2025 Edexcel IAL Biology (YBI11) 模擬試題連答案詳解

Sucrose is a disaccharide formed by a condensation reaction between a glucose molecule and a fructose molecule, linked by an \(\alpha\)-1,2-glycosidic bond. Complete hydrolysis breaks this bond to yield glucose and fructose.

1 mark: A - One \(\alpha\)-1,2-glycosidic bond to produce one glucose and one fructose molecule.

In semi-conservative replication, the first generation in \(^{14}\text{N}\) produces 100% hybrid DNA molecules (\(^{15}\text{N}/^{14}\text{N}\)). In the second generation, each hybrid molecule serves as a template. This results in 50% hybrid DNA molecules (\(^{15}\text{N}/^{14}\text{N}\)) and 50% light DNA molecules (\(^{14}\text{N}/^{14}\text{N}\)). Thus, 50% of the DNA molecules contain only \(^{14}\text{N}\).

1 mark: C - 50%

Because the ventricular pressure is higher than atrial pressure, the bicuspid (atrioventricular) valve is forced shut to prevent backflow of blood. However, because ventricular pressure has not yet exceeded aortic pressure, the aortic (semilunar) valve remains closed. This is the period of isovolumetric contraction where both valves are closed.

1 mark: D - Both bicuspid and aortic valves are closed

Oxygen is a small, non-polar molecule that easily dissolves in and diffuses through the hydrophobic lipid core of the bilayer. Ions (like \(\text{Ca}^{2+}\)) and polar/large molecules (like glucose and amino acids) cannot easily pass through the non-polar hydrophobic core and require transport proteins (facilitated diffusion or active transport).

1 mark: A - Oxygen (\(\text{O}_2\))

A triglyceride is synthesized by three condensation reactions, where each of the three fatty acids forms an ester bond with one of the three hydroxyl groups on a single glycerol molecule. Each condensation reaction releases one water molecule, resulting in a total of 3 water molecules released and 3 ester bonds formed.

1 mark: B - 3 water molecules released and 3 ester bonds formed

Transcription produces an mRNA strand that is complementary to the DNA template strand and runs in an antiparallel (5' to 3') direction. Under complementary base pairing rules: 3'-TAC-5' transcribes to 5'-AUG-3'; 3'-GGC-5' transcribes to 5'-CCG-3'; 3'-TTA-5' transcribes to 5'-AAU-3'. Therefore, the sequence is 5'- AUG CCG AAU -3'.

1 mark: A - 5'- AUG CCG AAU -3'

Arteries must withstand and maintain high blood pressure. They have thick walls containing abundant elastic fibres and muscle tissue, resulting in a narrower lumen. Veins carry blood under much lower pressure, so they have thinner walls with fewer elastic fibres and a wider lumen to minimize resistance to blood flow.

1 mark: A - The artery has a narrower lumen and more elastic fibres than the vein.

Since three nucleotides (one complete codon) are deleted, there is no frameshift mutation. The reading frame of the remaining mRNA codons is preserved, so the direct effect is simply the loss of a single amino acid (phenylalanine at position 508) while all other amino acids in the sequence remain unaffected.

1 mark: B - A single amino acid is deleted, but the rest of the sequence remains unchanged.

When a blood vessel is damaged, platelets and damaged tissue release thromboplastin. In the presence of calcium ions, thromboplastin converts the inactive plasma protein prothrombin into the active enzyme thrombin. Thrombin then catalyses the conversion of the soluble plasma protein fibrinogen into insoluble fibrin fibres, which form a mesh to trap blood cells and form a clot.

Correct Answer is A. 1 mark for selecting the option with: Substance released = Thromboplastin, Active enzyme = Thrombin, Insoluble protein = Fibrin. Reject B because prothrombin is not released directly and fibrinogen is soluble. Reject C because fibrin is the final insoluble protein, not the active enzyme. Reject D because thrombin is not released directly by damaged tissue.

First, calculate the total number of nucleotides: since the DNA is double-stranded and has 1600 base pairs, it contains 1600 * 2 = 3200 nucleotides. According to complementary base pairing, the percentage of adenine (A) is equal to the percentage of thymine (T), so A = T = 22%. The combined percentage of guanine (G) and cytosine (C) is 100% - (22% + 22%) = 56%. Since G = C, the percentage of guanine is 56% / 2 = 28%. The total number of guanine bases is 28% of 3200, which is calculated as 0.28 * 3200 = 896.

Correct Answer is C (896). 1 mark for the correct calculation. Distractor A (448) is incorrect because it uses 1600 total nucleotides instead of 3200. Distractor B (704) is incorrect because it is the number of adenine or thymine bases (22% of 3200). Distractor D (1792) is incorrect because it is the combined total of cytosine and guanine bases (56% of 3200).

Saturated fats are processed by the liver into low-density lipoproteins (LDLs), which circulate in the blood to transport cholesterol to body tissues. High dietary intake of saturated fats leads to an excess of LDLs, overloading LDL receptors. As a result, blood LDL levels remain elevated. These LDLs can penetrate the endothelium of arteries, particularly at sites of damage caused by high blood pressure. In the arterial wall, they undergo oxidation, initiating an inflammatory response that recruits macrophages and forms a fatty plaque called an atheroma. The atheroma narrows the arterial lumen, raising blood pressure and restricting blood flow to cardiac muscle or other tissues, increasing the risk of cardiovascular disease.

1. Saturated fats lead to increased synthesis of LDLs, elevating blood LDL/cholesterol levels. (1) 2. High LDL levels cause the accumulation of cholesterol in the damaged endothelium of arteries. (1) 3. This leads to the formation of an atheroma, which narrows the lumen and increases the risk of coronary heart disease or thrombosis. (1)

Phospholipid molecules are amphipathic, meaning they contain both hydrophilic and hydrophobic regions. The polar phosphate group forms a head that is hydrophilic (attracted to water), while the two non-polar fatty acid chains form tails that are hydrophobic (repelled by water). When placed in an aqueous environment, such as the intracellular and extracellular fluids, the hydrophilic heads spontaneously orient outwards to interact with water. The hydrophobic tails orient inwards, facing each other to minimize contact with water, creating a stable bilayer structure that acts as a barrier.

1. Phospholipids consist of a hydrophilic polar phosphate head and hydrophobic non-polar fatty acid tails. (1) 2. In an aqueous environment, the hydrophilic heads are attracted to water and face outwards towards the cytoplasm and extracellular fluid. (1) 3. The hydrophobic fatty acid tails are repelled by water and face inwards, pointing towards each other in the center of the bilayer. (1)

The acrosome is a specialized lysosome located at the tip of the sperm head. When the sperm reaches the egg and binds to the zona pellucida (the outer glycoprotein layer of the egg), the acrosome reaction is triggered. The acrosome membrane fuses with the sperm's cell surface membrane, releasing hydrolytic enzymes (such as acrosin) via exocytosis. These digestive enzymes break down the glycoproteins of the zona pellucida, creating a path for the sperm to traverse. This allows the sperm cell membrane to make contact and fuse with the egg cell surface membrane, enabling the sperm nucleus to enter the egg cytoplasm.

1. The acrosome undergoes the acrosome reaction, fusing with the sperm membrane to release digestive enzymes by exocytosis. (1) 2. These enzymes digest and break down the zona pellucida surrounding the egg. (1) 3. This allows the sperm cell membrane to fuse with the egg cell membrane, allowing the sperm nucleus to enter the egg. (1)

Xylem vessels are highly specialized for water transport and support. Their cell walls are thickened with lignin, a rigid, waterproof polymer. Lignin provides mechanical strength, preventing the vessels from collapsing inward under the high tension created by transpiration, while also supporting the stem. Additionally, xylem vessels are dead, hollow tubes that lack living contents such as cytoplasm or organelles, and their end walls have broken down completely. This structural modification creates a continuous column with minimal resistance to water flow. Finally, unlignified areas called pits allow the lateral movement of water between adjacent vessels.

1. Cell walls are reinforced with lignin, which provides mechanical strength to support the stem and prevents vessels from collapsing under tension. (1) 2. The vessels are dead, hollow tubes lacking cytoplasm and end walls, which creates a continuous column for uninterrupted water transport. (1) 3. Pits in the walls allow lateral transport of water to surrounding tissues or adjacent vessels. (1)

The atrioventricular (AV) node plays a critical role in regulating the cardiac cycle. When the sinoatrial node (SAN) initiates an electrical impulse, it spreads across the atria, causing atrial systole. Because of a non-conducting layer of fibrous tissue between the atria and ventricles, the impulse must pass through the AV node. The AV node introduces a short delay of approximately 0.1 seconds. This delay is essential as it allows time for the atria to finish contracting and fully empty blood into the ventricles. After this delay, the AV node transmits the electrical signal down the Bundle of His to the Purkyne fibers, which rapidly spread the depolarization wave to the apex, ensuring the ventricles contract from the bottom up to efficiently eject blood into the arteries.

1. The AV node receives the wave of electrical excitation from the sinoatrial node / atria. (1) 2. It delays the electrical impulse for about 0.1 seconds to allow the atria to complete contraction and empty blood into the ventricles. (1) 3. It conducts the impulse down the Bundle of His / Purkyne fibers to the apex, ensuring ventricles contract from the bottom up. (1)

During the elongation stage of translation, the ribosome brings two tRNA molecules carrying their respective amino acids close together. A condensation reaction occurs between the carboxyl group (\(-\text{COOH}\)) of the amino acid in the P site and the amine group (\(-\text{NH}_2\)) of the amino acid in the A site. This reaction is catalyzed by the ribozyme peptidyl transferase, which is an integral part of the large ribosomal subunit. The reaction results in the release of a water molecule (\(\text{H}_2\text{O}\)) and the creation of a covalent peptide bond (\(-\text{CO}-\text{NH}-\)) that links the amino acids together into a growing polypeptide chain.

1. A condensation reaction occurs between the amine group of one amino acid and the carboxyl group of the adjacent amino acid. (1) 2. This reaction is catalyzed by peptidyl transferase in the ribosome. (1) 3. A molecule of water (\(\text{H}_2\text{O}\)) is released, resulting in the formation of a covalent peptide bond. (1)

During meiosis I, specifically metaphase I, homologous pairs of chromosomes align along the equator of the spindle. The orientation of each homologous pair is completely random; the maternal chromosome of one pair may face one pole while the paternal chromosome faces the other, and this occurs independently for every other pair. When homologous chromosomes are separated during anaphase I and move to opposite poles, this random alignment ensures that each resulting haploid gamete receives a unique combination of maternal and paternal chromosomes. The mathematical number of potential combinations is \(2^n\), where \(n\) is the haploid number, contributing significantly to genetic diversity.

1. During metaphase I, homologous pairs of chromosomes align randomly at the equator of the spindle. (1) 2. The orientation of each homologous pair is independent of the orientation of any other pair. (1) 3. Separation of these chromosomes in anaphase I results in random and unique combinations of maternal and paternal chromosomes in the gametes. (1)

To prepare a temporary root tip squash, first cut a 5 to 10 mm section of a root tip, which contains the actively dividing meristematic cells. Place this root tip in warm dilute hydrochloric acid. This treatment hydrolyzes the middle lamella, softening the cell wall matrix so cells can be separated. Next, transfer the root tip to a clean slide, rinse it, and add a few drops of a stain such as acetic orcein, toluidine blue, or Feulgen's stain, which binds selectively to nucleic acids to make chromosomes visible. Gently warm the slide to intensify the stain. Finally, place a coverslip over the specimen and press down firmly and vertically with a thumb or pencil eraser to squash the tissue into a single-cell-thick layer, taking care not to slide the coverslip horizontally to avoid damaging the chromosomes.

1. Cut a small section of the root tip and heat/macerate it in hydrochloric acid to soften the tissue and break down the middle lamella. (1) 2. Place the root tip on a slide and apply a stain (such as acetic orcein or toluidine blue) to color the chromosomes. (1) 3. Cover with a coverslip and press down gently/firmly (squash) to flatten the cells into a single-cell-thick layer without moving the coverslip sideways. (1)

When the endothelium of an artery is damaged, collagen fibres are exposed. Platelets adhere to the exposed collagen and release clotting factors, specifically thromboplastin. Thromboplastin, in the presence of calcium ions \(Ca^{2+}\), catalyses the conversion of the inactive plasma protein prothrombin into the active enzyme thrombin. Thrombin then catalyses the conversion of soluble fibrinogen into insoluble fibrin. Fibrin fibres form a mesh that traps red blood cells and platelets, forming a blood clot.

1. Platelets adhere to exposed collagen and release thromboplastin. (1 mark) 2. Thromboplastin catalyses the conversion of prothrombin to thrombin in the presence of calcium ions. (1 mark) 3. Thrombin catalyses the conversion of soluble fibrinogen to insoluble fibrin to trap blood cells. (1 mark)

The phospholipid bilayer consists of hydrophilic phosphate heads facing outwards and hydrophobic fatty acid tails pointing inwards, creating a hydrophobic core. This hydrophobic core acts as a barrier to polar, hydrophilic, or charged substances (such as ions and glucose), preventing them from diffusing directly through. Small, non-polar, lipid-soluble molecules (such as oxygen and carbon dioxide) can easily dissolve in and diffuse directly through the hydrophobic core.

1. Identifies the hydrophobic core formed by fatty acid tails as a barrier to polar / charged / hydrophilic substances. (1 mark) 2. Explains that small, non-polar / lipid-soluble molecules can easily diffuse directly through the bilayer. (1 mark) 3. Explains that large polar molecules or ions require specific transport proteins to cross. (1 mark)

Water is a dipolar molecule because the oxygen atom has a slight negative charge (delta negative) and the hydrogen atoms have a slight positive charge (delta positive). This dipolar nature allows water to form hydrogen bonds with other polar substances and ionic solutes, enabling them to dissolve and be transported in blood plasma. Additionally, hydrogen bonding between water molecules creates cohesion, which allows water to flow as a continuous column (mass flow) under pressure through blood vessels.

1. Explains that water is dipolar with delta negative oxygen and delta positive hydrogen atoms. (1 mark) 2. States that water forms hydrogen bonds with polar / ionic solutes, allowing them to dissolve for transport. (1 mark) 3. Mentions cohesion between water molecules due to hydrogen bonding, which enables mass flow / continuous flow through blood vessels. (1 mark)

As temperature increases above the optimum, the excessive thermal energy breaks the weak hydrogen bonds and ionic bonds holding the tertiary structure of the enzyme together. This alters the specific 3D conformation of the enzyme, changing the shape of its active site. Consequently, the substrate is no longer complementary and cannot bind to the active site (the enzyme is denatured), leading to fewer enzyme-substrate complexes forming and a rapid decrease in the rate of reaction.

1. States that increased thermal energy breaks hydrogen / ionic bonds holding the tertiary structure of the enzyme. (1 mark) 2. Explains that this changes the shape of the active site, denaturing the enzyme. (1 mark) 3. Concludes that the substrate can no longer bind / fit, so fewer or no enzyme-substrate complexes form. (1 mark)

A disaccharide like sucrose is formed by a chemical reaction called a condensation reaction. This reaction occurs between two monosaccharides: glucose (specifically alpha-glucose) and fructose. During this reaction, a molecule of water \(H_2O\) is eliminated (released), and a covalent glycosidic bond (specifically a 1,2-glycosidic bond) is formed between the two monosaccharide units to join them together.

1. Identifies that it occurs via a condensation reaction between alpha-glucose and fructose. (1 mark) 2. Mentions the removal / elimination / release of a water molecule. (1 mark) 3. States that a glycosidic bond (or 1,2-glycosidic bond) is formed between the monosaccharides. (1 mark)

During semi-conservative DNA replication, DNA polymerase aligns free DNA nucleotides with their complementary bases on the template strand and catalyses the formation of phosphodiester bonds between adjacent nucleotides to synthesise a new strand. Because replication is discontinuous on the lagging strand, short segments called Okazaki fragments are produced. DNA ligase then catalyses the formation of phosphodiester bonds to join these Okazaki fragments together, completing the continuous sugar-phosphate backbone.

1. DNA polymerase catalyses the formation of phosphodiester bonds between adjacent nucleotides to form the new strand. (1 mark) 2. DNA polymerase works in a 5' to 3' direction using complementary base pairing. (1 mark) 3. DNA ligase joins Okazaki fragments together on the lagging strand by forming phosphodiester bonds. (1 mark)

The left ventricle has a much thicker muscular wall (thick myocardium) compared to both the right ventricle and the atria. This abundant cardiac muscle allows the left ventricle to contract with a very high force, generating high blood pressure during systole. This high pressure is essential to overcome the high resistance of the systemic circulation and successfully pump oxygenated blood all the way through the aorta to the rest of the body.

1. Identifies that the left ventricle has a thick muscular wall / thick myocardium. (1 mark) 2. Explains that this structure allows for powerful contractions that generate high blood pressure. (1 mark) 3. Relates this to the function of pumping blood a long distance through the aorta / around the entire systemic circulation. (1 mark)

1. Calculate Stroke Volume (SV) before training:
\( \text{SV} = \frac{\text{Cardiac Output}}{\text{Heart Rate}} = \frac{15.6}{145} = 0.1076 \text{ dm}^3 \) (or \( 107.6 \text{ cm}^3 \))

2. Calculate Stroke Volume (SV) after training:
\( \text{SV} = \frac{15.6}{120} = 0.1300 \text{ dm}^3 \) (or \( 130.0 \text{ cm}^3 \))

3. Calculate the percentage change:
\( \text{Percentage change} = \frac{0.1300 - 0.1076}{0.1076} \times 100 = +20.83\% \)

4. Structure-function relation:
- The left ventricle has a thick, muscular wall containing cardiac muscle fibres.
- These fibres stretch further during exercise due to increased venous return, resulting in a stronger contraction (Starling's Law of the Heart) to pump out more blood per beat.

Calculation (3 Marks):
- MP1: Correct calculation of stroke volume before training (\( 0.108 \text{ dm}^3 \) or \( 107.6 \text{ cm}^3 \)) [1]
- MP2: Correct calculation of stroke volume after training (\( 0.130 \text{ dm}^3 \) or \( 130.0 \text{ cm}^3 \)) [1]
- MP3: Correct percentage change calculation: \( +20.8\% \) or \( +20.83\% \) (Accept \( 21\% \)) [1]

Explanation (2 Marks):
- MP4: Thicker muscular wall of the left ventricle / contains cardiac muscle [1]
- MP5: Muscle fibres stretch more (due to venous return) leading to more powerful contraction / more blood expelled [1]

According to Fick's Law:
\( \text{Rate of Diffusion} \propto \frac{\text{Surface Area} \times \text{Concentration Difference}}{\text{Thickness}} \)

From Condition 1 to Condition 2:
- Surface area is doubled: \( \frac{5.0 \times 10^{-4}}{2.5 \times 10^{-4}} = 2 \) (increases rate by factor of 2)
- Concentration difference remains constant: factor of 1
- Thickness is increased by a factor of \( \frac{10.0}{7.5} = 1.333 \) (reduces rate by a factor of 1.333)

Therefore, the new rate is:
\( \text{New Rate} = 4.2 \times 10^{-6} \times \frac{2}{1.333} = 6.3 \times 10^{-6} \text{ mol s}^{-1} \)

Calculation (3 Marks):
- MP1: Recognises that surface area is doubled (\( \times 2 \)) [1]
- MP2: Recognises that thickness has increased by a factor of 1.33 (or ratio \( 7.5/10 = 0.75 \)) [1]
- MP3: Correct final calculated value of \( 6.3 \times 10^{-6}\text{ mol s}^{-1} \) (Accept standard form or decimal) [1]

Explanation (2 Marks):
- MP4: Membrane has a hydrophobic core / interior made of non-polar fatty acid tails [1]
- MP5: Polar molecules are hydrophilic and are repelled by / cannot dissolve in the non-polar hydrophobic core [1]

1. Calculate \( N \) (total number of individuals):
\( N = 15 + 5 + 10 + 2 = 32 \)

2. Calculate \( N(N-1) \):
\( N(N-1) = 32 \times 31 = 992 \)

3. Calculate \( \sum n(n-1) \) for all species:
- Meadow Buttercup: \( 15 \times 14 = 210 \)
- Red Clover: \( 5 \times 4 = 20 \)
- Ribwort Plantain: \( 10 \times 9 = 90 \)
- Common Dandelion: \( 2 \times 1 = 2 \)
\( \sum n(n-1) = 210 + 20 + 90 + 2 = 322 \)

4. Calculate \( D \):
\( D = \frac{992}{322} \approx 3.08 \) (or 3.1)

Calculation (3 Marks):
- MP1: Correct calculation of \( N = 32 \) and \( N(N-1) = 992 \) [1]
- MP2: Correct calculation of \( \sum n(n-1) = 322 \) [1]
- MP3: Correct calculation of \( D = 3.08 \) or \( 3.1 \) [1]

Analysis/Explanation (2 Marks):
- MP4: A site with lower biodiversity would have a lower value of \( D \) [1]
- MP5: Conserving these habitats maintains a large gene pool / variety of alleles which can be sourced for future breeding programmes / medicine development [1]

1. Since the DNA is double-stranded, cytosine (C) base-pairs with guanine (G). Therefore, if C = 34%, G must also be 34%.
2. Total percentage of C + G = 68%.
3. Total percentage of A + T = 100% - 68% = 32%.
4. Since adenine (A) base-pairs with thymine (T), A must make up half of this remaining percentage: 32% / 2 = 16% of the total bases.
5. Total number of bases in a 1500 base pair fragment = 1500 x 2 = 3000 bases.
6. Total number of adenine bases = 16% of 3000 = \( 0.16 \times 3000 = 480 \).

Calculation (3 Marks):
- MP1: Correct determination of the percentage of adenine (16%) [1]
- MP2: Correct determination of the total number of individual bases (3000 bases) [1]
- MP3: Correct final calculated number of adenine bases (480) [1]

Explanation (2 Marks):
- MP4: Weak hydrogen bonds between complementary base pairs are easily broken to allow separation of the two strands [1]
- MP5: Each single strand acts as a template for the synthesis of a new complementary strand (A-T, C-G) [1]

1. Calculate the number of divisions (\( n \)) in 3 hours:
\( 3 \text{ hours} = 180 \text{ minutes} \)
\( n = \frac{180}{20} = 9 \text{ generations} \)

2. Use the exponential growth formula:
\( N = N_0 \times 2^n \)
\( N = (2.0 \times 10^3) \times 2^9 \)
\( 2^9 = 512 \)
\( N = 2.0 \times 10^3 \times 512 = 1.024 \times 10^6 \text{ cells per cm}^3 \)

3. Explanation of antibiotics:
- Bactericidal antibiotics kill bacteria (e.g., by disrupting cell wall synthesis or membrane integrity, causing lysis).
- Bacteriostatic antibiotics prevent binary fission / cell division / growth of bacteria (e.g., by inhibiting protein synthesis), allowing host immune systems to clear the infection.

Calculation (3 Marks):
- MP1: Correctly calculates number of divisions/generations as 9 [1]
- MP2: Correct working using the exponential growth equation (e.g., \( 2.0 \times 10^3 \times 512 \)) [1]
- MP3: Correct final answer in standard form: \( 1.024 \times 10^6 \) (or \( 1.02 \times 10^6 \) or \( 1.0 \times 10^6 \)) [1]

Explanation (2 Marks):
- MP4: Bactericidal antibiotics cause bacterial death / destroy / lyse bacterial cells [1]
- MP5: Bacteriostatic antibiotics prevent/inhibit bacterial cell division/growth [1]

1. Endothelial damage: High blood pressure, toxins from smoking, or high blood cholesterol damage the endothelial lining of an artery.
2. Inflammatory response: White blood cells (macrophages) move into the artery wall and accumulate cholesterol/lipids, forming foam cells. This leads to the formation of a fatty streak and subsequently a fibrous plaque (atheroma).
3. Plaque hardening: Calcium salts and fibrous tissue deposit in the atheroma, causing it to harden, reducing the elasticity of the artery wall and narrowing the lumen, which increases blood pressure further.
4. Plaque rupture: The protective fibrous cap of the plaque can rupture, exposing the underlying collagen fibres to the blood.
5. Clotting cascade activation: Platelets bind to the exposed collagen and release thromboplastin. Thromboplastin, in the presence of calcium ions, catalyses the conversion of the inactive protein prothrombin into the active enzyme thrombin.
6. Clot formation and occlusion: Thrombin catalyses the conversion of soluble fibrinogen into insoluble fibrin fibres. These fibres form a mesh that traps blood cells, forming a blood clot (thrombus). If this occurs in a coronary artery, it blocks blood flow, preventing oxygen from reaching cardiac muscle tissue, leading to aerobic respiration stopping and cell death (myocardial infarction).

Indicative content:
- Damage to endothelial lining of artery (due to high blood pressure/toxins).
- Inflammatory response and accumulation of lipids/macrophages/foam cells forming an atheroma/plaque.
- Hardening of plaque (calcium deposits) and narrowing of the lumen.
- Rupture of plaque exposes collagen fibres.
- Platelets adhere to collagen and release thromboplastin.
- Thromboplastin converts prothrombin to thrombin (requires Ca2+).
- Thrombin converts soluble fibrinogen to insoluble fibrin.
- Fibrin mesh traps blood cells to form a clot (thrombus).
- Blockage of coronary artery restricts oxygen supply to cardiac muscle, causing myocardial infarction.

Level 1 (1-2 marks): Explains either the formation of plaque OR the clotting cascade, but lacks detail or logical sequence.
Level 2 (3-4 marks): Explains both plaque formation and blood clotting, showing a clear connection, but may omit some intermediate steps (e.g. role of calcium ions or details of the coronary blockage).
Level 3 (5-6 marks): A detailed and logically structured explanation covering endothelial damage, plaque formation, the full clotting cascade (thromboplastin, prothrombin/thrombin, fibrinogen/fibrin), and how this blocks coronary blood flow to cause myocardial infarction.

1. Change in protein structure: The deletion of three nucleotides results in the loss of a single amino acid (phenylalanine) from the polypeptide chain. This alters the primary structure, which in turn alters the folding and tertiary structure of the CFTR protein.
2. Non-functional channel: Because of the altered shape, the CFTR channel protein is either misfolded and degraded before reaching the cell surface membrane, or it is present but cannot function correctly as a chloride ion channel.
3. Lack of chloride transport: Since the CFTR channel is absent or non-functional, chloride ions (\(\text{Cl}^-\)) cannot be actively transported out of the epithelial cells into the mucus lining the airways.
4. Sodium and water movement: The failure to secrete chloride ions, combined with the unregulated absorption of sodium ions (\(\text{Na}^+\)) into the cells, creates a concentration gradient. Water is drawn out of the mucus and into the epithelial cells by osmosis.
5. Thick, sticky mucus: The loss of water from the mucus makes it highly viscous, thick, and sticky.
6. Effect on lungs: The cilia lining the airways are unable to beat effectively to move this thick mucus upwards. This leads to blocked airways, reduced gas exchange surface area, and trapped bacteria, causing recurring lung infections.

Indicative content:
- Mutation (deletion of 3 bases) leads to loss of one amino acid.
- Alters primary structure and consequently the tertiary structure / 3D shape of the CFTR protein.
- CFTR protein is degraded / does not reach the cell membrane / cannot open.
- Chloride ions (\(\text{Cl}^-\)) cannot be transported out of the cell into the mucus.
- Sodium ions (\(\text{Na}^+\)) move into the cell.
- Water moves out of the mucus into the cells by osmosis down a water potential gradient.
- Mucus becomes dehydrated, viscous, thick and sticky.
- Cilia cannot clear the thick mucus, leading to blocked airways / infections.

Level 1 (1-2 marks): Identifies that the protein structure is altered and that mucus becomes thick, but fails to explain the mechanism of ion and water movement.
Level 2 (3-4 marks): Explains the link between the mutation, the altered CFTR protein, and the lack of chloride transport, but may lack detail on osmosis or the physical effect on cilia.
Level 3 (5-6 marks): Gives a complete, logically structured explanation connecting the genetic mutation to protein structure, the detailed mechanism of ion and water transport (osmosis), and the resulting physical properties of the mucus and its impact on cilia.