2023 Edexcel IAS-Level Biology (XBI11) Practice Paper with Answers

Lactose is a disaccharide composed of the monosaccharides glucose and galactose joined by a glycosidic bond. The reaction that breaks down a disaccharide into its constituent monosaccharide monomers by the addition of water is a hydrolysis reaction.

1 mark for the correct answer B (glucose and galactose; hydrolysis reaction).

Arteries have a narrower lumen than veins to help maintain high blood pressure. They have a thicker wall with a larger amount of elastic fibres and smooth muscle to withstand and recoil to maintain high blood pressure, and they do not have valves along their length (except for the semilunar valves at the exit of the heart).

1 mark for the correct answer B.

When a blood vessel wall is damaged, platelets release the clotting factor thromboplastin. In the presence of calcium ions, thromboplastin catalyses the conversion of 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.

1 mark for the correct answer A.

The leveling off of the rate of uptake at higher concentrations indicates that transport is mediated by membrane proteins (carriers or channels) which have become saturated. Because the metabolic inhibitor (which blocks ATP synthesis) does not affect the rate of uptake, the process must be passive and does not require ATP. Therefore, the mechanism of transport is facilitated diffusion.

1 mark for the correct answer C.

Triglycerides are formed by a condensation reaction between one glycerol molecule and three fatty acid molecules. Each fatty acid binds to glycerol via an ester bond, releasing a total of three molecules of water. They are non-polar hydrophobic molecules that are insoluble in water, and saturated fatty acids contain no carbon-to-carbon double bonds.

1 mark for the correct answer A.

Thrombin is an active protease enzyme that plays a critical role in the coagulation cascade. It is converted from its inactive precursor prothrombin by thromboplastin and calcium ions. Once active, thrombin catalyses the conversion of the soluble plasma protein fibrinogen into insoluble fibrin fibers. These fibrin fibers polymerise and form a structural mesh that traps platelets and red blood cells, sealing the damaged blood vessel and forming a stable blood clot.

1. Reference to thrombin acting as an enzyme or catalysing a reaction (1 mark); 2. Catalyses the conversion of soluble fibrinogen into insoluble fibrin (1 mark); 3. Reference to fibrin forming a mesh or network that traps platelets or red blood cells to form a clot (1 mark).

Lactose is a disaccharide carbohydrate. It is formed from a condensation reaction between two hexose monosaccharides: glucose and galactose. During this reaction, a molecule of water is eliminated, and a covalent bond called a 1,4-glycosidic bond is established between carbon-1 of the galactose molecule and carbon-4 of the glucose molecule.

1. Identifies the constituent monosaccharides as glucose and galactose (1 mark); 2. States that the monosaccharides are joined by a 1,4-glycosidic bond (1 mark); 3. Identifies the reaction type as a condensation reaction / involves the removal of a water molecule (1 mark).

Carrier proteins span the cell membrane and possess highly specific binding sites that are complementary to a particular molecule or ion. Unlike simple diffusion, which is passive and occurs down a concentration gradient through the lipid bilayer, active transport requires energy. The binding of ATP to the carrier protein causes it to change its tertiary shape (conformation), thereby carrying the solute across the membrane against its concentration gradient.

1. Reference to carrier protein having a specific binding site for a molecule or ion (1 mark); 2. ATP is required to provide energy for a conformational / shape change in the protein (1 mark); 3. This allows transport against the concentration gradient, whereas simple diffusion is passive / down a concentration gradient (1 mark).

The artery wall has a thick layer of elastic fibres (elastin) in the tunica media. These fibres stretch when high-pressure blood is pumped out of the heart during systole, and then recoil during diastole, which maintains a high and continuous blood pressure. Additionally, the smooth muscle layer can contract to constrict the lumen and increase pressure, while a thick layer of collagen provides structural strength to prevent the artery from bursting under high pressure.

1. Elastic fibres stretch and recoil to maintain blood pressure or smooth out blood flow (1 mark); 2. Smooth muscle contracts to constrict the lumen, increasing resistance and pressure (1 mark); 3. Collagen fibres provide high tensile strength to withstand high blood pressure and prevent bursting (1 mark).

High levels of circulating Low-Density Lipoproteins (LDLs) transport cholesterol in the blood. If there is damage to the endothelial lining of an artery (e.g., due to high blood pressure), LDL cholesterol accumulates within the artery wall. These LDLs undergo oxidation, triggering an inflammatory response where white blood cells (macrophages) engulf the lipids to become foam cells. The accumulation of these foam cells and cholesterol leads to the formation of a fatty deposit or plaque (atheroma) beneath the endothelium.

1. High LDL level leads to accumulation of cholesterol/LDL in the damaged endothelial lining of the artery (1 mark); 2. LDL is oxidized and engulfed by macrophages/white blood cells to form foam cells (1 mark); 3. Accumulation of foam cells and lipid deposits forms an atheroma / fatty plaque (1 mark).

The primary structure of an enzyme is the specific linear sequence of amino acids in its polypeptide chain. This sequence determines the types and positions of the R-groups. If the primary structure is altered, different chemical bonds (such as hydrogen, ionic, or disulfide bonds) will form, leading to a change in the tertiary folding of the protein. Consequently, the specific 3D shape of the active site is altered, meaning the substrate can no longer fit or bind to it, preventing the formation of enzyme-substrate complexes.

1. Change in primary structure / amino acid sequence changes the position/type of R-groups (1 mark); 2. This alters the tertiary structure / 3D folding due to different hydrogen/ionic/disulfide bonds forming (1 mark); 3. This deforms the active site, so it is no longer complementary to the substrate / enzyme-substrate complexes cannot form (1 mark).

Fick's Law states that the rate of diffusion is proportional to (surface area x difference in concentration) / thickness of exchange surface. The mammalian lung is highly adapted to maximize this rate: 1) It has millions of alveoli which provide a very large surface area. 2) The alveolar and capillary walls are both only one cell thick, consisting of flat squamous epithelial cells, which minimizes the thickness of the diffusion pathway. 3) Ventilation (breathing) and continuous capillary blood flow maintain a steep concentration gradient for oxygen and carbon dioxide.

1. Large surface area provided by many/millions of alveoli (1 mark); 2. Extremely short diffusion distance because alveolar and capillary walls are only one cell thick (1 mark); 3. Steep concentration gradient maintained by continuous ventilation and/or blood flow (1 mark).

During semi-conservative DNA replication, DNA polymerase plays a crucial role. First, it aligns free activated mononucleotides with their complementary bases on the exposed single-stranded DNA template (A with T, C with G). Second, it catalyses condensation reactions to form covalent phosphodiester bonds between the 5' phosphate group of an incoming nucleotide and the 3' hydroxyl group of the existing nucleotide chain, thus synthesizing the new complementary DNA strand in a 5' to 3' direction.

1. Aligns free/activated nucleotides with complementary bases on the template strand (1 mark); 2. Catalyses the formation of phosphodiester bonds between adjacent nucleotides (1 mark); 3. Synthesizes the new DNA strand / sugar-phosphate backbone in a 5' to 3' direction (1 mark).

1. Glycogen is composed of \(\alpha\)-glucose monomers linked by 1,4-glycosidic bonds, forming a compact shape that allows large amounts of energy to be stored in a small volume. 2. It contains frequent 1,6-glycosidic bonds which create a highly branched structure. This provides many terminal ends, allowing rapid hydrolysis by glycogen phosphorylase to release glucose when energy demand is high. 3. It is a large, insoluble molecule, meaning it does not dissolve in the cytoplasm and has no osmotic effect on the cell (preventing water from entering by osmosis).

1. Identifies glycogen as a polymer of \(\alpha\)-glucose with 1,4-glycosidic bonds making it compact (1 mark).
2. Explains that the highly branched structure (due to 1,6-glycosidic bonds) provides many terminal ends for rapid hydrolysis to release glucose (1 mark).
3. Explains that glycogen is insoluble, meaning it has no osmotic effect on the cell (1 mark).

1. High dietary salt intake increases the concentration of solutes in the blood, causing more water to be retained in the circulatory system by osmosis. This increases blood volume and leads to high blood pressure (hypertension). 2. Persistent high blood pressure exerts physical stress on the arterial walls, causing microscopic inflammatory damage to the delicate endothelial lining. 3. This endothelial damage initiates an inflammatory response, leading to the accumulation of white blood cells, lipids, and cholesterol to form an atheroma (plaque), which narrows the lumen and restricts blood flow.

1. Explains that a high-salt diet increases blood volume, leading to high blood pressure / hypertension (1 mark).
2. Mentions that high blood pressure causes physical / inflammatory damage to the endothelium / endothelial lining of arteries (1 mark).
3. Links endothelial damage to subsequent plaque / atheroma formation, narrowing the arterial lumen (1 mark).

1. The muscular wall of the left ventricle is much thicker and contains significantly more cardiac muscle tissue compared to the thin wall of the left atrium. 2. When the ventricle contracts (ventricular systole), it generates a much more powerful contraction, resulting in a higher peak pressure. 3. This high pressure is necessary to overcome the high resistance of the systemic circulation and pump blood all around the body, whereas the left atrium only needs to generate enough pressure to pump blood a short distance into the left ventricle.

1. States that the left ventricle has a much thicker muscular wall / more cardiac muscle than the left atrium (1 mark).
2. Explains that ventricular contraction is much more forceful / generates higher pressure than atrial contraction (1 mark).
3. Explains that the left ventricle must pump blood against high resistance all around the body (systemic circulation), whereas the left atrium only pumps blood a short distance into the ventricle (1 mark).

1. During the clotting cascade, the release of thromboplastin triggers the conversion of inactive prothrombin into the active enzyme thrombin. 2. Thrombin then acts as a protease, catalysing the conversion of the soluble plasma protein fibrinogen into insoluble fibrin fibres. 3. These insoluble fibrin fibres polymerise to form a sticky meshwork over the damaged area, trapping red blood cells and platelets to form a stable blood clot.

1. Identifies thrombin as an active enzyme / protease (1 mark).
2. Explains that thrombin catalyses the conversion of the soluble protein fibrinogen into insoluble fibrin (1 mark).
3. Explains that insoluble fibrin forms a mesh / network that traps red blood cells / platelets to form a clot (1 mark).

1. Similarity: Both processes require specific transmembrane proteins (carrier proteins) to transport large, polar, or charged molecules across the selectively permeable membrane. 2. Difference: Facilitated diffusion is a passive process that does not require metabolic energy (ATP) and moves substances down their concentration gradient (from high to low concentration). 3. Difference: Active transport requires active metabolic energy in the form of ATP to pump substances against their concentration gradient (from low to high concentration), and only utilises carrier proteins, whereas facilitated diffusion can use both channel and carrier proteins.

1. Similarity: Both require specific membrane proteins / transport proteins to move substances across the membrane (1 mark).
2. Difference: Facilitated diffusion is passive / does not require ATP, whereas active transport requires energy in the form of ATP (1 mark).
3. Difference: Facilitated diffusion moves substances down a concentration gradient, whereas active transport moves substances against a concentration gradient (1 mark).
[Accept as alternative difference: Facilitated diffusion can use channel or carrier proteins, whereas active transport only uses carrier proteins.]

1. A gene mutation alters the sequence of DNA bases. This changes the mRNA codon sequence during transcription, which alters the specific sequence of amino acids (primary structure) translationally. 2. A change in the primary structure changes the positions of the R-groups, meaning different ionic, hydrogen, and disulfide bonds form during folding. This alters the overall 3D tertiary structure of the protein. 3. Consequently, the shape of the active site is changed, meaning the substrate is no longer complementary in shape, cannot bind to the active site, and no enzyme-substrate complexes can form.

1. Explains that a mutation changes the DNA base sequence, altering the sequence of amino acids / primary structure of the protein (1 mark).
2. Explains that this alters the positions of R-groups, changing the bonds (e.g. hydrogen, ionic, disulfide) and changing the 3D tertiary structure (1 mark).
3. Explains that this changes the shape of the active site, so the substrate is no longer complementary / cannot bind to form enzyme-substrate complexes (1 mark).

1. During semi-conservative DNA replication, DNA polymerase matches free activated mononucleotides to their complementary bases on the exposed template strand (A with T, and C with G). 2. It catalyses condensation reactions that form covalent phosphodiester bonds between the sugar and phosphate groups of adjacent nucleotides. 3. This builds the new sugar-phosphate backbone, synthesising the complementary strand in a strict \(5'\) to \(3'\) direction.

1. States that DNA polymerase aligns free nucleotides along the template strand via complementary base pairing (1 mark).
2. Explains that it catalyses the formation of phosphodiester bonds / condensation reactions between adjacent nucleotides to form the sugar-phosphate backbone (1 mark).
3. Mentions that synthesis occurs specifically in a \(5'\) to \(3'\) direction (1 mark).

1. A mutation in the CFTR gene results in a non-functional or absent CFTR membrane protein, preventing the active transport of chloride ions out of the epithelial cells into the mucus. 2. As chloride ions remain inside the cells, water does not move out of the cells by osmosis to dilute the mucus, leaving the mucus abnormally thick and sticky. 3. This thick mucus blocks the narrow bronchioles and covers the alveoli. This reduces ventilation (air flow), decreases the surface area available for gas exchange, and increases the diffusion distance for oxygen and carbon dioxide.

1. Explains that a mutation in the CFTR gene leads to non-functional / absent chloride channels, so chloride ions cannot be transported out of epithelial cells (1 mark).
2. Explains that water does not move out of the cells by osmosis, resulting in the production of thick, sticky mucus (1 mark).
3. Explains that this mucus blocks airways / covers alveoli, reducing ventilation, increasing diffusion distance, and reducing the surface area available for gas exchange (1 mark).

During the blood clotting cascade, prothrombin is converted into the active enzyme thrombin. Thrombin then acts as a catalyst to convert the soluble plasma protein fibrinogen into insoluble fibrin fibers. These fibrin fibers entangle platelets and red blood cells, forming a meshwork that seals the damaged blood vessel and prevents further blood loss and entry of pathogens.

1. Thrombin catalyses the conversion of soluble fibrinogen into insoluble fibrin (1 mark). 2. Fibrin forms a mesh/network of fibers (1 mark). 3. This mesh traps platelets and red blood cells to form a clot (1 mark).

Amylopectin is a polysaccharide made of alpha-glucose molecules. It contains 1,4-glycosidic bonds and frequent 1,6-glycosidic bonds, which create a highly branched structure. This branching provides many terminal ends, allowing rapid hydrolysis by enzymes to release glucose when energy is needed. Furthermore, because it is a very large molecule, it is insoluble in water, ensuring it does not alter the water potential of the plant cell, while its compact shape allows efficient storage.

1. Branched structure (due to 1,6-glycosidic bonds) allows rapid hydrolysis or release of glucose (1 mark). 2. Large and insoluble, so it does not affect the water potential or osmotic balance of the cell (1 mark). 3. Compact structure allows a high density of glucose or energy to be stored in a small space (1 mark).

Saturated fatty acids have hydrocarbon chains with only single covalent bonds between carbon atoms (\(\text{C}-\text{C}\)), allowing the chains to be straight. Unsaturated fatty acids contain at least one double bond (\(\text{C}=\text{C}\)), which introduces a kink in the hydrocarbon chain. This kink prevents the unsaturated fatty acid molecules from packing closely together. Consequently, the intermolecular forces (van der Waals forces) between unsaturated molecules are weaker, requiring less thermal energy to break, resulting in a lower melting point (often liquid at room temperature) compared to saturated fatty acids (often solid).

1. Saturated fatty acids have only single \(\text{C}-\text{C}\) bonds, whereas unsaturated fatty acids contain at least one \(\text{C}=\text{C}\) double bond (1 mark). 2. Double bonds cause kinks in the chain, preventing close packing of molecules (1 mark). 3. This reduces intermolecular forces, resulting in a lower melting point for unsaturated fatty acids (1 mark).

The left ventricle has a much thicker muscular wall than the left atrium. When it contracts, it generates high pressure to force blood out of the heart, through the aorta, and around the entire body (systemic circulation), which has high resistance. The left atrium only needs to pump blood a very short distance into the left ventricle, requiring much less force. Additionally, the high pressure in the ventricle relative to the atrium forces the atrioventricular (bicuspid) valve to close tightly, preventing backflow of blood into the atrium.

1. High ventricular pressure is needed to pump blood against high resistance through the systemic circulation or to the rest of the body (1 mark). 2. Left atrium only pumps blood a short distance into the ventricle, requiring low pressure (1 mark). 3. The pressure gradient forces the atrioventricular valve to close to prevent backflow (1 mark).

Water is a polar molecule due to the unequal sharing of electrons between oxygen and hydrogen. Oxygen is more electronegative, gaining a slight negative charge (\(\delta^-\)), while the hydrogen atoms have a slight positive charge (\(\delta^+\)). When ionic compounds (like inorganic salts) enter water, the negative oxygen ends of water molecules are attracted to positive ions (cations), and the positive hydrogen ends are attracted to negative ions (anions). This causes water molecules to cluster around the ions, separating and dissolving them, which allows them to be transported in the blood or cytoplasm.

1. Water is dipole/polar, with \(\delta^-\)\ oxygen and \(\delta^+\)\ hydrogen atoms (1 mark). 2. Oppositely charged parts of water molecules attract and surround positive/negative ions (1 mark). 3. This separates/dissolves the ions, allowing them to be transported in aqueous media (1 mark).

A diet rich in saturated fats increases the concentration of low-density lipoproteins (LDLs) in the blood, which transport cholesterol to arteries, leading to lipid deposition and plaque (atheroma) formation. Meanwhile, high salt intake increases the osmotic pressure of blood, leading to increased water retention and high blood pressure (hypertension). Hypertension puts physical stress on the arterial endothelium, causing damage that triggers an inflammatory response. This inflammation promotes further atheroma development, narrowing the lumen of coronary arteries and increasing the risk of myocardial infarction or stroke.

1. High saturated fat increases LDL cholesterol, leading to atheroma or plaque formation in arteries (1 mark). 2. High salt intake increases blood pressure or hypertension (1 mark). 3. High blood pressure damages endothelial lining, accelerating inflammatory response and plaque formation (1 mark).

To compare the vitamin C concentration, use a titration method with the redox indicator DCPIP (dichlorophenolindophenol). 1. Standardisation: Pipette a fixed volume (e.g., 1.0 cm³) of 1% DCPIP solution into a test tube. Fill a syringe or micro-burette with a standard 0.1% vitamin C solution. Add the standard solution drop by drop to the DCPIP, shaking gently, until the blue colour of the DCPIP just disappears (decolourises). Record the volume of standard solution used. 2. Testing the juices: Repeat this exact process using fresh orange juice, and then with the boiled orange juice (which must be cooled to room temperature first). Record the volume of each juice required to decolourise the DCPIP. 3. Variables to control for validity: Keep the volume and concentration of DCPIP constant. Ensure both juices are at the same temperature when tested. Use orange juice from the same carton/source to control for initial concentration. 4. Reliability: Repeat the titration at least three times for each type of juice to obtain concordant results (within 0.1 cm³ of each other) and calculate a mean volume. 5. Analysis: Calculate the concentration of vitamin C in each sample using the relationship: Concentration of juice = (Volume of standard vitamin C / Volume of juice) * Concentration of standard.

Level 1 (1-2 marks): Describes a simple titration method where juice is added to DCPIP. Mentions a colour change but lacks detail on precise measurement or control of variables. Level 2 (3-4 marks): Explains how to determine the endpoint (blue to colourless). Identifies at least two key control variables (e.g., same volume of DCPIP, same temperature of juices). Mentions the need to repeat the experiment to calculate a mean. Level 3 (5-6 marks): Provides a fully valid and reliable method. Includes a standardisation step using a known concentration of vitamin C to allow quantitative calculation of concentrations. Explicitly describes how to ensure validity (e.g., cooling boiled juice, same batch of juice, identical mixing technique) and reliability (at least three repeats to identify anomalies and calculate a mean).

The nucleolus is a dense structure located within the eukaryotic nucleus. It is the site of ribosomal RNA (rRNA) transcription and the assembly of ribosomal subunits from rRNA and proteins.

Award 1 mark for the correct option B. Reject options A, C, and D as they do not specifically perform the synthesis and assembly of ribosomal subunits.

Xylem vessels form continuous, hollow tubes because their end walls break down entirely during maturation to allow efficient water transport. In contrast, sclerenchyma fibres have closed, tapered ends and function solely for mechanical support rather than transport.

Award 1 mark for the correct option C. Reject A, B, and D because both cell types are dead at maturity, undergo secondary thickening, and contain lignin in their cell walls.

To calculate the mitotic index, divide the total number of cells in mitosis by the total number of cells counted. Cells in mitosis = \(15 + 8 + 5 + 12 = 40\). Mitotic index = \(40 / 250 = 0.160\).

Award 1 mark for the correct option B. Option A is incorrect as it divides only telophase cells by total. Options C and D are incorrect calculations.

The heterozygosity index is calculated as the number of heterozygotes divided by the total population size: \(120 / 500 = 0.24\).

Award 1 mark for the correct option B. Reject A, C, and D which represent incorrect calculations of the heterozygosity index.

Pluripotent stem cells can differentiate into almost any cell type representing the three embryonic germ layers, but they cannot differentiate into extraembryonic tissues, such as the placenta.

Award 1 mark for the correct option B. Option A describes totipotent cells. Option C describes multipotent cells. Option D is incorrect because stem cells are undifferentiated.

Magnesium ions (\(Mg^{2+}\)) are a core component of chlorophyll. A deficiency prevents chlorophyll synthesis, leading to chlorosis, starting in the older leaves because magnesium is mobile and is remobilized to younger leaves.

Award 1 mark for the correct option B. Calcium deficiency typically affects growing tips first. Nitrate deficiency leads to overall stunting and pale green color. Sodium is not an essential major nutrient.

The acrosome reaction begins when the sperm contacts the zona pellucida of the oocyte. This stimulates an influx of calcium ions into the sperm head, triggering fusion of the outer acrosomal membrane with the sperm's cell surface membrane, leading to exocytosis of hydrolytic enzymes.

Award 1 mark for the correct option A. Option B describes the cortical reaction. Options C and D describe incorrect order of physiological processes.

The acrosome reaction is initiated when receptors on the sperm head bind to glycoproteins in the zona pellucida of the oocyte. This binding triggers the fusion of the acrosomal membrane with the sperm cell surface membrane, releasing hydrolytic enzymes (such as acrosin) via exocytosis into the surrounding area. These enzymes digest a pathway through the glycoprotein matrix of the zona pellucida, enabling the sperm cell to move forward and fuse with the oocyte cell membrane.

1. Reference to sperm receptors binding to glycoproteins on the oocyte's zona pellucida. 2. Fusion of the acrosomal membrane with the sperm cell surface membrane to release hydrolytic enzymes (e.g. acrosin) via exocytosis. 3. Digestion of the glycoprotein matrix / zona pellucida to allow the sperm to reach and fuse with the oocyte membrane.

Independent assortment occurs during metaphase I of meiosis, where homologous chromosomes align randomly along the equator of the spindle. The alignment and orientation of one homologous pair is entirely independent of any other pair. When the homologous chromosomes are separated in anaphase I, they are distributed into daughter cells in random combinations. This results in \(2^n\) possible maternal and paternal chromosome combinations in the gametes (where n is the haploid number).

1. Homologous chromosome pairs align randomly at the spindle equator during metaphase I. 2. The alignment / orientation of one homologous pair is independent of all other homologous pairs. 3. Separation of these chromosomes in anaphase I distributes maternal and paternal chromosomes randomly into gametes, creating new allele combinations.

The transport of water through xylem vessels occurs under high tension due to transpiration. To prevent the vessels from collapsing inward under this negative pressure, the cell walls of xylem are heavily thickened and reinforced with a tough, waterproof polymer called lignin. Additionally, the cells are dead and completely empty, lacking cytoplasm and end walls to form hollow, continuous tubes, which minimizes resistance to the upward flow of the water column. Unlignified areas called pits allow water to move laterally between adjacent vessels.

1. Cell walls are thickened / reinforced with lignin to resist tension and prevent the vessels collapsing inward. 2. Vessels are hollow tubes with no cytoplasm or end walls to allow a continuous, uninterrupted column of water. 3. Pits are present in the walls to allow lateral movement of water between adjacent vessels.

Embryonic stem cells are pluripotent, meaning they retain the ability to differentiate into almost any cell type of the adult organism (all specialized body cells derived from the three germ layers, excluding extra-embryonic tissues). In contrast, tissue (adult) stem cells are multipotent, meaning they are partially specialized and can only differentiate into a limited, specific range of cell types related to their tissue of origin (for example, bone marrow stem cells can only form blood cells). This is because embryonic stem cells have more of their genome active and fewer genes permanently silenced than tissue stem cells.

1. Embryonic stem cells are pluripotent and can differentiate into almost any specialized cell type in the organism. 2. Tissue (adult) stem cells are multipotent and can only differentiate into a limited range of cell types associated with that tissue. 3. Embryonic stem cells have more active genes / fewer permanently silenced genes than tissue stem cells.

Cellulose is a polysaccharide composed of long, straight, unbranched chains of \(\beta\)-glucose monomers linked by \(\beta\)-1,4-glycosidic bonds. To allow these bonds to form, alternate \(\beta\)-glucose molecules must be inverted by 180 degrees. Many of these parallel straight chains form hydrogen bonds between the hydroxyl groups of adjacent chains. This extensive cross-linking groups the chains into tough bundles called microfibrils, which possess high tensile strength to prevent the plant cell from bursting under high turgor pressure.

1. Consists of \(\beta\)-glucose monomers joined by \(\beta\)-1,4-glycosidic bonds, where alternate glucose molecules are rotated 180 degrees to form straight, unbranched chains. 2. Many hydrogen bonds form between parallel adjacent cellulose chains. 3. This cross-linking groups the chains together into microfibrils, providing high tensile strength to resist turgor pressure.

The synthesis of digestive enzymes (which are proteins) begins on ribosomes bound to the rough endoplasmic reticulum (rER). The synthesized polypeptide chain enters the lumen of the rER, where it is folded into its specific 3D tertiary structure. The folded proteins are then packaged into transport vesicles that bud off the rER and fuse with the Golgi apparatus. Inside the Golgi apparatus, the proteins are chemically modified (such as by adding carbohydrate chains to form glycoproteins) and sorted. Finally, they are packaged into secretory vesicles that move to and fuse with the cell surface membrane, releasing the enzymes by exocytosis.

1. Ribosomes on the rER synthesize proteins which enter the rER lumen to be folded into their tertiary structure. 2. Transport vesicles transport the folded proteins from the rER to the Golgi apparatus. 3. Golgi apparatus modifies proteins (e.g., adding carbohydrate groups) and packages them into secretory vesicles for release via exocytosis.

To preserve plant genetic diversity for long periods, seeds are dried to remove moisture and stored at low temperatures (typically around -20 degrees Celsius). These conditions significantly reduce the rate of enzyme activity and respiration within the seed. This slows down metabolic processes, prevents premature germination, and conserves the seed's stored nutrient reserves, thereby extending its viability and lifespan. Furthermore, the lack of moisture and low temperatures prevent the growth of fungal and bacterial decomposers, which would otherwise rot the seeds.

1. Low moisture and low temperature reduce the rate of enzyme activity and respiration in seeds. 2. This prevents germination and conserves stored nutrient reserves to extend seed viability. 3. These conditions inhibit the growth of decomposers (bacteria and fungi) that could cause decay.

DNA methylation involves the covalent addition of methyl groups \(-\text{CH}_3\) to cytosine bases within CpG islands, which are typically located in the promoter regions of genes. The physical presence of these methyl groups prevents the binding of transcription factors and RNA polymerase to the promoter, preventing the initiation of transcription. Additionally, DNA methylation can recruit proteins such as methyl-CpG-binding domain proteins, which recruit histone deacetylases. This causes chromatin to condense tightly into heterochromatin, further preventing transcription machinery from accessing the gene.

1. Methyl groups are added to cytosine bases at CpG islands in the promoter region of the gene. 2. This prevents the binding of transcription factors / RNA polymerase to the promoter. 3. It triggers chromatin condensation into tightly packed heterochromatin, making the gene inaccessible for transcription.

1. Proteins synthesized by the rER are transported to the Golgi apparatus where they are chemically modified (e.g., folded, or carbohydrate chains are added to form glycoproteins).
2. The modified proteins are then sorted and packaged into secretory vesicles.
3. These vesicles move towards and fuse with the cell surface membrane, releasing the enzyme (amylase) out of the cell via exocytosis.

- Mark 1: Reference to the chemical modification of proteins / folding / addition of carbohydrate chains (to form glycoproteins) (1)
- Mark 2: Reference to packaging the modified proteins into (secretory) vesicles (1)
- Mark 3: Reference to transport of vesicles to, and fusion with, the cell surface membrane / secretion via exocytosis (1)
[Do not accept: simple transport of protein without mention of vesicle packaging or exocytosis.]

1. Cellulose is a polymer of \(\beta\)-glucose monomers, whereas starch (amylose) is a polymer of \(\alpha\)-glucose monomers.
2. In cellulose, alternate glucose molecules are rotated \(180^\circ\), resulting in straight, unbranched chains, whereas amylose forms a coiled, helical structure.
3. Straight cellulose chains lie parallel to each other, allowing hydrogen bonds to cross-link adjacent chains to form strong microfibrils, which provide high tensile strength to the cell wall.

- Mark 1: Cellulose consists of \(\beta\)-glucose (and 1,4-glycosidic bonds) whereas amylose consists of \(\alpha\)-glucose (1)
- Mark 2: Cellulose forms straight, unbranched chains whereas amylose forms coiled/helical chains (1)
- Mark 3: Many hydrogen bonds form between parallel cellulose chains to produce microfibrils (providing strength) (1)
[Accept: reference to inverted glucose monomers for cellulose. Reject: 1,6-glycosidic bonds in cellulose or amylose.]

1. Totipotent stem cells have the potential to differentiate into all specialized cell types, including extraembryonic cells/placenta, whereas pluripotent stem cells can differentiate into all body cell types but not extraembryonic tissues.
2. Totipotent cells are only present for a limited time in the early embryo (zygote up to the 8-cell stage).
3. Pluripotent stem cells can be obtained from the inner cell mass of the blastocyst (early embryo).

- Mark 1: Totipotent cells can give rise to all body cell types AND extraembryonic tissues / placenta (1)
- Mark 2: Pluripotent cells can give rise to all cell types of the embryo / cannot give rise to extraembryonic tissues (1)
- Mark 3: Pluripotent stem cells are found in the inner cell mass / blastocyst (1)
[Accept: blastula for blastocyst. Reject: adult stem cells as an example of pluripotent cells (as they are multipotent).]

1. Seeds are collected from many different plants and geographically diverse populations to maximize the size and variety of the gene pool (capturing different alleles).
2. Seeds are dehydrated and stored at sub-zero temperatures (around \(-20^\circ\text{C}\)) to slow down metabolic activity, preventing decay and maintaining long-term viability.
3. Samples are periodically germinated to test viability; if viability drops below a threshold, the seeds are grown into plants to harvest fresh, genetically identical seeds for restocking.

- Mark 1: Seeds are collected from multiple individuals / different populations to maximize the range of alleles / gene pool (1)
- Mark 2: Stored in cold and dry conditions to reduce enzyme activity / prevent germination / prevent decay (1)
- Mark 3: Periodic germination tests are carried out to check viability, and seeds are grown to replenish stock if viability decreases (1)
[Accept: reference to maintaining genetic variation / reducing genetic drift.]

1. During prophase/metaphase, spindle fibres attach to the centromeres of the chromosomes.
2. The contraction and alignment of these fibres position the chromosomes along the equator (metaphase plate) of the cell.
3. During anaphase, the spindle fibres contract and shorten, pulling the sister chromatids apart to opposite poles of the cell, ensuring each new nucleus receives an identical set of chromosomes.

- Mark 1: Attachment of spindle fibres to centromeres (of chromosomes / chromatids) (1)
- Mark 2: Alignment of chromosomes along the equator / metaphase plate (1)
- Mark 3: Contraction / shortening of fibres to pull sister chromatids to opposite poles (during anaphase) (1)
[Accept: separation of chromatids. Reject: separation of homologous chromosomes (this is meiosis).]

1. Acetyl groups are added to the positively charged lysine residues on histone tails, neutralizing their charge.
2. This weakens the attraction between histones and the negatively charged DNA backbone, causing the chromatin to decondense (become looser/euchromatin).
3. The relaxed chromatin structure allows transcription factors and RNA polymerase to bind to the promoter region of the DNA, leading to gene transcription.

- Mark 1: Addition of acetyl groups neutralizes positive charges on histones / reduces binding between histones and DNA (1)
- Mark 2: Chromatin becomes less condensed / open / forms euchromatin (1)
- Mark 3: Allows RNA polymerase / transcription factors access to bind to DNA / promoter, initiating transcription / gene expression (1)
[Reject: reference to DNA methylation instead of acetylation.]

1. Magnesium is a vital component of the chlorophyll molecule.
2. A deficiency in magnesium ions prevents chlorophyll synthesis, causing the leaves to become yellow (chlorosis).
3. This leads to a decrease in light absorption, which reduces the rate of photosynthesis, meaning fewer organic molecules (such as glucose/sucrose) are produced for respiration, cell division, and growth.

- Mark 1: Magnesium is needed for the synthesis / production of chlorophyll (1)
- Mark 2: Deficiency causes chlorosis / yellowing of leaves (1)
- Mark 3: Less photosynthesis occurs, resulting in fewer organic compounds / sugars for respiration / cell division / cell wall synthesis / growth (1)
[Accept: less glucose/sucrose produced for energy/growth. Reject: magnesium is part of the cell wall (this is calcium).]

1. Upon contact with the receptors of the zona pellucida (jelly layer), the acrosome membrane fuses with the cell surface membrane of the sperm.
2. This triggers the release of hydrolytic/digestive enzymes (such as acrosin) from the acrosome via exocytosis.
3. These enzymes digest a pathway through the follicular cells and the zona pellucida, allowing the sperm head to reach and fuse with the egg cell surface membrane.

- Mark 1: Contact with the zona pellucida triggers fusion of the acrosome membrane with the sperm cell membrane (1)
- Mark 2: Release of hydrolytic / digestive enzymes (e.g., acrosin) by exocytosis (1)
- Mark 3: Digestion / breakdown of the zona pellucida to allow passage of the sperm to the egg cell membrane (1)
[Accept: digestion of follicle cells / jelly layer. Reject: reference to the cortical reaction (which prevents polyspermy, rather than helping the sperm reach the membrane).]

The acrosome reaction is triggered when the sperm contacts the jelly-like layer of the egg. The outer membrane of the acrosome fuses with the plasma membrane of the sperm, releasing stored hydrolytic enzymes such as acrosin. These enzymes digest the proteins in the zona pellucida, clearing a path so the sperm can pass through and fuse with the egg cell membrane.

1. Fusion of the acrosome membrane with the sperm cell surface membrane (1 mark). 2. Release of hydrolytic enzymes / acrosin / digest enzymes by exocytosis (1 mark). 3. Digestion of the zona pellucida / follicle cell layer to allow penetration (1 mark).

Sclerenchyma fibres have cell walls heavily thickened with lignin, which provides high compressive strength and rigidity. The cells are dead and lose their cytoplasm, forming hollow, strong tubes. Additionally, the cellulose microfibrils in their walls are laid down in a criss-cross pattern, which increases tensile strength and prevents buckling.

1. Cell walls are thickened with lignin to provide rigidity / compressive strength (1 mark). 2. Cells are dead / hollow to form rigid structural columns (1 mark). 3. Cellulose microfibrils are arranged in a criss-cross pattern to resist tensile forces (1 mark).

DNA methylation involves the attachment of methyl groups to CpG sites on the DNA molecule. This modification prevents transcription factors and RNA polymerase from binding to the promoter regions of specific genes, effectively silencing them. As a result, certain proteins are not produced, while other active genes are transcribed and translated, determining the unique structure and function of the differentiated cell.

1. Addition of methyl groups to CpG sites / cytosine bases of DNA (1 mark). 2. Prevents binding of transcription factors / RNA polymerase, switching genes off (1 mark). 3. Only specific genes are transcribed into mRNA / translated into proteins, changing cell structure/function (1 mark).

Proteins are synthesised by ribosomes bound to the outer surface of the rER, then pass into the rER lumen where they fold. These proteins are packaged into transport vesicles that bud off the rER and travel to, and fuse with, the membranes of the Golgi apparatus. Inside the Golgi apparatus, the proteins are modified by the addition of carbohydrate groups to form glycoproteins, which are then packaged into secretory vesicles that move to the cell surface membrane for release via exocytosis.

1. Ribosomes on the rER synthesise proteins which enter the rER lumen to fold (1 mark). 2. Transport vesicles bud off the rER and transport the proteins to the Golgi apparatus (1 mark). 3. The Golgi apparatus modifies the proteins by adding carbohydrates / packaging them into secretory vesicles for exocytosis (1 mark).

Drying and storing seeds at very low temperatures dramatically decreases water potential and kinetic energy, which reduces metabolic rate and respiration to a near-zero level. This prevents the seeds from prematurely germinating during storage. Additionally, these extreme conditions prevent the growth and reproduction of decomposers, such as bacteria and fungi, which would otherwise decay and destroy the seeds.

1. Reduces enzyme activity / rate of respiration / metabolic rate in the seeds (1 mark). 2. Prevents the seeds from germinating during storage (1 mark). 3. Inhibits the growth of decay-causing microorganisms / bacteria / fungi, extending viability (1 mark).

During prophase I of meiosis, homologous chromosomes pair up to form bivalents. Non-sister chromatids can cross over, breaking and rejoining at sites called chiasmata. This process exchanges segments of DNA between maternal and paternal chromosomes, resulting in new combinations of alleles on the recombinant chromatids, which increases genetic diversity in the resulting gametes.

1. Homologous chromosomes pair up / form bivalents during prophase I (1 mark). 2. Non-sister chromatids break and rejoin at chiasmata (1 mark). 3. This leads to exchange of alleles / genetic material, producing recombinant chromatids / new combinations of alleles (1 mark).

Studbooks act as a comprehensive database containing the pedigree, parental history, and location of every individual of an endangered species kept in captivity. Conservationists use this data to carefully select breeding pairs that are as unrelated as possible. This prevents inbreeding depression and ensures that genetic diversity is maximised within the captive population, maintaining its long-term viability.

1. Studbooks maintain a record of the pedigree / ancestry / genetic relationships of all captive individuals (1 mark). 2. Used to select mating pairs that are unrelated to prevent inbreeding / inbreeding depression (1 mark). 3. Helps to maintain high genetic diversity / prevent genetic drift within the captive population (1 mark).

The middle lamella is the outermost layer of the plant cell wall and is primarily made of the polysaccharide pectin. Pectin binds with calcium ions to form a stable gel-like compound called calcium pectate. This calcium pectate acts as a strong glue or cement that adheres the primary cell walls of neighbouring cells together, providing stability and strength to the overall plant tissue.

1. Middle lamella is composed of pectin (1 mark). 2. Pectin interacts/binds with calcium ions to form calcium pectate (1 mark). 3. Calcium pectate acts as an adhesive / cement to hold adjacent cell walls together, providing tissue stability (1 mark).

A detailed structural comparison of a mature sieve tube element and a mature xylem vessel element highlights several key similarities and differences:

**Similarities:**
1. **Tubular arrangement:** Both consist of cells joined end-to-end to form long, continuous columns adapted for transport.
2. **Lack of nucleus:** Both cell types lack a nucleus at maturity to facilitate the efficient flow of substances (sap or water) with minimal resistance.
3. **Cellulose present:** Both structures have cell walls containing cellulose.

**Differences:**
1. **Viability and cytoplasm:** Sieve tube elements are living cells containing a thin layer of cytoplasm and a cell membrane, whereas xylem vessels are dead cells completely devoid of cytoplasm, organelles, and membranes.
2. **End walls:** Sieve tube elements have perforated end walls called sieve plates, whereas xylem vessels have no end walls, forming completely open tubes.
3. **Lignification:** Xylem vessel walls are thickened with lignin, which provides structural reinforcement, whereas sieve tube walls are non-lignified.
4. **Associated cells:** Sieve tube elements are closely associated with companion cells via plasmodesmata, whereas xylem vessel elements do not have companion cells.
5. **Pits:** Xylem vessel walls contain pits (areas of unlignified, thin cellulose wall) to allow lateral water movement, which are absent in sieve tubes.

**Marking Scheme (6 Marks Maximum):**

*Award 1 mark per correct comparative point, up to a maximum of 6 marks. To achieve full marks, the answer must include at least one similarity and at least one difference.*

**Similarities (Max 3 marks):**
- **S1:** Both are composed of cells joined end-to-end to form a long / tubular / column-like structure.
- **S2:** Both lack a nucleus / lack most organelles (such as ribosomes or vacuoles) at maturity.
- **S3:** Both have cell walls containing cellulose.

**Differences (Max 5 marks - must be comparative):**
- **D1:** Sieve tube elements are living / have cytoplasm / have a cell membrane whereas xylem vessels are dead / hollow / lack cytoplasm / lack cell membranes.
- **D2:** Sieve tube elements have end walls containing pores / sieve plates whereas xylem vessels have no end walls (open ends).
- **D3:** Xylem vessel walls are lignified / contain lignin whereas sieve tube walls are non-lignified.
- **D4:** Sieve tube elements are associated with companion cells (connected by plasmodesmata) whereas xylem vessels are not associated with companion cells.
- **D5:** Xylem vessels have pits in their walls (for lateral transport) whereas sieve tubes do not have pits.

Total number of cells in mitosis = 18 (prophase) + 12 (metaphase) + 6 (anaphase) + 9 (telophase) = 45 cells. Total number of cells counted = 240. Mitotic Index = (Number of cells in mitosis / Total number of cells) * 100 = (45 / 240) * 100 = 18.75%.

1 mark: Correctly calculating the total number of dividing cells as 45. 1 mark: Correct formula or substitution showing 45 divided by 240. 1 mark: Final correct answer of 18.75% (accept 18.75 without the percentage sign).

Magnification = Image size / Actual size. Therefore, Actual size = Image size / Magnification. Image size = 36 mm = 36,000 \(\mu\text{m}\). Actual size = 36,000 / 450 = 80 \(\mu\text{m}\). Alternatively, 36 mm / 450 = 0.08 mm; 0.08 mm * 1000 = 80 \(\mu\text{m}\).

1 mark: Correct conversion of mm to micrometres (36 mm = 36,000 \(\mu\text{m}\)) or equivalent in mm. 1 mark: Correct arrangement of the magnification formula showing division of image size by magnification. 1 mark: Correct final calculation of 80 \(\mu\text{m}\) (accept 80).

To ensure a valid comparison of tensile strength between different plant species, variables affecting fiber breaking point must be controlled: 1. Length of the fibre (as longer fibres may have more weak points), 2. Diameter/thickness of the fibre (thicker fibres can withstand more force), 3. Temperature or humidity of the testing environment, or the rate at which mass/force is added to the fibre.

1 mark: Same/constant length of the fibre. 1 mark: Same/constant diameter or thickness of the fibre (accept cross-sectional area). 1 mark: Same environmental temperature/humidity OR same rate of loading (adding masses).

Mass of vitamin C required to decolourise 1.0 cm\(^{3}\) of DCPIP = Volume of standard * Concentration of standard = 2.0 cm\(^{3}\) * 1.0 mg cm\(^{-3}\) = 2.0 mg. This same mass of vitamin C (2.0 mg) is present in the 1.6 cm\(^{3}\) of grapefruit juice. Therefore, Concentration of grapefruit juice = Mass / Volume = 2.0 mg / 1.6 cm\(^{3}\) = 1.25 mg cm\(^{-3}\).

1 mark: Correctly calculating the mass of vitamin C in the standard titration as 2.0 mg. 1 mark: Setting up the correct ratio showing concentration of juice = 2.0 mg / 1.6 cm\(^{3}\). 1 mark: Final correct answer of 1.25 mg cm\(^{-3}\) (accept 1.25).

1. Calibrate the colorimeter using a reference cuvette containing distilled water to set the absorbance to 0 (or transmission to 100%). 2. Select a green filter (around 520-550 nm) as betalain is a red pigment and absorbs green light most strongly. 3. Ensure the outside of each cuvette is dry and free of fingerprints before placing it in the colorimeter.

1 mark: Calibrating the colorimeter using a blank / cuvette of distilled water (to 0 absorbance / 100% transmission). 1 mark: Selection of a green/blue-green filter (500 to 550 nm range). 1 mark: Wiping the cuvette clean/dry before reading to prevent scattering of light.

Rate at 1.0% = 12.0 cm\(^{3}\) / 30 s = 0.40 cm\(^{3}\) s\(^{-1}\). Rate at 2.0% = 21.0 cm\(^{3}\) / 30 s = 0.70 cm\(^{3}\) s\(^{-1}\). Percentage increase in rate = ((Rate at 2.0% - Rate at 1.0%) / Rate at 1.0%) * 100 = ((0.70 - 0.40) / 0.40) * 100 = (0.30 / 0.40) * 100 = 75.0%.

1 mark: Correct calculation of the rates at both concentrations (0.40 and 0.70 cm\(^{3}\) s\(^{-1}\)). 1 mark: Correct formula for percentage change showing ((0.70 - 0.40) / 0.40) or ((21 - 12) / 12). 1 mark: Correct calculation of 75% or 75.0%.

A low-power tissue plan diagram shows the outline of the different tissues (e.g., epidermis, cortex, vascular bundles) but does not show individual cells, whereas a high-power drawing shows individual cell walls and cellular detail. Two rules for plan diagrams: 1. Draw only tissue boundaries (no individual cells). 2. Do not use shading or hatching to represent different layers.

1 mark: Explanation that low-power plan shows only tissue boundaries/regions while high-power shows individual cells. 1 mark: Rule 1: No individual cells should be drawn in the plan. 1 mark: Rule 2: No shading/colouring/hatching is allowed.

To perform a two-fold serial dilution: 1. Mix 10 cm\(^{3}\) of the 10.0% stock solution with 10 cm\(^{3}\) of distilled water to create 20 cm\(^{3}\) of a 5.0% solution. 2. Take 10 cm\(^{3}\) of this 5.0% solution and mix it with 10 cm\(^{3}\) of distilled water to produce 20 cm\(^{3}\) of a 2.5% solution. 3. Take 10 cm\(^{3}\) of this 2.5% solution and mix it with 10 cm\(^{3}\) of distilled water to produce 20 cm\(^{3}\) of a 1.25% solution. This leaves exactly 10 cm\(^{3}\) of each solution for the test (except the final one, which can be discarded down to 10 cm\(^{3}\)).

1 mark: Correct dilution ratio (1:1 ratio / equal volumes of solution and water, i.e., 10 cm\(^{3}\) of each). 1 mark: Describes transferring the newly made dilution to the next tube of water (serial transfer). 1 mark: Specifying that solutions must be thoroughly mixed at each step before the next dilution.

1. Calculate distance of stage micrometer divisions: 2 divisions = 2 * 0.1 mm = 0.2 mm = 200 \(\mu\text{m}\). 2. Calibrate eyepiece graticule: 40 eyepiece units = 200 \(\mu\text{m}\), so 1 eyepiece unit = 200 / 40 = 5 \(\mu\text{m}\). 3. Calculate actual length of cell: 18 eyepiece units * 5 \(\mu\text{m}\) per unit = 90 \(\mu\text{m}\).

[1 mark] Calculating the equivalent of 1 eyepiece unit to be 5 \(\mu\text{m}\) (or stage micrometer distance of 200 \(\mu\text{m}\)). [1 mark] Multiplying measured units (18) by calibration factor (5). [1.33 marks] Correct final answer of 90.

Controlled variables are critical to ensure that any change in absorbance is only due to the independent variable (ethanol concentration). Controlled variables include: 1. Volume of ethanol (e.g., 10 \(\text{cm}^3\) in each tube). 2. Temperature (all tubes kept in a water bath at 25 degrees Celsius). 3. Surface area/dimensions of beetroot pieces (using a cork borer to get same diameter and cutting to same length with a scalpel and ruler). 4. Washing time/process to remove excess pigment from cut surfaces. 5. Source/variety of beetroot. Controlling temperature ensures that thermal damage to cell membranes does not confound the results.

[1.5 marks] 0.5 marks for each valid controlled variable stated (up to three). [1 mark] Explaining how one chosen variable is controlled. [0.83 marks] Explaining why controlling this variable ensures validity (e.g., prevents thermal damage to membranes from confounding results).

1. Convert the time to seconds: 2 minutes and 15 seconds = (2 * 60) + 15 = 135 seconds. 2. Calculate rate: Rate = Volume of oxygen collected / time in seconds = 18.0 \(\text{cm}^3\) / 135 s = 0.13333... \(\text{cm}^3 \text{ s}^{-1}\). 3. Round to 3 significant figures: 0.133 \(\text{cm}^3 \text{ s}^{-1}\).

[1 mark] Conversion of time to seconds (135 s). [1 mark] Correct formula application (volume divided by time). [1.33 marks] Correct calculation of rate to 3 significant figures (0.133).

1. Sum the number of cells undergoing mitosis (active stages): 24 (prophase) + 8 (metaphase) + 6 (anaphase) + 12 (telophase) = 50 cells. 2. Calculate the total number of cells: 50 + 150 (interphase) = 200 cells. 3. Calculate mitotic index: Mitotic Index = (Number of cells in mitosis / Total number of cells) * 100 = (50 / 200) * 100 = 25%. 4. Express to 3 significant figures: 25.0%.

[1 mark] Calculating total number of dividing cells (50) or total cells (200). [1 mark] Setting up the correct ratio: (50 / 200). [1.33 marks] Correct final percentage value to 3 significant figures (25.0).

1. Use the titration relationship: Concentration of Sample = (Volume of Standard / Volume of Sample) * Concentration of Standard. 2. Substitute values: Concentration of orange juice = (0.85 / 2.15) * 10.0 \(\text{mg cm}^{-3}\). 3. Calculate: Concentration = 0.3953488... * 10.0 = 3.953488... \(\text{mg cm}^{-3}\). 4. Express to 2 decimal places: 3.95 \(\text{mg cm}^{-3}\).

[1 mark] Correctly stating or using the inverse relationship between volume required and concentration. [1 mark] Calculation setup (0.85 / 2.15) * 10. [1.33 marks] Final concentration calculated to 2 decimal places (3.95).

1. Find the change in mass: Final Mass - Initial Mass = 2.87 g - 3.42 g = -0.55 g (a loss of 0.55 g). 2. Calculate percentage change: (Change in Mass / Initial Mass) * 100 = (-0.55 / 3.42) * 100 = -16.08187...%. 3. Round to one decimal place: -16.1%.

[1 mark] Correct calculation of change in mass (-0.55 g). [1 mark] Correct division by initial mass (3.42 g) rather than final mass. [1.33 marks] Correct percentage change to one decimal place including the correct sign (-16.1).

1. Convert both measurements to the same unit. Convert drawing size to micrometres: 5.2 cm = 52 mm = 52,000 \(\mu\text{m}\). 2. Use the magnification formula: Magnification = Drawing Size / Actual Size. 3. Calculate magnification: Magnification = 52,000 / 65 = 800. Therefore, the magnification of the drawing is x800.

[1 mark] Convert drawing size to micrometres (52,000 \(\mu\text{m}\)) or actual size to centimetres (0.0065 cm). [1 mark] Setting up the correct equation (Drawing Size / Actual Size). [1.33 marks] Correct final magnification of 800 (accept x800).