2022 AQA GCSE Biology 8461 Practice Paper with Answers

a) The three main structures found in plant cells but absent in animal cells are the cellulose cell wall, chloroplasts, and a large permanent vacuole.

b)(i) The cell wall is made of cellulose and strengthens and supports the cell, maintaining its shape.
(ii) The permanent vacuole contains cell sap (a weak solution of sugar and salts) and keeps the cell rigid and turgid to support the plant.

c) Using the magnification formula:
\(\text{Magnification} = \frac{\text{size of image}}{\text{real size of object}}\)
\(\text{Magnification} = \frac{15\text{ mm}}{0.05\text{ mm}} = 300\)
Therefore, the magnification is \(\times 300\).

d) Electron microscopes have a much higher resolution and higher magnification compared to light microscopes, allowing scientists to see very small sub-cellular structures (like ribosomes and mitochondria) in detail.

- (a) [3 marks] 1 mark for each correct structure: cell wall, chloroplasts, permanent vacuole/vacuole. (Accept starch grain/plastid; reject cell membrane, cytoplasm, nucleus).
- (b)(i) [2 marks] To strengthen the cell (1 mark) and support/maintain shape (1 mark).
- (b)(ii) [2 marks] Contains cell sap (1 mark) and keeps the cell turgid/firm (1 mark).
- (c) [3.5 marks]
- Correct substitution: \(\frac{15}{0.05}\) (1.5 marks)
- Correct calculation: \(300\) (2 marks)
- (Award full 3.5 marks for correct final answer of \(\times 300\) or 300 without working).
- (d) [2 marks] Any one from: higher magnification (2 marks) OR higher resolution / can see smaller organelles like ribosomes (2 marks). (Accept: more detail).

a) Osmosis is defined as the diffusion of water molecules from a dilute solution (high water concentration) to a concentrated solution (low water concentration) through a partially permeable membrane.

b)(i) Change in mass = Final mass - Starting mass = \(3.2\text{ g} - 4.0\text{ g} = -0.8\text{ g}\) (a decrease of 0.8 g).
(ii) \(\text{Percentage change} = \frac{-0.8}{4.0} \times 100 = -20\%\) (a 20% decrease).

c) The concentrated salt solution has a lower concentration of water (more concentrated solute) than the cytoplasm inside the potato cells. Consequently, water moves out of the potato cells by osmosis across the partially permeable cell membranes, reducing the mass of the cylinder.

d) The student must blot the cylinders dry to remove any excess liquid resting on the outside surface of the potato. If left on, this water would increase the measured final mass, leading to inaccurate results.

- (a) [3 marks] Diffusion of water (1 mark) from a dilute to a concentrated solution (1 mark) through a partially permeable membrane (1 mark).
- (b)(i) [1 mark] \(-0.8\) g (or a decrease of \(0.8\) g).
- (b)(ii) [3 marks]
- Substitution: \(\frac{-0.8}{4.0} \times 100\) (1.5 marks)
- Correct calculation: \(-20\)% or a \(20\)% decrease (1.5 marks).
- (c) [3.5 marks]
- Water moved out of the potato cells (1 mark)
- by osmosis (1 mark)
- because the external salt solution had a lower concentration of water / higher concentration of solutes than the potato cytoplasm (1.5 marks).
- (d) [2 marks] To remove excess surface liquid/water (1 mark) so it does not add to the final mass of the potato cylinder (1 mark).

a) Chromosomes are made of DNA (deoxyribonucleic acid).

b) In eukaryotic cells, chromosomes are located inside the nucleus.

c) Mitosis is necessary for the growth of multicellular organisms, for repairing damaged tissues by replacing dead or damaged cells, and for asexual reproduction.

d)(i) Before a cell can divide, it must grow and increase its sub-cellular structures. Crucially, the DNA replicates to produce two identical copies of each chromosome.
(ii) During mitosis, the chromosomes line up along the center of the cell. Spindle fibers pull one set of chromosomes to each opposite pole (end) of the cell. Finally, the nucleus divides to form two new nuclei containing identical genetic material.

e) Mitosis produces two genetically identical daughter cells. Therefore, each new skin cell will contain exactly the same number of chromosomes as the parent cell, which is 46.

- (a) [1 mark] DNA (deoxyribonucleic acid).
- (b) [1 mark] Nucleus (reject nucleolus).
- (c) [2 marks] Any two from: Growth (1 mark); repair of tissues / replacement of damaged cells (1 mark); asexual reproduction (1 mark).
- (d)(i) [2 marks] The DNA/chromosomes replicate / copy (1 mark) and the cell grows / increases sub-cellular structures (1 mark).
- (d)(ii) [4 marks]
- Chromosomes line up in the center (1 mark)
- One set of chromosomes is pulled to opposite ends of the cell (2 marks)
- The nucleus divides (1 mark).
- (e) [2.5 marks] 46 (2.5 marks).

a) The palisade mesophyll layer contains cells packed with chloroplasts. Its position near the top of the leaf ensures maximum light absorption for photosynthesis.

b) Stomata are tiny pores on the underside of leaves surrounded by guard cells. When water is plentiful, guard cells swell (become turgid) and curve, opening the stomata to allow gas exchange. In dry or hot conditions, the guard cells lose water (become flaccid) and come together, closing the stomatal pore to prevent excessive water loss via transpiration.

c) Increase in rate of water loss = Rate in warm, windy air - Rate in cool, still air
\(\text{Increase} = 4.5\text{ g/hour} - 1.2\text{ g/hour} = 3.3\text{ g/hour}\).

d) Water is transported up the plant through xylem tissue, which consists of hollow tubes of dead cells reinforced with lignin.

- (a) [3 marks] Palisade mesophyll (1 mark). Function: Photosynthesis / absorbs maximum light (2 marks).
- (b) [4 marks] Guard cells open and close stomata (1 mark). In dry/hot/dark conditions, guard cells close the stomata (1 mark) which prevents water vapor from escaping (1 mark) by evaporation/transpiration (1 mark).
- (c) [2.5 marks]
- Correct calculation: \(4.5 - 1.2 = 3.3\) (1.5 marks)
- Correct unit: g/hour or g per hour (1 mark).
- (d) [3 marks] Xylem (3 marks) (reject phloem).

a) Proteins are large polymers made of amino acids. During digestion, proteases break down proteins into individual amino acids.

b)(i) Amylase works in the mouth (where it is secreted in saliva) and in the small intestine (where pancreatic amylase is added).
(ii) The stomach contains hydrochloric acid, giving it a very low pH (around pH 1.5 - 2). This acidic environment denatures amylase because the highly acidic conditions disrupt the chemical bonds keeping the active site in shape. Once denatured, amylase cannot bind to starch.

c)(i) Bile is produced in the liver and stored in the gall bladder until it is needed.
(ii) Bile helps fat digestion in two ways:
1. It is alkaline, so it neutralises the acidic mixture coming from the stomach, creating optimum pH conditions for lipase to work in the small intestine.
2. It emulsifies fats, breaking large globules of fat into many tiny droplets. This greatly increases the surface area for lipase enzymes to interact with, speeding up digestion.

- (a) [2 marks] Amino acids (2 marks).
- (b)(i) [2 marks] Mouth (1 mark) and small intestine (1 mark).
- (b)(ii) [3 marks]
- Stomach contains hydrochloric acid / is acidic / has low pH (1 mark)
- Amylase is denatured (1 mark)
- because the shape of its active site is changed (1 mark).
- (c)(i) [1.5 marks] Gall bladder (1.5 marks).
- (c)(ii) [4 marks]
- Emulsifies fats / breaks fat into tiny droplets (1 mark) to increase the surface area for lipase to work on (1 mark).
- Neutralises hydrochloric acid from the stomach (1 mark) to provide the optimum alkaline pH for lipase enzymes to work (1 mark).

a)(i) Malaria is transmitted by female Anopheles mosquitoes acting as vectors. When they bite a human, they pass the protist into the bloodstream.
(ii) Preventing mosquito bites (using insect repellent, mosquito nets) or destroying mosquito breeding sites (draining stagnant water) reduces transmission.

b) The skin forms a continuous physical barrier that blocks entry, and it secretes antimicrobial lipids that kill bacteria. The trachea has goblet cells that produce sticky mucus to trap inhaled pathogens, and ciliated cells with hair-like structures (cilia) that sweep the mucus up and out of the airways to the throat to be swallowed and destroyed by stomach acid.

c) White blood cells protect the body by:
1. Phagocytosis, where they engulf and digest foreign cells.
2. Producing antibodies, which are proteins custom-designed to target and destroy specific antigens on pathogens.
3. Producing antitoxins, which counteract and neutralise the toxic substances produced by bacteria.

d) A vaccine contains dead or weakened forms of a pathogen. When injected, it stimulates white blood cells to produce specific antibodies against the antigens. If the person meets the real pathogen later, memory cells recognize it and produce antibodies rapidly and in large quantities, destroying the pathogen before they fall ill.

- (a)(i) [1.5 marks] Via a mosquito vector / mosquito bites (1.5 marks).
- (a)(ii) [1 mark] Any one from: Use mosquito nets (1 mark); use insect repellent (1 mark); drain standing water to stop breeding (1 mark); take antimalarial drugs (1 mark).
- (b) [4 marks]
- Skin: Acts as a physical barrier (1 mark) / secretes antimicrobial substances to kill bacteria (1 mark).
- Trachea: Secretes mucus to trap pathogens (1 mark) and has cilia (hairs) to sweep mucus up to the throat to be swallowed (1 mark).
- (c) [4 marks] Any two from:
- Phagocytosis / engulfing and digesting pathogens (2 marks)
- Producing specific antibodies to target/destroy pathogens (2 marks)
- Producing antitoxins to neutralise toxins released by bacteria (2 marks).
- (d) [2 marks] Introduces dead or inactive pathogens (1 mark), which stimulates white blood cells to produce antibodies / create memory cells for a rapid immune response if infected later (1 mark).

a) The missing product is glucose.

b)(i) The student can use an LED bulb, which emits very little heat, or place a large glass container of water between the lamp and the pondweed to absorb heat energy without blocking light.
(ii) As the distance from the light source increases, the rate of photosynthesis (measured by the number of bubbles per minute) decreases. This is an inverse relationship.

c) At larger distances (like 40 cm), the light intensity is very low. Light energy is necessary to power the chemical reaction of photosynthesis. Because there is insufficient light energy, the reaction cannot happen at its maximum speed. Thus, light intensity is the factor restricting the overall rate.

d) By heating the greenhouse and adding carbon dioxide, growers remove temperature and carbon dioxide concentration as limiting factors. This maximizes the rate of photosynthesis, allowing plants to grow faster and produce larger crop yields, leading to greater profit.

- (a) [1.5 marks] Glucose (1.5 marks) (accept sugar/starch).
- (b)(i) [2 marks] Use an LED lamp (1 mark) OR place a glass tank/beaker of water between the lamp and the pondweed to absorb heat (1 mark).
- (b)(ii) [2 marks] As distance increases, the rate of photosynthesis/bubble count decreases (2 marks). (Accept: inverse relationship).
- (c) [4 marks]
- Increasing distance reduces light intensity (1 mark)
- Light energy is required for photosynthesis to take place (1 mark)
- At low light intensity, there is not enough energy to drive the reaction at its maximum rate (1 mark)
- Therefore, light intensity limits the overall rate (1 mark).
- (d) [3 marks]
- To increase temperature and carbon dioxide concentration so they are no longer limiting factors (1 mark)
- This increases the rate of photosynthesis (1 mark)
- Resulting in faster plant growth / higher crop yields / more profit (1 mark).

a) The balanced chemical symbol equation is:
\(\text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \rightarrow 6\text{CO}_2 + 6\text{H}_2\text{O}\)

b) Two differences are:
1. Aerobic respiration uses oxygen, whereas anaerobic respiration in muscles takes place in the absence of oxygen.
2. Aerobic respiration fully breaks down glucose to produce carbon dioxide and water, while anaerobic respiration only partially breaks down glucose, producing lactic acid. Additionally, aerobic respiration yields much more energy per glucose molecule.

c)(i) The substance produced is lactic acid.
(ii) After exercise, lactic acid must be removed. It is transported by the blood from the muscles to the liver. In the liver, lactic acid is reacted with oxygen to convert it back into glucose. This requirement for oxygen to clear the lactic acid is known as the 'oxygen debt', which is why breathing rate remains high after exercise.

- (a) [3 marks]
- Reactants: \(C_6H_{12}O_6 + 6O_2\) (1 mark)
- Products: \(6CO_2 + 6H_2O\) (1 mark)
- Correctly balanced (1 mark)
- (Award 1 mark total if word equation is given instead: Glucose + Oxygen -> Carbon Dioxide + Water).
- (b) [4 marks] Any two differences from:
- Aerobic uses oxygen, anaerobic does not (2 marks).
- Aerobic produces carbon dioxide and water, anaerobic produces lactic acid (2 marks).
- Aerobic releases a large amount of energy, anaerobic releases a small amount of energy (2 marks).
- (c)(i) [1.5 marks] Lactic acid (1.5 marks).
- (c)(ii) [4 marks]
- Blood carries lactic acid from muscles to the liver (1 mark)
- In the liver, lactic acid is converted back to glucose (1 mark)
- This reaction requires oxygen (1 mark)
- This is why an oxygen debt is paid / breathing and heart rate remain high after exercise to supply oxygen to the liver (1 mark).

a) Homeostasis refers to the regulation of internal conditions of a cell or organism to maintain optimum conditions for function in response to internal and external changes. This is vital to keep enzymes from denaturing and to ensure cells function efficiently.

b) Two other conditions controlled by homeostasis are: 1. Blood glucose concentration, 2. Water levels (or carbon dioxide levels/pH).

c) Parts of a reflex arc involved include receptors (which detect the stimulus/change) and coordination centres (such as the brain, spinal cord, or pancreas which receive and process information).

d) The two main mechanisms to lower body temperature are:
1. Sweat glands produce sweat, which evaporates from the skin surface, transferring heat energy away from the body.
2. Blood vessels supplying the skin capillaries dilate (vasodilation), which allows more blood to flow close to the skin surface, increasing heat loss by radiation.

a)
- Regulation of internal conditions of a cell or organism [1 mark]
- To maintain optimum conditions for function / enzyme activity [1 mark]

b)
- Blood glucose concentration [1 mark]
- Water levels [1 mark] (accept: salt levels / pH / urea concentration)

c)
- Receptors [1 mark]
- Coordination centre / brain / spinal cord [1 mark] (also accept sensory neurone / motor neurone / effector)

d)
- Sweat glands produce sweat [1 mark]
- Sweat evaporates, transferring thermal energy (cooling the body) [1 mark]
- Blood vessels supplying the skin dilate (vasodilation) [1 mark] (do not accept 'capillaries dilate' or 'blood vessels move')
- More blood flows close to the skin surface, losing heat by radiation [1 mark]

a) A reflex action is an automatic and rapid response to a stimulus that does not involve the conscious part of the brain. It is important because it prevents or minimises damage to the body (protects from harm).

b)
- (i) receptor
- (ii) sensory
- (iii) synapse
- (iv) effector (or muscle / gland)

c)
- Decrease in reaction time: \( 0.25\text{ s} - 0.20\text{ s} = 0.05\text{ s} \)
- Percentage decrease: \( \frac{0.05}{0.25} \times 100 = 20\% \)

a)
- Rapid / automatic response [1 mark]
- Does not involve the conscious part of the brain [1 mark]
- Protects the body from harm / injury [1 mark]

b)
- (i) receptor [1 mark]
- (ii) sensory [1 mark]
- (iii) synapse [1 mark]
- (iv) effector [1 mark] (accept muscle / gland)

c)
- Correct calculation of decrease: \( 0.05\text{ s} \) [1 mark]
- Correct division by original value: \( \frac{0.05}{0.25} \) [1 mark]
- Correct final percentage: \( 20\% \) [1 mark]
- (Award 3 marks for correct final answer with or without working)

a) The pancreas is the gland that monitors and controls blood glucose levels.

b) When blood glucose is too high, the pancreas releases the hormone insulin. Insulin causes glucose to move from the blood into body cells. In liver and muscle cells, excess glucose is converted into glycogen for storage.

c)
- (i) In Type 1 diabetes, the pancreas fails to produce sufficient insulin.
- (ii) It is managed/treated with insulin injections.
- (iii) A major risk factor for Type 2 diabetes is obesity.

d) In a healthy person, insulin keeps blood glucose within a normal range, so all glucose filtered by the kidneys is reabsorbed back into the blood. In a diabetic person, blood glucose is so high that the kidneys cannot reabsorb all of it, so some glucose is excreted in the urine.

a)
- Pancreas [1 mark]

b)
- Insulin [1 mark]
- (Released from the) pancreas [1 mark]
- Causes glucose to move from blood into cells [1 mark]
- In liver / muscle cells, glucose is converted to glycogen [1 mark]

c)
- (i) insulin [1 mark]
- (ii) insulin injections / insulin therapy [1 mark]
- (iii) obesity / lack of exercise / poor diet [1 mark]

d)
- Blood glucose concentration is extremely high [1 mark]
- Not all glucose can be reabsorbed by the kidneys (so it is excreted in urine) [1 mark]

a)
- (i) Auxin is the hormone that coordinates growth in response to light.
- (ii) Auxin accumulates / moves to the shaded side of the shoot. This unequal distribution causes cells on the shaded side to elongate more than cells on the lit side. Consequently, the shoot bends towards the light source.

b) This response is called phototropism (specifically positive phototropism).

c)
- (i) Auxins are used as weedkillers, rooting powders, or in tissue culture to promote growth.
- (ii) Gibberellins are used to end seed dormancy, promote flowering, or increase fruit size.
- (iii) Ethene is used to control the ripening of fruit.

d) Gravitropism causes roots to grow downwards. This is useful because it anchors the plant securely in the ground and ensures the roots grow towards water and essential mineral ions in the soil.

a)
- (i) Auxin [1 mark]
- (ii) Auxin moves to / accumulates on the shaded side [1 mark]
- Cells on the shaded side elongate / grow more [1 mark]
- Causing the shoot to bend towards the light [1 mark]

b)
- Phototropism [1 mark] (accept positive phototropism)

c)
- (i) Weedkiller / rooting powder / plant tissue culture [1 mark] (any one)
- (ii) Ending seed dormancy / promoting flowering / increasing fruit size [1 mark] (any one)
- (iii) Ethene [1 mark]

d)
- Anchors the plant in the soil [1 mark]
- Helps the plant absorb water / mineral ions [1 mark]

a)
- (i) Sexual reproduction requires 2 parents.
- (ii) Asexual reproduction involves mitosis.
- (iii) Offspring of asexual reproduction are genetically identical (clones) to the parent.

b)
- (i) Meiosis in human males takes place in the testes.
- (ii) A human sperm cell contains 23 chromosomes (which is half the number found in a normal body cell, which has 46).

c)
- (i) The genetic material is DNA.
- (ii) A section of DNA coding for a protein is a gene.
-
d) Understanding the human genome helps medicine by: 1. Identifying genes linked to different types of diseases (such as cancer or heart disease), which allows early treatment. 2. Helping us understand and develop therapies for inherited disorders (such as cystic fibrosis).

a)
- (i) Two / 2 [1 mark]
- (ii) Mitosis [1 mark]
- (iii) Identical [1 mark] (accept: clones)

b)
- (i) Testes [1 mark] (accept: testis)
- (ii) Half / 23 (compared to 46) [1 mark]

c)
- (i) DNA / Deoxyribonucleic acid [1 mark]
- (ii) Gene [1 mark]

d)
- Search for genes linked to different types of disease [1 mark]
- Understand / treat inherited disorders [1 mark]
- Explaining detail: allows targeted treatments / genetic screening / early intervention [1 mark]

a)
- (i) Mendel suggested that 'hereditary units' (now known as genes) were passed on.
- (ii) Mendel's work was not accepted because scientists did not know about chromosomes, DNA, or genes, and they did not understand how cell division worked.

b)
- (i) Darwin's theory challenged the belief that God created all individual species of plants and animals, and there was lack of direct evidence to convince everyone immediately.
- (ii) Lamarck suggested that characteristics acquired by an organism during its lifetime are passed on to its offspring (e.g., a giraffe stretching its neck). This is incorrect because acquired characteristics do not alter an organism's DNA/genes, so they cannot be inherited.

c)
- (i) Fossils form when hard parts of organisms (like bones) are replaced by minerals as they decay, or as preserved traces like footprints, burrows, or impressions in mud.
- (ii) The fossil record is incomplete because many early organisms were soft-bodied (so they decayed completely without leaving fossils), and many fossils have been destroyed by geological activity (e.g. tectonic plate movements, volcanic activity).

a)
- (i) Hereditary units / factors [1 mark]
- (ii) DNA / chromosomes / genes were not yet discovered [1 mark]
- Cell division / meiosis was not yet understood [1 mark]

b)
- (i) Challenged the idea of divine creation / lack of sufficient evidence at the time / mechanism of inheritance (genes) was unknown [1 mark] (any one)
- (ii) Description: Acquired characteristics / changes during a lifetime are inherited [1 mark]
- Explanation: Acquired traits do not affect genes/DNA (which are passed to offspring) [1 mark]

c)
- (i) From parts of organisms that have not decayed [1 mark]
- When parts are replaced by minerals [1 mark] (or as preserved traces/footprints)
- (ii) Many early organisms were soft-bodied / left no fossil trace [1 mark]
- Fossils have been destroyed by geological activity / volcanic activity / earthquakes [1 mark]

a)
- (i) The correct sequence from largest to smallest is: Kingdom \(\rightarrow\) Phylum \(\rightarrow\) Class \(\rightarrow\) Order \(\rightarrow\) Family \(\rightarrow\) Genus \(\rightarrow\) Species.
- (ii) The first word, Vulpes, represents the Genus.
- (iii) This naming system is called the binomial system.

b)
- (i) The three domains are: Archaea, Bacteria, and Eukaryota (or Eukarya).
- (ii) New techniques in biochemistry and chemical analysis, along with DNA and RNA sequencing, allowed scientists to look inside cells and compare genetic codes, showing differences that were invisible under a microscope.

a)
- (i) Class [1 mark]
- Order [1 mark]
- Genus [1 mark]
- (ii) Genus [1 mark]
- (iii) Binomial (system) [1 mark]

b)
- (i) Archaea [1 mark]
- Bacteria [1 mark]
- Eukaryota [1 mark]
- (ii) DNA sequencing / RNA sequencing [1 mark]
- Advances in biochemistry / chemical analysis of cells [1 mark]

a)
- (i) A community refers to all the populations of different species living and interacting together in a specific habitat.
- (ii) An ecosystem is the interaction of a community of living organisms (biotic) with the non-living (abiotic) parts of their environment.

b)
- (i) Plants compete for light, space, water, and mineral ions.
- (ii) Animals compete for food, mates, and territory.

c)
- (i) Temperature is an abiotic (non-living) factor.
- (ii) Pathogens are biotic (living) factors.

d) A cactus has several adaptations:
1. Spines instead of leaves, which reduces the surface area to decrease water loss through transpiration.
2. A thick, waxy cuticle, which acts as a barrier to reduce water evaporation.
3. Deep or very wide root systems, which allow the cactus to absorb maximum water from the dry soil.

a)
- (i) All the populations of different species living in a habitat together [1 mark]
- (ii) Interaction of living organisms (community) with the non-living (abiotic) parts of their environment [1 mark]

b)
- (i) Light / space / water / mineral ions [2 marks for any two]
- (ii) Food / mates / territory [2 marks for any two]

c)
- (i) Abiotic [1 mark]
- (ii) Biotic [1 mark]

d)
- Description of feature + survival benefit [1 mark each, up to 2 marks]
- e.g., Spines/needles [1 mark] to reduce water loss / transpiration [1 mark]
- e.g., Deep/wide roots [1 mark] to absorb maximum water [1 mark]
- e.g., Thick waxy cuticle [1 mark] to prevent water evaporation [1 mark]

a) The organ that monitors and controls blood glucose is the pancreas.

b) When blood glucose levels are too high, the pancreas releases insulin. This hormone causes glucose to move from the blood into cells, particularly liver and muscle cells. In these cells, the excess glucose is converted into glycogen for storage.

c) (i) Type 1 diabetes is treated using daily insulin injections or an insulin pump.

(ii) To calculate BMI:
$$\text{BMI} = \frac{95}{1.75^2} = \frac{95}{3.0625} \approx 31.0204$$
Rounding to 1 decimal place gives 31.0.

d) Two lifestyle changes to manage Type 2 diabetes without medication are increasing physical exercise and adopting a diet that is lower in simple carbohydrates and sugars.

a) Pancreas [1 mark]

b)
- pancreas releases the hormone insulin [1 mark]
- (insulin causes) glucose to move from the blood into cells / liver / muscle cells [1 mark]
- glucose is converted into glycogen (for storage) [1 mark]

c) (i)
- insulin injection(s) / insulin therapy / insulin pump [1 mark] (do not accept 'insulin' alone)

c) (ii)
- correct substitution of values: 95 / 1.75^2 or 95 / 3.0625 [1 mark]
- raw calculation of BMI as 31.02... [1 mark]
- correct rounding to 1 decimal place: 31.0 [1 mark]
*Note: Allow 3 marks for correct final answer 31.0 (or 31) with or without working.*

d) Any two from:
- eat a low-carbohydrate / low-sugar / balanced diet [1 mark]
- exercise regularly / increase physical activity [1 mark]
- lose weight / reduce body mass [1 mark]

a) (i) Large ears provide a large surface area relative to body volume, and contain many blood capillaries. This allows heat to be transferred efficiently from the blood to the cooler surrounding air, helping the fox stay cool.
(ii) Large ears are a physical part of the animal's body, so this is a structural adaptation.

b) (i) An animal that hunts and eats other animals is a predator (or carnivore).
(ii) At night, air temperatures are much lower. Hunting in cooler temperatures means the fox loses less water through sweating, panting, and respiration, conserving vital body water in the dry desert.

c) (i) Plants are called producers because they produce glucose and other complex organic nutrients using energy from sunlight via photosynthesis. They do not need to consume other organisms.
(ii) Deep roots grow deep down into the soil to access water tables that other shallow-rooted plants cannot reach. This helps them compete successfully for scarce water resources in arid conditions.

a) (i)
- large surface area / many capillaries near the surface [1 mark]
- to lose / transfer heat to the surroundings [1 mark] (accept: to keep the fox cool)

a) (ii)
- Structural [1 mark]

b) (i)
- Predator / carnivore [1 mark]

b) (ii)
- air temperature is lower at night / it is cooler [1 mark]
- reduces water loss / saves energy / prevents overheating [1 mark]
*OR*
- prey animals are active at night [1 mark]
- making it easier to find food [1 mark]

c) (i)
- they make / produce their own food / glucose [1 mark]
- by photosynthesis / using light energy [1 mark]

c) (ii)
- can reach / absorb water stored deep underground [1 mark]
- allows them to survive / obtain water when surface soil is dry / compete successfully for water [1 mark]