2023 Edexcel GCSE Biology (1BI0) 模擬試題連答案詳解

At high temperatures, the chemical bonds holding the enzyme's specific three-dimensional structure together are broken. This alters the shape of the active site, meaning the substrate molecule can no longer fit into it. This irreversible process is known as denaturation.

1 mark for the correct answer A.

To calculate magnification, use the formula: \(\text{Magnification} = \frac{\text{Image size}}{\text{Actual size}}\). Substituting the given values: \(\text{Magnification} = \frac{30\text{ mm}}{0.15\text{ mm}} = 200\). Therefore, the magnification is \(\times 200\).

1 mark for the correct answer C.

During metaphase, the chromosomes line up along the equator (middle) of the spindle fibres in preparation for separation.

1 mark for the correct answer B.

A cross between a homozygous dominant parent (\(RR\)) and a homozygous recessive parent (\(rr\)) produces only heterozygous offspring (\(Rr\)). Since the round allele (\(R\)) is dominant, all offspring will have the round seed phenotype. Therefore, 0% of the offspring will have wrinkled seeds.

1 mark for the correct answer A.

Ardi (Ardipithecus ramidus) lived approximately 4.4 million years ago, making her older than Lucy (Australopithecus afarensis, approx 3.2 million years old), Homo habilis (approx 2.4-1.4 million years old), and Homo erectus (approx 1.8-0.1 million years old).

1 mark for the correct answer A.

Selective breeding allows humans to choose individual plants or animals with desirable characteristics (such as high milk yield, resistance to disease, or high crop yield) and breed them together to ensure those useful traits are passed on to the offspring.

1 mark for the correct answer B.

Malaria is a disease caused by Plasmodium, which is a single-celled eukaryotic organism classified as a protist. It is transmitted to humans by female Anopheles mosquitoes acting as vectors.

1 mark for the correct answer B.

Phase 1 clinical trials are primarily designed to test the safety, toxicity, and side effects of a new drug in a small group of healthy human volunteers before administering it to patients who have the disease.

1 mark for the correct answer C.

To calculate the magnification, use the formula: \( \text{Magnification} = \text{Image size} / \text{Actual size} \). Substituting the given values: \( \text{Magnification} = 15 / 0.05 = 300 \). Therefore, the magnification is x300.

1 mark for the correct substitution into the formula: \( 15 / 0.05 \). 1 mark for the correct calculation: 300 (or x300).

Mitosis is a type of cell division that produces genetically identical diploid cells. It is essential for the growth of multicellular organisms and for repairing damaged tissues by replacing damaged or old cells.

1 mark for growth (or increasing cell number). 1 mark for repair of damaged tissues (or replacing dead/damaged cells / asexual reproduction).

The phenotype is tall because the dominant allele (T) masks the effect of the recessive allele (t). The genotype is heterozygous because the plant has two different alleles (one dominant T and one recessive t) for this gene.

1 mark for identifying the phenotype as tall. 1 mark for identifying the genotype as heterozygous. 1 mark for explaining that heterozygous means containing two different alleles.

In a bacterial population, random genetic mutations can create an allele that confers resistance to a specific antibiotic. When antibiotics are overused, they create a selection pressure that kills non-resistant bacteria. The resistant bacteria survive and reproduce by binary fission, passing the resistance allele to their offspring, which increases the proportion of resistant bacteria over time.

1 mark for stating that genetic mutations produce variation/resistance. 1 mark for explaining that resistant bacteria survive the antibiotic exposure (natural selection). 1 mark for stating that the survivors reproduce and pass on the resistance allele to their offspring.

Communicable diseases are caused by pathogens and can be passed or transmitted from one individual to another (such as tuberculosis, flu, or cholera). Non-communicable diseases cannot be transmitted between individuals and are caused by genetic factors, lifestyle, or environmental factors (such as cardiovascular disease, diabetes, or cancer).

1 mark for explaining that communicable diseases can be transmitted between organisms, while non-communicable diseases cannot. 1 mark for a correct example of a communicable disease (e.g., flu, tuberculosis, cholera). 1 mark for a correct example of a non-communicable disease (e.g., cardiovascular disease, diabetes, cancer).

At high temperatures, the chemical bonds holding the enzyme's protein structure together break. This changes the shape of the enzyme's active site permanently, which is called denaturation. As a result, the substrate is no longer complementary to the active site and cannot fit, so no enzyme-substrate complexes can form.

1 mark for explaining that high temperatures break chemical bonds, causing the active site to change shape (denaturation). 1 mark for stating that the substrate can no longer fit into the active site. 1 mark for explaining that no enzyme-substrate complexes can form (or no reaction occurs).

When a stimulus is detected by a receptor, an electrical impulse is generated. The impulse travels along the sensory neurone to the central nervous system (CNS). In the spinal cord, the impulse is passed across a synapse to a relay neurone, and then across another synapse to a motor neurone. The motor neurone carries the impulse to the effector (such as a muscle or gland) to produce a response.

1 mark for stating the impulse travels from the receptor along the sensory neurone to the relay neurone (in the CNS/spinal cord). 1 mark for stating the impulse travels from the relay neurone along the motor neurone to the effector.

Ardi (approx 4.4 million years old) had a divergent big toe for climbing trees and longer arms. Lucy (approx 3.2 million years old) had arches in her feet more adapted for walking upright (bipedalism) and shorter, more human-like arms, indicating a transition towards walking upright.

1 mark for identifying that Ardi had features adapted for climbing/tree-dwelling (e.g., divergent big toe / longer arms). 1 mark for identifying that Lucy had features adapted for walking upright (e.g., arched feet / shorter arms).

Amylase is a protein enzyme. When heated to high temperatures such as 80 degrees C, the bonds holding its structure together break. This causes the enzyme to denature, changing the specific shape of its active site. Because the active site is no longer complementary to the starch substrate, the starch cannot bind to it, and digestion stops.

1 mark: enzyme is denatured. 1 mark: active site changes shape. 0.5 marks: starch substrate can no longer fit into the active site.

During anaphase, the spindle fibres contract and shorten. This pulls the sister chromatids (or chromosomes) apart at the centromere, moving them to opposite ends (poles) of the dividing cell.

1 mark: spindle fibres contract or shorten. 1 mark: sister chromatids or chromosomes are pulled apart or separated. 0.5 marks: chromosomes move to opposite ends or poles of the cell.

The parental genotypes are Tt (heterozygous tall) and tt (short). Crossing these gives offspring genotypes: Tt, Tt, tt, and tt. Two out of four offspring are homozygous recessive (tt), which corresponds to a probability of 0.5 or 50%.

1 mark: identification of the parental genotypes and gametes as Tt and tt. 1 mark: correct Punnett square showing offspring genotypes Tt and tt in a 1:1 ratio. 0.5 marks: correct probability of 0.5 (or 50%).

Genetic engineering allows the direct insertion of a desired gene into the wheat plant, making the process much faster than selective breeding, which takes many generations. However, a disadvantage is that genetically modified seeds can be expensive for farmers to buy, and there may be concerns about their long-term environmental impact.

1 mark: stating a valid advantage (e.g., faster process, can introduce genes from other species). 1 mark: stating a valid disadvantage (e.g., higher cost of seeds, potential ecological/health concerns, ethical issues). 0.5 marks: comparing the two methods clearly in the response.

Consuming high levels of saturated fats leads to cholesterol and fatty deposits building up inside the walls of the coronary arteries. This narrows the lumen of the arteries, restricting blood flow. As a result, less oxygen and glucose reach the heart muscle cells, preventing aerobic respiration and potentially leading to a heart attack.

1 mark: fatty deposits or cholesterol build up inside the coronary arteries. 1 mark: coronary arteries narrow, reducing blood flow. 0.5 marks: less oxygen or glucose reaches the heart muscle cells.

To find the magnification, divide the image size by the real size: Magnification = 40 mm / 0.08 mm = 500. Therefore, the magnification is 500.

1 mark: setting up the calculation correctly as 40 divided by 0.08. 1 mark: correct calculation of the value 500. 0.5 marks: expressing the final answer with a times symbol (x500 or times 500) or clearly as a ratio.

The stimulus is detected by receptors in the skin. An electrical impulse travels along the sensory neurone to the central nervous system (spinal cord). The impulse is passed across a synapse to a relay neurone, and then across another synapse to a motor neurone. The motor neurone carries the impulse to the effector, which is a muscle in the arm, causing it to contract and pull the hand away.

1 mark: impulse travels from receptor along the sensory neurone to the relay neurone (in the spinal cord / CNS). 1 mark: impulse travels from the relay neurone along the motor neurone to the effector. 0.5 marks: identifying the effector as a muscle that contracts.

When a patient stops taking antibiotics early, the weaker bacteria have been killed, but the more resistant bacteria survive. These surviving resistant bacteria then reproduce rapidly because they have less competition. They pass on their genes for antibiotic resistance to their offspring, leading to a population of bacteria that are fully resistant to that antibiotic.

1 mark: stating that stopping early allows the most resistant bacteria to survive. 1 mark: explaining that these surviving bacteria reproduce and pass on their resistance genes to offspring. 0.5 marks: stating that the overall population of resistant bacteria increases.

To calculate magnification, use the formula: Magnification = Image size / Actual size. Convert both measurements to the same units if needed, but here both are in millimetres (mm). Magnification = 24 mm / 0.06 mm = 400. Therefore, the magnification is x400.

1 mark for using the correct formula: Magnification = Image size / Actual size. 1 mark for correct substitution: 24 / 0.06. 0.5 marks for the correct final answer of 400 (or x400).

Mitosis produces two genetically identical daughter cells. To achieve this, the cell must replicate its DNA so that each new cell receives a full and identical set of genetic information. The chromosome number of the resulting daughter cells remains the same as the original parent cell.

1 mark for explaining that replication ensures each daughter cell gets a full/complete set of DNA. 1 mark for stating that the daughter cells are genetically identical to each other/parent cell. 0.5 marks for stating that the chromosome number remains the same (diploid).

The heterozygous tall plant has the genotype Tt. The short plant, being recessive, must be homozygous recessive with the genotype tt. Crossing Tt and tt yields offspring with genotypes Tt, Tt, tt, and tt. Two out of four offspring have short stems (tt), which is a probability of 50%.

1 mark for identifying the parent genotypes correctly as Tt and tt. 1 mark for calculating/deducing the offspring genotypes correctly as 2 Tt and 2 tt. 0.5 marks for the correct probability of 50% (or 0.5 or 1/2).

Lucy has a larger skull/brain capacity than the older Ardi fossil. Additionally, Lucy's feet and leg structure (such as arched feet and pelvic shape) show adaptation for walking upright (bipedalism), whereas Ardi has a grasping big toe and longer arms, indicating adaptation for climbing trees.

1 mark for identifying that Lucy had a larger brain capacity / skull size than Ardi. 1 mark for identifying that Lucy was better adapted to upright walking / bipedalism than Ardi (e.g., foot shape, pelvic structure). 0.5 marks for indicating that this shows an evolutionary trend of increasing brain size and transition from tree-climbing to walking upright.

Tuberculosis is caused by a bacterium (specifically Mycobacterium tuberculosis). It is spread through airborne transmission when an infected person coughs or sneezes, releasing droplets containing the bacteria into the air, which are then inhaled by others.

1 mark for identifying the pathogen group as bacteria / bacterium. 1 mark for stating transmission is airborne / via droplets. 0.5 marks for mentioning that these droplets are released by coughing/sneezing and inhaled.

To test for reducing sugars, Benedict's reagent is added to the food sample. The mixture must then be heated in a hot water bath (typically at 80 degrees Celsius) for a few minutes. If a high concentration of reducing sugar is present, the colour changes from blue to brick-red.

1 mark for naming Benedict's reagent. 0.5 marks for stating that heating / a hot water bath is required. 1 mark for stating the positive result colour change to brick-red (accept red or orange-red).

Sensory neurones have a myelin sheath, which acts as an electrical insulator and speeds up the transmission of electrical impulses by allowing them to jump from node to node. They also have a very long dendron and axon, which allows electrical impulses to travel long distances through a single cell without having to cross multiple slow synapses.

1 mark for mentioning the myelin sheath (or myelination). 0.5 marks for explaining that the myelin sheath insulates the neurone to increase transmission speed. 1 mark for identifying the long axon / dendron which carries impulses over long distances (reducing the number of synapses).

BMI is calculated using the formula: BMI = mass / (height squared). Substituting the given values: BMI = 80 / (1.6 * 1.6) = 80 / 2.56 = 31.25. A BMI of 31.25 is over 30, which classifies this individual as obese. Obesity is a major risk factor for cardiovascular disease, so yes, this person has an increased risk.

1 mark for correct calculation of height squared (1.6 * 1.6 = 2.56). 1 mark for the correct calculation of BMI (31.25 or 31). 0.5 marks for stating that this BMI (above 30) indicates obesity / high risk of cardiovascular disease.

Cancer begins when mutations in DNA cause cells to lose control over their cell cycle. This leads to rapid and uncontrolled cell division by mitosis, resulting in a growing mass of abnormal cells called a tumor.

Award up to 2 marks:
- MP1: For identifying that the cells undergo uncontrolled cell division / rapid mitosis [1 mark].
- MP2: For stating that this results in a mass / clump / group of abnormal cells [1 mark].
- Reject: any reference to meiosis.

Antibiotics are drugs designed to target and destroy bacteria (for example, by breaking down their cell walls). They have no effect on viruses, which have different structures and reproduce inside host cells. Prescribing antibiotics for viral infections is ineffective and contributes to the development of antibiotic-resistant bacteria.

Award up to 3 marks:
- MP1: Antibiotics only kill/affect bacteria OR do not kill viruses [1 mark].
- MP2: Explanation that viruses have a different structure / reproduce inside host cells [1 mark].
- MP3: State that unnecessary use leads to the development of antibiotic resistance in bacteria [1 mark].

At a highly acidic pH of 2, the chemical bonds holding the amylase protein together are disrupted, causing the active site of the enzyme to change shape. The enzyme is denatured. As a result, the starch substrate can no longer fit into the active site, and the enzyme-controlled reaction stops completely.

Award up to 2 marks:
- MP1: The active site of the enzyme changes shape / denatures [1 mark].
- MP2: (Therefore) the substrate can no longer fit/bind into the active site / no enzyme-substrate complexes can form [1 mark].
- Reject: 'enzyme is killed' or 'enzyme dies' for MP1.

To selectively breed cows for high milk yield, the farmer must first choose the female cows in the herd that produce the most milk and breed them with bulls from high-yielding mothers. Once offspring are produced, the farmer selects the highest-yielding female offspring and breeds them. This selection and breeding process is repeated over many generations until the high milk yield characteristic is consistently established in the herd.

Award up to 3 marks:
- MP1: Select the cows (and bulls) with the highest milk yield / desired characteristic [1 mark].
- MP2: Breed these selected individuals together [1 mark].
- MP3: Select the offspring with the highest yields and repeat the process over many generations [1 mark].

Physical barriers prevent pathogens from entering the body. The skin forms a continuous, protective outer layer that blocks pathogens; if damaged, blood clots to seal cuts. Mucus in the respiratory tract traps inhaled pathogens and dust. Cilia are tiny hair-like structures on ciliated cells that beat to sweep the trapped mucus up to the back of the throat, where it can be swallowed and destroyed. Chemical defences destroy pathogens. Hydrochloric acid in the stomach has a very low pH, which kills most pathogens that enter on food or drink. Lysozyme is an enzyme found in tears, saliva, and mucus that destroys bacterial cells by breaking down their cell walls.

Level 1 (1-2 marks): Simple description of at least one physical barrier or chemical defence (e.g., skin blocks pathogens, or stomach acid kills bacteria) with limited detail. Level 2 (3-4 marks): Explains how at least one physical barrier and one chemical defence work, using some correct scientific terms. Level 3 (5-6 marks): Comprehensive explanation of multiple physical barriers (skin, mucus, cilia) and chemical defences (stomach acid, lysozyme), clearly detailing how each acts to protect the body. Scientific terminology is used accurately throughout.

Natural selection in bacteria begins with a random mutation in the DNA of a bacterium. This mutation creates a new allele that makes the bacterium resistant to a specific antibiotic. This introduces variation into the bacterial population, where some bacteria are resistant and others are not. When antibiotics are introduced to the environment, they act as a selection pressure. The antibiotic kills the non-resistant bacteria, but the resistant bacteria survive. The surviving resistant bacteria then reproduce rapidly by binary fission. They pass on the allele for antibiotic resistance to their offspring. Over generations, the proportion of resistant bacteria increases until the entire population is resistant.

Level 1 (1-2 marks): Mentions key terms such as mutation or survival, but lacks clear sequencing or explanation of how resistance spreads. Level 2 (3-4 marks): Explains how mutation causes resistance and how antibiotics kill the non-resistant bacteria, leaving the resistant ones to survive and reproduce. Level 3 (5-6 marks): Provides a clear, logical, and complete explanation showing how variation arises from mutation, antibiotics act as a selection pressure killing non-resistant bacteria, and the resistant survivors reproduce and pass on the resistance allele, increasing its frequency in the population.

Root hair cells have long projections that grow out into the soil. This greatly increases the surface area of the cell, allowing osmosis to happen more rapidly to absorb water from the soil.

1 mark for the correct option (b).

The pancreas is the organ that monitors blood glucose concentration. If the concentration is too high, it secretes insulin into the blood.

1 mark for the correct option (c).

Veins transport blood back to the heart under low pressure. They have relatively thin walls compared to arteries and contain valves to prevent blood from flowing backwards.

1 mark for the correct option (b).

During photosynthesis, plants take in carbon dioxide from the atmosphere and water from the soil to produce glucose and oxygen, utilizing light energy.

1 mark for the correct option (b).

The active site is the specific region on an enzyme molecule where the substrate binds and where the chemical reaction takes place.

1 mark for the correct option (a).

Phloem tissue is responsible for translocation, which is the transport of dissolved sugars (such as sucrose) and amino acids around the plant.

1 mark for the correct option (b).

When body temperature falls, skeletal muscles rapidly contract and relax (shivering) to generate heat from respiration. Sweating decreases, vasoconstriction occurs, and hairs stand upright.

1 mark for the correct option (c).

In a mutualistic relationship, both organisms benefit. Nitrogen-fixing bacteria provide the plant with nitrogen compounds, while the plant provides the bacteria with sugars from photosynthesis. Head lice, tapeworms, and mistletoe are examples of parasites where only one organism benefits at the expense of the host.

1 mark for the correct option (c).

Decomposers feed on and break down dead plant and animal remains. As they digest this organic material, they perform aerobic respiration to obtain energy, which releases carbon dioxide into the air.

Award 1 mark for stating that decomposers break down, digest, or feed on dead organic matter. Award 1 mark for stating that decomposers respire, releasing carbon dioxide.

Companion cells contain a large number of mitochondria. These mitochondria carry out aerobic respiration to release energy (ATP). This energy is used to actively transport sucrose into the phloem sieve tubes against a concentration gradient.

Award 1 mark for identifying that companion cells have many mitochondria. Award 1 mark for stating that mitochondria release energy or ATP via respiration. Award 1 mark for linking this energy to the active transport of sucrose.

Vasoconstriction causes the arterioles near the surface of the skin to narrow. This reduces the volume of blood flowing through the capillaries near the skin surface, which decreases the amount of heat energy lost to the surroundings by radiation.

Award 1 mark for stating that arterioles or blood vessels near the skin surface narrow or constrict (do not accept 'capillaries constrict'). Award 1 mark for stating that less blood flows near the skin surface, reducing heat loss by radiation.

Alveoli have thin walls (only one cell thick) and a large total surface area (as well as a rich capillary blood supply). Thin walls speed up gas exchange by providing a very short diffusion pathway for oxygen and carbon dioxide. A large surface area allows more gas molecules to diffuse at the same time.

Award 1 mark each for any two correct adaptations: thin walls / one cell thick, large surface area, good blood supply / capillary network, or moist surface (maximum 2 marks). Award 1 mark for a correct explanation linked to one of the chosen adaptations (e.g., thin walls provide a short diffusion distance, large surface area increases the rate of diffusion, or good blood supply maintains a steep concentration gradient).

At a low temperature of \(5^\circ\text{C}\), the enzyme and starch molecules have very little kinetic energy. This means they move slowly, resulting in fewer successful collisions between the active site of the enzyme and the substrate, forming fewer enzyme-substrate complexes.

Award 1 mark for stating that molecules have low kinetic energy or move slowly at low temperatures. Award 1 mark for stating there are fewer successful collisions or fewer enzyme-substrate complexes formed.

When blood glucose levels are low, the pancreas detects this change and secretes the hormone glucagon. Glucagon travels through the blood to the liver, where it triggers liver cells to break down stored glycogen into glucose. This glucose is then released into the bloodstream, raising the blood glucose concentration back to normal.

Award 1 mark for stating that glucagon is released by the pancreas. Award 1 mark for stating that glucagon acts on the liver (or muscle cells) to convert glycogen. Award 1 mark for stating that glycogen is converted to glucose, which is released into the blood.

When fertilisers wash into the pond, the high levels of nutrients (nitrates and phosphates) cause rapid growth of algae on the surface, known as an algal bloom. This thick layer of algae blocks sunlight from reaching aquatic plants deeper in the water, preventing them from photosynthesising so they die.

Award 1 mark for stating that fertilisers cause rapid algal growth or an algal bloom. Award 1 mark for stating that the algae block sunlight, preventing photosynthesis of plants below the surface (or causing underwater plants to die).

An increase in light intensity increases the rate of transpiration. This occurs because guard cells respond to light by absorbing water and becoming turgid, which causes the stomata to open wider to allow carbon dioxide in for photosynthesis. Wider stomata allow more water vapour to diffuse out of the leaf into the air.

Award 1 mark for stating that transpiration rate increases. Award 1 mark for explaining that stomata open wider (in response to light for photosynthesis). Award 1 mark for stating that this allows more water vapour to diffuse out of the leaf.

First, calculate the change in mass: \(1.92\text{ g} - 2.40\text{ g} = -0.48\text{ g}\). Next, divide this change by the initial mass: \(\frac{-0.48\text{ g}}{2.40\text{ g}} = -0.20\). Finally, multiply by 100 to find the percentage change: \(-0.20 \times 100 = -20\%\). A negative value indicates a 20% decrease in mass.

MP1: Calculates the correct change in mass: \(-0.48\text{ g}\) (allow \(0.48\text{ g}\)) [1 mark]. MP2: Divides change in mass by initial mass: \(\frac{-0.48}{2.40}\) (or \(\frac{0.48}{2.40}\)) [1 mark]. MP3: Correctly calculates percentage change as \(-20\%\) (accept \(20\%\) decrease or \(-20\)) [1 mark]. Award 3 marks for correct final answer with no working.

When water is plentiful, it enters the guard cells by osmosis, causing them to become turgid. Because of their thick inner walls and thin outer walls, they swell and curve outwards, opening the stomatal pore. When water is scarce, the guard cells lose water by osmosis, become flaccid, and the stomatal pore closes.

MP1: Guard cells absorb water by osmosis and become turgid (to open the stoma) [1 mark]. MP2: Guard cells lose water and become flaccid (to close the stoma) [1 mark].

During exercise, the body temperature rises. In response, blood vessels (arterioles) that supply blood to the capillaries in the skin dilate (widen). This process is vasodilation. It allows more blood to flow near the skin surface, meaning more heat energy is radiated from the blood into the surroundings, cooling the body down.

MP1: Arterioles / blood vessels supplying the skin capillaries dilate / widen [1 mark]. (Reject 'capillaries dilate'). MP2: This increases blood flow through the skin capillaries / close to the skin surface [1 mark]. MP3: Resulting in more heat being lost by radiation (into the surrounding environment) [1 mark].

When blood glucose levels rise, the pancreas detects this and secretes the hormone insulin. Insulin travels in the bloodstream to target organs, primarily the liver and muscle cells. It triggers these cells to take up excess glucose from the blood and convert it into insoluble glycogen, which is stored, thereby lowering blood glucose levels back to normal.

MP1: Insulin is released by the pancreas (into the blood) [1 mark]. MP2: It causes liver or muscle cells to take up glucose / convert glucose to glycogen (for storage) [1 mark].

Capillaries are adapted for exchange in several ways: 1) Their walls are extremely thin (one cell thick), providing a very short diffusion pathway for substances like oxygen and glucose. 2) They have highly permeable walls, allowing easy passage of substances. 3) They have a very narrow lumen, which slows down blood flow, giving more time for diffusion to occur.

Award up to 3 marks. Must link structural adaptation to its functional benefit. MP1: Capillary wall is only one cell thick [1 mark]. MP2: This provides a short diffusion distance / pathway [1 mark]. MP3: Highly permeable walls (to allow substances to pass) OR narrow lumen (which slows blood flow to allow more time for exchange) [1 mark].

Decomposers feed on dead plant and animal material, as well as waste products. They break down this organic matter to obtain nutrients. During this process, the decomposers carry out cellular respiration to release energy, which produces carbon dioxide as a waste product. This carbon dioxide is then released into the atmosphere.

MP1: Decomposers digest / break down dead organic matter or animal waste [1 mark]. MP2: Decomposers carry out respiration, which releases carbon dioxide [1 mark].

Mutualism is a close relationship between two different species where both organisms benefit from the interaction. A clear example is nitrogen-fixing bacteria living in the root nodules of leguminous plants. The bacteria gain carbohydrates and protection from the plant, while the plant receives nitrates produced by the bacteria, which are essential for making proteins.

MP1: Mutualism is defined as a relationship between two species where both organisms benefit [1 mark]. MP2: Identifies a valid mutualistic pair (e.g., nitrogen-fixing bacteria and legumes OR cleaner fish and sharks OR flowers and bees) [1 mark]. MP3: Explains how both organisms benefit in the chosen example (e.g., bacteria get sugars, plant gets nitrates; or bees get food/nectar, flowers get pollinated) [1 mark].

Enzymes are protein molecules with a highly specific three-dimensional shape, including an active site. When heated above their optimum temperature (such as above \(45^\circ\text{C}\)), the high thermal energy breaks the chemical bonds holding the protein structure together. This changes the shape of the active site permanently. The substrate can no longer fit into the active site, meaning the enzyme is denatured and cannot catalyze the reaction.

MP1: High temperatures cause chemical bonds in the enzyme to break, changing the shape of the active site [1 mark]. MP2: The substrate can no longer fit into the active site / is no longer complementary (meaning the enzyme is denatured) [1 mark]. (Reject 'enzyme is killed / dies').

An increase in wind speed increases the rate of transpiration because it moves water molecules away from the surface of the leaf. This maintains a steep concentration gradient of water vapour between the inside of the leaf and the outside air, leading to faster diffusion of water out through the stomata.

1 mark: State that increased wind speed increases transpiration rate. 1 mark: Explain that wind removes water vapour from the leaf surface. 0.5 marks: State that this maintains a steep diffusion or concentration gradient.

Capillaries have walls that are only one cell thick, which reduces the distance that substances (like oxygen and glucose) need to diffuse. Their narrow lumen slows down blood flow and brings red blood cells close to the capillary wall, maximizing the efficiency of exchange.

1 mark: Wall is one cell thick / very thin. 1 mark: Short diffusion distance / pathway. 0.5 marks: Narrow lumen slows blood flow / increases surface-area-to-volume ratio.

Decomposers (such as bacteria and fungi) feed on dead organisms and waste products. They secrete enzymes to digest this organic matter and absorb the nutrients. During respiration, decomposers release carbon dioxide \(CO_2\) into the atmosphere, making it available again for photosynthesis.

1 mark: Break down / digest dead organic matter / waste. 1 mark: Decomposers respire. 0.5 marks: Release carbon dioxide into the atmosphere.

When core body temperature decreases, vasoconstriction occurs. The muscles in the walls of arterioles near the skin constrict, reducing blood flow to the skin capillaries and minimizing heat loss to the surroundings. Additionally, shivering occurs, which is rapid muscle contraction that increases the rate of respiration, releasing thermal energy to warm the body.

1 mark: Vasoconstriction / blood vessels near skin surface narrow. 1 mark: Less blood flows near skin surface / less heat lost by radiation. 0.5 marks: Shivering / rapid muscle contraction to generate heat.

Increasing temperature up to the optimum increases the kinetic energy of both the enzyme and substrate molecules. This causes them to move faster, leading to more frequent and successful collisions between substrates and the enzyme's active site, increasing the rate of reaction.

1 mark: Rate of reaction increases. 1 mark: Molecules gain kinetic energy and move faster. 0.5 marks: Leading to more frequent / successful collisions.

Xylem tissue transports water and mineral ions from the roots, up the stem, and to the leaves. Phloem tissue transports dissolved sugars (specifically sucrose) and other soluble food molecules from the leaves to other parts of the plant (such as growing regions and storage organs) in a process called translocation.

1 mark: Xylem transports water and minerals up the plant. 1 mark: Phloem transports dissolved sugars/sucrose. 0.5 marks: Mention of translocation / transport in phloem is bidirectional (up and down).

When blood glucose levels drop, the pancreas detects this change and secretes the hormone glucagon. Glucagon travels in the blood to target cells, mainly in the liver. It triggers the liver cells to break down stored glycogen into glucose, which is then released back into the bloodstream to raise blood glucose levels.

1 mark: Released by the pancreas. 1 mark: Travels to the liver to convert glycogen into glucose. 0.5 marks: Glucose is released into the blood to raise levels back to normal.

Plants absorb nitrogen in the form of nitrates from the soil through active transport. They use this to make proteins. Animals obtain their nitrogen by eating plants or other animals. When plants and animals die or produce waste, decomposers (bacteria and fungi) break down the nitrogen compounds (like proteins and urea) back into ammonia, which is eventually converted back to nitrates.

1 mark: Plants absorb nitrates from the soil to make proteins. 1 mark: Animals obtain nitrogen by feeding / eating. 0.5 marks: Decomposers break down waste / dead material to return nitrogen compounds to the soil.

1. The biconcave shape creates a larger surface area compared to the volume of the cell (a high surface area to volume ratio). 2. This larger surface area increases the rate at which oxygen can diffuse across the cell membrane. 3. Additionally, the shape ensures a short diffusion distance to the centre of the cell, further increasing the efficiency of oxygen transport.

[1 mark] for stating that the biconcave shape increases the surface area to volume ratio of the red blood cell. [1 mark] for explaining that this increases the rate of diffusion of oxygen. [0.5 marks] for noting that it reduces the diffusion distance to the centre of the cell.

1. Stonefly nymphs are indicator species for clean, unpolluted water because they require high concentrations of dissolved oxygen. 2. Bloodworms are indicator species for polluted water because they can survive in low oxygen environments. 3. Therefore, finding many bloodworms and no stonefly nymphs indicates that the river is highly polluted, leading to low dissolved oxygen levels.

[1 mark] for identifying that stonefly nymphs are indicator species for clean water or require high dissolved oxygen levels. [1 mark] for identifying that bloodworms are indicator species for polluted water or can tolerate low oxygen levels. [0.5 marks] for concluding that the river has high levels of pollution.

1. Water moves into the guard cells from surrounding epidermal cells by osmosis. 2. This movement of water causes the guard cells to swell and become turgid. 3. The inner cell walls of guard cells are thicker and less flexible than the outer walls, causing the cells to curve outward and pull apart, which opens the stoma.

[1 mark] for stating that water enters the guard cells by osmosis. [1 mark] for stating that the guard cells swell and become turgid. [0.5 marks] for explaining that the thicker inner walls cause the guard cells to curve outwards, opening the pore.

1. The blood vessels (arterioles) supplying the skin capillaries dilate (widen). 2. This increases the volume of blood flowing close to the surface of the skin. 3. As a result, more thermal energy (heat) is transferred from the blood to the surrounding environment via radiation.

[1 mark] for explaining that blood vessels/arterioles supplying skin capillaries dilate or widen (do not accept 'capillaries dilate'). [1 mark] for explaining that this increases blood flow closer to the skin surface. [0.5 marks] for stating that more heat is lost to the environment by radiation.

When blood glucose levels rise too high (for example, after eating):
- The pancreas detects the rise and secretes the hormone insulin into the blood.
- Insulin travels to the liver and muscle cells.
- Insulin causes these cells to take in glucose from the blood and convert it into insoluble glycogen for storage.
- This reduces the blood glucose level back to normal.

When blood glucose levels fall too low (for example, during exercise):
- The pancreas detects the fall and secretes the hormone glucagon into the blood.
- Glucagon travels to the liver.
- Glucagon causes the stored glycogen to be broken down back into glucose.
- This glucose is released back into the blood, raising the blood glucose level back to normal.

Indicative content:
- (When high) Pancreas detects high blood glucose levels
- (When high) Pancreas releases insulin
- Insulin travels in the blood to the liver/muscles
- Insulin causes liver/muscle cells to absorb glucose
- Glucose is converted into glycogen
- Blood glucose levels return to normal
- (When low) Pancreas detects low blood glucose levels
- (When low) Pancreas releases glucagon
- Glucagon travels in the blood to the liver
- Glucagon causes glycogen to be broken down into glucose
- Glucose is released into the blood
- Blood glucose levels return to normal

Marking levels:
- Level 1 (1-2 marks): Simple description of one process, e.g., states that insulin lowers blood glucose or glucagon raises it. The response lacks detail and key biological terms.
- Level 2 (3-4 marks): Clear explanation of either what happens when glucose is high OR when glucose is low, using some correct scientific terms (such as insulin, liver, or glycogen), or a brief outline of both.
- Level 3 (5-6 marks): Detailed and balanced explanation of both processes (high and low blood glucose regulation), correctly identifying the role of the pancreas, both hormones (insulin and glucagon), and the conversion of glucose to/from glycogen in the liver.

To set up and use the potometer:
1. Cut the leafy shoot under water to prevent air bubbles entering the xylem vessels.
2. Assemble the potometer under water and fit the shoot tightly into the stopper.
3. Use petroleum jelly (Vaseline) to seal the joints and make the apparatus completely airtight.
4. Introduce a single air bubble into the capillary tube.
5. Record the starting position of the bubble on the scale/ruler.
6. Measure the distance the bubble travels in a set period of time (e.g. 10 minutes) to calculate the rate of water uptake.

To investigate wind speed:
- Use a fan at different distances (or speed settings) to vary the wind speed, keeping temperature and light intensity constant.

Explanation of the effect of wind speed:
- An increase in wind speed will increase the rate of transpiration.
- This is because the moving air blows away water vapour that has accumulated around the outside of the leaves/stomata.
- This maintains a steep concentration gradient for water vapour between the inside and the outside of the leaf, causing water to diffuse out more rapidly.

Indicative content:
- Cut shoot under water
- Seal joints / make airtight using petroleum jelly
- Introduce an air bubble into the capillary tube
- Measure distance bubble moves using a ruler/scale
- Record time taken
- Use a fan to change wind speed
- Control other variables (e.g., light, temperature)
- Increased wind speed increases the rate of transpiration / water uptake
- Moving air removes water vapour from near the leaf/stomata
- Maintains a steep concentration gradient (allowing faster diffusion of water)

Marking levels:
- Level 1 (1-2 marks): Simple description of how to use a potometer or a basic statement that wind speeds up transpiration. Little or no scientific explanation.
- Level 2 (3-4 marks): Detailed description of how to set up/use the potometer OR a clear explanation of how wind speed increases transpiration in terms of water vapour and diffusion.
- Level 3 (5-6 marks): Comprehensive answer covering both a detailed, practical method to measure water uptake using a potometer (including airtight setup/bubble measurement) and a clear scientific explanation of how wind speed maintains a steep concentration gradient to increase diffusion.