2023 Edexcel AS Level Biology A (Salters-Nuffield) (8BN0) Practice Paper with Answers

When a blood vessel is damaged, platelets and damaged tissue release thromboplastin. In the presence of calcium ions, thromboplastin acts as an enzyme catalyst to convert the inactive plasma protein prothrombin into the active enzyme thrombin.

Award 1 mark for the correct option (b). Correct answer explains that thromboplastin catalyses the conversion of prothrombin to thrombin in the presence of calcium ions.

Since three nucleotides form a single codon, deleting exactly three nucleotides results in the loss of one amino acid (phenylalanine at position 508) from the polypeptide chain. Because the deletion is a multiple of three, it does not cause a frameshift mutation, meaning downstream amino acids remain unaffected.

Award 1 mark for selecting option (b). The candidate must recognize that deleting three bases removes one complete codon without causing a frameshift.

Sucrose is a disaccharide formed by a condensation reaction between a molecule of \(\alpha\)-glucose and a molecule of fructose. A water molecule is released during this reaction, forming a 1,2-glycosidic bond.

Award 1 mark for selecting option (c), which correctly identifies the two constituent monosaccharides (glucose and fructose) and the reaction type (condensation).

Phospholipids are amphipathic molecules. The phosphate heads are hydrophilic (polar) and orient outward towards the aqueous environments of the cytoplasm and intercellular space, while the fatty acid tails are hydrophobic (non-polar) and orient inward, away from water, to form the hydrophobic core of the bilayer.

Award 1 mark for selecting option (b). Candidate must correctly identify the hydrophilic/hydrophobic nature of the head and tail, along with their correct orientation.

During ventricular systole, the left ventricle contracts, rapidly increasing the intraventricular pressure. Once this pressure exceeds the pressure in the aorta, the semilunar valve is forced open, and blood is ejected from the ventricle into the aorta.

Award 1 mark for selecting option (d). The candidate must identify that ventricular contraction (systole) generates the high pressure required to exceed aortic pressure and open the semilunar valve.

After the first division in \(^{14}\text{N}\), all DNA molecules consist of one heavy and one light strand (100% hybrid). During the second division in \(^{14}\text{N}\), these strands separate and act as templates. The heavy strands form hybrid DNA, while the light strands form completely light DNA. This results in 50% hybrid DNA and 50% light DNA.

Award 1 mark for selecting option (c), representing the correct percentage of hybrid molecules based on the semi-conservative model of replication.

HDLs have a higher protein-to-lipid ratio (making them denser) and function to transport cholesterol away from body tissues to the liver for excretion or recycling, thereby reducing the risk of atherosclerosis. LDLs have more lipid and transport cholesterol to the body tissues.

Award 1 mark for selecting option (a). Candidate must correctly identify that HDLs have a higher protein proportion and carry cholesterol to the liver.

Since the parents have normal pigmentation but have produced an offspring with the recessive condition (aa), both parents must be heterozygous carriers (Aa). A cross between two carriers (Aa x Aa) yields a genotype probability of 25% AA, 50% Aa, and 25% aa. Therefore, the probability of having a carrier child (Aa) is 0.50 (or 50%).

Award 1 mark for selecting option (b), demonstrating understanding of autosomal recessive monohybrid inheritance and carrier probability.

First, calculate the absolute reduction in the mortality rate: \(120 - 42 = 78\) per 10,000 person-years. Next, calculate this reduction as a percentage of the rate for inactive individuals: \((78 / 120) \times 100 = 65\%\).

1.00 mark for calculating the correct absolute difference of 78. 1.75 marks for calculating the correct percentage reduction of 65%. Award full marks for correct answer without working.

1. Probability that the mother is a carrier = \(1/25 = 0.04\). 2. If both parents are carriers, the probability of them passing the recessive mutant allele to their offspring is \(0.5\) for each parent. 3. Total probability of an affected child = \(\text{P(mother is carrier)} \times \text{P(mother passes allele)} \times \text{P(father passes allele)}\) which equals \(1/25 \times 1/2 \times 1/2 = 1/100 = 0.01\) (or 1%).

1.00 mark for identifying the correct probabilities of allele transmission (0.5 for mother, 0.5 for father). 1.75 marks for calculating the final correct probability of 0.01 (or 1% / 1/100). Accept equivalent decimal, fraction, or percentage.

1. Calculate pulse pressure: \(\text{systolic} - \text{diastolic} = 122 - 78 = 44\text{ mmHg}\). 2. Use the MAP formula: \(\text{MAP} = 78 + \frac{1}{3}(44) = 78 + 14.67 = 92.67\text{ mmHg}\). 3. Round to one decimal place: \(92.7\text{ mmHg}\).

1.00 mark for finding the correct pulse pressure of 44 mmHg. 1.75 marks for the correct final MAP of 92.7 mmHg (accept 92.7, reject 92.67 or 93 due to rounding instructions).

Substitute the values directly into Fick's Law equation: \(\text{Relative Rate} = \frac{(1.2 \times 10^{-2}) \times 4.0}{0.5} = \frac{0.048}{0.5} = 0.096\).

1.00 mark for calculating correct numerator (0.048). 1.75 marks for final calculated relative rate of 0.096.

1. Calculate total fat mass: \(24 + 36 + 15 = 75\text{ g}\). 2. Calculate total unsaturated fat mass (monounsaturated + polyunsaturated): \(36 + 15 = 51\text{ g}\). 3. Calculate percentage: \((51 / 75) \times 100 = 68.0\%\).

1.00 mark for identifying total fat mass as 75g and total unsaturated fat mass as 51g. 1.75 marks for calculating the correct percentage to 3 significant figures (68.0%).

1. Calculate absolute change in rate: \(4.5 - 1.2 = 3.3\) picomoles per second. 2. Calculate percentage increase: \((3.3 / 1.2) \times 100 = 275\%\).

1.00 mark for correct absolute change in rate of 3.3. 1.75 marks for correct final percentage increase of 275%.

1. Calculate BMI: \(\text{BMI} = \frac{\text{mass in kg}}{(\text{height in m})^2} = \frac{82.5}{1.72^2} = \frac{82.5}{2.9584} = 27.886\dots\text{ kg m}^{-2}\). 2. Rounded to 1 decimal place: 27.9. 3. Classification: Overweight (defined as a BMI between 25.0 and 29.9).

1.00 mark for correct calculation of BMI to 27.9 (accept 27.89). 1.75 marks for linking the correct BMI to the classification of 'Overweight' (accept 'overweight').

1. Total bases = \(2400 \times 2 = 4800\) bases. 2. Since Cytosine = 35%, Guanine = 35% (total CG content = 70%). 3. This leaves 30% for Adenine + Thymine. 4. Since Adenine equals Thymine, Adenine makes up 15% of the total bases. 5. Total Adenine bases = \(0.15 \times 4800 = 720\).

1.00 mark for calculating total bases as 4800 OR identifying that Adenine constitutes 15% of total bases. 1.75 marks for calculating the correct number of adenine bases (720).

First, state Fick's Law equation incorporating the diffusion coefficient: Rate = (Surface Area \times Concentration Difference \times Diffusion Coefficient) / Thickness. Substitute the values: Rate = (1.5 \times 10^{-5} \times 0.040 \times 2.5 \times 10^{-11}) / (7.5 \times 10^{-9}). First calculate the concentration gradient term: (1.5 \times 10^{-5} \times 0.040) / (7.5 \times 10^{-9}) = 6.0 \times 10^{-7} / 7.5 \times 10^{-9} = 80. Then multiply by the diffusion coefficient: 80 \times 2.5 \times 10^{-11} = 2.0 \times 10^{-9} mol s^{-1}.

- 1 mark for correct substitution of values into the formula: (1.5 \times 10^{-5} \times 0.040 / 7.5 \times 10^{-9}) \times 2.5 \times 10^{-11}.
- 1 mark for calculating intermediate value of 80 (gradient term).
- 0.75 marks for the final correct answer in standard form with correct units: 2.0 \times 10^{-9} mol s^{-1} (accept 2 \times 10^{-9} mol s^{-1}, reject other units or non-standard form).

When the endothelium is damaged, collagen is exposed, causing platelets to adhere and release thromboplastin. Thromboplastin catalyzes the conversion of prothrombin into active thrombin in the presence of calcium ions. Thrombin then acts as a protease to catalyze the hydrolysis of soluble fibrinogen into insoluble fibrin, which forms a mesh that traps blood cells and forms a secure clot.

- 1 mark for stating that platelets aggregate at the damaged endothelium/exposed collagen and release thromboplastin.
- 1 mark for explaining that thromboplastin converts inactive prothrombin into active thrombin (in the presence of calcium ions).
- 0.75 marks for stating that thrombin catalyzes the conversion of soluble fibrinogen into insoluble fibrin, trapping blood cells to form the clot.

1. Calculate BMI: BMI = mass / height^2 = 94.5 / (1.78)^2 = 94.5 / 3.1684 = 29.826... = 29.8 kg m^-2.
2. Calculate WHR: WHR = waist / hip = 102 / 96 = 1.0625 = 1.06.
3. Identify risk category: His BMI of 29.8 is classified as overweight but not obese (threshold 30.0). However, his WHR of 1.06 is well above the high-risk threshold of 0.90 for males. Thus, WHR places him in a higher risk category.

- 1 mark for correct BMI calculation and unit: 29.8 kg m^-2 (accept 29.8, reject 30 or 29.83).
- 1 mark for correct WHR calculation: 1.06 (accept 1.063, reject 1.1).
- 0.75 marks for identifying that WHR represents a higher risk category because 1.06 is greater than the threshold of 0.90 (abdominal obesity), whereas his BMI (29.8) is below the obesity threshold (30.0).

In cystic fibrosis, impaired chloride transport leads to thick, sticky mucus that blocks the pancreatic duct. This prevents digestive enzymes, including pancreatic lipase, from being secreted into the duodenum. Consequently, dietary lipids are not hydrolyzed into fatty acids and monoglycerides. The absence of these breakdown products lowers their concentration in the intestinal lumen, reducing the concentration gradient and severely decreasing their absorption across the epithelial lining of the small intestine.

- 1 mark for stating that thick mucus blocks pancreatic secretions, preventing pancreatic lipase from reaching the duodenum.
- 1 mark for explaining that this leads to reduced hydrolysis of lipids into fatty acids and monoglycerides.
- 0.75 marks for explaining that this decreases the concentration gradient of digestion products across the ileum wall, leading to malabsorption of lipids.

1. In double-stranded DNA, Cytosine (C) = Guanine (G). Given C = 544, G must also be 544. Therefore, total C + G = 1088 nucleotides. The remaining nucleotides are Adenine (A) and Thymine (T): 3400 - 1088 = 2312. Since A = T, the number of adenine nucleotides = 2312 / 2 = 1156.
2. Purine bases are Adenine (A) and Guanine (G). Total purines = 1156 + 544 = 1700. The percentage of purines is (1700 / 3400) * 100 = 50%. (Alternatively, double-stranded DNA always contains exactly 50% purine bases due to complementary pairing with pyrimidines).

- 1 mark for calculating the correct number of A and T nucleotides combined (2312) or stating C=G=544.
- 1 mark for calculating the correct number of adenine nucleotides: 1156.
- 0.75 marks for stating that the percentage of purines is 50% (accept correct working showing 1700/3400 = 50%).

A single amino acid substitution changes the primary structure of the polypeptide chain, which alters the specific R-groups present. This can prevent the formation of critical bonds (such as hydrogen, ionic, or disulfide bonds) that stabilize the protein's overall folding. The alteration in folding changes the tertiary structure (3D shape) of the entire enzyme. Since the precise shape of the active site is determined by this overall tertiary conformation, the active site becomes deformed and is no longer complementary to the substrate. Consequently, the substrate cannot bind, no enzyme-substrate complexes can form, and catalytic activity is lost.

- 1 mark for explaining that a change in the primary structure alters the R-groups, preventing the formation of specific ionic, hydrogen, or disulfide bonds.
- 1 mark for stating that this alters the folding and overall tertiary structure (3D shape) of the enzyme.
- 0.75 marks for explaining that this changes the shape of the active site so that it is no longer complementary, preventing the formation of enzyme-substrate complexes.

1. Calculate Stroke Volume (SV): SV = End-diastolic volume - End-systolic volume = 138 cm^3 - 53 cm^3 = 85 cm^3.
2. Calculate Heart Rate (HR): One cycle lasts 0.80 seconds. HR = 60 seconds / 0.80 seconds/beat = 75 beats per minute (bpm).
3. Calculate Cardiac Output (CO): CO = SV * HR = 85 cm^3 * 75 bpm = 6375 cm^3 min^-1.
4. Convert units to dm^3 min^-1: 6375 / 1000 = 6.375 dm^3 min^-1 (accept 6.4 dm^3 min^-1 rounded to 2 significant figures).

- 1 mark for calculating correct Stroke Volume: 85 cm^3.
- 1 mark for calculating correct Heart Rate: 75 bpm.
- 0.75 marks for correct final cardiac output with correct units: 6.375 dm^3 min^-1 or 6.4 dm^3 min^-1 (deduct 1 mark if calculated as 6375 but unit conversion to dm^3 min^-1 is incorrect or missing).

Water is a polar molecule with a slight negative charge on the oxygen atom and a slight positive charge on the hydrogen atoms. This dipolar nature allows water to form hydrogen bonds with other water molecules (causing cohesion) and with other polar or ionic substances. Because of this, key nutrients (like glucose and amino acids) and ions (like sodium) easily dissolve in water because hydration shells form around them. This excellent solvent property allows the blood plasma to transport a wide range of dissolved substances. Furthermore, the strong cohesive forces between water molecules allow blood to flow as a continuous column under pressure during mass transport through the blood vessels.

- 1 mark for explaining that water is dipolar because oxygen has a slight negative charge and hydrogen has a slight positive charge, allowing it to form hydrogen bonds.
- 1 mark for explaining that this polarity allows water to act as a solvent by surrounding and dissolving polar molecules/ionic substances for transport.
- 0.75 marks for explaining that cohesion between water molecules allows blood to flow under pressure / as a continuous stream through blood vessels (mass flow).

1. First, calculate the total person-years of exposure: \(12,500 \text{ participants} \times 8 \text{ years} = 100,000 \text{ person-years}\). 2. Next, calculate the incidence rate per person-year: \(\frac{320 \text{ cases}}{100,000 \text{ person-years}} = 0.0032 \text{ cases per person-year}\). 3. Convert this rate to the rate per 1,000 person-years: \(0.0032 \times 1,000 = 3.2 \text{ cases per 1,000 person-years}\).

1 mark: Correct calculation of total person-years (100,000 person-years). 1 mark: Correct calculation of incidence rate (3.2). 0.75 mark: Correct units (per 1,000 person-years).

1. Convert the cell volume from \(\text{cm}^3\) to \(\text{dm}^3\): \(4.2 \times 10^{-9} \text{ cm}^3 \times 10^{-3} \text{ dm}^3 \text{ cm}^{-3} = 4.2 \times 10^{-12} \text{ dm}^3\). 2. Calculate the concentration in \(\text{mol dm}^{-3}\): \(\text{Concentration} = \frac{2.52 \times 10^{-14} \text{ moles}}{4.2 \times 10^{-12} \text{ dm}^3} = 0.006 \text{ mol dm}^{-3}\). 3. Convert \(\text{mol dm}^{-3}\) to \(\text{mmol dm}^{-3}\): \(0.006 \text{ mol dm}^{-3} \times 1000 = 6.0 \text{ mmol dm}^{-3}\).

1 mark: Correct conversion of volume to \(\text{dm}^3\) (\(4.2 \times 10^{-12}\)). 1 mark: Correct concentration calculation in \(\text{mol dm}^{-3}\) (\(0.006\)). 0.75 mark: Correct conversion to \(\text{mmol dm}^{-3}\) (\(6.0\)) with appropriate units.

1. Calculate target mass using the formula \(\text{BMI} = \frac{\text{mass}}{\text{height}^2}\): \(24.0 = \frac{\text{mass}}{1.75^2} \Rightarrow \text{mass} = 24.0 \times 3.0625 = 73.5 \text{ kg}\). 2. Calculate mass to lose: \(88.0 \text{ kg} - 73.5 \text{ kg} = 14.5 \text{ kg}\). 3. Calculate energy deficit in kJ: \(14.5 \text{ kg} \times 30,000 \text{ kJ kg}^{-1} = 435,000 \text{ kJ}\). 4. Convert to MJ: \(\frac{435,000 \text{ kJ}}{1,000} = 435 \text{ MJ}\).

1 mark: Correct calculation of target weight (73.5 kg) or mass loss (14.5 kg). 1 mark: Correct calculation of energy deficit in kJ (435,000 kJ). 0.75 mark: Correct conversion to MJ (435 MJ) with units.

1. Total bases = \(450 \times 2 = 900\) bases. 2. Cytosine (C) = \(32\%\) of 900 = \(288\) bases. Therefore, Guanine (G) = \(288\) bases. This gives \(288\) C-G base pairs. 3. Remaining percentage for Adenine (A) and Thymine (T) = \(100\% - (32\% + 32\%) = 36\%\). 4. Number of A-T bases = \(36\%\) of 900 = \(324\) bases, which gives \(162\) A-T base pairs. 5. C-G pairs form 3 hydrogen bonds each: \(288 \times 3 = 864\). 6. A-T pairs form 2 hydrogen bonds each: \(162 \times 2 = 324\). 7. Total hydrogen bonds = \(864 + 324 = 1188\).

1 mark: Correct identification of the number of C-G pairs (288) and A-T pairs (162). 1 mark: Correct application of hydrogen bond rules (3 for C-G, 2 for A-T). 0.75 mark: Correct total of 1188.

1. Convert the plasma sample volume to \(\text{dm}^3\): \(2.5 \text{ cm}^3 = 0.0025 \text{ dm}^3\). 2. Calculate the mass of fibrinogen in the sample: \(3.0 \text{ g dm}^{-3} \times 0.0025 \text{ dm}^3 = 0.0075 \text{ g}\), which is \(7.5 \text{ mg}\). 3. Calculate the mass of fibrin produced using the 85% conversion efficiency: \(7.5 \text{ mg} \times 0.85 = 6.375 \text{ mg}\). 4. Round to 3 significant figures: \(6.38 \text{ mg}\).

1 mark: Correct calculation of total fibrinogen mass in the sample (7.5 mg or 0.0075 g). 1 mark: Correct application of 85% conversion efficiency (6.375 mg). 0.75 mark: Correct rounding to 3 significant figures with units (6.38 mg).

1. Each amino acid is coded by a triplet of nucleotides (codon). For \(340\) amino acids, this requires: \(340 \times 3 = 1020\) nucleotides. 2. The stop codon does not code for an amino acid, but is still present in the mRNA molecule as a triplet: \(1 \text{ stop codon} \times 3 = 3\) nucleotides. 3. Total minimum nucleotides = \(1020 + 3 = 1023\) nucleotides.

1 mark: Correct calculation of nucleotides required for the amino acids (1020). 1 mark: Correct addition of stop codon nucleotides (+3). 0.75 mark: Correct final answer of 1023.

1. Calculate resting cardiac output: \(\text{Resting Cardiac Output} = 78 \text{ bpm} \times 65 \text{ cm}^3 = 5070 \text{ cm}^3 \text{ min}^{-1}\). 2. Calculate exercise cardiac output (110% increase means multiplying by 2.10): \(5070 \times 2.10 = 10647 \text{ cm}^3 \text{ min}^{-1}\). 3. Calculate exercise heart rate: \(\text{Exercise HR} = \frac{10647 \text{ cm}^3 \text{ min}^{-1}}{85 \text{ cm}^3} \approx 125.26 \text{ bpm}\). 4. Round to the nearest whole number: \(125 \text{ bpm}\).

1 mark: Correct calculation of resting cardiac output (5070 cm3 min-1). 1 mark: Correct calculation of exercise cardiac output (10,647 cm3 min-1). 0.75 mark: Correct heart rate calculation rounded to nearest whole number (125 bpm).

1. Calculate the initial number of red blood cells in the 0.5 mm3 aliquot: \(4.8 \times 10^6 \text{ cells mm}^{-3} \times 0.5 \text{ mm}^3 = 2.4 \times 10^6 \text{ cells}\). 2. Find the percentage of remaining intact cells: \(100\% - 45\% = 55\%\). 3. Calculate the number of intact cells: \(2.4 \times 10^6 \times 0.55 = 1.32 \times 10^6\) intact cells.

1 mark: Correct calculation of initial cell count in the aliquot (2.4 x 10^6). 1 mark: Multiplication by the remaining percentage (0.55). 0.75 mark: Correct final value in scientific notation (1.32 x 10^6).

When an atheroma ruptures, the endothelial lining of the coronary artery is damaged, exposing collagen fibres in the blood vessel wall to the blood. Platelets immediately bind to this exposed collagen and become activated. The damaged tissue and activated platelets release an enzyme called thromboplastin. In the presence of calcium ions and clotting factors, 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. Fibrin forms a mesh of fibres that traps red blood cells and more platelets, forming a blood clot (thrombus). If this clot blocks the lumen of the coronary artery, blood flow to the downstream heart muscle (myocardium) is severely restricted or stopped. Consequently, the heart muscle cells are deprived of oxygen and glucose, which are essential for aerobic respiration. The cells are forced to respire anaerobically, leading to lactic acid accumulation, which lowers the pH and damages enzymes. Ultimately, the lack of ATP and build-up of toxic waste causes cell death (necrosis) of the cardiac tissue, resulting in a myocardial infarction.

Level 1 (1-2 marks): Answer describes some parts of the clotting cascade or mentions coronary blockage/heart attack. The explanation is fragmented, lacks detail, and has limited logical structure. Level 2 (3-4 marks): Answer describes the sequential steps of the clotting cascade (e.g., thromboplastin, prothrombin to thrombin, fibrinogen to fibrin) OR explains how a blockage leads to lack of oxygen and death of heart muscle. The explanation shows a reasonable logical flow with some scientific terms used correctly. Level 3 (5-6 marks): Answer provides a complete, detailed, and logically structured account of BOTH the blood clotting cascade (including roles of collagen, thromboplastin, calcium ions, thrombin, and fibrin) AND the progression to myocardial infarction (occlusion of coronary artery, lack of oxygen/aerobic respiration, anaerobic respiration, cell death). Indicative content points: 1. Rupture of atheroma exposes collagen. 2. Platelets release thromboplastin. 3. Thromboplastin catalyses conversion of prothrombin to thrombin (requires calcium ions). 4. Thrombin catalyses conversion of soluble fibrinogen to insoluble fibrin, forming a mesh/clot. 5. Clot blocks/occludes the coronary artery, cutting off blood flow to heart muscle. 6. Muscle cells lack oxygen/glucose for aerobic respiration. 7. Anaerobic respiration leads to lactic acid build-up, cell damage, and cell death/myocardial infarction.

Calcium ions are essential for the formation of calcium pectate, which cements adjacent plant cell walls together in the middle lamella. Without calcium, plants show stunted growth and weak cell walls.

1 mark for correct identification of calcium pectate and its location in the middle lamella (B).

Pluripotent stem cells can differentiate into almost all cell types of the body (from the three germ layers), but unlike totipotent cells, they cannot form extra-embryonic tissues such as the placenta.

1 mark for distinguishing pluripotency from totipotency and multipotency (B).

Cellulose is a linear, unbranched polymer of \(\beta\)-glucose molecules linked by 1,4-glycosidic bonds, whereas starch is made of \(\alpha\)-glucose monomers and contains branches in its amylopectin component.

1 mark for identifying the polymer and bond type specific to cellulose (A).

DNA replication (synthesis) occurs during the S phase of interphase. During this phase, the mass of DNA in the cell increases continuously until it has doubled.

1 mark for identifying S phase as the period of DNA replication (B).

To prevent germination, slow down decay, and reduce cellular respiration, seeds are dried (low moisture) and kept frozen (low temperature, typically around \(-20^\circ\text{C}\)).

1 mark for identifying the correct low moisture and low temperature combination (B).

Increased acetylation of histone proteins reduces their positive charge, weakening their interaction with DNA. This relaxes the chromatin structure, making the DNA more accessible to RNA polymerase and transcription factors, thereby increasing transcription.

1 mark for identifying histone acetylation as a promoter of transcription (C).

The zone of inhibition is the clear area around the disk where bacteria cannot grow. A larger zone indicates that the active antimicrobial substance in the garlic extract is more effective at preventing bacterial growth.

1 mark for linking a larger zone of inhibition to greater antimicrobial effectiveness (B).

Continuous variation (such as human height) is typically polygenic (controlled by many genes) and exhibits a wide range of phenotypes that can be modified by environmental factors (e.g., nutrition).

1 mark for recognizing that continuous variation is polygenic and environmentally influenced (C).

First, calculate the total number of cells undergoing mitosis: \( 28 \text{ (prophase)} + 12 \text{ (metaphase)} + 8 \text{ (anaphase)} + 15 \text{ (telophase)} = 63 \text{ dividing cells} \).
Next, divide the number of dividing cells by the total number of cells: \( \frac{63}{342} \approx 0.18421 \).
Finally, multiply by 100 to convert to a percentage and round to three significant figures: \( 0.18421 \times 100 = 18.421\% \rightarrow 18.4\% \).

1. **Method Mark (1.0 mark):** Correct sum of active mitotic cells: \( 28 + 12 + 8 + 15 = 63 \) cells.
2. **Accuracy Mark (1.0 mark):** Correct division by total cells to find fraction: \( \frac{63}{342} \) or \( 0.184 \).
3. **Precision and Units Mark (0.75 marks):** Final answer expressed as a percentage to 3 significant figures: \( 18.4\% \) (Accept 18.4 without the percentage sign if units are assumed, but reject 18% or 18.42%).

Tensile strength is calculated using the formula:
\( \text{Tensile Strength} = \frac{\text{Force (N)}}{\text{Cross-sectional Area (mm}^2\text{)}} \)
Substitute the given values into the equation:
\( \text{Tensile Strength} = \frac{13.5 \text{ N}}{0.045 \text{ mm}^2} = 300 \text{ N mm}^{-2} \).

1. **Formula Mark (1.0 mark):** Use of correct formula \( \text{Tensile Strength} = \frac{\text{Force}}{\text{Area}} \).
2. **Calculation Mark (1.0 mark):** Correct substitution of values: \( \frac{13.5}{0.045} \).
3. **Accuracy Mark (0.75 marks):** Correct final answer: \( 300 \text{ N mm}^{-2} \) (Accept 300).

Pluripotent stem cells (such as embryonic stem cells) have the capacity to give rise to any cell type derived from the three germ layers (ectoderm, endoderm, mesoderm) but cannot form extraembryonic tissues like the placenta. Multipotent stem cells (such as adult stem cells in bone marrow) are more restricted and can only differentiate into cell types of a closely related family of cells (e.g., hematopoietic stem cells into red blood cells, white blood cells, and platelets).

1. **Pluripotency description (1.0 mark):** State that pluripotent cells can differentiate into almost all cell types / any cell type in the body (except placenta/extraembryonic tissue).
2. **Multipotency description (1.0 mark):** State that multipotent cells are limited to a specific range or lineage of specialized cell types (e.g., blood cells from bone marrow stem cells).
3. **Comparative clarity (0.75 marks):** Correctly contrast the hierarchy of potency, emphasizing that pluripotent cells have a wider/less restricted differentiation potential than multipotent cells.

1. Calculate total number of individuals, \( N \):
\( N = 15 + 8 + 7 = 30 \)
\( N(N-1) = 30 \times 29 = 870 \)

2. Calculate \( n(n-1) \) for each species:
- Silver birch: \( 15 \times 14 = 210 \)
- English oak: \( 8 \times 7 = 56 \)
- Rowan: \( 7 \times 6 = 42 \)

3. Sum these values: \( \sum n(n-1) = 210 + 56 + 42 = 308 \)

4. Divide \( N(N-1) \) by \( \sum n(n-1) \):
\( D = \frac{870}{308} \approx 2.8247 \)
To 2 decimal places, \( D = 2.82 \).

1. **Method Mark (1.0 mark):** Correctly calculated numerator value: \( N(N-1) = 870 \).
2. **Method Mark (1.0 mark):** Correctly calculated denominator sum: \( \sum n(n-1) = 308 \) (working shown for individual species products: 210, 56, 42).
3. **Accuracy Mark (0.75 marks):** Final correct value of \( 2.82 \) to two decimal places. (Reject 2.8 or 2.825).

Proteins synthesized on the rough endoplasmic reticulum (RER) are transported to the Golgi apparatus in transition vesicles. Within the cisternae of the Golgi, these proteins are chemically modified (e.g., by adding carbohydrate groups to form glycoproteins, or lipid groups to form lipoproteins). The modified proteins are then sorted and packaged into secretory vesicles, which bud off from the Golgi and move along microtubules to fuse with the cell surface membrane, releasing the proteins via exocytosis.

1. **Modification Mark (1.0 mark):** Mention modification of proteins (e.g., addition of carbohydrate groups / glycosylation / lipid groups).
2. **Packaging Mark (1.0 mark):** Mention packaging of these proteins into (secretory) vesicles.
3. **Transport/Destination Mark (0.75 marks):** Reference to sending these vesicles to the cell surface membrane for release / exocytosis.

Magnesium ions are a key central component of the chlorophyll molecule. When magnesium is deficient in the soil, the plant cannot synthesize adequate chlorophyll, which results in the loss of green pigmentation (chlorosis), causing leaves to turn yellow. With less chlorophyll, the plant absorbs less light energy, decreasing the rate of photosynthesis. This leads to a reduction in the production of carbohydrates (such as glucose), which are vital substrates for cellular respiration and the synthesis of structural materials (like cellulose) and proteins needed for cell division and growth.

1. **Biochemical link (1.0 mark):** Explain that magnesium ions are required for the synthesis of chlorophyll.
2. **Functional effect (1.0 mark):** Link chlorophyll deficiency to reduced light absorption and a lower rate of photosynthesis.
3. **Growth impact (0.75 marks):** Connect reduced photosynthesis to decreased production of glucose/sugars, resulting in less respiration/energy or less structural material (cellulose/proteins) for plant growth.

DNA methylation is an epigenetic mechanism where methyl groups are covalently added to cytosine bases in the DNA, typically at CpG islands located in promoter regions of genes. The presence of these methyl groups alters the DNA conformation and physically prevents transcription factors and RNA polymerase from binding to the promoter region. Consequently, the gene is silenced because transcription cannot occur, preventing the production of mRNA and the corresponding protein.

1. **Mechanism Location (1.0 mark):** State that methyl groups are added to DNA/cytosine bases at CpG sites/islands in the promoter region of a gene.
2. **Transcription Prevention (1.0 mark):** Explain that this prevents the binding of transcription factors / RNA polymerase.
3. **Consequence (0.75 marks):** State that this inhibits transcription / silences the gene, preventing mRNA synthesis or protein production.

Magnification is calculated using the formula:
\( \text{Magnification} = \frac{\text{Image Size}}{\text{Actual Size}} \)

1. Convert units to be consistent. Convert the image size from millimetres (mm) to micrometres (\(\mu\text{m}\)):
\( 18 \text{ mm} \times 1000 = 18,000\ \mu\text{m} \).

2. Divide the image size in micrometres by the actual size in micrometres:
\( \text{Magnification} = \frac{18,000\ \mu\text{m}}{3.6\ \mu\text{m}} = 5000 \).

Therefore, the magnification is \( \times 5000 \).

1. **Unit Conversion Mark (1.0 mark):** Correct conversion of image units: \( 18 \text{ mm} = 18,000\ \mu\text{m} \) OR actual size: \( 3.6\ \mu\text{m} = 0.0036 \text{ mm} \).
2. **Formula Application Mark (1.0 mark):** Division of image size by actual size: \( \frac{18,000}{3.6} \) or \( \frac{18}{0.0036} \).
3. **Accuracy Mark (0.75 marks):** Correct final magnification value of 5000 (Accept \( \times 5000 \)).

Step 1: Calculate the total number of cells currently undergoing mitosis (dividing cells). Dividing cells = Prophase + Metaphase + Anaphase + Telophase = 32 + 12 + 8 + 14 = 66 cells. Step 2: Calculate the mitotic index as a percentage of the total observed cells. Mitotic Index = (Number of dividing cells / Total number of cells) * 100 = (66 / 250) * 100 = 26.4%.

1 mark: Correct calculation of the total number of dividing cells (66). 1 mark: Correct substitution into the mitotic index formula (66 / 250) * 100. 0.75 mark: Correct final answer of 26.4% with unit (%).

Step 1: Calculate the radius of the fibre. Radius (r) = diameter / 2 = 0.40 mm / 2 = 0.20 mm. Step 2: Calculate the cross-sectional area of the cylindrical fibre. Area = \(\pi r^2\) = \(\pi \times (0.20)^2\) = 0.12566 mm^2. Step 3: Calculate the tensile strength. Tensile Strength = Force / Area = 12.0 N / 0.12566 mm^2 = 95.493 N mm^-2 (or MPa). Step 4: Round the final answer to 3 significant figures: 95.5 MPa.

1 mark: Correct calculation of the cross-sectional area as 0.126 mm^2 (or 0.1257 mm^2) based on a radius of 0.20 mm. 1 mark: Correct substitution of force and area into the formula (12.0 / 0.12566). 0.75 mark: Correct final answer of 95.5 MPa (accept 95.2 to 96.0 depending on rounding of pi) with correct unit.

When using a patient's own stem cells, the cells are genetically identical to the patient's body cells. This means they express the same self-antigens (MHC/HLA glycoproteins) on their cell surface membrane, preventing the patient's immune system from recognizing them as foreign and initiating an immune response (rejection). Consequently, the patient does not need to take lifelong immunosuppressant drugs, which have severe side effects. Additionally, because these cells are generated from somatic adult cells rather than blastocysts, their use does not involve the ethical controversy of destroying human embryos.

1 mark: Reference to no risk of immune rejection / no immune response because the stem cells are genetically identical / have the same self-antigens. 1 mark: Reference to no need for immunosuppressant drugs (which leave the patient vulnerable to infections). 0.75 mark: Reference to the avoidance of ethical issues/concerns associated with the destruction of human embryos.

Step 1: Calculate N, the total number of individuals of all species. N = 12 + 8 + 4 + 2 + 1 = 27. Therefore, N(N-1) = 27 * 26 = 702. Step 2: Calculate n(n-1) for each species: Species A: 12 * 11 = 132; Species B: 8 * 7 = 56; Species C: 4 * 3 = 12; Species D: 2 * 1 = 2; Species E: 1 * 0 = 0. Step 3: Sum the n(n-1) values. \(\sum n(n-1)\) = 132 + 56 + 12 + 2 + 0 = 202. Step 4: Calculate the index of diversity (d). d = 702 / 202 = 3.4752... Step 5: Round to 2 decimal places, giving 3.48.

1 mark: Correct calculation of N = 27 and N(N-1) = 702. 1 mark: Correct calculation of the sum of n(n-1) as 202. 0.75 mark: Correct final value of 3.48 (must be rounded to 2 decimal places).

DNA methylation involves the addition of methyl groups to cytosine bases, typically at CpG sites within the promoter region of a gene. This modification physically blocks the binding of transcription factors and RNA polymerase to the promoter, preventing transcription from being initiated. Furthermore, DNA methylation can recruit histone-modifying proteins that lead to chromatin condensation (forming heterochromatin). Because the DNA becomes tightly packed, the transcription machinery cannot access the gene, effectively silencing its expression.

1 mark: Reference to the addition of methyl groups to DNA cytosine bases / CpG islands in the promoter region of the gene. 1 mark: Reference to the prevention of transcription factor / RNA polymerase binding to the promoter. 0.75 mark: Reference to chromatin condensation / forming tightly packed heterochromatin, which prevents transcription of the gene.

Seeds are dried to reduce their water content and stored at sub-zero temperatures to drastically slow down metabolic processes and enzyme-controlled reactions (such as respiration) within the seed. This prevents premature germination and keeps the embryo alive in a dormant state for a long period. Additionally, low moisture and cold temperatures inhibit the growth and reproduction of decay-causing microorganisms, such as fungi and bacteria, which would otherwise decompose and destroy the seeds.

1 mark: Reference to drying and low temperatures reducing enzyme activity / rate of respiration / metabolic rate within the seeds to keep them dormant / prevent germination. 1 mark: Reference to preventing the growth / reproduction of decay-causing microorganisms (such as bacteria / fungi). 0.75 mark: Reference to increasing the viability / lifespan of the seeds for long-term storage.

Plant height is a polygenic characteristic, meaning it is controlled by several genes at different loci (polygenic inheritance). Each gene has an additive effect on the phenotype, creating a range of potential heights (continuous variation) rather than distinct categories. This genotype sets the maximum possible height the plant can achieve. However, environmental factors such as light intensity, temperature, water availability, and soil mineral ions (like nitrates for protein synthesis or magnesium for chlorophyll) are necessary for growth. If these resources are limiting, the plant cannot perform photosynthesis or build tissues effectively, meaning it will not reach its genetic potential height.

1 mark: Reference to height being polygenic / controlled by multiple genes at different loci, which have an additive effect on the phenotype (giving continuous variation / setting genetic potential). 1 mark: Reference to environmental factors (such as light intensity, water, or nutrient availability, e.g., nitrates / magnesium) being needed for growth / protein synthesis / photosynthesis. 0.75 mark: Reference to the fact that if environmental resources are limiting, the plant cannot reach its maximum genetic potential height.

The primary plant cell wall is composed of cellulose microfibrils. Cellulose is a polysaccharide made of beta-glucose monomers joined by beta-1,4-glycosidic bonds, where alternate glucose molecules are rotated 180 degrees. These linear, unbranched chains run parallel to each other and are linked by hydrogen bonds to form tough bundles called microfibrils. The microfibrils are arranged in a criss-cross net-like structure and are held together by a matrix of hemicelluloses and pectin. This composition gives the cell wall a very high tensile strength, allowing it to resist the high internal turgor pressure when water enters by osmosis, thereby preventing the cell from bursting and providing mechanical support to the plant.

1 mark: Description of cellulose as beta-glucose chains linked by beta-1,4-glycosidic bonds, forming linear chains. 1 mark: Cellulose chains are held together by hydrogen bonds to form microfibrils, which are arranged in a criss-cross / net-like structure embedded in pectin / hemicelluloses. 0.75 mark: This arrangement/composition provides high tensile strength to resist turgor pressure / prevent the cell bursting.

To calculate the heterozygosity index (\(H\)), use the formula: \(H = \frac{\text{Number of heterozygous loci}}{\text{Total number of loci}}\). Substituting the given values: \(H = \frac{44}{160} = 0.275\). Rounding to two decimal places gives 0.28.

1. Correct formula or identification of values: \(44 / 160\) (1 mark).
2. Correct calculation of 0.275 (1 mark).
3. Correct rounding to 2 decimal places: 0.28 (0.75 marks).

DNA methylation involves the attachment of methyl groups to cytosine bases (often at CpG sites) in the promoter region of a gene. This modification changes the structure of the DNA and physically blocks transcription factors and RNA polymerase from binding to the promoter. Consequently, transcription is inhibited, meaning no mRNA is produced, and the translation of the encoded protein cannot occur.

1. Reference to the addition of methyl groups to CpG islands / promoter region of DNA (1 mark).
2. Prevents the binding of RNA polymerase / transcription factors (1 mark).
3. Inhibits transcription, so no mRNA is synthesized for translation (0.75 marks).

First, calculate the total number of cells in mitosis (prophase + metaphase + anaphase + telophase): \(18 + 11 + 6 + 5 = 40\text{ cells}\). Next, find the total number of cells observed: \(40 + 160 = 200\text{ cells}\). Finally, calculate the mitotic index as a percentage: \(\text{Mitotic Index} = \left( \frac{40}{200} \right) \times 100 = 20.0\%\).

1. Correctly calculates cells in mitosis (40) and total cells (200) (1 mark).
2. Divides mitotic cells by total cells and multiplies by 100 (1 mark).
3. Correct answer to 1 decimal place with percentage sign: 20.0% (0.75 marks).

In the plant cell wall, cellulose microfibrils (held together by hydrogen bonds) are not aligned in a single direction; instead, they are laid down in a criss-cross, multi-layered meshwork. This structural arrangement is embedded in a sticky matrix of hemicelluloses and pectins. The cross-hatched meshwork provides high tensile strength in all directions, preventing the cell wall from stretching too much and bursting when the vacuole fills with water and exerts turgor pressure.

1. Description of cellulose microfibrils arranged in a criss-cross / mesh-like pattern (1 mark).
2. Microfibrils are bound together by a matrix containing hemicelluloses / pectins (1 mark).
3. This arrangement provides high tensile strength to resist bursting under high turgor pressure (0.75 marks).

1. Calculate \(N\) (total individuals): \(N = 8 + 12 + 10 = 30\). \(N(N-1) = 30 \times 29 = 870\).
2. Calculate \(n(n-1)\) for each species:
- Species A: \(8 \times 7 = 56\)
- Species B: \(12 \times 11 = 132\)
- Species C: \(10 \times 9 = 90\)

3. Sum the values: \(\sum n(n-1) = 56 + 132 + 90 = 278\).
4. Calculate \(D\): \(D = 1 - \frac{278}{870} = 1 - 0.3195 = 0.6805\). Rounding to two decimal places gives 0.68.

1. Correct calculations of \(N = 30\) and \(N(N-1) = 870\) (1 mark).
2. Correct sum of \(n(n-1) = 278\) (1 mark).
3. Correct evaluation of \(D = 0.68\) rounded to two decimal places (0.75 marks).

Pluripotent stem cells can divide and differentiate into almost any specialized cell type in the body (belonging to any of the three germ layers), but they cannot form extraembryonic tissues such as the placenta. Multipotent stem cells are more restricted, as they can only differentiate into a limited range of cell types related to their tissue of origin. An example of a multipotent stem cell is a hematopoietic stem cell found in human bone marrow, which can differentiate into various blood cell types.

1. Clear definition of pluripotent stem cells (can differentiate into almost any cell type except extraembryonic tissues) (1 mark).
2. Clear definition of multipotent stem cells (can only differentiate into a limited/restricted range of cell types) (1 mark).
3. Correct location of human multipotent stem cells, e.g., bone marrow / skin / brain / umbilical cord (0.75 marks).

Use the formula: \(\text{Actual Size} = \frac{\text{Image Size}}{\text{Magnification}}\).
1. Convert the image size from millimetres to micrometres: \(24\text{ mm} = 24 \times 1000 = 24,000\ \mu\text{m}\).
2. Divide by the magnification: \(\text{Actual Size} = \frac{24,000}{800} = 30\ \mu\text{m}\).
3. Expressed to two significant figures, the answer is 30.

1. Correctly converts millimetres to micrometres: \(24\text{ mm} = 24,000\ \mu\text{m}\) (1 mark).
2. Correct substitution into the formula: \(24,000 / 800\) (1 mark).
3. Correct final answer of 30 (or \(30\ \mu\text{m}\)) expressed to two significant figures (0.75 marks).

Translation initiation begins when the small ribosomal subunit binds to the mRNA molecule and scans for the start codon (usually AUG). A tRNA molecule carrying the amino acid methionine, with the complementary anticodon (UAC), binds to this codon. The large ribosomal subunit then associates to form the functional ribosome. Translation termination is triggered when the ribosome encounters a stop codon (UAA, UAG, or UGA) on the mRNA. Since there are no tRNA molecules with complementary anticodons for these codons, a protein called a release factor binds to the ribosome instead, causing the completed polypeptide chain to be cleaved and released.

1. Initiation involves ribosome assembly at the start codon (AUG) and binding of tRNA carrying methionine / complementary anticodon (1 mark).
2. Termination occurs when the ribosome reaches a stop codon (UAA/UAG/UGA) (1 mark).
3. No tRNA binds to the stop codon; instead, a release factor binds to release the polypeptide chain (0.75 marks).

Similarities:
- Both methods aim to identify a safe and effective therapeutic dosage.
- Both methods ultimately test the drug on patients who suffer from the disease the drug is intended to treat (with digitalis for Withering, and in Phase 2/3 of modern trials).
- Both historically derived their active ingredients from natural plant sources.

Differences:
- Pre-clinical stage: Modern testing involves testing on isolated cells, tissues, and live animals to evaluate toxicity and efficacy before human administration. Withering did not perform pre-clinical animal testing.
- Healthy volunteers: Modern trials feature Phase 1, where the drug is tested on a small group of healthy volunteers to determine safety and pharmacokinetics. Withering tested directly on patients with dropsy.
- Bias control (Double-blind & Placebo): Modern Phase 2 and 3 trials use placebos and double-blind techniques (where neither patients nor doctors know who receives the drug) to eliminate psychological bias and ensure objectivity. Withering used an open-label method, slowly increasing the dose on his patients while observing symptoms.
- Sample size and progression: Modern trials use standardized, multi-phase scaling (Phase 1, 2, and 3) with progressively larger, statistically significant sample sizes. Withering used a smaller, less systematic group of individuals and adjusted doses trial-by-error.
- Regulation: Modern trials are strictly regulated by ethical committees and national agencies (e.g., MHRA), requiring rigorous protocol approval, whereas Withering operated independently without regulatory oversight.

This is a Quality of Written Communication (QWC) question marked using a level-of-response grid.

Level 1 (1-2 marks):
- Mentions simple, isolated points of comparison or contrast (e.g., Withering tested on sick people, modern trials test on animals or healthy people).
- Communication is basic with little structure.

Level 2 (3-4 marks):
- Provides a structured comparison containing at least one similarity and several clear differences (e.g., pre-clinical trials, healthy volunteers in Phase 1, or placebos).
- Explains the scientific purpose of at least one difference (e.g., placebos prevent bias, Phase 1 tests safety).
- Writing is structured and mostly clear.

Level 3 (5-6 marks):
- Explains a comprehensive range of both similarities and differences.
- Explicitly compares pre-clinical stages, Phase 1 (healthy volunteers), Phase 2/3 (patients, placebos, double-blind trials), and regulatory/ethical controls.
- Argument is highly coherent, using precise biological terminology throughout.