0.3%):** Correct calculation and rounding of 8 and 12 bubbles per minute. [1]
- **M5 Mean (0.4% & 0.5%):** Correct calculation and rounding of 15 and 16 bubbles per minute. [1]

**(a)(ii)**
- **M1:** Independent variable: Concentration of sodium hydrogencarbonate (Accept: carbon dioxide concentration). [1]
- **M2:** Dependent variable: Rate of photosynthesis / number of bubbles released per minute (or per unit time). [1]

**(a)(iii)**
- **M1 & M2:** Any two controlled variables correctly identified (e.g., light intensity/distance of lamp, temperature, plant species, plant length, volume of solution). [2]

**(b)**
- **M1:** State formula: \(\text{Magnification} = \frac{\text{size of image}}{\text{actual size}}\). [1]
- **M2:** Correct substitution: \(\frac{42}{0.07}\). [1]
- **M3:** Correct final answer: \(\times 600\) or 600. (Reject units in final value, e.g., 600 mm is wrong). [1]

**(c) Plan Investigation (Max 8 marks):**
- **P1 (IV):** Test at least 5 different temperatures. [1]
- **P2 (IV Method):** Use of water baths to maintain constant temperatures. [1]
- **P3 (DV):** Count number of bubbles or measure volume of gas produced in a specific time. [1]
- **P4 (CV1):** Keep light intensity constant by maintaining lamp at a fixed distance. [1]
- **P5 (CV2):** Use a heat shield (beaker of water) to prevent lamp heating. [1]
- **P6 (CV3):** Use the same concentration / volume of sodium hydrogencarbonate solution. [1]
- **P7 (CV4):** Use the same species and length of *Cabomba* pondweed. [1]
- **P8 (Equilibration):** Allow plant to adjust to the temperature for a set time (e.g., 5 mins) before counting. [1]
- **P9 (Reliability):** Repeat each temperature trial at least 3 times and calculate a mean. [1]
- **P10 (Safety):** Wear safety goggles or use caution with hot water. [1]

**(a)(i) Graph Plotting Details:**
- **Axes:** x-axis is labeled 'Time / minutes' (scaled 0 to 6); y-axis is labeled 'Breathing rate / breaths per minute' (scaled 0 to 50).
- **Scale:** Plotted points occupy more than half of the grid in both directions.
- **Plots:** All 7 points accurately plotted with a small 'x' or a circled dot.
- **Line:** Points joined with clean, ruled straight lines. No extrapolation beyond the coordinates (0,14) and (6,14).

**(a)(ii)**
- Breathing rate increases rapidly during the exercise period (from 14 breaths per minute at 0 minutes to a maximum peak of 44 breaths per minute at 3 minutes).
- Breathing rate decreases rapidly once exercise stops (from 44 breaths per minute at 3 minutes to 14 breaths per minute at 6 minutes).
- The athlete's breathing rate completely returns to its initial resting level (14 breaths per minute) by the 6th minute (which is 3 minutes after exercise ended).

**(a)(iii)**
- **During exercise:** Muscles contract more frequently and powerfully, requiring more energy. Aerobic respiration increases, which produces more carbon dioxide as a waste product. The brain detects the elevated carbon dioxide concentration in the blood and triggers an increase in breathing rate to increase oxygen uptake and carbon dioxide removal.
- **After exercise:** The breathing rate remains higher than the resting rate at 4 and 5 minutes to pay off the oxygen debt. Extra oxygen is required to break down lactic acid, which accumulated in the muscles due to anaerobic respiration during the high-intensity exercise, into carbon dioxide and water.

**(b)** Any one from:
- Use a freshly disinfected or sterile mouthpiece for each subject.
- Check that the soda lime (which absorbs carbon dioxide) is fresh and working properly.
- Ensure the subject does not have a medical history of asthma or heart disease.
- Stop the test immediately if the subject feels dizzy or faint.

**(c) Plan Investigation:**
- **Independent Variable:** Type of exercise performed (e.g., walking, jogging, sprinting).
- **Dependent Variable:** Skin temperature, measured on the forehead or inner wrist using a digital infrared thermometer.
- **Control Variables:**
- Duration of each exercise type (e.g., exactly 5 minutes of exercise).
- Environmental conditions (e.g., perform the test in the same room with constant temperature and humidity).
- Participant characteristics (e.g., same age group, similar fitness level, wearing identical athletic clothing).
- **Method:**
1. Measure and record the participant's resting skin temperature before starting exercise.
2. Have the participant perform the chosen exercise (e.g., walking) for 5 minutes.
3. Immediately measure and record the skin temperature at the exact same location.
4. Calculate the change in skin temperature (final temperature minus initial temperature).
5. Allow the participant to rest completely until heart and breathing rates return to resting levels before testing the next type of exercise.
- **Reliability:** Repeat the test with at least 5 different participants and calculate the mean temperature change for each type of exercise.
- **Safety:** Ensure participants have warmed up, are properly hydrated, and stop exercising if they experience discomfort.

**(a)(i)**
- **M1 Axes:** 'Time / minutes' on x-axis and 'Breathing rate / breaths per minute' on y-axis (with units). [1]
- **M2 Scale:** Linear, easy-to-read scale where plotted points cover at least half of the grid in both dimensions. [1]
- **M3 Plotting:** All 7 points plotted accurately within half a small square. [1]
- **M4 Line:** Points joined with clean, ruled straight lines or a smooth curve. No thick double lines or sketchy lines. [1]
- **M5 Extrapolation:** No extrapolation beyond 0 min or 6 min. [1]

**(a)(ii)**
- **M1:** Breathing rate increases during exercise (0 to 3 min) AND reaches a maximum / peak of 44 breaths per minute. [1]
- **M2:** Breathing rate decreases after exercise (3 to 6 min). [1]
- **M3:** Breathing rate returns to the original resting value of 14 breaths per minute by 6 minutes / after 3 minutes of rest. [1]

**(a)(iii)**
- **M1:** Muscle contraction increases during exercise, requiring more energy / ATP. [1]
- **M2:** Rate of respiration increases, releasing more carbon dioxide into the blood. [1]
- **M3:** Brain detects increase in blood carbon dioxide concentration and increases breathing rate (to supply oxygen / remove carbon dioxide). [1]
- **M4:** After exercise, extra oxygen is needed to repay oxygen debt / break down lactic acid. [1]

**(b)**
- **M1:** Any one from: sterile mouthpiece for each participant, ensure soda lime is fresh, do not use with asthmatic/unwell individuals, limit duration of test. [1]

**(c) Plan Investigation (Max 7 marks):**
- **P1 (IV):** Test at least three distinct types of exercise (e.g., walking, jogging, sprinting). [1]
- **P2 (DV):** Measure skin temperature using a digital thermometer / infrared sensor / thermal camera. [1]
- **P3 (CV1):** Constant duration of exercise (e.g., 5 minutes for each type). [1]
- **P4 (CV2):** Perform exercise in the same environmental temperature / room. [1]
- **P5 (CV3):** Use subjects of similar characteristics (age / fitness / sex) OR use the same subject with adequate rest periods. [1]
- **P6 (Method):** Measure temperature before AND immediately after exercise (to calculate temperature difference). [1]
- **P7 (Reliability):** Repeat with at least 5 different participants (or repeat trials on the same participant) and calculate a mean. [1]
- **P8 (Safety):** Ensure participant hydration / warm-up / monitor for fatigue. [1]