2022 Edexcel GCSE Biology (1BI0) 模擬試題連答案詳解

Mitochondria are the organelles where aerobic respiration takes place, releasing energy for the cell. Ribosomes are responsible for protein synthesis, chloroplasts are the site of photosynthesis, and the nucleus contains DNA.

1 mark for B (Mitochondrion). Reject all other options.

Embryonic stem cells are undifferentiated and can differentiate into most cell types. Therefore, they have the potential to replace damaged cells, such as nerve cells, in people who are paralysed.

1 mark for B (Replacing damaged nerve cells to treat paralysis). Reject other options as they do not represent medical uses of embryonic stem cells.

In DNA, the complementary base pairing rules state that adenine (A) always pairs with thymine (T), and cytosine (C) always pairs with guanine (G). Therefore, the complementary sequence to A-G-T-C-C-A is T-C-A-G-G-T.

1 mark for A (T-C-A-G-G-T). Reject options containing Uracil (U) as this is a DNA strand, not RNA.

Ardi is a famous fossilised hominid skeleton dating back to approximately 4.4 million years ago. 'Lucy' (Australopithecus afarensis) lived around 3.2 million years ago.

1 mark for C (4.4 million years ago). Reject all other options.

Stents are tiny, expandable metal mesh tubes inserted into narrowed or blocked arteries to keep them open, which restores blood flow to the heart muscle. Statins are cholesterol-lowering drugs, and heart bypass surgery uses a blood vessel from another part of the body to bypass blocked arteries.

1 mark for A (Stents). Reject all other options.

Motor neurones carry electrical impulses from the central nervous system (brain and spinal cord) to the effectors (muscles or glands) to bring about a response. Sensory neurones carry impulses from receptors to the CNS, while relay neurones connect sensory and motor neurones within the CNS.

1 mark for C (Motor neurone). Reject all other options.

Cardiovascular disease is a non-communicable disease because it is caused by lifestyle factors or genetics and cannot be passed from person to person. Tuberculosis, cholera, and malaria are all communicable diseases caused by pathogens.

1 mark for B (Cardiovascular disease). Reject other options.

Ligase is the enzyme used to join DNA strands together by catalyzing the formation of phosphodiester bonds. Restriction enzymes are used to cut DNA at specific base sequences, while protease and amylase are digestive enzymes.

1 mark for A (Ligase). Reject all other options.

To find the total magnification of a microscope, you multiply the magnification of the eyepiece lens by the magnification of the objective lens: \(\text{Total magnification} = 10 \times 40 = 400\).

D - \(\times 400\) (1 mark). Any other option selected scores 0 marks.

Cholera is caused by the bacterium Vibrio cholerae. Therefore, the pathogen responsible is a bacterium.

A - bacterium (1 mark). Any other option selected scores 0 marks.

To ensure a valid test (a fair test), all variables other than the independent variable (pH) must be kept constant. These are the control variables. In an enzyme-controlled reaction such as amylase breaking down starch, the control variables include temperature, amylase concentration, amylase volume, starch concentration, and starch volume.

1 mark for identifying any valid control variable (e.g. temperature). 1 mark for identifying a second valid control variable (e.g. starch concentration). Accept: volume of solutions or concentration of solutions. Reject: amount of starch/amylase (must specify volume or concentration to be precise).

During mitosis, specifically in metaphase, spindle fibres attach to the centromeres of the chromosomes. In anaphase, these spindle fibres contract, pulling the sister chromatids apart towards opposite ends (poles) of the dividing cell. This ensures that each new daughter cell receives an identical set of chromosomes.

1 mark: for mentioning that spindle fibres attach to chromosomes or chromatids. 1 mark: for mentioning they pull chromosomes or chromatids apart to opposite ends or poles of the cell. Accept: separate chromosomes.

Cardiovascular disease (CVD) can be influenced by lifestyle factors. Key changes that reduce the risk include: eating a balanced diet lower in saturated fats and salt (which lowers blood cholesterol and blood pressure), stopping smoking (which avoids damage to blood vessels from tobacco smoke chemicals), and doing regular exercise (which strengthens the heart muscle and helps maintain a healthy weight).

1 mark for each correct lifestyle change up to a maximum of 2 marks. Accept: losing weight, reducing alcohol consumption, reducing salt intake, regular exercise. Reject: eating healthy on its own (must be specific, e.g. low saturated fat, less salt).

Selective breeding and genetic engineering both aim to produce organisms with characteristics that are useful to humans (such as high crop yield or disease resistance). However, selective breeding takes many generations and relies on natural reproduction, whereas genetic engineering is a rapid process where a specific gene is directly transferred into the genome of another organism, often of a different species.

1 mark for a correct similarity (e.g. both produce organisms with desirable characteristics, both involve human intervention). 1 mark for a correct difference (e.g. genetic engineering directly transfers genes/DNA, whereas selective breeding involves choosing parents to breed naturally).

The genotype is the collection of alleles or the genetic code that determines a trait (for example, Bb). The phenotype is how these alleles are expressed as physical characteristics in the organism (for example, brown eyes).

1 mark: Genotype defined as the alleles or genetic makeup or genes of an organism. 1 mark: Phenotype defined as the physical expression or characteristics or features of an organism.

In a reflex arc, a stimulus is first detected by a receptor. This generates an electrical impulse that 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), which brings about the response.

1 mark for the correct sequence of neurones: sensory neurone to relay neurone to motor neurone. 1 mark for correctly linking the receptor at the start and the effector at the end of this neurone sequence.

First, find the radius from the diameter: radius \(r = \text{diameter} / 2 = 14 / 2 = 7\text{ mm}\). Next, calculate the area using the formula: \(A = \pi r^2 = 3.14 \times 7^2 = 3.14 \times 49 = 153.86\text{ mm}^2\). Rounding to 1 decimal place gives \(153.9\text{ mm}^2\).

1 mark for calculating the radius correctly: \(r = 7\text{ mm}\). 1 mark for calculating the area: \(3.14 \times 7^2 = 153.86\) and rounding correctly to 1 decimal place: \(153.9\text{ mm}^2\). Accept correct answer with no working for full marks.

Fossils are the preserved remains or traces of organisms from many years ago. By arranging fossils in chronological order based on the depth of the rock layers they are found in, scientists can observe gradual anatomical changes over time. This provides direct evidence of evolutionary pathways and shows how modern species share common ancestors with extinct species.

1 mark: for stating that fossils show anatomical changes in organisms over long periods of time. 1 mark: for mentioning that they allow scientists to compare extinct species with modern species or trace common ancestors.

1. The stage of mitosis described where chromosomes are pulled apart is anaphase. 2. In the next stage (telophase), the chromosomes reach the opposite poles of the cell. 3. A new nuclear membrane reforms around each of the two sets of chromosomes.

1 mark for identifying the stage as anaphase. 1 mark for stating that chromosomes reach the opposite poles of the cell. 1 mark for stating that new nuclear membranes/envelopes reform around each group of chromosomes.

1. It has a long dendron which carries electrical impulses from sensory receptor cells to the cell body. 2. It has an axon that carries impulses away from the cell body towards the central nervous system. 3. The axon and dendron are surrounded by a myelin sheath, which acts as an electrical insulator and speeds up the transmission of the impulses.

Any three points from: Has a long dendron to carry impulses from receptors to the cell body (1 mark); Has an axon to carry impulses from the cell body to the CNS / relay neurone (1 mark); Has a myelin sheath to insulate the neurone (1 mark); Myelin sheath increases the speed of electrical impulse transmission (1 mark).

1. Convert both measurements to the same unit: Actual width = \(0.008\text{ mm}\), Image width = \(2.4\text{ cm} = 24\text{ mm}\). 2. Use the magnification formula: Magnification = Image size / Actual size. 3. Substitute the values: Magnification = \(24\text{ mm} / 0.008\text{ mm} = 3000\).

1 mark for converting units correctly (e.g. \(2.4\text{ cm} = 24\text{ mm}\) or \(0.008\text{ mm} = 0.0008\text{ cm}\)). 1 mark for setting up the calculation: \(24 / 0.008\) (or equivalent with consistent units). 1 mark for the correct final answer of 3000 (accept \(\times 3000\)). Award 3 marks for correct final answer with no working shown.

1. Amylase has an optimum pH (around pH 7) at which it works best. 2. At a highly acidic pH like pH 2, the shape of the active site of the amylase enzyme is altered / denatured. 3. As a result, the substrate (starch) can no longer fit into the active site, meaning fewer or no enzyme-substrate complexes can form, decreasing the rate of reaction.

1 mark for identifying that the enzyme is denatured / active site changes shape. 1 mark for explaining that the substrate can no longer fit into the active site / is no longer complementary. 1 mark for explaining that fewer / no enzyme-substrate complexes can form.

1. The parents are heterozygous (\(\text{Bb}\)), meaning they both have one dominant allele (\(\text{B}\)) and one recessive allele (\(\text{b}\)). Since black fur is dominant, both parents will have the black fur phenotype. 2. A genetic cross (\(\text{Bb} \times \text{Bb}\)) produces offspring with genotypes: \(\text{BB}\) (black), \(\text{Bb}\) (black), \(\text{Bb}\) (black), and \(\text{bb}\) (brown). 3. The probability of offspring with brown fur (\(\text{bb}\)) is 1 out of 4, which is \(25\%\).

1 mark for identifying the parent phenotype as black fur. 1 mark for showing or describing the genetic cross results (e.g., genotypes \(\text{BB}\), \(\text{Bb}\), \(\text{Bb}\), \(\text{bb}\)). 1 mark for stating the probability is 25%.

1. The gene responsible for beta-carotene production is cut from the donor DNA using a restriction enzyme. 2. The same restriction enzyme is used to cut the plasmid/vector DNA, leaving complementary sticky ends. Ligase enzymes are then used to join the gene into the plasmid/vector. 3. The vector is used to insert the gene into the genome of the rice plant cells at an early stage of development.

1 mark for stating restriction enzymes are used to cut out the gene (leaving sticky ends). 1 mark for stating ligase enzymes are used to join the gene into a vector / plasmid. 1 mark for stating that the vector is used to insert the gene into the rice plant genome or cells.

1. A stent is a small, expandable metal mesh tube. 2. It is inserted into a narrowed or blocked coronary artery and expanded. 3. This keeps the artery open, allowing blood to flow freely and ensuring the heart muscle receives sufficient oxygen and glucose.

1 mark for defining a stent as a mesh tube or metal tube placed inside an artery. 1 mark for explaining that it holds/keeps the artery open / widens the lumen. 1 mark for explaining that this restores/increases blood flow or oxygen delivery to the heart muscle.

1. Physical barrier: Skin (or mucus/cilia in airways). 2. Chemical barrier: Hydrochloric acid in the stomach (or lysozymes in tears/saliva). 3. Explanation: For example, the skin forms a waterproof protective outer layer that physically blocks pathogens from entering the blood, or stomach acid has a very low pH which kills most swallowed pathogens.

1 mark for identifying a correct physical barrier (such as skin / mucus / cilia). 1 mark for identifying a correct chemical barrier (such as stomach acid / lysozymes). 1 mark for a correct explanation of how either barrier protects (such as skin blocking pathogen entry, or stomach acid destroying swallowed pathogens).

Before a cell can undergo mitosis, it must prepare during interphase. This involves two major events: 1) DNA replication, ensuring that each of the two new daughter cells receives a full copy of the genetic material, and 2) Cell growth, where the cell synthesizes proteins and duplicates organelles like mitochondria and ribosomes so both cells can function.

Mark 1: DNA is replicated / duplicated / copied (1).
Mark 2: Cell grows in size / synthesizes more sub-cellular structures (such as mitochondria or ribosomes) (1).

Using the formula: \( \text{Magnification} = \frac{\text{Image size}}{\text{Actual size}} \).
Given: \( \text{Image size} = 15 \text{ mm} \) and \( \text{Actual size} = 0.05 \text{ mm} \).
\( \text{Magnification} = \frac{15}{0.05} = 300 \). Therefore, the magnification is x300.

Mark 1: Correct rearrangement or substitution of numbers into formula: \( \frac{15}{0.05} \) (1).
Mark 2: Correct calculation of magnification: 300 or x300 (1).

A heterozygous tall plant has the genotype Tt. A short plant must be homozygous recessive, so its genotype is tt. When crossed (Tt x tt), the potential offspring genotypes are Tt, Tt, tt, and tt. Two out of the four possible offspring have the genotype tt (short). Therefore, the probability is 2/4, which is equal to 50%.

Mark 1: Identification of correct parental genotypes: Tt and tt (or shown correctly in a Punnett square) (1).
Mark 2: Correct probability given as 50% (1). Accept 0.5 or 1/2.

Selective breeding involves breeding closely related individuals with desired traits over many generations. This process reduces the gene pool (genetic variation) of the crop. If all plants are genetically similar, they will all share the same vulnerabilities. For example, if a new disease pathogen emerges or climate conditions change, the entire crop could be destroyed because none of the plants carry resistant alleles.

Mark 1: Identifies that selective breeding reduces genetic variation / reduces the gene pool / causes inbreeding (1).
Mark 2: Explains that this makes the crop highly vulnerable to being completely wiped out by a single pathogen/disease OR an environmental change (1).

The body has external physical defenses. One major physical barrier is the skin, which forms an unbroken protective layer over the entire body preventing microbes from penetrating deeper tissues. Another example is mucus in the respiratory tract, which physically traps inhaled pathogens; cilia then sweep this mucus upwards and away from the lungs.

Mark 1: Identifies a valid physical barrier, e.g., skin, mucus, or cilia (1).
Mark 2: Describes how it blocks/traps pathogens, e.g., skin forms a continuous protective layer / mucus traps pathogens / cilia sweep mucus containing pathogens away (1).
Note: Reject chemical barriers like stomach acid or tears.

Embryonic stem cells are pluripotent, meaning they have the potential to differentiate into almost any cell type found in the human body. In contrast, adult stem cells are multipotent and can only differentiate into a narrow range of cell types related to the tissue they are sourced from (e.g., bone marrow stem cells can only differentiate into different blood cells).

Mark 1: States that embryonic stem cells can differentiate into any/all types of specialized cells (1).
Mark 2: States that adult stem cells can only differentiate into a limited/few types of cells (1).

The age of stone tools can be estimated in two main ways: 1) Stratigraphy, which is dating the sedimentary rock layers in which the tool was found (older tools are deeper down), and 2) Radiometric dating (such as carbon-14 dating) of any organic materials (like wooden handles or bone fragments) found in association with the stone tool.

Mark 1: Mentioning stratigraphy / rock layers (tools found in deeper rock layers are older) (1).
Mark 2: Mentioning carbon dating / radiometric dating of organic materials (such as wood or bone) found in the same layer as the tool (1).

If a patient stops taking antibiotics early, only the least resistant bacteria will have been killed. The bacteria that are more resistant to the antibiotic will survive. These surviving resistant bacteria can then reproduce, pass on their resistance genes, and cause a new, harder-to-treat infection that can spread to others.

Mark 1: Explains that stopping early leaves the most resistant bacteria alive (1).
Mark 2: These surviving resistant bacteria will multiply/reproduce, leading to antibiotic resistance (1).

(a) The stage of mitosis where chromosomes line up at the equator is metaphase.
(b) In the next stage (anaphase), the chromosomes (or sister chromatids) are separated/split and pulled to opposite ends (poles) of the cell by spindle fibres.

Part (a):
- Metaphase [1 mark]

Part (b):
- Chromatids / chromosomes are separated / split / pulled apart [1 mark]
- To opposite poles / ends of the cell (by spindle fibres) [1 mark]

(a) A stent (or balloon angioplasty) can be inserted into the narrowed coronary artery to keep it open.
(b) By keeping the artery open, it increases blood flow to the heart muscle. This ensures that the heart muscle cells receive sufficient oxygen and glucose for aerobic respiration, preventing them from dying and thus reducing the risk of a heart attack.

Part (a):
- Stent (accept balloon angioplasty / bypass surgery) [1 mark]

Part (b):
- Increases/restores blood flow / oxygen / glucose supply to the heart muscle [1 mark]
- (So heart muscle can continue to) respire / prevent heart muscle cells from dying [1 mark]

(a) Since both parents are heterozygous carriers, their genotype must be \(Ff\).
(b) Crossing two heterozygous parents (\(Ff \times Ff\)) results in the following offspring genotypes:
- \(FF\) (healthy, non-carrier) - 25%
- \(Ff\) (healthy carrier) - 50%
- \(ff\) (has cystic fibrosis) - 25%
Therefore, the probability of the child having cystic fibrosis is 25%, 0.25, or \(\frac{1}{4}\).

Part (a):
- \(Ff\) (accept both having \(Ff\)) [1 mark]

Part (b):
- Correct genetic diagram / Punnett square showing offspring genotypes (\(FF\), \(Ff\), \(Ff\), \(ff\)) OR correct gametes identified [1 mark]
- Correct probability: 25% / 0.25 / \(\frac{1}{4}\) / 1 in 4 [1 mark]

(a) Scientists can estimate the age of a stone tool by identifying the rock or sediment layer (stratum) in which it was found, as deeper layers are generally older (stratigraphy). They can also date fossils or organic remains found in the same layer as the tool.
(b) Over evolutionary time, stone tools became more complex, sophisticated, and finely worked/specialised.

Part (a):
- Reference to looking at the rock/sediment layer / stratigraphy (deeper layers are older) [1 mark]
- Dating fossils / organic material found in the same layer (e.g. using carbon dating) [1 mark]

Part (b):
- Tools became more complex / sophisticated / detailed / worked more finely [1 mark]

1. The stimulus (heat from the hot pan) is detected by temperature/pain receptors in the skin of the hand. 2. This triggers an electrical impulse that travels along the sensory neurone to the central nervous system (specifically the spinal cord). 3. At the end of the sensory neurone, the impulse reaches a synapse (a gap between two neurones). 4. A chemical neurotransmitter is released, which diffuses across the synapse and triggers a new electrical impulse in the relay neurone. 5. The impulse travels through the relay neurone and crosses another synapse to reach the motor neurone. 6. The electrical impulse travels along the motor neurone to the effector, which is the muscle in the arm. 7. The muscle contracts, causing the hand to be pulled away from the hot pan. This pathway is fast and automatic because it does not involve the conscious part of the brain, thereby minimizing tissue damage.

Level 1 (1-2 marks): Simple description of the reflex action. Mentions some components (e.g. hand, hot pan, muscles, brain/spinal cord) but without a clear chronological sequence of the pathway.
Level 2 (3-4 marks): Clear description of the pathway of the impulse, showing the correct sequence of most neurones (sensory -> relay -> motor) and linking the effector to the final movement.
Level 3 (5-6 marks): Detailed and fully ordered explanation of the reflex arc. Must include the role of receptors, all three neurones in order, the role of synapses/neurotransmitters, the effector's response, and a clear explanation of why the action is rapid/protective (e.g., bypasses conscious thought to prevent tissue damage).

To reduce the risk of CVD, lifestyle changes are essential: 1) Reducing intake of saturated fats to prevent high blood cholesterol levels and fatty plaque buildup in arteries. 2) Regular exercise to strengthen the heart muscle and lower blood pressure. 3) Stopping smoking to avoid chemicals like nicotine and carbon monoxide which damage blood vessels and increase blood pressure.
If a person has developed CVD, treatments include: 1) Life-long medications: Statins can be prescribed to reduce blood cholesterol levels; Antihypertensives can lower high blood pressure; Beta-blockers can regulate heart rate. 2) Surgical interventions: Stents can be inserted into narrowed coronary arteries to keep them open and restore blood flow to the heart muscle; Coronary artery bypass graft (CABG) surgery can be performed to bypass blocked arteries using a healthy blood vessel from another part of the body.

Level 1 (1-2 marks): Identifies at least one lifestyle change or one treatment method for CVD with little or no explanation of how they work.
Level 2 (3-4 marks): Explains how specific lifestyle changes reduce the risk of CVD OR describes at least two different medical/surgical treatments for someone with CVD.
Level 3 (5-6 marks): A comprehensive discussion that explains both how multiple lifestyle changes reduce CVD risk (e.g. diet, exercise, smoking) AND details both medical (e.g. statins, antihypertensives) and surgical (e.g. stents, bypass) treatments, showing a clear, balanced structure.

Root hair cells have a long, hair-like extension that projects into the soil. This structural adaptation greatly increases the surface area of the cell membrane in contact with the soil water, allowing water to be absorbed more rapidly by osmosis. Root hair cells do not contain chloroplasts because they are located underground where there is no light for photosynthesis.

1 mark for correct option A. Reject all other options.

Arteries carry blood under high pressure away from the heart. To withstand and maintain this high pressure, they have thick walls composed of muscle and elastic fibres, and a narrow lumen. Unlike veins, arteries do not have valves to prevent the backflow of blood because the high pressure keeps the blood flowing in one direction.

1 mark for correct option A. Reject all other options.

Biotic factors are living parts of an ecosystem that affect other living organisms. The number of herbivores is a biotic factor because herbivores are living organisms that eat plants, directly affecting plant survival and distribution. Temperature, light intensity, and soil pH are all non-living, physical factors, which are classified as abiotic factors.

1 mark for correct option C. Reject all other options.

When blood glucose levels rise, such as after a meal, the pancreas detects this change and secretes the hormone insulin. Insulin causes body cells, particularly muscle and liver cells, to take up more glucose from the blood and convert it into glycogen for storage, thereby lowering blood glucose levels. Glucagon is released when blood glucose is too low.

1 mark for correct option A. Reject all other options.

Osmosis is defined specifically as the diffusion of water molecules from a dilute solution (higher water concentration) to a more concentrated solution (lower water concentration) across a partially permeable membrane. A partially permeable membrane has microscopic pores that allow small water molecules to pass through but block larger solute molecules.

1 mark for correct option B. Reject all other options.

In this experiment, the independent variable is the light intensity, which is manipulated by moving the light source. The dependent variable is the rate of photosynthesis, measured by counting oxygen bubbles. Temperature is a key control variable that must be kept constant because temperature affects the enzymes involved in photosynthesis. If temperature changes, it would affect the rate of bubble production, making the results invalid.

1 mark for correct option B. Reject all other options.

Photosynthesis is the only process among the options that removes carbon dioxide from the atmosphere. Green plants and algae take in atmospheric carbon dioxide and use light energy to react it with water to produce glucose and oxygen. Combustion, respiration, and decomposition all release carbon dioxide back into the atmosphere.

1 mark for correct option C. Reject all other options.

When body temperature is too high, blood vessels (arterioles) supplying the skin capillaries dilate (widen) in a process called vasodilation. This increases the volume of blood flowing close to the skin surface, allowing more thermal energy to be radiated and lost to the surrounding environment, cooling the body down.

1 mark for correct option B. Reject all other options.

(a) Biotic factors include living elements that affect other organisms, such as competition with other plant species for nutrients/water, herbivory (being eaten), and disease/pathogens. (b) Abiotic factors are non-living chemical and physical parts of the environment, such as soil pH, soil moisture, temperature, or mineral ion concentration.

(a) 1 mark for each correct biotic factor up to a maximum of 2 marks: e.g., competition with grass (1), grazing by animals/pests (1), disease/fungal infections (1). (b) 0.5 marks for any correct abiotic factor: e.g., soil pH (0.5), temperature (0.5), soil moisture/water levels (0.5).

(a) Nitrogen-fixing bacteria live in soil or root nodules and convert atmospheric nitrogen into ammonium ions or nitrates. (b) In waterlogged soil, oxygen levels are very low. This encourages denitrifying bacteria (which are anaerobic) to actively convert nitrate ions from the soil into nitrogen gas, reducing the nitrate content of the soil.

(a) 0.5 marks for nitrogen-fixing bacteria / nitrogen-fixers. (b) 2 marks total: 1 mark for identifying that waterlogged soils are anaerobic/lacking in oxygen which allows denitrifying bacteria to thrive (1); 1 mark for explaining that these bacteria convert nitrates in the soil back into nitrogen gas (1).

(a) Osmosis is the passive movement of water molecules from a region of higher water potential to a region of lower water potential across a partially permeable membrane. (b) The long extension increases the surface area of the root cell in contact with the soil, allowing more water and mineral ions to be absorbed per unit time. The thin cell wall ensures a shorter pathway for faster diffusion and osmosis.

(a) 0.5 marks for osmosis. (b) 2 marks total: 1 mark for identifying the structural adaptation (e.g., long projection/extension/hair-like shape) (1); 1 mark for explaining how this increases the rate of absorption (e.g., increases surface area to volume ratio) (1). Alternatively, accept: thin cell wall (1) which provides a short diffusion distance (1).

(a) Phloem is the specialized plant tissue responsible for the transport of organic nutrients (sucrose, amino acids) during translocation. (b) Phloem sieve tubes are made of cells joined end-to-end with perforated sieve plates, allowing continuous flow. They lack a nucleus to minimize resistance to flow. Companion cells next to them are packed with mitochondria to generate the ATP required to actively pump sucrose into the sieve tubes.

(a) 0.5 marks for phloem. (b) 2 marks total: 1 mark for describing a structural adaptation (e.g., sieve tubes have no nucleus/very little cytoplasm or have sieve plates) (1); 1 mark for explaining how this aids transport (e.g., allows free flow of sap) (1). OR 1 mark for companion cells having many mitochondria (1) and 1 mark for explaining that this provides ATP/energy for active transport of sucrose (1).

(a) Homeostasis is the regulation of internal conditions of a cell or organism to maintain optimum conditions for function. (b) When the temperature rises, the hypothalamus triggers mechanisms to lose heat: vasodilation increases blood flow near the skin surface, radiating heat away; sweat glands secrete sweat, and the evaporation of water from the skin removes thermal energy from the body.

(a) 0.5 marks for homeostasis. (b) 2 marks total: 1 mark for describing vasodilation (blood vessels dilate to increase blood flow to skin surface) (1) or sweating (sweat glands release sweat) (1); 1 mark for explaining how this cools the body (e.g., heat is lost by radiation from blood / evaporation of sweat removes heat) (1).

(a) The pancreas detects changes in blood glucose levels and secretes insulin directly into the blood. (b) Insulin travels to the liver and muscle cells. It binds to receptors and stimulates these cells to absorb glucose from the bloodstream, where it is converted into glycogen and stored, thus lowering the concentration of glucose in the blood.

(a) 0.5 marks for pancreas. (b) 2 marks total: 1 mark for stating that insulin causes liver/muscle cells to absorb glucose from the blood (1); 1 mark for stating that glucose is converted into glycogen for storage (1).

(a) Haemoglobin is a iron-containing protein in red blood cells that reversibly binds to oxygen to form oxyhaemoglobin. (b) The biconcave disc shape creates a larger surface area compared to a simple sphere, which increases the rate of oxygen diffusion. Having no nucleus maximizes the internal volume of the cell, meaning more haemoglobin molecules can fit inside to transport a greater volume of oxygen.

(a) 0.5 marks for haemoglobin. (b) 2 marks total: 1 mark for explaining that the biconcave shape increases the surface area (to volume ratio) for faster diffusion of oxygen (1); 1 mark for explaining that the lack of a nucleus leaves more space to carry more haemoglobin/oxygen (1).

(a) The active site of an enzyme is a highly specific cleft where substrate molecules bind and undergo a chemical reaction. (b) High temperatures supply excess kinetic energy, causing the enzyme's structure to vibrate excessively until the weak bonds (like hydrogen bonds) holding its tertiary structure break. This denatures the enzyme, permanently altering the shape of the active site so that the substrate is no longer complementary, halting the reaction.

(a) 0.5 marks for active site. (b) 2 marks total: 1 mark for stating that high temperatures break bonds, changing the shape of the active site / denaturing the enzyme (1); 1 mark for explaining that the substrate is no longer complementary / cannot fit into the active site, so no reaction occurs (1). (Reject 'enzyme dies/kills').

Nitrogen-fixing bacteria have a mutualistic relationship with leguminous plants. 1. The bacteria convert atmospheric nitrogen into nitrates/ammonium, which the plant needs to build amino acids and proteins. 2. The host plant provides the bacteria with glucose/carbohydrates (produced via photosynthesis) and a protected environment in the root nodules.

Award 1 mark for describing the benefit to the plant (obtains nitrates/nitrogen compounds to make amino acids/proteins). Award 1 mark for describing the benefit to the bacteria (obtains glucose/carbohydrates/sugars/energy/shelter from the plant).

Alveoli have several adaptations for gas exchange: 1. Thin walls (one cell thick) which minimize the distance gases have to diffuse. 2. A massive total surface area which maximizes the amount of oxygen and carbon dioxide that can diffuse at once. 3. An excellent blood supply (capillaries) to continuously carry oxygen away and bring carbon dioxide, maintaining a steep concentration gradient.

Award up to 3 marks. Max 2 marks for naming adaptations, and max 2 marks for explaining their functions: One-cell thick / thin walls (1 mark) which reduces diffusion distance (1 mark); Large surface area (1 mark) to increase rate of diffusion (1 mark); Good blood supply / capillaries (1 mark) to maintain the concentration gradient (1 mark).

1. High blood glucose levels are detected by the pancreas, which secretes the hormone insulin into the blood. 2. Insulin targets the liver and muscle cells. 3. It causes these cells to take up excess glucose from the blood and convert it into glycogen, which is an insoluble storage carbohydrate.

Award 1 mark for stating that insulin is released by the pancreas. Award 1 mark for stating that insulin causes liver/muscle cells to absorb glucose from the blood. Award 1 mark for stating that glucose is converted into glycogen for storage.

Xylem and phloem have different transport directions and carry different materials. Xylem transports water and minerals unidirectionally (upwards only). Phloem transports sucrose and organic nutrients bidirectionally (both up and down the plant).

Award 1 mark for identifying the difference in substances transported (water/minerals in xylem vs sugars/sucrose/amino acids in phloem). Award 1 mark for identifying the difference in direction of flow (one-way/upwards in xylem vs two-way/up and down in phloem).

1. Calculate the change in mass: 3.40 g - 4.00 g = -0.60 g. 2. Calculate the percentage change: (-0.60 / 4.00) * 100 = -15%.

Award 1 mark for the correct calculation of change in mass (-0.60 g) or setting up the calculation correctly: ((3.40 - 4.00) / 4.00) * 100. Award 1 mark for the correct final answer of -15% (allow 15% decrease/loss).

Decomposers such as bacteria and fungi require specific conditions to survive, grow, and respire. The three main abiotic factors affecting their rate of activity are: 1. Temperature (warmer temperatures speed up enzyme reactions). 2. Moisture / water availability (decomposers need water to carry out biological processes). 3. Oxygen levels (needed for aerobic respiration).

Award 1 mark for each of the following up to a maximum of 3 marks: Temperature (accept warmth), water availability (accept moisture / humidity), oxygen availability (accept aeration).

Red blood cells are adapted by having: 1. A biconcave disc shape, which gives them a high surface area to volume ratio, speeding up diffusion of oxygen. 2. No nucleus, providing extra space inside the cell to carry more haemoglobin (and thus more oxygen). 3. Containing haemoglobin, a protein that binds to oxygen.

Award 1 mark for naming any valid structural feature and 1 mark for linking it to its function, up to a maximum of 2 marks: Biconcave shape (1) to increase surface area / speed up diffusion (1); No nucleus (1) to provide more space for haemoglobin/oxygen (1); Contains haemoglobin (1) to bind/carry oxygen (1).

1. Receptors in the skin detect the thermal stimulus and generate an electrical impulse. 2. The impulse travels along the sensory neurone to the spinal cord (CNS). 3. It crosses a chemical synapse to a relay neurone, and then another synapse to a motor neurone. 4. The motor neurone carries the impulse to the effector (muscle), which contracts to move the hand away.

Award 1 mark for mentioning that the impulse travels along a sensory neurone from the receptor/stimulus. Award 1 mark for mentioning that the impulse passes through a relay neurone (in the CNS/spinal cord) and across synapses. Award 1 mark for mentioning that the impulse travels along a motor neurone to the effector/muscle.

During exercise, muscles contract more frequently and require more energy. The heart rate increases to pump more blood to these muscles. This blood delivers oxygen and glucose, which are needed for aerobic respiration. It also helps to remove carbon dioxide, a waste product of respiration, more quickly.

1 mark for identifying that increased heart rate pumps more blood to deliver oxygen or glucose to the working muscles. 1 mark for linking this to increased aerobic respiration or energy release in the muscles. 0.5 marks for mentioning the faster removal of carbon dioxide or waste products.

Decomposers are microorganisms that break down dead organic matter. They secrete digestive enzymes onto the dead leaves to break down complex organic molecules into simpler soluble nutrients. These nutrients are absorbed by the decomposers, and mineral ions (such as nitrates) are released into the soil. Plants can then absorb these mineral ions through their roots for growth.

1 mark for stating that decomposers secrete enzymes to break down organic matter or dead leaves. 1 mark for stating that mineral ions or nitrates are released into the soil. 0.5 marks for stating that plants absorb these mineral ions through their roots.

For part (a), decomposers require oxygen for aerobic respiration to break down organic matter efficiently. They also require moisture (water) because biological reactions occur in solution. For part (b), the main groups of decomposers are microscopic organisms such as bacteria and fungi.

For part (a):
- Moisture / water / humidity [1 mark]
- Oxygen [1 mark]
(Ignore: temperature)

For part (b):
- Bacteria / Fungi / Fungus [0.5 marks]
(Do not accept: worms / insects / detritivores)

(a) The red blood cell has a biconcave shape (it is dented on both sides). This specific shape increases the cell's surface area to volume ratio, allowing oxygen to diffuse into and out of the cell much faster. (b) Haemoglobin is the protein pigment that chemically binds with oxygen in the lungs and releases it in the tissues.

For part (a):
- Biconcave (disc/shape) [0.5 marks]
- Provides a large surface area to volume ratio (for faster diffusion) [1.0 mark]

For part (b):
- Haemoglobin (accept hemoglobin) [1.0 mark]

(a) The pancreas is the endocrine organ that senses glucose levels and releases hormones to control them. (b) When blood glucose is high, insulin is secreted. It travels in the blood to target organs, prompting liver and muscle cells to absorb excess glucose and convert it into the insoluble storage carbohydrate called glycogen.

For part (a):
- Pancreas [1.0 mark]

For part (b) (any two points from the following, up to a maximum of 1.5 marks):
- (Insulin causes) glucose to move from blood into cells / liver / muscle [1.0 mark]
- (Glucose is) converted into glycogen [0.5 marks]
- Glycogen is stored [0.5 marks]

(a) Root hair cells have long projections that provide a large surface area to absorb water efficiently by osmosis. (b) Transpiration is the process where water evaporates from the leaf surfaces and pulls a column of water up through the xylem vessels. (c) Factors like high temperature, high wind speed, or high light intensity increase transpiration rate, while high humidity decreases it.

For part (a):
- Root hair cell(s) [1.0 mark]

For part (b):
- Transpiration [0.5 marks]

For part (c):
- Any one environmental factor from: (increased) temperature / (increased) wind (speed) / (increased) light (intensity) / (decreased) humidity [1.0 mark]

(a) Osmosis is a passive transport mechanism specifically defined as the net movement of water molecules from a solution with a higher water concentration to one with a lower water concentration through a membrane that only lets certain molecules pass (partially permeable). (b) The independent variable is the factor that is changed by the experimenter, which is the concentration of sucrose (sugar) solution.

For part (a):
- (Movement of) water (molecules) [0.5 marks]
- From high water concentration to low water concentration (or down a water concentration gradient) [0.5 marks]
- Through / across a partially permeable membrane (accept semi-permeable membrane) [0.5 marks]

For part (b):
- Concentration of sucrose / sugar solution [1.0 mark]
(Accept: concentration of solution / strength of solution)

(a) The area of the quadrat is \(\text{width} \times \text{length} = 0.5\text{ m} \times 0.5\text{ m} = 0.25\text{ m}^2\).
(b) First, find how many times the quadrat fits into the total field area: \(200\text{ m}^2 / 0.25\text{ m}^2 = 800\). Then, multiply this factor by the average number of dandelions per quadrat: \(800 \times 4 = 3200\) dandelions in total. (Alternatively, dandelions per \(\text{m}^2 = 4 / 0.25 = 16\); total dandelions = \(16 \times 200 = 3200\)).

For part (a):
- \(0.25\) (\(\text{m}^2\)) [1.0 mark]

For part (b):
- Correct method shown to scale up from quadrat area to total area (e.g. \(200 / 0.25 = 800\) or \(4 / 0.25 = 16\)) [0.5 marks]
- \(3200\) [1.0 mark]
(Award 1.5 marks for correct final answer of \(3200\) with or without working)

1. Runoff/Leaching: The heavy rain washes the highly soluble nitrates from the fertiliser into the nearby pond.
2. Algal Bloom: The high level of nutrients causes rapid multiplication of algae at the surface of the pond, forming a thick green layer.
3. Light Blockage: The algal bloom blocks sunlight from penetrating the deeper layers of the water.
4. Plant Death: Submerged plants deeper in the pond cannot absorb light, meaning they cannot photosynthesise and subsequently die.
5. Decomposition: Bacteria and other decomposers break down the dead plant tissues. Due to the abundant food source, the bacterial population increases rapidly.
6. Oxygen Depletion: The bacteria respire aerobically, consuming vast amounts of dissolved oxygen from the water.
7. Fish Suffocation: The concentration of dissolved oxygen falls so low that fish and other aquatic animals cannot respire and die.

Indicative content:
- Nitrates/fertiliser leach/wash into the pond due to rain.
- This causes rapid growth of algae on the water surface (algal bloom).
- The algae block sunlight from reaching plants deeper in the water.
- Deeper plants cannot photosynthesise and die.
- Bacteria/decomposers feed on the dead plants and multiply.
- Bacteria respire aerobically, using up dissolved oxygen in the water.
- Oxygen levels drop, leading to the death of fish due to lack of oxygen.

Level Descriptors:
- Level 1 (1-2 marks): The explanation is simple and fragmented. Identifies a few key points (e.g., fertiliser enters water, plants die, fish die) but lacks logical sequencing or detail on decomposition/oxygen.
- Level 2 (3-4 marks): The explanation is mostly structured and covers several key stages. Clearly links fertiliser runoff to algal growth and plant death, or links decomposition to oxygen loss and fish death.
- Level 3 (5-6 marks): A detailed, fully coherent and logically structured explanation. Covers the complete process from fertiliser runoff, algal bloom, plant death, bacterial decomposition, aerobic respiration/oxygen depletion, to the final death of the fish.

1. Detection: Thermoreceptors in the skin detect changes in the external temperature, and receptors in the hypothalamus of the brain monitor the temperature of the blood flowing through it.
2. Coordination: The hypothalamus acts as a control center and sends nervous impulses to effectors in the skin to trigger cooling mechanisms.
3. Sweating: Sweat glands are stimulated to secrete sweat onto the surface of the skin. As the water in sweat evaporates, it absorbs and carries away heat energy from the body, cooling it down.
4. Vasodilation: Arterioles supplying skin capillaries dilate (widen). This allows more blood to flow close to the skin surface, increasing the rate of heat transfer to the environment by radiation.
5. Flattening of Hairs: Hair erector muscles relax, causing hairs to lie flat against the skin. This prevents a layer of warm air from being trapped next to the skin.
6. Negative Feedback: Once body temperature returns to the set point (approx. 37 degrees Celsius), these mechanisms are reduced.

Indicative content:
- Temperature changes are detected by receptors in the skin.
- Receptors in the hypothalamus monitor blood temperature.
- Hypothalamus coordinates the response (acts as control center).
- Sweat glands produce sweat, which evaporates to transfer heat energy away from the skin.
- Vasodilation occurs (arterioles supplying capillaries near the skin surface dilate/widen).
- More blood flows near the skin surface, increasing heat loss by radiation.
- Hair erector muscles relax so hairs lie flat, preventing insulating air from being trapped.
- This is an example of negative feedback.

Level Descriptors:
- Level 1 (1-2 marks): Identifies simple cooling mechanisms (e.g., sweating/vasodilation) but with limited explanation of how they lower temperature or how they are detected.
- Level 2 (3-4 marks): Explains at least two mechanisms (e.g., sweating and vasodilation) showing how they help lose heat, or links detection (hypothalamus/skin) to a response.
- Level 3 (5-6 marks): Provides a comprehensive explanation including detection by receptors (hypothalamus/skin) and a detailed description of both sweating (evaporation) and vasodilation (blood flow and radiation) to lower body temperature.