2025 Edexcel IAS-Level Biology (XBI11) Practice Paper with Answers

Part (a) Platelets release thromboplastin in response to damage to blood vessel walls. Thromboplastin, along with calcium ions and clotting factors, catalyses the conversion of inactive prothrombin into the active enzyme thrombin. Thrombin then catalyses the conversion of soluble fibrinogen into insoluble fibrin. The fibrin fibres form a mesh that traps blood cells to form a clot. Part (b) Inhibiting the coagulation cascade prevents the production of thrombin, which in turn prevents the conversion of fibrinogen to fibrin. This reduces the formation of blood clots (thrombi) within blood vessels, such as coronary arteries. This maintains blood flow and oxygen supply to cardiac muscle, reducing the risk of myocardial infarction. Part (c) Decrease in events = \(84 - 52 = 32\). Percentage decrease = \((32 / 84) \times 100 = 38.1\%\) (or 38.09%).

Part (a) [4 marks] 1. Platelets release thromboplastin (1 mark) 2. Thromboplastin converts prothrombin to thrombin (1 mark) 3. Calcium ions or clotting factors required (1 mark) 4. Thrombin converts soluble fibrinogen to insoluble fibrin (1 mark) Part (b) [3 marks] 1. Inhibitor prevents thrombin or fibrin formation (1 mark) 2. Reduces risk of blood vessel occlusion or thrombosis (1 mark) 3. Maintains blood flow or oxygen delivery to heart tissue (1 mark) Part (c) [3 marks] 1. Calculation of difference: 84 - 52 = 32 (1 mark) 2. Division by original value: 32 / 84 (1 mark) 3. Correct percentage with rounding: 38.1% or 38.09% (1 mark)

Part (a) Both amylose and amylopectin are polysaccharides composed of alpha-glucose monomers linked by 1,4-glycosidic bonds. However, amylose is unbranched and forms a coiled helical structure, whereas amylopectin is branched and contains both 1,4-glycosidic and 1,6-glycosidic bonds. Part (b) Glycogen is highly branched due to many 1,6-glycosidic bonds, which provides many terminal ends for rapid hydrolysis by enzymes to release glucose for respiration. It is compact, allowing a high amount of energy to be stored in a small volume. It is a large, insoluble molecule, so it does not affect the osmotic potential of the cell and cannot diffuse out across the cell membrane. Part (c) Glucose is a small monomer with multiple polar hydroxyl (-OH) groups that readily form hydrogen bonds with water molecules. Starch is a massive polymer where most hydroxyl groups are involved in internal bonding or are tucked inside the complex 3D helical structure, preventing them from interacting with water.

Part (a) [4 marks] 1. Both contain alpha-glucose or 1,4-glycosidic bonds (1 mark) 2. Amylose is unbranched or helical (1 mark) 3. Amylopectin is branched (1 mark) 4. Amylopectin contains 1,6-glycosidic bonds (1 mark) Part (b) [4 marks] 1. Highly branched or many terminal ends for rapid hydrolysis to release glucose (1 mark) 2. Compact structure stores more glucose or energy in a small space (1 mark) 3. Insoluble so has no osmotic effect or does not cause water to enter cell (1 mark) 4. Large molecule so cannot diffuse out of the cell (1 mark) Part (c) [2 marks] 1. Glucose has many exposed polar hydroxyl groups that form hydrogen bonds with water (1 mark) 2. Starch is a large macromolecule with fewer exposed hydroxyl groups or cannot form hydrogen bonds with water (1 mark)

Part (a) The CFTR protein is a channel protein that actively transports chloride ions out of epithelial cells into the mucus lining the airways. This accumulation of chloride ions lowers the water potential of the mucus. Consequently, water moves out of the cells and into the mucus by osmosis, which dilutes the mucus and keeps it thin and runny. Part (b) The F508del mutation causes the CFTR protein to fold incorrectly. As a result, the misfolded protein is recognized by the cell's quality control system in the endoplasmic reticulum and is degraded rather than being transported to the cell surface membrane. Because there are no CFTR channel proteins in the membrane, chloride ions cannot be transported out of the cells. Water does not move out of the cells by osmosis, leaving the mucus dehydrated, thick, and sticky. Part (c) Facilitated diffusion. Chloride ions are charged or polar and cannot pass through the hydrophobic phospholipid bilayer directly, so they require a transport protein channel to move down their electrochemical gradient.

Part (a) [4 marks] 1. CFTR acts as a channel for chloride ions (1 mark) 2. Chloride ions move out of epithelial cells into mucus (1 mark) 3. This lowers the water potential of mucus (1 mark) 4. Water moves into mucus by osmosis, keeping it thin/runny (1 mark) Part (b) [4 marks] 1. Mutation causes misfolding of CFTR protein (1 mark) 2. Misfolded protein is degraded in the rough endoplasmic reticulum or does not reach membrane (1 mark) 3. No chloride transport out of cell (1 mark) 4. Water does not leave cells, making mucus thick and sticky (1 mark) Part (c) [2 marks] 1. Facilitated diffusion (1 mark) 2. Ions are charged/polar and must pass through a channel protein down a gradient (1 mark)

Part (a) DNA helicase unwinds and unzips the double helix by breaking the hydrogen bonds between complementary base pairs. Both separated strands act as templates. Free DNA nucleotides align with their complementary bases on the template strands via hydrogen bonding (A to T, C to G). DNA polymerase then joins the adjacent nucleotides together by forming phosphodiester bonds, creating a new sugar-phosphate backbone. Part (b) DNA polymerase is involved in DNA replication, whereas RNA polymerase is involved in transcription. DNA polymerase uses DNA nucleotides (containing thymine), while RNA polymerase uses RNA nucleotides (containing uracil). DNA polymerase synthesizes a double-stranded DNA molecule from two templates, whereas RNA polymerase synthesizes a single-stranded RNA molecule from a single template strand. Part (c) Since cytosine is 28%, guanine must also be 28% due to complementary base pairing. Total C + G = 28% + 28% = 56%. Therefore, the remaining bases (A + T) must make up 100% - 56% = 44%. Since adenine pairs with thymine, adenine must be 44% / 2 = 22%.

Part (a) [4 marks] 1. DNA helicase breaks hydrogen bonds or unwinds double helix (1 mark) 2. Both strands act as templates (1 mark) 3. Complementary base pairing of free nucleotides (A-T, C-G) (1 mark) 4. DNA polymerase forms phosphodiester bonds between nucleotides (1 mark) Part (b) [3 marks] 1. DNA polymerase for replication vs RNA polymerase for transcription (1 mark) 2. DNA polymerase incorporates T vs RNA polymerase incorporates U (1 mark) 3. DNA polymerase replicates both strands vs RNA polymerase transcribes one strand (1 mark) Part (c) [3 marks] 1. Identifies C = G, so G = 28% and C+G = 56% (1 mark) 2. Calculates remaining bases A+T = 44% (1 mark) 3. Correctly divides by 2 to get 22% adenine (1 mark)

Part (a) During prophase, chromatin condenses and shortens to form visible chromosomes, which appear as two sister chromatids joined at a centromere. During metaphase, the chromosomes align individually along the equator (metaphase plate) of the cell, and spindle fibres attach to the centromere of each chromosome. Part (b) Total number of dividing cells = \(35 + 15 + 10 + 5 = 65\) cells. Total number of cells observed = 250 cells. Mitotic index = \(65 / 250 = 0.26\) (or 26%). Part (c) Mitosis produces genetically identical cells, which is essential for: 1) growth of tissues by increasing the number of cells; 2) repair of damaged tissues by replacing dead or damaged cells; and 3) asexual reproduction, ensuring that the offspring have the exact same genetic information as the parent.

Part (a) [4 marks] 1. Prophase: Chromosomes condense or become visible (1 mark) 2. Chromosomes consist of two chromatids joined at centromere (1 mark) 3. Metaphase: Chromosomes align along the equator of cell (1 mark) 4. Spindle fibres attach to centromeres (1 mark) Part (b) [3 marks] 1. Sums dividing cells: 35 + 15 + 10 + 5 = 65 (1 mark) 2. Divides by total cells: 65 / 250 (1 mark) 3. Correct final answer of 0.26 or 26% (1 mark) Part (c) [3 marks] 1. Produces genetically identical cells (1 mark) 2. Required for growth (increasing cell number) (1 mark) 3. Required for tissue repair or cell replacement (1 mark)

Part (a) The mammalian sperm cell has a flagellum (tail) that provides motility, enabling it to swim towards the secondary oocyte. The middle piece contains numerous mitochondria to produce ATP via respiration to power the movement of the flagellum. The head contains an acrosome, which is a specialized lysosome containing hydrolytic enzymes to digest the zona pellucida. It also has a haploid nucleus containing 23 chromosomes to restore the diploid number upon fertilization. Part (b) The acrosome reaction is initiated when the sperm contacts the zona pellucida of the egg. The acrosome membrane fuses with the sperm cell surface membrane, releasing hydrolytic enzymes by exocytosis. These enzymes digest the zona pellucida, allowing the sperm to reach and fuse with the egg cell membrane. The fusion of the sperm and egg membranes triggers the cortical reaction. Cortical granules in the egg cytoplasm fuse with the egg cell membrane and release their contents by exocytosis into the space between the membrane and the zona pellucida. These chemicals harden the zona pellucida, creating a fertilization membrane that prevents any further sperm from entering (polyspermy).

Part (a) [4 marks] 1. Flagellum/tail for swimming/motility (1 mark) 2. Mitochondria in middle piece to supply ATP (1 mark) 3. Acrosome containing enzymes to digest zona pellucida (1 mark) 4. Haploid nucleus to restore diploid state (1 mark) Part (b) [6 marks] 1. Acrosome reaction: sperm binds to zona pellucida (1 mark) 2. Acrosomal enzymes released via exocytosis (1 mark) 3. Enzymes digest jelly coat or zona pellucida (1 mark) 4. Cortical reaction: fusion of sperm and egg membranes (1 mark) 5. Cortical granules release contents via exocytosis (1 mark) 6. Zona pellucida hardens or fertilisation membrane forms to prevent polyspermy (1 mark)

Part (a) Both xylem vessels and sclerenchyma fibres consist of dead cells with secondary cell walls heavily thickened with lignin to provide structural support. However, xylem vessels are hollow, continuous tubes with no end walls (open ended) to transport water and mineral ions, whereas sclerenchyma fibres have closed, tapered ends and function solely for support. Part (b) Cellulose is a polymer of beta-glucose where alternate monomers are rotated 180 degrees, forming straight, unbranched chains. Many parallel cellulose chains form hydrogen bonds between adjacent hydroxyl groups, grouping them together into strong microfibrils. These microfibrils are arranged in a criss-cross mesh embedded in a matrix of hemicelluloses and pectins, providing strength in multiple planes. Part (c) Nitrate ions are essential for the synthesis of amino acids, which are the building blocks of proteins, and nucleic acids (DNA/RNA) required for cell division and growth. Magnesium ions are needed to make chlorophyll, which absorbs light for photosynthesis. Without photosynthesis, the plant cannot produce glucose for respiration or cellulose synthesis, resulting in severely stunted growth.

Part (a) [4 marks] 1. Both have lignified or secondary cell walls or both are dead (1 mark) 2. Both provide support (1 mark) 3. Xylem has no end walls or forms continuous tubes, sclerenchyma has closed/tapered ends (1 mark) 4. Xylem transports water/minerals, sclerenchyma only supports (1 mark) Part (b) [3 marks] 1. Beta-glucose chains are straight/unbranched (1 mark) 2. Hydrogen bonds form between adjacent chains to form microfibrils (1 mark) 3. Criss-cross arrangement of microfibrils provides strength in multiple directions (1 mark) Part (c) [3 marks] 1. Nitrate needed for amino acids or proteins or nucleic acids (1 mark) 2. Magnesium needed for chlorophyll synthesis (1 mark) 3. Lack of chlorophyll reduces photosynthesis, leading to less glucose/energy for growth (1 mark)

Part (a) Endemism refers to a species that is native to and restricted to a unique, defined geographical location, such as an island, and is not found naturally anywhere else on Earth. Genetic diversity can be measured by calculating the heterozygosity index (H) using the formula: H = number of heterozygotes / total number of individuals in the population. A higher index indicates greater genetic diversity. Alternatively, researchers can determine the proportion of polymorphic gene loci by dividing the number of polymorphic loci by the total number of loci analyzed. Part (b) First, seeds are harvested from a wide range of individual plants to maximize genetic diversity. They are cleaned and x-rayed to ensure only healthy seeds containing viable embryos are selected. The seeds are then dehydrated in a controlled environment to lower their water content to around 5%. Finally, they are kept in sterile containers at low temperatures (typically -20 degrees Celsius). This drastically reduces metabolic rate and prevents decay by fungi or bacteria. Part (c) Seed banks take up significantly less physical space, allowing thousands of species to be kept in a single facility. They are also much cheaper to maintain than keeping live plants in a botanic garden, and seeds are protected from environmental hazards such as pests, plant diseases, or extreme weather events.

Part (a) [4 marks] 1. Endemism definition: species found only in one specific geographic area (1 mark) 2. Measure by calculating heterozygosity index: H = heterozygotes / total population (1 mark) 3. Alternatively, calculate proportion of polymorphic gene loci (1 mark) 4. Or use gel electrophoresis or DNA sequencing to compare base sequences (1 mark) Part (b) [4 marks] 1. Cleaned or X-rayed to ensure viable embryos (1 mark) 2. Dried to reduce water content or prevent germination (1 mark) 3. Stored at very low temperatures or -20 degrees C (1 mark) 4. Low temperature/dry conditions slow down respiration or prevent microbial decay (1 mark) Part (c) [2 marks] Any two from: 1. Takes up less space (1 mark) 2. Lower maintenance cost or less labor intensive (1 mark) 3. Less vulnerable to disasters or disease or pests (1 mark) 4. Can store larger numbers of individuals to preserve more genetic diversity (1 mark)

(a) 1. Cut the terminal 2 to 5 mm of the garlic root tip. 2. Place the root tip in warm hydrochloric acid to soften the cell walls and break down the middle lamella. 3. Rinse with water and place on a microscope slide. 4. Add a few drops of a stain such as acetic orcein or toluidine blue to stain the chromosomes. 5. Gently squash the root tip under a coverslip using a thumb or paper towel to obtain a single layer of cells. (b) Total dividing cells = \(18 + 12 + 6 + 4 = 40\) cells. Total cells observed = 120 cells. Mitotic index = \(40 / 120 = 0.33\) (or 33.3%). (c) 1. Spindle fibers cannot form or attach to the centromeres of chromosomes. 2. Chromosomes cannot align at the equator during metaphase, and sister chromatids cannot separate during anaphase. 3. This arrests the cell cycle at metaphase and prevents completion of mitosis. 4. Since mitosis cannot be completed, cytokinesis does not occur, stopping cell division.

(a) [Max 4 marks] 1. Treat root tip with hydrochloric acid to soften cell walls (1) 2. Rinse in water (1) 3. Use of a suitable stain, e.g., acetic orcein or toluidine blue (1) 4. Macerate or squash gently under coverslip to obtain a single cell layer (1). (b) [2 marks] 1. Correct calculation of total dividing cells (40) or correct formula (1) 2. Correct final answer of 0.33 / 33.3% / 1/3 (1). (c) [Max 4 marks] 1. No spindle fibers formed to attach to centromeres (1) 2. Chromosomes cannot align at the equator or chromatids cannot separate (1) 3. Cell cycle arrested or stopped at metaphase or anaphase (1) 4. Prevents division of nucleus, so cell division cannot proceed (1).

(a) Both consist of dead cells and both have secondary cell walls thickened with lignin. Both lack living cytoplasm and organelles, leaving a hollow central lumen. (b) 1. Lignified walls provide structural strength to prevent the vessels from collapsing under the negative pressure of tension. 2. The lack of end walls between cells forms a continuous, uninterrupted hollow tube for water flow. 3. The absence of cytoplasm and organelles minimizes resistance to the transport of water. 4. Pits in the lignified walls allow the lateral movement of water to adjacent vessels. (c) 1. Use a micrometer screw gauge or Vernier caliper to measure the diameter of the fibre. 2. Take multiple measurements along the length of the fibre and at different angles to calculate a mean diameter. 3. Use the mathematical formula for the area of a circle, \(\text{Area} = \pi r^2\), where r is the radius. 4. Ensure that the measurements are focused at the thinnest point of the fibre, as this represents the weakest point where the fibre is most likely to break under tension.

(a) [2 marks] Any two from: both contain lignin or lignified walls (1); both are dead cells or lack living contents (1); both have a hollow central lumen (1). (b) [Max 4 marks] 1. Lignified walls resist negative pressure and collapse (1) 2. Open ends or hollow tube allows continuous column of water (1) 3. Lack of cytoplasm reduces resistance to water flow (1) 4. Pits allow lateral flow of water (1). (c) [Max 4 marks] 1. Use micrometer or Vernier calipers (1) 2. Measure diameter in multiple positions and orientations and calculate mean (1) 3. Calculate cross-sectional area using \(\pi r^2\) (where r is half of the mean diameter) (1) 4. Ensure measurement is taken at the point of break or thinnest point to ensure a valid comparison (1).

(a) 1. Damaged endothelial cells and platelets release the enzyme clotting factor called thromboplastin. 2. Thromboplastin, in the presence of calcium ions and vitamin K, catalyzes the conversion of the inactive plasma protein prothrombin into the active enzyme thrombin. 3. Thrombin then acts as an enzyme to catalyze the conversion of the soluble plasma protein fibrinogen into insoluble fibrin fibers. 4. Fibrin fibers form a mesh or network across the damaged area. 5. This mesh traps platelets and red blood cells, forming a solid blood clot. (b) 1. Arteries have a much thicker tunica media containing more muscle fibers and elastic tissue than veins, allowing them to withstand high blood pressure. 2. Arteries have a narrow lumen to maintain high pressure, whereas veins have a much wider lumen to minimize resistance to blood flow. 3. Veins contain semi-lunar valves to prevent the backflow of blood under low pressure, which are absent in arteries. (c) Risk in the drug group = \(150 / 12500 = 0.012\) (or 1.2%). Risk in the placebo group = \(240 / 12500 = 0.0192\) (or 1.92%). Relative Risk Reduction = \(\frac{0.0192 - 0.012}{0.0192} \times 100 = 37.5\%\).

(a) [5 marks] 1. Release of thromboplastin from damaged tissue and platelets (1) 2. Thromboplastin converts prothrombin to thrombin (1) 3. Calcium ions / Vitamin K required for this conversion (1) 4. Thrombin converts soluble fibrinogen to insoluble fibrin (1) 5. Fibrin forms a mesh that traps red blood cells and platelets (1). (b) [Max 3 marks] 1. Artery has thicker muscular and elastic wall / vein has thinner wall (1) 2. Artery has narrower lumen / vein has wider lumen (1) 3. Veins have valves, arteries do not (1) 4. Artery has more collagen/tough outer layer (1). (c) [2 marks] 1. Correct calculation of risk rates: 1.2% and 1.92% (or 0.012 and 0.0192) (1) 2. Correct calculation of RRR as 37.5% (1).

(a) 1. DNA replication involves the enzyme DNA polymerase, whereas transcription involves RNA polymerase. 2. DNA replication copies both strands of the DNA molecule, whereas transcription uses only one template strand of DNA to construct mRNA. 3. DNA replication produces a double-stranded DNA molecule, whereas transcription produces a single-stranded mRNA molecule. 4. DNA replication uses free nucleotides containing deoxyribose sugar and thymine, whereas transcription uses nucleotides containing ribose sugar and uracil. (b) A DNA mononucleotide is made of three key components: a pentose sugar called deoxyribose, a phosphate group, and a nitrogenous base. The phosphate group is bonded to carbon-5 of the deoxyribose sugar, and the nitrogenous base (which can be adenine, thymine, cytosine, or guanine) is bonded to carbon-1. (c) 1. After one round of replication in a medium containing light \(^{14}\text{N}\), the extracted DNA formed a single, intermediate density band. 2. This ruled out the conservative replication theory, which would have produced two separate bands (one heavy band at the bottom and one light band at the top). 3. The single intermediate band proved that each new DNA molecule contained one heavy original strand and one newly synthesized light strand.

(a) [Max 4 marks] 1. DNA replication involves DNA polymerase whereas transcription involves RNA polymerase (1) 2. DNA replication makes a double-stranded DNA molecule whereas transcription makes single-stranded mRNA (1) 3. DNA replication copies both strands whereas transcription copies only one/template strand (1) 4. Replication uses thymine/deoxyribose whereas transcription uses uracil/ribose (1). (b) [3 marks] 1. Deoxyribose sugar, phosphate group, and nitrogenous base (1) 2. Phosphate attached to carbon 5 and base attached to carbon 1 of the sugar (1) 3. Correct naming of at least two bases: adenine, thymine, cytosine, guanine (1). (c) [3 marks] 1. DNA formed a single, intermediate band / hybrid density band (1) 2. This rules out conservative replication (which would show two bands) (1) 3. Because each DNA molecule contains one heavy template strand and one new light strand (1).

(a) Totipotent stem cells can differentiate into any cell type of the body, including extra-embryonic tissues like the placenta, whereas pluripotent stem cells can differentiate into any body cell type but cannot form extra-embryonic tissues. (b) 1. Transcription factors bind to specific promoter regions on DNA. 2. This binding either activates (turns on) or represses (turns off) the transcription of specific genes. 3. Active genes are transcribed to produce specific mRNA molecules, which are then translated into specific proteins. 4. These proteins modify the cell's structure and biochemical pathways permanently, causing it to differentiate. (c) Advantages: 1. No destruction of human embryos is required, avoiding major ethical and moral concerns. 2. iPSCs can be generated from the patient's own adult body cells, reducing the risk of immune rejection during transplantation. Disadvantages/Ethical issues: 1. There is a risk of genetic abnormalities or tumor/cancer formation as the genes used to induce pluripotency can act as oncogenes. 2. Long-term safety and efficiency of reprogramming somatic cells is not fully understood, posing unpredictable health risks.

(a) [2 marks] 1. Totipotent can differentiate into any cell type AND extra-embryonic tissues/placenta (1) 2. Pluripotent can differentiate into any body cell type but NOT extra-embryonic tissues (1). (b) [Max 4 marks] 1. Transcription factors bind to promoter regions of genes (1) 2. Stimulate or prevent transcription / RNA polymerase binding (1) 3. Specific mRNA is produced only from active genes (1) 4. Translation of mRNA produces specific proteins (1) 5. These proteins determine the cell's structure / function (1). (c) [Max 4 marks] Max 4 marks total, allowing up to 3 marks for advantages or disadvantages. Must include at least one of each: Advantages: 1. No embryo destroyed / fewer ethical objections (1) 2. Patient's own cells used so no immune rejection (1). Disadvantages/Ethical: 3. Risk of cancer / tumor formation / oncogene activation (1) 4. Safety/side effects not fully known (1).

(a) Biodiversity is the variety of living organisms in an area, which includes species richness (the number of different species present) and species evenness (the relative abundance of each species). (b) Total population size \(N = 15 + 25 + 10 = 50\). For Species A: \(n / N = 15 / 50 = 0.3\), giving \((n / N)^2 = 0.09\). For Species B: \(n / N = 25 / 50 = 0.5\), giving \((n / N)^2 = 0.25\). For Species C: \(n / N = 10 / 50 = 0.2\), giving \((n / N)^2 = 0.04\). Sum of \((n / N)^2 = 0.09 + 0.25 + 0.04 = 0.38\). Thus, Simpson's Index of Diversity \(D = 1 - 0.38 = 0.62\). (c) 1. Seed banks store a wide variety of seeds under cool, dry conditions to maintain viability and prevent germination or decay. 2. This preserves the genetic diversity (gene pool) of endangered plant species. 3. Seeds take up much less space than growing adult plants, allowing many species to be stored at a lower cost. 4. Stored seeds can be germinated to reintroduce plant species back into the wild or for scientific research and education.

(a) [2 marks] 1. Variety / range of living organisms / species in a habitat (1) 2. Reference to both species richness (number of species) and genetic diversity / species evenness (1). (b) [4 marks] 1. Calculate N = 50 (1) 2. Calculate n/N for each species: 0.3, 0.5, 0.2 (1) 3. Calculate sum of (n/N)^2 = 0.09 + 0.25 + 0.04 = 0.38 (1) 4. Calculate D = 1 - 0.38 = 0.62 (1). (c) [Max 4 marks] 1. Seeds stored in cold, dry conditions to prevent germination / decay / fungal growth (1) 2. Conserves genetic diversity / large sample size stored (1) 3. Less space / lower cost than conserving adult plants (1) 4. Viability tested regularly (by germinating samples) (1) 5. Can be used to reintroduce plants to wild / restore habitats (1).

(a) 1. Glycogen is a polymer of \(\alpha\)-glucose, which is easily hydrolyzed to release glucose for respiration. 2. It has many 1,6-glycosidic bonds, making it highly branched. This provides many terminal ends for rapid enzyme hydrolysis. 3. It is large and insoluble in water, so it does not affect the water potential or osmotic balance of the cell. 4. It is compact, allowing a large amount of energy to be stored in a small space. (b) Similarities: 1. Both contain a glycerol backbone. 2. Both contain fatty acid chains attached by ester bonds. 3. Both contain carbon, hydrogen, and oxygen. Differences: 1. Triglycerides have three fatty acids, whereas phospholipids have two fatty acids and a phosphate group. 2. Triglycerides are completely hydrophobic/non-polar, whereas phospholipids have a hydrophilic/polar head and hydrophobic/non-polar tails. (c) 1. High solvent power (polar nature allows it to dissolve ionic and polar substances like glucose and salts). 2. High specific heat capacity (prevents rapid temperature changes, maintaining stable conditions for transport).

(a) [Max 4 marks] 1. Highly branched / many 1,6-glycosidic bonds allows rapid hydrolysis (1) 2. Insoluble so has no osmotic effect / does not affect water potential of cell (1) 3. Compact molecule so large amount can be stored in small volume (1) 4. Made of \(\alpha\)-glucose monomers which are easily released for respiration (1). (b) [Max 4 marks] Max 4 marks (must have at least one similarity and one difference): Similarities: 1. Both contain glycerol (and fatty acids) (1) 2. Both contain ester bonds (1) 3. Both contain C, H, O (1). Differences: 4. Triglyceride has 3 fatty acids whereas phospholipid has 2 fatty acids + phosphate group (1) 5. Triglyceride is hydrophobic/non-polar whereas phospholipid has hydrophilic/polar head and hydrophobic tails (1). (c) [2 marks] Any two from: 1. Liquid at room temperature / high cohesive forces allow bulk flow (1) 2. Excellent solvent for polar/ionic substances (1) 3. High specific heat capacity (minimizes temperature fluctuations) (1).

(a) 1. The membrane consists of a double layer (bilayer) of phospholipid molecules. 2. Phospholipids are 'fluid' because they can move laterally relative to one another. 3. Proteins are embedded within the bilayer in a random, patchy arrangement resembling a 'mosaic'. 4. These include intrinsic (transmembrane) proteins and extrinsic (peripheral) proteins, along with glycoproteins, glycolipids, and cholesterol. (b) 1. A mutation in the CFTR gene leads to a non-functional or absent CFTR channel protein in the cell membrane. 2. Chloride ions cannot be actively transported out of the epithelial cells into the mucus. 3. Consequently, sodium ions do not move out of the cells, and water is not drawn into the mucus by osmosis. 4. The lack of water causes the mucus to become dehydrated, thick, and sticky. (c) 1. Active transport requires ATP/energy, whereas facilitated diffusion is passive and does not require ATP. 2. Active transport moves substances against their concentration gradient (from low to high concentration), whereas facilitated diffusion moves substances down their concentration gradient (from high to low concentration).

(a) [Max 4 marks] 1. Phospholipid bilayer structure with hydrophobic tails inside and hydrophilic heads outside (1) 2. Fluidity allows lateral movement of phospholipids/proteins (1) 3. Proteins scattered/embedded like a mosaic (1) 4. Details of components: glycolipids, glycoproteins, cholesterol, channel/carrier proteins (1). (b) [Max 4 marks] 1. CFTR channel protein is absent / non-functional (1) 2. Chloride ions cannot leave epithelial cells (into mucus) (1) 3. Sodium ions move into cells / water moves out of mucus into cells (by osmosis) (1) 4. Mucus becomes dehydrated / viscous / sticky (1). (c) [2 marks] 1. Active transport requires ATP / energy, facilitated diffusion does not (1) 2. Active transport is against concentration gradient, facilitated diffusion is down/with concentration gradient (1).