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

Part A: Water is a polar molecule with a slight negative charge on the oxygen atom (\delta^{-}) and a slight positive charge on the hydrogen atoms (\delta^{+}). This allows water to form hydrogen bonds with other water molecules, resulting in high cohesion, which helps water flow in continuous columns (mass flow) through blood vessels. The polar nature of water allows it to dissolve polar and ionic solutes (such as glucose, amino acids, and salts) by forming hydration shells around them, facilitating their transport. Part B: Arteries have a thick wall containing a large amount of collagen fibers to provide high tensile strength and withstand high pressure without bursting. They have a thick layer of elastic fibers (elastin) in the tunica media which stretch during systole to accommodate high-pressure pulses and recoil during diastole to maintain blood pressure. They contain a thick layer of smooth muscle that can contract (vasoconstriction) or relax (vasodilation) to regulate lumen diameter and control pressure. The folded endothelium allows the lumen to expand to accommodate pulses of high-pressure blood without tearing.

Part A (Max 4 marks): 1. Reference to oxygen being delta negative (\delta^{-}) and hydrogen being delta positive (\delta^{+}) / water is a dipole (1 mark). 2. Hydrogen bonds form between water molecules (1 mark). 3. This creates cohesive forces that hold water molecules together to allow mass flow / continuous transport column (1 mark). 4. Water acts as a solvent because polar / ionic substances can dissolve in it / form hydration shells (1 mark). 5. Hydrophilic molecules are transported in solution (1 mark). Part B (Max 4 marks): 1. Thick layer of collagen to provide high tensile strength / prevent bursting (1 mark). 2. Elastic fibers (elastin) stretch to accommodate high pressure during systole (1 mark). 3. Elastic recoil of elastic fibers maintains high blood pressure during diastole (1 mark). 4. Smooth muscle contracts to constrict the lumen / help maintain high pressure (1 mark). 5. Folded endothelium prevents damage during stretching / allows expansion (1 mark).

Part A: Similarities: Both processes use DNA as a template, involve complementary base pairing, and occur in the nucleus. Differences: Transcription uses RNA polymerase while replication uses DNA polymerase. Transcription produces a single-stranded mRNA using one template strand, whereas replication produces two double-stranded DNA molecules using both strands. Uracil is used in transcription, whereas thymine is used in replication. Part B: A single base deletion is a frameshift mutation, which alters the reading frame of all subsequent codons from the point of mutation. This changes the sequence of amino acids in the primary structure. Consequently, different R-group interactions (such as hydrogen, ionic, or disulfide bonds) form, altering the overall tertiary structure and three-dimensional shape, resulting in a non-functional protein.

Part A (Max 4 marks, must include at least one similarity and one difference): Similarities (Max 2): 1. Both involve helicase / unwinding of DNA / use of DNA as a template (1 mark). 2. Both involve complementary base pairing / formation of phosphodiester bonds (1 mark). 3. Both occur in the nucleus (1 mark). Differences (Max 2): 4. Transcription produces single-stranded RNA whereas replication produces double-stranded DNA (1 mark). 5. Transcription uses RNA polymerase whereas replication uses DNA polymerase (1 mark). 6. Transcription incorporates uracil whereas replication incorporates thymine (1 mark). 7. Only one DNA template strand is copied in transcription, whereas both are copied in replication (1 mark). Part B (Max 4 marks): 1. Reference to frameshift mutation / shift in the reading frame (1 mark). 2. Alters all subsequent codons / triplets from the point of mutation (1 mark). 3. Leading to a change in the primary structure / amino acid sequence of the polypeptide (1 mark). 4. This alters the position of R-groups / changes the hydrogen, ionic, or disulfide bonds (1 mark). 5. Resulting in a different tertiary structure / three-dimensional shape, leading to a non-functional hemoglobin (1 mark).

Part A: Energy budget is the balance between energy intake (from food) and energy expenditure (through respiration and physical activity). An imbalance where energy intake exceeds energy expenditure (positive energy balance) means the excess energy is stored as fat in adipose tissue, leading to an increase in body mass index (BMI) and obesity. Part B: High saturated lipid intake is associated with elevated levels of low-density lipoproteins (LDLs) in the blood. LDLs transport cholesterol and deposit it in the endothelium of arteries. This triggers an inflammatory response, leading to plaque (atheroma) formation (atherosclerosis). High salt intake causes water retention, increasing blood volume and raising blood pressure (hypertension). High blood pressure damages the endothelial lining of arteries, making them more susceptible to further inflammatory responses, plaque formation, and blood clots (thrombosis), reducing oxygen delivery to the heart muscle (myocardial infarction) or brain (stroke).

Part A (Max 3 marks): 1. Energy budget is the balance between energy intake and energy expenditure (1 mark). 2. If intake is greater than expenditure, there is a positive energy balance (1 mark). 3. Excess energy is converted to / stored as fat/adipose tissue leading to obesity (1 mark). Part B (Max 5 marks): 1. High saturated fat increases blood cholesterol / LDL levels (1 mark). 2. LDLs transport cholesterol and deposit it in the artery walls (endothelium) (1 mark). 3. This leads to the formation of an atheroma / plaque (1 mark). 4. High salt intake causes water retention in blood / increases blood volume (1 mark). 5. This leads to high blood pressure / hypertension (1 mark). 6. High blood pressure damages the endothelium, promoting further plaque / clot formation (1 mark). 7. Plaque narrows the lumen of arteries, reducing oxygen delivery to heart muscle (myocardial infarction) / brain (stroke) (1 mark).

Part A: According to the fluid mosaic model, the cell membrane is a double layer (bilayer) of phospholipids, where the hydrophilic phosphate heads face outwards and the hydrophobic fatty acid tails face inwards. Proteins of various shapes and sizes are scattered (mosaic) throughout this bilayer. Some proteins are transmembrane (integral), while others are peripheral. The phospholipids and proteins can move laterally, giving the membrane a fluid quality. The membrane also contains cholesterol, glycolipids, and glycoproteins. Part B: Facilitated diffusion is a passive process that does not require ATP, moving substances down their concentration gradient (from high to low concentration). Active transport is an active process requiring ATP to move substances against their concentration gradient (from low to high concentration). Facilitated diffusion uses both channel and carrier proteins, whereas active transport uses carrier proteins/pumps only.

Part A (Max 4 marks): 1. Double layer / bilayer of phospholipids with hydrophilic heads facing outwards and hydrophobic tails pointing inwards (1 mark). 2. Phospholipids can move laterally within the monolayer / fluid property (1 mark). 3. Proteins are scattered / embedded randomly within the bilayer / mosaic appearance (1 mark). 4. Reference to intrinsic / transmembrane proteins AND extrinsic / peripheral proteins (1 mark). 5. Presence of other components: cholesterol, glycoproteins, or glycolipids (1 mark). Part B (Max 4 marks): 1. Facilitated diffusion is passive / does not require ATP, whereas active transport requires ATP (1 mark). 2. Facilitated diffusion occurs down a concentration gradient, whereas active transport occurs against a concentration gradient (1 mark). 3. Facilitated diffusion uses channel or carrier proteins, whereas active transport uses carrier proteins / pumps only (1 mark). 4. Both processes involve membrane proteins (similarity - max 1 mark if clearly stated as comparison) (1 mark).

Part A: When a blood vessel wall is damaged, platelets and damaged tissue release the enzyme thromboplastin. In the presence of calcium ions (\text{Ca}^{2+}) and vitamin K, thromboplastin catalyzes the conversion of the inactive plasma protein prothrombin into the active enzyme thrombin. Thrombin then acts as an enzyme to catalyze the conversion of the soluble plasma protein fibrinogen into insoluble fibrin fibers. These fibrin fibers form a meshwork/net at the site of damage, which traps red blood cells and platelets to form a blood clot (thrombus). Part B: If a clot forms in an undamaged blood vessel (thrombosis), it can restrict or completely block blood flow. This reduces the supply of oxygen and glucose to tissues. If this occurs in a coronary artery, it causes a myocardial infarction (heart attack). If it occurs in a cerebral artery, it deprives brain cells of oxygen, causing a stroke.

Part A (Max 5 marks): 1. Damaged tissue / platelets release thromboplastin (1 mark). 2. Thromboplastin catalyzes the conversion of prothrombin to thrombin (1 mark). 3. Calcium ions / Vitamin K are required for this conversion (1 mark). 4. Thrombin catalyzes the conversion of soluble fibrinogen into insoluble fibrin (1 mark). 5. Fibrin forms a mesh of fibers (1 mark). 6. Red blood cells / platelets are trapped in the mesh to form a clot (1 mark). Part B (Max 3 marks): 1. A clot in undamaged vessels blocks the lumen / restricts blood flow (1 mark). 2. Reducing supply of oxygen / glucose to tissues (1 mark). 3. Can lead to myocardial infarction if in coronary arteries / stroke if in cerebral arteries / tissue death (1 mark).

Part A: DNA is double-stranded, which allows both strands to act as templates for the synthesis of complementary strands. The hydrogen bonds holding the base pairs together are relatively weak, allowing the strands to easily unzip. Specific complementary base pairing (A with T, C with G) ensures that the new nucleotides are aligned in a highly specific, accurate sequence. The strong covalent phosphodiester bonds in the sugar-phosphate backbone keep the order of the bases stable and protected. Part B: Bacteria were grown initially in heavy nitrogen (^{15}\text{N}). After one round of replication in light nitrogen (^{14}\text{N}), the DNA was found to be of intermediate density, representing a hybrid (one strand ^{15}\text{N} and one strand ^{14}\text{N}). This ruled out the conservative model of replication, which would have produced two distinct bands (one heavy and one light). After two rounds of replication, there were two distinct bands: one intermediate and one light. This ruled out the dispersive model (which would only produce a single hybrid band of decreasing density) and confirmed the semi-conservative model, showing that each DNA molecule conserved one template strand and made one new light strand.

Part A (Max 4 marks): 1. DNA has a double-stranded structure / two strands, allowing both to act as templates (1 mark). 2. Hydrogen bonds between bases are weak, allowing easy unzipping / separation of strands (1 mark). 3. Complementail base pairing / A-T and C-G ensures accurate alignment of nucleotides (1 mark). 4. Strong covalent phosphodiester bonds in sugar-phosphate backbone prevent alteration of base sequence during replication (1 mark). Part B (Max 4 marks): 1. After one generation in ^{14}\text{N}, there was a single band of intermediate density, containing one ^{15}\text{N} strand and one ^{14}\text{N} strand (1 mark). 2. This ruled out conservative replication because there was no purely heavy band or light band (1 mark). 3. After two generations in ^{14}\text{N}, there were two bands: one intermediate and one light (1 mark). 4. This ruled out dispersive replication because a light band was present (dispersive would only produce one hybrid band of decreasing density) (1 mark). 5. This confirmed semi-conservative replication where each new DNA molecule consists of one original/conserved strand and one newly synthesized strand (1 mark).

### Part (a)
1. **Calculate the LDL:HDL ratio for Group A:**
\(\text{Ratio}_A = \frac{4.20}{1.10} = 3.8182\)

2. **Calculate the LDL:HDL ratio for Group B:**
\(\text{Ratio}_B = \frac{2.80}{1.40} = 2.0000\)

3. **Calculate the difference in ratio:**
\(\text{Difference} = 3.8182 - 2.0000 = 1.8182\)

4. **Calculate the percentage decrease:**
\(\text{Percentage Decrease} = \left( \frac{1.8182}{3.8182} \right) \times 100\% = 47.619\%\)
Rounding to three significant figures gives **47.6%**.

### Part (b)
- LDLs transport cholesterol in the blood and can deposit it in the walls of the arteries.
- High blood pressure or toxins cause damage to the endothelial lining of an artery.
- This damage triggers an inflammatory response, attracting white blood cells (macrophages) to the site.
- Macrophages ingest the excess cholesterol/LDLs, transforming into foam cells.
- These foam cells accumulate to form a fatty streak, which develops into an atheroma (plaque) beneath the endothelium.
- Calcium salts and fibrous tissue build up, causing the plaque to harden, narrowing the lumen of the artery and reducing blood flow.

### Part (c)
- **Lifestyle Changes:**
- Reducing dietary saturated fat lowers circulating LDL levels, reducing plaque progression.
- Regular aerobic exercise helps lower resting blood pressure, increases HDL levels, and aids weight management.
- Ceasing smoking removes nicotine (a vasoconstrictor) and carbon monoxide (which damages endothelium), reducing arterial damage.
- **Medical Treatments:**
- **Statins:** Inhibit the enzyme in the liver responsible for cholesterol synthesis, lowering LDL concentration.
- **Antihypertensives (e.g., ACE inhibitors, beta-blockers, calcium channel blockers):** Lower blood pressure, reducing the mechanical stress on arterial walls and preventing further endothelial damage.
- **Platelet inhibitors (e.g., low-dose aspirin):** Prevent blood clotting (thrombosis) at the site of existing plaques, reducing the risk of myocardial infarction or stroke.

### Part (a) [4 Marks]
* **MP1:** Correct calculation of LDL:HDL ratio for Group A as 3.82 (or 3.818) [1 Mark].
* **MP2:** Correct calculation of LDL:HDL ratio for Group B as 2.00 [1 Mark].
* **MP3:** Correct mathematical setup for percentage decrease: \(\frac{3.818 - 2.00}{3.818} \times 100\) [1 Mark].
* **MP4:** Correct final value calculated to 3 significant figures: **47.6%** (Accept range 47.6% to 47.7%) [1 Mark].
* *Note: If no working is shown, a correct final answer of 47.6% gains all 4 marks.*

### Part (b) [6 Marks]
* **MP1:** LDLs transport cholesterol in the blood to the tissues/arteries [1 Mark].
* **MP2:** Damage occurs to the endothelium/endothelial lining of the artery (due to high blood pressure, smoking toxins, etc.) [1 Mark].
* **MP3:** Inflammatory response is triggered, leading to accumulation of white blood cells / macrophages [1 Mark].
* **MP4:** Macrophages engulf LDL/cholesterol to form foam cells [1 Mark].
* **MP5:** Accumulation of foam cells forms a fatty streak / atheroma under the endothelium [1 Mark].
* **MP6:** Deposition of calcium salts / fibrous tissue hardens the plaque, narrowing the lumen and restricting blood flow [1 Mark].

### Part (c) [6 Marks]
* **MP1:** Diet low in saturated fat / high in fiber lowers blood LDL levels / reduces further plaque build-up [1 Mark].
* **MP2:** Regular exercise lowers blood pressure / increases HDL levels / reduces body mass [1 Mark].
* **MP3:** Smoking cessation reduces arterial damage / vasoconstriction caused by nicotine/carbon monoxide [1 Mark].
* **MP4:** Statins lower blood cholesterol by inhibiting liver synthesis [1 Mark].
* **MP5:** Antihypertensives (accept specific examples like beta-blockers, ACE inhibitors, diuretics) lower blood pressure, reducing mechanical strain on arterial walls [1 Mark].
* **MP6:** Platelet inhibitors / anticoagulants (e.g., aspirin) reduce the risk of blood clot formation (thrombosis) on existing plaques [1 Mark].

### Part (a)
- **Primary Structure:** The deletion of three nucleotides (one codon) results in the loss of one specific amino acid (phenylalanine) at position 508. This directly changes the sequence of amino acids.
- **Secondary Structure:** The loss of this amino acid alters local hydrogen bonding, which disrupts the normal alpha-helix or beta-pleated sheet configurations.
- **Tertiary Structure:** The modified secondary structure changes how the polypeptide folds into its 3D shape. Hydrophobic and hydrophilic interactions, ionic bonds, or disulfide bridges are formed incorrectly, leading to a misfolded protein.
- **Consequences for Function:** The cell's quality control system in the endoplasmic reticulum (ER) recognizes the CFTR protein as misfolded and targets it for degradation. Consequently, little or no CFTR protein reaches the apical cell membrane. Any CFTR that does reach the membrane fails to open/close correctly, preventing the passage of chloride ions.

### Part (b)
- In healthy individuals, CFTR channels pump chloride ions (\(\text{Cl}^-\)) out of epithelial cells into the mucus lining the airways.
- In cystic fibrosis, because CFTR is missing or non-functional, chloride ions cannot leave the cells.
- Consequently, sodium ions (\(\text{Na}^+\)) are actively absorbed into the epithelial cells rather than being regulated.
- This creates a high solute concentration inside the cells, resulting in a lower water potential inside the cells relative to the mucus.
- Water moves out of the mucus and into the epithelial cells by osmosis.
- The mucus becomes dehydrated, highly viscous, thick, and sticky.
- The cilia lining the airways cannot beat effectively to clear this thick mucus.
- Trapped bacteria, dust, and pathogens are not removed; the stagnant, nutrient-rich mucus provides ideal conditions for bacterial growth, leading to chronic lung infections.

### Part (c)
- **Amniocentesis procedure:**
- An ultrasound scan is used to locate the fetus and placenta.
- A fine needle is inserted through the abdomen and uterine wall of the mother.
- A sample of amniotic fluid (which contains shed fetal cells) is extracted.
- The fetal cells are cultured and genetically tested for the CFTR mutation.
- **Key differences between Amniocentesis and CVS (any one):**
- **Timing:** Amniocentesis is performed later in pregnancy (typically weeks 15–20), whereas CVS is performed earlier (weeks 10–14).
- **Sample source:** Amniocentesis extracts amniotic fluid containing fetal cells, whereas CVS samples tissue from the chorionic villi of the placenta.
- **Miscarriage risk:** CVS has a slightly higher risk of miscarriage (~1%) compared to amniocentesis (~0.5%).
- **Turnaround time:** CVS results are often available faster because more fetal DNA can be directly extracted without waiting for cell cultures.

### Part (a) [6 Marks]
* **MP1:** The deletion mutation removes one codon, changing the primary structure / amino acid sequence (missing phenylalanine) [1 Mark].
* **MP2:** Altered amino acid sequence disrupts hydrogen bonding, altering the secondary structure (alpha-helices/beta-sheets) [1 Mark].
* **MP3:** Changes the 3D folding / tertiary structure because ionic bonds, disulfide bridges, or hydrophobic interactions form differently [1 Mark].
* **MP4:** CFTR protein is misfolded [1 Mark].
* **MP5:** Misfolded protein is recognized as faulty and degraded in the endoplasmic reticulum / fails to be transported to the cell membrane [1 Mark].
* **MP6:** If any protein reaches the membrane, the channel gate cannot open, preventing chloride ion transport [1 Mark].

### Part (b) [6 Marks]
* **MP1:** Chloride ions (\(\text{Cl}^-\)) are not transported out of the epithelial cells into the mucus [1 Mark].
* **MP2:** Sodium ions (\(\text{Na}^+\)) are continuously absorbed into the cells [1 Mark].
* **MP3:** Water potential inside the epithelial cells becomes lower than in the mucus / water potential gradient is created [1 Mark].
* **MP4:** Water moves out of the mucus and into the cells by osmosis [1 Mark].
* **MP5:** Mucus becomes dehydrated, viscous, and sticky [1 Mark].
* **MP6:** Cilia are unable to beat/clear the thick mucus, trapping pathogens and providing an environment for bacterial growth/infection [1 Mark].

### Part (c) [4 Marks]
* **MP1:** Needle is inserted through the abdomen/uterus under ultrasound guidance [1 Mark].
* **MP2:** A sample of amniotic fluid containing fetal cells is extracted [1 Mark].
* **MP3:** Cells are cultured and analyzed for the presence of the CF mutation [1 Mark].
* **MP4:** State one valid difference between amniocentesis and CVS (e.g., Amniocentesis is performed at 15–20 weeks vs CVS at 10–14 weeks OR CVS takes placental tissue vs amniocentesis takes fluid OR CVS has a slightly higher miscarriage risk) [1 Mark].

Amylase is synthesized on ribosomes attached to the rough endoplasmic reticulum (rER), folded into its tertiary structure in the rER lumen, and transported via vesicles to the Golgi apparatus. In the Golgi apparatus, it is modified (e.g., glycosylated) and packaged into secretory vesicles. These vesicles travel along microtubules and fuse with the cell surface membrane, releasing the enzyme by exocytosis. The nucleolus produces the ribosomal subunits required for translation, while the Golgi apparatus modifies and packages the synthesized protein.

(a) Maximum of 5 marks:
1. Translation of mRNA on ribosomes attached to the rER [1]
2. Polypeptide enters the lumen of the rER and folds into its 3D/tertiary structure [1]
3. Transported from rER to Golgi apparatus in transport vesicles [1]
4. Vesicles fuse with the cis-face of the Golgi apparatus [1]
5. Golgi modifies the protein (e.g., glycosylation/addition of carbohydrate) [1]
6. Packaged into secretory vesicles from the trans-face [1]
7. Vesicles move along the cytoskeleton and fuse with the cell surface membrane via exocytosis [1]

(b) Maximum of 3 marks:
1. Nucleolus: site of rRNA synthesis / assembly of ribosomal subunits [1]
2. Ribosomes: site of translation / polypeptide synthesis [1]
3. Golgi apparatus: modifies proteins AND packages them into secretory vesicles [1]

Spermatogenesis and oogenesis both use meiosis to produce haploid cells, but spermatogenesis results in four small, motile sperm continuously from puberty, whereas oogenesis results in one large, non-motile ovum plus polar bodies cyclically starting before birth. Genetic variation is introduced via crossing over in prophase I and independent assortment in metaphase I/II.

(a) Maximum of 5 marks:
Similarities (max 2):
1. Both processes involve meiosis to produce haploid cells [1]
2. Both processes involve early mitotic divisions of germ cells [1]
3. Both involve growth phases before division [1]
Differences (max 4):
4. Spermatogenesis produces four functional spermatozoa, oogenesis produces one functional ovum and polar bodies [1]
5. Spermatogenesis is continuous from puberty, oogenesis is cyclic / paused during development [1]
6. Spermatogenesis begins at puberty, oogenesis begins during fetal development [1]
7. Spermatogenesis produces motile gametes, oogenesis produces non-motile gametes [1]

(b) Maximum of 3 marks:
1. Crossing over: Homologous chromosomes pair up and exchange genetic material during prophase I [1]
2. Creates recombinant chromatids / new combinations of maternal and paternal alleles [1]
3. Independent assortment: Homologous chromosome pairs/chromatids align randomly at the spindle equator during metaphase I/II [1]
4. Leads to random distribution of maternal and paternal chromosomes into gametes [1]

Pluripotent stem cells can differentiate into almost any cell type (excluding extra-embryonic tissue), whereas multipotent stem cells can only differentiate into a limited range of cells of a closely related family. Differentiation is regulated by transcription factors which activate specific genes, while epigenetic modifications like DNA methylation silence genes by blocking transcription factor binding.

(a) Maximum of 2 marks:
1. Pluripotent stem cells can differentiate into almost any cell type (except extra-embryonic tissue) [1]
2. Multipotent stem cells can differentiate into a limited range of cell types / specific cell lineage only [1]

(b) Maximum of 6 marks:
1. Transcription factors bind to specific promoter regions on DNA [1]
2. This stimulates (or represses) the transcription of target genes / synthesis of mRNA [1]
3. Active genes are transcribed and translated to produce specific structural or functional proteins [1]
4. These proteins permanently alter the cell's structure and function (differentiation) [1]
5. DNA methylation involves the addition of methyl groups to cytosine bases in DNA [1]
6. High levels of methylation prevent transcription factors from binding / silence genes [1]
7. Histone acetylation opens chromatin allowing transcription / deacetylation condenses chromatin preventing transcription [1]
8. These modifications ensure only tissue-specific genes are expressed [1]

Xylem and sclerenchyma are both dead, lignified, hollow structural tissues. However, xylem vessels form continuous open-ended tubes for water transport and have pits, whereas sclerenchyma fibres have tapered closed ends and function purely for support. Cellulose is a straight polymer of \(\beta\)-glucose with hydrogen bonds forming microfibrils, which criss-cross in the cell wall to provide strength.

(a) Maximum of 5 marks:
Similarities (max 2):
1. Both consist of dead cells at maturity [1]
2. Both are hollow / lack cytoplasm / organelles [1]
3. Both have secondary cell walls thickened with lignin [1]
4. Both provide structural support to the plant [1]
Differences (max 3):
5. Xylem vessels have completely open end walls / form continuous tubes, whereas sclerenchyma fibres have closed / tapered ends [1]
6. Xylem vessels have pits in their cell walls (for lateral water transport), whereas sclerenchyma fibres do not [1]
7. Xylem vessels function in transport of water and mineral ions, whereas sclerenchyma fibres function solely in support [1]

(b) Maximum of 3 marks:
1. Cellulose is a linear, unbranched polymer of \(\beta\)-glucose joined by \(\beta\)-1,4-glycosidic bonds [1]
2. Alternate glucose molecules are rotated by 180 degrees to form straight chains [1]
3. Hydrogen bonds form between adjacent parallel chains, grouping them into strong microfibrils [1]
4. Microfibrils are arranged in a criss-cross mesh embedded in a matrix of pectin/hemicellulose in the cell wall [1]

Drying and freezing seeds reduces moisture and enzyme activity, preventing premature germination and bacterial or fungal decay to maximize longevity. Captive breeding increases population sizes and controls inbreeding using studbooks, but faces challenges like a small gene pool, loss of survival behaviors, high costs, and difficulties in successful reintroduction.

(a) Maximum of 3 marks:
1. Drying reduces moisture content / water potential of seeds [1]
2. Low temperatures and low moisture reduce / slow down enzyme activity / metabolic rate [1]
3. Prevents / delays germination of seeds [1]
4. Inhibits the growth of bacteria / fungi / decay microorganisms [1]
5. Extends the period of viability / survival time of the seeds [1]

(b) Maximum of 5 marks:
Advantages (max 3):
1. Increases population size of endangered species in a safe / predator-free environment [1]
2. Breeding can be managed (using studbooks/pedigrees) to maintain genetic diversity / minimize inbreeding [1]
3. Protects animals from poaching, habitat fragmentation, or extreme weather [1]
4. Raises public awareness and funding for conservation [1]
Challenges (max 3):
5. High cost and intensive resources required [1]
6. Genetic drift / inbreeding depression due to small initial gene pool [1]
7. Loss of wild instincts / natural behaviors (e.g., hunting, predator avoidance) due to habituation [1]
8. Reintroduction may fail due to ongoing threats in the wild (diseases, habitat loss, poaching) [1]

Both trials aimed to find safe, effective doses for diseased patients, but Withering tested directly on diseased humans without pre-clinical animal testing or healthy Phase I volunteers. Modern testing uses large sample sizes, placebo groups, and double-blind setups to scientifically establish safety and efficacy and remove bias.

(a) Maximum of 5 marks:
Similarities (max 2):
1. Both tested the active substance on patients suffering from the target disease [1]
2. Both aimed to find the effective therapeutic dose [1]
3. Both monitored patients for toxic side effects [1]
Differences (max 4):
4. Withering did not perform pre-clinical testing on cells/animals, whereas modern testing does [1]
5. Withering did not test on healthy volunteers (Phase I), whereas modern protocols always do first [1]
6. Withering had a very small sample size, whereas modern clinical trials use hundreds/thousands of subjects [1]
7. Modern testing uses randomized control groups with placebos, which Withering did not use [1]
8. Modern testing is highly regulated, standardizing safety protocols and ethical approvals [1]

(b) Maximum of 3 marks:
1. Placebo: An inactive treatment / dummy pill identical in appearance to the drug [1]
2. Used as a control to measure the baseline psychological / placebo effect against the drug's physiological effect [1]
3. Double-blind: Neither patients nor researchers know who is receiving the active drug or placebo [1]
4. Eliminates bias / subconscious influence from doctors during evaluation or from patients reporting symptoms [1]

Part (a):
- Seed banks require much less space than animal enclosures as seeds are small (1 mark).
- Seeds can be kept dormant for long periods under controlled dry and cold conditions, whereas animals must be continuously fed and managed throughout their lifespans (1 mark).
- Lower maintenance costs overall (less labour, food, and veterinary care required for seed banks) (1 mark).
- It is easier to store large numbers of seeds, representing high genetic diversity, compared to maintaining a viable breeding population of animals (1 mark).
- Ethical concerns are minimal with plant gametes/seeds compared to animal captivity and breeding (1 mark).
- Easier and cheaper to transport seeds than live animals for reintroduction programmes (1 mark).

Part (b)(i):
- Formula: \(H = \frac{\text{Number of heterozygotes}}{\text{Total number of individuals}}\)
- Working: \(H = \frac{14}{50} = 0.28\) (1 mark for working, 1 mark for correct final answer of 0.28).

Part (b)(ii):
- A high heterozygosity index indicates a high level of genetic variation/diversity within the gene pool (1 mark).
- This increases the adaptability of the population to changing environmental selection pressures (e.g., disease, climate change), reducing the risk of extinction (1 mark).

Part (c):
- Use of DNA profiling / sequencing / molecular markers to identify genetic relationships and measure existing heterozygosity (1 mark).
- Keeping detailed records / pedigree databases / studbooks to plan matings or crosses between unrelated individuals (1 mark).
- Regular exchange of genetic material (e.g., swapping animals between zoos, transferring seeds or pollen between seed banks) to prevent isolated inbred populations (1 mark).
- Use of cryopreservation of gametes / artificial insemination / IVF / hand-pollination to introduce alleles from underrepresented or deceased individuals (1 mark).
- Avoidance of selective breeding for specific traits to preserve wild-type diversity (1 mark).
- Large population size maintained to minimize the effects of genetic drift (1 mark).

Part (a) [Max 6 marks]:
- Accept any six valid comparative points comparing seed banks to captive breeding (space, cost, lifespan/dormancy, scale of genetic diversity, ethics, ease of reintroduction).

Part (b)(i) [2 marks]:
- 1 mark for correct division of 14 by 50.
- 1 mark for correct value: 0.28 (allow 28% if working is shown, but index is ideally expressed as decimal).

Part (b)(ii) [2 marks]:
- 1 mark for linking high index to high genetic diversity/variation.
- 1 mark for linking high diversity to survival/adaptation advantage.

Part (c) [Max 6 marks]:
- Award 1 mark per valid scientific mechanism (e.g., studbooks, DNA profiling, exchange of individuals, cryopreservation, large gene pool maintenance, avoiding artificial selection).

Part (a):
- Flagellum / tail present which beats / moves to provide motility/propulsion to swim towards the egg (1 mark).
- Streamlined shape to reduce resistance/drag during movement (1 mark).
- Middle piece packed with mitochondria to produce ATP via aerobic respiration to power the flagellum (1 mark).
- Acrosome (specialized lysosome) at the head containing hydrolytic/acrosin enzymes to digest the jelly coat/zona pellucida of the egg (1 mark).
- Haploid nucleus containing 23 chromosomes to restore diploidy (46) upon fusion with the haploid egg nucleus (1 mark).
- Specialized receptors on the head membrane to bind to the glycoproteins of the zona pellucida (1 mark).

Part (b):
- In plants, two male gamete nuclei are involved in the fertilization process, whereas in mammals only one sperm gamete nucleus is involved (1 mark).
- In plants, one male nucleus fuses with the female egg cell to form a diploid (2n) zygote AND the second male nucleus fuses with two polar nuclei to form a triploid (3n) endosperm nucleus (double fertilization) (1 mark).
- In mammals, the sperm nucleus fuses only with the egg nucleus to form a diploid (2n) zygote; no endosperm or triploid tissue is formed (1 mark).
- Plant fertilization occurs inside the embryo sac within an ovule, whereas mammalian fertilization occurs in the oviduct / fallopian tube (1 mark).
- Mammals use a cortical reaction (hardening of the zona pellucida) to prevent polyspermy, whereas plants use chemical guidance/self-incompatibility mechanisms or degeneration of synergids to prevent multiple tube entry (1 mark).
- Plant gametes are delivered via a growing pollen tube, whereas animal gametes rely on sperm motility (swimming) to reach the egg (1 mark).

Part (c):
- Convert length from mm to \(\mu\text{m}\): \(4.8 \text{ mm} \times 1000 = 4800 \ \mu\text{m}\) (1 mark).
- Divide by the time duration (4 hours): \(\frac{4800 \ \mu\text{m}}{4 \text{ hours}}\) (1 mark).
- Calculate final rate: \(1200\) (1 mark).
- Provide correct units: \(\mu\text{m hr}^{-1}\) or \(\mu\text{m/hr}\) (1 mark).
- Alternative working: \(4.8 \text{ mm} / 4 \text{ hours} = 1.2 \text{ mm hr}^{-1}\) (1 mark) followed by conversion \(1.2 \times 1000 = 1200 \ \mu\text{m hr}^{-1}\) (2 marks) + units (1 mark).

Part (a) [Max 6 marks]:
- 1 mark for structure linked to 1 mark for function. Maximum 6 marks total.
- Examples: Middle piece with mitochondria (structure) -> ATP production for swimming (function); Acrosome with enzymes (structure) -> digestion of zona pellucida (function).

Part (b) [Max 6 marks]:
- Must be contrasts (comparisons of differences). Award marks for parallel points describing both plants and mammals.

Part (c) [4 marks]:
- 1 mark for correct conversion of unit (e.g., 4.8 mm to 4800 micrometers).
- 1 mark for dividing by 4.
- 1 mark for correct numerical answer: 1200.
- 1 mark for correct units: \(\mu\text{m hr}^{-1}\) / \(\mu\text{m/hr}\).

Step-by-step calculations for Part (b):

1. Calculate mean breaking mass (\(m\)) in kilograms:
- For \(Phormium\;tenax\): \((450+490+420+510+480)/5 = 470\text{ g} = 0.470\text{ kg}\).
- For \(Agave\;sisalana\): \((820+890+780+910+850)/5 = 850\text{ g} = 0.850\text{ kg}\).

2. Calculate breaking force (\(F = m \times g\)):
- \(Phormium\): \(0.470\text{ kg} \times 9.81\text{ N kg}^{-1} = 4.6107\text{ N}\).
- \(Agave\): \(0.850\text{ kg} \times 9.81\text{ N kg}^{-1} = 8.3385\text{ N}\).

3. Calculate cross-sectional area (\(A = \pi r^2\)):
- \(Phormium\) radius \(r = 0.24 / 2 = 0.12\text{ mm}\).
\(A = \pi \times (0.12)^2 = 0.045239\text{ mm}^2\).
- \(Agave\) radius \(r = 0.36 / 2 = 0.18\text{ mm}\).
\(A = \pi \times (0.18)^2 = 0.101788\text{ mm}^2\).

4. Calculate tensile strength (\(\text{Strength} = F / A\)):
- \(Phormium\): \(4.6107\text{ N} / 0.045239\text{ mm}^2 = 101.92\text{ N mm}^{-2}\).
- \(Agave\): \(8.3385\text{ N} / 0.101788\text{ mm}^2 = 81.92\text{ N mm}^{-2}\).

5. Round final answers to consistent and appropriate significant figures (typically 3 sig figs):
- \(Phormium\;tenax\) Tensile Strength = \(102\text{ N mm}^{-2}\)
- \(Agave\;sisalana\) Tensile Strength = \(81.9\text{ N mm}^{-2}\)

Part (a) [Maximum 6 marks]
- MP1: Soak leaves/stems in water for a defined duration (retting) to soften surrounding tissues (1)
- MP2: Select fibres of the same length and inspect for consistent physical condition (1)
- MP3: Secure fibre ends in clamp stands using padding/cardboard to prevent physical shearing at the clamps (1)
- MP4: Add standard masses systematically/at fixed intervals until failure (1)
- MP5: Record final mass supported prior to breaking (1)
- MP6: Replicate trial at least 5 times per species and outline a suitable safety precaution (e.g., cushioned landing zone, safety glasses) (1)

Part (b) [Maximum 6 marks]
- MP1: Correctly calculates mean breaking masses as \(0.470\text{ kg}\) and \(0.850\text{ kg}\) (1)
- MP2: Correctly calculates mean breaking forces as \(4.61\text{ N}\) (or \(4.6\text{ N}\)) and \(8.34\text{ N}\) (or \(8.3\text{ N}\)) (1)
- MP3: Correctly determines radius of both fibres (\(0.12\text{ mm}\) and \(0.18\text{ mm}\)) (1)
- MP4: Correctly calculates cross-sectional areas as \(0.0452\text{ mm}^2\) and \(0.102\text{ mm}^2\) (1)
- MP5: Correctly calculates both tensile strengths: \(Phormium\) in range \(101.9 - 102.0\text{ N mm}^{-2}\) AND \(Agave\) in range \(81.8 - 82.0\text{ N mm}^{-2}\) (1)
- MP6: Provides correct units (\(N mm^{-2}\) or \(MPa\)) and quotes answers to 3 significant figures (1)

Part (c) [Maximum 3 marks]
- MP1: Identifies that different species (or individual fibres) have different diameters/cross-sectional areas (1)
- MP2: Explains that a thicker fibre has more mechanical cells and breaks at a higher mass purely due to size, not material property (1)
- MP3: Concludes that dividing force by area standardises the size variable, allowing a fair comparison of intrinsic fibre strength (1)

Part (d) [Maximum 3 marks]
- MP1: Identifies one correct variable (e.g., retting duration, environmental humidity, rate of mass addition, plant tissue source) (1)
- MP2: Explains how this variable is controlled (e.g., keeping in a temperature-controlled room, using a timer to add weights every 10 seconds) (1)
- MP3: Explains that standardising this variable ensures the validity of the comparison (i.e. ensures only species differences affect the outcome) (1)

(a) A suitable table must have clearly structured headers with units, individual replicate data, and the calculated mean values correct to 2 decimal places:
- Mean at 20 °C: \(\frac{0.05 + 0.07 + 0.06}{3} = 0.06\) a.u.
- Mean at 35 °C: \(\frac{0.12 + 0.15 + 0.14}{3} = 0.137 \approx 0.14\) a.u.
- Mean at 50 °C: \(\frac{0.38 + 0.41 + 0.39}{3} = 0.393 \approx 0.39\) a.u.
- Mean at 65 °C: \(\frac{0.72 + 0.76 + 0.74}{3} = 0.74\) a.u.
- Mean at 80 °C: \(\frac{0.94 + 0.98 + 0.96}{3} = 0.96\) a.u.

(b) The graph should be plotted with:
- Temperature (°C) on the x-axis and Mean Absorbance (a.u.) on the y-axis.
- Sensible linear scales covering at least 50% of the grid.
- Points plotted accurately and connected with straight, dot-to-dot lines or a smooth curve.
- Range bars plotted vertically through each mean point, extending from the lowest replicate value to the highest replicate value (e.g., at 20 °C, the bar ranges from 0.05 to 0.07).

(c)
- Independent Variable: Temperature / °C
- Dependent Variable: Absorbance of the solution / a.u.
- Controlled Variable: e.g., Volume of distilled water in the tubes. This is achieved by using a graduated pipette or measuring cylinder to add exactly 10.0 cm³ of water to each test tube.

(d)
- Washing: Cutting the beetroot cylinders ruptures the cell membranes and tonoplast of cells along the cut edge, releasing betalain pigment onto the surface. Thorough washing removes this surface-bound pigment so that any pigment detected in the water baths during the experiment is solely due to the temperature treatment damaging cell membranes.
- Control Experiment: Run a control set of beetroot cylinders incubated in distilled water at room temperature (e.g., 20 °C) for the same duration to show that membrane leakage does not happen significantly in the absence of elevated temperatures, or use distilled water with no beetroot to calibrate/blank the colorimeter.

**Part (a) [4 Marks maximum]**
- **1 mark**: Table has complete headers with physical quantities and units: "Temperature / °C" and "Absorbance / arbitrary units (a.u.)" (or similar).
- **1 mark**: Raw data (Replicates 1, 2, 3) are clearly separated and correctly transcribed into the table.
- **1 mark**: A column for "Mean absorbance" is present and calculations are correct.
- **1 mark**: All raw and mean data values are recorded consistently to 2 decimal places.

**Part (b) [5 Marks maximum]**
- **1 mark (A - Axes)**: Correct axes chosen (Temperature on x-axis, Mean Absorbance on y-axis) with complete labels and units.
- **1 mark (S - Scale)**: Scales on both axes are linear, easy to read, and occupy at least half the available grid size (no awkward scale values like 3 or 7 squares per unit).
- **1 mark (P - Plotting)**: All 5 mean data points are plotted accurately (within half a small grid square).
- **1 mark (L - Line)**: Points joined with clean, ruled, straight lines (dot-to-dot) OR a smooth curve of best fit. Reject a thick or multiple-sketched line.
- **1 mark (R - Range bars)**: Correctly drawn vertical range bars showing the maximum and minimum values for each of the five temperatures (e.g., at 65 °C, a bar from 0.72 to 0.76).

**Part (c) [3 Marks maximum]**
- **1 mark**: Correctly identifies both Independent Variable (Temperature) AND Dependent Variable (Absorbance of the solution).
- **1 mark**: Identifies one relevant control variable (e.g., beetroot surface area/dimensions, volume of water, source of beetroot, incubation time).
- **1 mark**: Describes how this variable is controlled (e.g., cut cylinders to exactly 2.0 cm length using a ruler and scalpel; use a pipette to measure exactly 10.0 cm³ of distilled water; use a stopwatch to time exactly 20 minutes).

**Part (d) [4 Marks maximum]**
- **1 mark**: Washing removes pigment released from damaged cells along the cut edges of the beetroot cylinder.
- **1 mark**: This ensures that any measured absorbance is due solely to the temperature treatment, ensuring the validity of the results.
- **1 mark**: A suitable control experiment involves incubating beetroot cylinders under identical conditions but keeping them at a constant reference low temperature (e.g., 4 °C or 20 °C) / or using water without beetroot.
- **1 mark**: To show that no pigment leaks / no change in absorbance occurs without the elevated temperature treatment.

**Part (a)**
A null hypothesis states that there is no relationship or significant difference between the variables.
- Null Hypothesis: "There is no significant difference in the mean fresh mass of radish seedlings grown in different concentrations of nitrate ions."

**Part (b)**
A suitable table must have independent variable in the first column, raw data replicates in the middle columns, and the calculated mean in the final column. Units must only be in the column headers.

| Nitrate concentration / % | Fresh mass of individual seedling / g | | | | | Mean fresh mass / g |
|---|---|---|---|---|---|---|
| | Seedling 1 | Seedling 2 | Seedling 3 | Seedling 4 | Seedling 5 | |
| 0.00 | | | | | | |
| 0.01 | | | | | | |
| 0.05 | | | | | | |
| 0.10 | | | | | | |
| 0.20 | | | | | | |

**Part (c)**
An effective, valid, and reliable method must include:
1. Seed germination under uniform conditions, selecting seedlings of the same age and initial length/mass.
2. Supporting each seedling in a container (e.g., tube/beaker) of the respective solution using cotton wool around the stem, ensuring roots are fully submerged.
3. Controlling temperature (e.g., in a thermostatically controlled room/incubator) and light (e.g., using a growth lamp with a fixed 16-hour light/8-hour dark photoperiod).
4. Providing constant aeration (e.g., bubbling air into the solutions with an aquarium pump) to prevent anaerobic root respiration.
5. Topping up the solutions with distilled water regularly to replace water lost by evaporation/transpiration, ensuring solute concentration remains constant.
6. At 4 weeks, removing seedlings, blotting them gently with a paper towel to remove surface water, and weighing them on a balance to the nearest 0.01 g.
7. Replicating the entire set-up at least three times to obtain a mean and identify anomalies.

**Part (d)**
- Anomalous result: \(0.45\text{ g}\)
- Justification: It is significantly lower than and lies completely outside the tight cluster/range of the other four replicates (which range from \(1.24\) to \(1.35\text{ g}\)).
- Mean calculation:
$$\text{Mean} = \frac{1.24 + 1.35 + 1.30 + 1.28}{4} = \frac{5.17}{4} = 1.2925\text{ g} \approx 1.29\text{ g}$$

**Part (e)**
- Nitrate ions are absorbed by plants and combined with organic acids (from respiration) to synthesise amino acids, which are the building blocks of proteins/enzymes needed for structural growth.
- They are also components of chlorophyll and nucleic acids (DNA/RNA); without them, photosynthesis, cell division, and protein synthesis are severely restricted, causing chlorosis and stunted growth.

**Part (a) [2 Marks]**
- **MP1:** States "no significant difference" or "no correlation/effect". (1)
- **MP2:** Correctly identifies both independent variable (nitrate ion concentration) and dependent variable (fresh mass / growth of radish seedlings). (1)
*Accept:* "The concentration of nitrate ions has no significant effect on the fresh mass of radish seedlings."
*Reject:* Hypotheses that predict a direction (e.g. "higher nitrate increases growth").

**Part (b) [3 Marks]**
- **MP1:** Column heading for independent variable with correct unit: "Nitrate concentration / %" (or "(w/v)"). (1)
- **MP2:** Headings for raw data replicates with correct unit: "Fresh mass of individual seedling / g" (or equivalent) showing space for 5 replicates. (1)
- **MP3:** Headings for calculated mean with correct unit: "Mean fresh mass / g". (1)
*Note:* Penalise 1 mark if units are written inside the data cells.

**Part (c) [6 Marks]**
- **MP1 (Initial standardization):** Germinate seeds and select seedlings of similar size/mass to start. (1)
- **MP2 (Apparatus setup):** Use of cotton wool to suspend seedlings over the culture solution. (1)
- **MP3 (Abiotic control):** Keep all treatments at the same temperature OR light intensity/duration. (1)
- **MP4 (Aeration/Water control):** Aerate the solutions using a pump OR top up solutions with distilled water to maintain concentration. (1)
- **MP5 (Measurement method):** Blot seedlings dry before weighing on an electronic balance (to 2 decimal places / 0.01 g). (1)
- **MP6 (Reliability):** Repeat the entire experiment at least 3 times to calculate a mean / identify anomalies. (1)

**Part (d) [3 Marks]**
- **MP1:** Identifies \(0.45\text{ g}\) as the anomaly. (1)
- **MP2 (Justification):** Explains that this value is much lower than the others / does not fit the pattern / is far outside the range of \(1.24\text{ g}\) to \(1.35\text{ g}\). (1)
- **MP3 (Calculation):** Correctly calculates the mean as \(1.29\text{ g}\) (accept \(1.293\text{ g}\) or \(1.2925\text{ g}\)). (1)

**Part (e) [2 Marks]**
- **MP1:** Nitrates are used to produce amino acids / proteins / nucleic acids / chlorophyll. (1)
- **MP2:** Lack of these biological molecules prevents cell division / translation / photosynthesis, leading to reduced cell growth/chlorosis. (1)