2024 Edexcel AS Level Biology B (8BI0) Practice Paper with Answers

Eukaryotic cells contain 80S ribosomes in their cytoplasm. Organelles of endosymbiotic origin, such as mitochondria, contain 70S ribosomes (similar to their ancestral bacterial counterparts). Prokaryotic cells contain 70S ribosomes in their cytoplasm. Therefore, the correct combination is eukaryotic cytoplasm: 80S, mitochondrial matrix: 70S, and prokaryotic cytoplasm: 70S.

Award 1 mark for the correct option (A).
Reject all other options.

A triglyceride is synthesised by esterifying each of the three hydroxyl (-OH) groups of a single glycerol molecule with a fatty acid chain (stearic acid in this case). Each ester bond is formed via a condensation reaction, which releases one molecule of water. With three fatty acids, three ester bonds are formed, releasing three water molecules.

Award 1 mark for the correct option (B).
Reject all other options.

Human Immunodeficiency Virus (HIV) is an enveloped retrovirus. Its genome consists of single-stranded RNA, which is contained within a protein capsid. This capsid is surrounded by an outer envelope composed of a lipid bilayer containing glycoproteins.

Award 1 mark for the correct option (C).
Reject all other options.

During the S phase of interphase, DNA replication takes place, which duplicates every chromosome. However, the duplicated DNA strands remain joined together as sister chromatids at the centromere, so they are still counted as single chromosomes. Consequently, in the subsequent \(G_2\) phase, the cell still has 24 chromosomes, but each chromosome now consists of 2 sister chromatids, leading to a total of \(24 \times 2 = 48\) chromatids.

Award 1 mark for the correct option (C).
Reject all other options.

First, calculate the water potential of the plant cell using the formula: \(\psi = ̈\psi_s + \psi_p = -0.9\text{ MPa} + 0.2\text{ MPa} = -0.7\text{ MPa}\). Since the surrounding solution has a higher water potential (\(-0.4\text{ MPa}\)) than the cell cytoplasm (\(-0.7\text{ MPa}\)), water moves into the cell down a water potential gradient by osmosis. This influx of water increases protoplast volume, which increases the pressure potential against the cell wall, making the cell more turgid. Plant cells do not lyse (burst) due to the mechanical strength of the cellulose cell wall.

Award 1 mark for the correct option (C).
Reject all other options.

First, calculate the surface area: \(SA = 4 \times \pi \times 15^2 = 900\pi \approx 2827.43\ \mu\text{m}^2\). Next, calculate the volume: \(V = \frac{4}{3} \times \pi \times 15^3 = 4500\pi \approx 14137.17\ \mu\text{m}^3\). Divide surface area by volume to find the ratio: \(\text{SA:V} = 2827.43 / 14137.17 = 0.2\ \mu\text{m}^{-1}\). Alternatively, simplify the ratio formula: \(\text{SA:V} = \frac{4\pi r^2}{\frac{4}{3}\pi r^3} = \frac{3}{r}\). Substituting the radius: \(\frac{3}{15} = 0.2\ \mu\text{m}^{-1}\).

1 mark for correct method of substituting values into the formulas or simplifying the ratio expression to 3/r. 1.2 marks for the correct final numerical answer of 0.2 (with or without units of micrometers^-1 or ratio format 0.2:1).

Use the magnification formula: \(\text{Magnification} = \frac{\text{Image size}}{\text{Actual size}}\). First, convert the image size from millimeters to micrometers: \(80\text{ mm} = 80,000\ \mu\text{m}\). Then, calculate the magnification: \(\text{Magnification} = \frac{80,000}{2.5} = 32,000\).

1 mark for correctly converting the unit of measurement to ensure both sizes use the same units (e.g., 80,000 micrometers or 0.0025 millimeters). 1.2 marks for the correct final calculation of 32,000 (accept x32000 or 32000 times).

The formation of a triglyceride is a condensation reaction. Each fatty acid reacts with one of the hydroxyl groups of glycerol to form an ester bond, releasing one molecule of water per bond. Because three fatty acids are joined to one glycerol molecule, three ester bonds are formed and three water molecules are released.

1 mark for identifying the reaction as condensation and the bond formed as an ester bond. 1.2 marks for stating that exactly 3 water molecules are released.

First, calculate the volume of water lost (the volume of the capillary tube the bubble moved through): \(\text{Volume} = \text{cross-sectional area} \times \text{distance} = 0.8\text{ mm}^2 \times 45\text{ mm} = 36\text{ mm}^3\). Next, calculate the rate of transpiration per minute: \(\text{Rate} = \frac{\text{Volume}}{\text{Time}} = \frac{36\text{ mm}^3}{15\text{ min}} = 2.4\text{ mm}^3\text{ min}^{-1}\).

1 mark for correctly calculating the volume of water uptake as 36 mm^3. 1.2 marks for the correct final rate of 2.4 with appropriate units (mm^3 min^-1).

The mitotic index is the ratio of cells undergoing mitosis to the total number of cells observed. Total cells in mitosis = 18 (prophase) + 12 (metaphase) + 6 (anaphase) + 4 (telophase) = 40 cells. Total cells observed = 250. Mitotic index = \(\frac{40}{250} \times 100\% = 16\%\).

1 mark for calculating the total number of dividing cells (40) or showing the fraction 40/250. 1.2 marks for the correct final percentage of 16% (accept 16).

First, convert the cardiac output from dm^3 min^-1 to cm^3 min^-1: \(5.4\text{ dm}^3\text{ min}^{-1} = 5400\text{ cm}^3\text{ min}^{-1}\). The formula for cardiac output is: \(\text{Cardiac Output} = \text{Heart Rate} \times \text{Stroke Volume}\). Rearrange to solve for stroke volume: \(\text{Stroke Volume} = \frac{\text{Cardiac Output}}{\text{Heart Rate}} = \frac{5400}{75} = 72\text{ cm}^3\).

1 mark for converting 5.4 dm^3 to 5400 cm^3 or for correctly rearranging the equation. 1.2 marks for the correct stroke volume of 72 cm^3 (accept 72).

Each amino acid is coded for by a triplet of three DNA bases (a codon in mRNA). For 144 amino acids, the number of nucleotides needed is \(144 \times 3 = 432\) nucleotides. An additional stop codon, which does not code for an amino acid but signals the end of translation, is also represented by 3 nucleotides on the template strand. Thus, the total minimum number of nucleotides is \(432 + 3 = 435\).

1 mark for multiplying the number of amino acids by 3 to get 432 nucleotides. 1.2 marks for adding 3 additional nucleotides for the stop codon to reach the correct total of 435.

To break down maltose into two molecules of alpha-glucose without enzymes, a hydrolysis reaction is carried out. This involves adding water to break the 1,4-glycosidic bond. In a laboratory setting, this chemical reaction is catalyzed by heating the maltose solution with dilute hydrochloric acid.

1 mark for identifying that a hydrolysis reaction occurs, involving the addition of water to break the glycosidic bond. 1.2 marks for stating that the reaction requires heating with dilute acid (such as hydrochloric acid).

Identify the two nearest data points: 0.4 mol dm\(^{-3}\) (+3.0%) and 0.6 mol dm\(^{-3}\) (-2.0%). The percentage change decreases by 5.0% over a concentration increase of 0.2 mol dm\(^{-3}\). To find the concentration where the change is 0.0%, calculate the fraction of change from +3.0% to 0.0%: \((3.0 / 5.0) \times 0.2\) mol dm\(^{-3}\) = 0.12 mol dm\(^{-3}\). Add this to the starting concentration: 0.4 + 0.12 = 0.52 mol dm\(^{-3}\).

1 mark for calculating the rate of change or fraction of concentration change: \((3.0 / 5.0) \times 0.2 = 0.12\) mol dm\(^{-3}\). 1.2 marks for the correct final concentration of 0.52 mol dm\(^{-3}\) (accept 0.52).

Convert both lengths to the same unit, micrometres (\(\mu\)m): 48 mm = 48,000 \(\mu\)m. Use the formula: Magnification = Image size / Actual size. Magnification = 48,000 / 1.5 = 32,000. Therefore, the magnification is x32,000.

1 mark for converting units correctly to 48,000 \(\mu\)m or 0.0015 mm. 1.2 marks for correct final magnification of x32,000 or 32,000 (accept with or without the 'x').

The total time of 120 minutes allows for exactly 3 lytic cycles (120 / 40 = 3). After cycle 1 (40 min), 120 phages are produced. After cycle 2 (80 min), \(120 \times 120 = 14,400\) phages are produced. After cycle 3 (120 min), \(14,400 \times 120 = 1,728,000\) phages are produced.

1 mark for identifying that 3 lytic cycles occur or for calculating 14,400 after cycle 2. 1.2 marks for the correct final answer of 1,728,000 bacteriophages.

Add Benedict's reagent to samples of both solutions and heat them in a water bath. Maltose is a reducing sugar and will turn the blue reagent green, yellow, orange, or red. Sucrose is a non-reducing sugar and will remain blue. To confirm sucrose, heat a fresh sample with dilute hydrochloric acid, neutralize with sodium hydrogencarbonate, and repeat the Benedict's test to get a positive red precipitate.

1 mark for adding Benedict's reagent, heating, and identifying that maltose gives a positive result (red/orange precipitate) while sucrose remains blue. 1.2 marks for describing the hydrolysis of sucrose with acid, heating, neutralization, and re-testing to get a positive result.

In double-stranded DNA, the percentage of cytosine (C) equals the percentage of guanine (G), so G = 22%. Together, C + G = 44%. The remaining bases must be adenine (A) and thymine (T), which make up \(100\% - 44\% = 56\%\). Since A pairs with T, the percentage of adenine is \(56\% / 2 = 28\%\).

1 mark for showing that C + G = 44% or that A + T = 56%. 1.2 marks for the correct final percentage of 28% (accept 28).

A triglyceride is formed by joining one glycerol molecule to three fatty acid molecules. Each fatty acid reacts with one hydroxyl group of the glycerol in a condensation reaction, releasing one molecule of water and forming one ester bond. Therefore, 3 water molecules are produced and 3 ester bonds are formed.

1 mark for stating that 3 water molecules are produced. 1.2 marks for naming the ester bond.

The mitotic index is the proportion of cells undergoing mitosis. Sum of cells in mitosis = 28 (prophase) + 12 (metaphase) + 8 (anaphase) + 6 (telophase) = 54 cells. Total cells counted = 350. Mitotic Index = \((54 / 350) \times 100 = 15.428...\%\). To one decimal place, this is 15.4%.

1 mark for calculating the total number of cells in mitosis as 54. 1.2 marks for the correct percentage of 15.4% (accept 15.4 without % symbol, reject 15 or 15.43).

First find the total number of individuals, N = 15 + 30 + 5 = 50. Calculate \((n/N)^2\) for each species: Species A = \((15/50)^2\) = 0.09. Species B = \((30/50)^2\) = 0.36. Species C = \((5/50)^2\) = 0.01. Sum of these terms = 0.09 + 0.36 + 0.01 = 0.46. Simpson's Index of Diversity (d) = 1 - 0.46 = 0.54.

1 mark for calculating N = 50 and showing the sum of \((n/N)^2\) as 0.46. 1.2 marks for the correct index value of 0.54.

Each triglyceride molecule contains three ester bonds. Complete hydrolysis of one triglyceride molecule yields one glycerol molecule and three fatty acids. Therefore, the number of glycerol molecules produced is equal to the number of triglyceride molecules hydrolyzed. Since each triglyceride has 3 ester bonds, the calculation is: \(4.5 \times 10^{18} / 3 = 1.5 \times 10^{18}\).

1 mark for dividing the number of ester bonds by 3. 1 mark for the correct final value in standard form: \(1.5 \times 10^{18}\).

Magnification = Image size / Actual size. Convert units to be consistent: \(18\text{ mm} = 18000\ \mu\text{m}\). Then, Magnification = \(18000\ / 2.4 = 7500\). Therefore, the magnification is 7500.

1 mark for converting \(18\text{ mm}\) to \(18000\ \mu\text{m}\) (or converting \(2.4\ \mu\text{m}\) to \(0.0024\text{ mm}\)). 1 mark for calculating the correct magnification of 7500 (accept \(\times 7500\) or 7,500).

1. HIV contains RNA as its genetic material, whereas T4 bacteriophage contains DNA. 2. HIV is enveloped (surrounded by a lipid membrane), whereas T4 bacteriophage is non-enveloped and has a protein-only outer structure including a complex tail.

1 mark for identifying the difference in genetic material (RNA vs DNA). 1 mark for identifying the presence of a lipid envelope in HIV vs its absence in T4 bacteriophage.

Oxygen is a small, non-polar molecule, which allows it to dissolve in and freely pass through the hydrophobic core of the phospholipid bilayer. In contrast, sodium ions (\(\text{Na}^+\)) are charged and hydrophilic, meaning they are repelled by the hydrophobic fatty acid tails and cannot cross without the hydrophilic pathway provided by channel proteins.

1 mark for explaining that oxygen is non-polar/hydrophobic and can interact with or pass through the fatty acid tails. 1 mark for explaining that sodium ions are charged/hydrophilic and are repelled by the hydrophobic core.

1. Amylose consists of monomers of \(\alpha\)-glucose, whereas cellulose is made of monomers of \(\beta\)-glucose. 2. Amylose has a coiled/helical structure, whereas cellulose consists of straight, unbranched chains. 3. In cellulose, alternate glucose monomers are rotated/inverted by \(180^\circ\) to form glycosidic bonds, whereas all monomers in amylose are in the same orientation.

1 mark for each correct structural difference stated (up to 2 marks). Accept: \(\alpha\)-glucose vs \(\beta\)-glucose; coiled/helical vs straight/unbranched; alternate monomers rotated \(180^\circ\) vs same orientation.

Volume of water taken up = cross-sectional area * distance moved = \(0.80\text{ mm}^2 * 45.0\text{ mm} = 36.0\text{ mm}^3\). Rate of water uptake = Volume of water / Time = \(36.0\text{ mm}^3 / 15.0\text{ minutes} = 2.4\text{ mm}^3\text{ min}^{-1}\).

1 mark for calculating the volume of water uptake as \(36.0\text{ mm}^3\) (method: \(0.80 * 45.0\)). 1 mark for the correct rate of \(2.4\) or \(2.40\text{ mm}^3\text{ min}^{-1}\) (accuracy).

1. Ice-cold: Reduces the activity of hydrolytic enzymes (e.g., proteases) that could break down and damage organelle structures.
2. Isotonic: Ensures there is no net movement of water into or out of the organelles by osmosis, preventing chloroplasts from swelling and bursting (osmotic lysis) or shriveling.
3. Buffered: Maintains a constant pH to prevent the denaturation of membrane proteins and enzymes within the organelles.
4. Centrifugation Step 1: Centrifuge the filtered homogenate at a low speed (e.g., \(1000 \times g\)) for a short period to pellet heavy cell debris, cell walls, and nuclei. Keep the liquid supernatant.
5. Centrifugation Step 2: Pour the supernatant into a clean tube and centrifuge at a higher speed (e.g., \(3000 \times g\)) to pellet the chloroplasts, leaving smaller organelles (mitochondria, ribosomes) in the supernatant.

• **Mark 1 (Ice-cold)**: Low temperature reduces/prevents hydrolytic enzyme activity (which would otherwise digest cell organelles). [Accept: reduces denaturation of proteins/enzymes].
• **Mark 2 (Isotonic)**: Water potential of the solution matches that of organelles to prevent net osmosis / prevents swelling and bursting / lysis of organelles.
• **Mark 3 (Buffered)**: Maintains constant pH to prevent protein/enzyme denaturation.
• **Mark 4 (First centrifugation)**: Centrifuge homogenate at low speed to remove heavy debris/nuclei/cell walls as a pellet and retain the supernatant.
• **Mark 5 (Second centrifugation)**: Centrifuge supernatant at higher speed to sediment chloroplasts as a pellet (and discard new supernatant).

1. **Cut**: Use a scalpel to cut a small (1-2 mm) section from the very tip of an actively growing onion root, as this contains the meristematic region where cells are actively dividing.
2. **Acid Hydrolysis**: Place the root tip in dilute hydrochloric acid (HCl) and gently warm it. This breaks down the pectins in the middle lamella, allowing the cells to separate easily during squashing.
3. **Rinse and Stain**: Rinse the root tip thoroughly with water, then transfer it to a microscope slide and add a few drops of an appropriate stain (e.g., acetic orcein, toluidine blue, or Feulgen's stain) to bind to and color the DNA/chromosomes.
4. **Maceration**: Use a mounted needle to gently break up/macerate the root tissue on the slide.
5. **Squash**: Place a coverslip over the specimen and press down firmly and vertically with a thumb or paper towel (without rolling or sliding) to spread the cells out into a single, thin layer so light can pass through.

• **Mark 1**: Cut the meristematic region / apical 1-2 mm of the root tip.
• **Mark 2**: Heat/treat root tip in hydrochloric acid (HCl) to break down the middle lamella / cell wall (allowing cells to separate).
• **Mark 3**: Add a suitable stain that binds to DNA (e.g., acetic orcein, toluidine blue, Feulgen's stain) [Reject: iodine, Benedict's].
• **Mark 4**: Macerate/break up tissue using a mounted needle.
• **Mark 5**: Push down firmly and vertically on the coverslip (using a thumb/paper towel) to obtain a single layer of cells (without rolling/moving coverslip sideways).

1. **Dilution series**: Prepare a dilution series of catalase to produce at least five different concentrations (e.g., 20%, 40%, 60%, 80%, 100%) using distilled water.
2. **Set up reaction**: Mix a fixed volume and concentration of hydrogen peroxide with a fixed volume of one of the catalase concentrations in a sealed reaction vessel.
3. **Measure product**: Connect the vessel to a gas syringe (or inverted measuring cylinder) and record the volume of oxygen gas produced at regular short intervals (e.g., every 10 seconds) for the first 1-2 minutes.
4. **Calculate initial rate**: Plot a graph of oxygen volume against time for each concentration. Draw a tangent at time \(t = 0\) seconds and calculate its gradient (volume/time) to determine the initial rate of reaction.
5. **Control variables**: Keep temperature constant using a water bath, and maintain a constant pH by using a buffer solution in all trials.

• **Mark 1**: Prepare a dilution series of catalase to produce at least 5 different concentrations (using distilled water).
• **Mark 2**: Keep concentration and volume of hydrogen peroxide substrate constant.
• **Mark 3**: Measure gas volume produced at short, regular intervals (e.g., every 10 seconds) using a gas syringe or inverted measuring cylinder.
• **Mark 4**: Plot gas volume against time and draw a tangent at \(t=0\) to calculate the initial rate of reaction.
• **Mark 5**: Control variables: use a water bath to maintain constant temperature AND use a buffer to maintain constant pH.

1. **Structural Differences (Capsid/Envelope)**: HIV is surrounded by an outer phospholipid envelope containing glycoproteins (gp120), whereas bacteriophages are non-enveloped and have an external protein capsid head and a tail assembly.
2. **Structural Differences (Genetic Material)**: HIV contains single-stranded RNA (along with the enzymes reverse transcriptase and integrase), whereas bacteriophages typically contain double-stranded DNA and do not carry reverse transcriptase.
3. **Attachment structures**: HIV has glycoprotein spikes (gp120) for attachment, while bacteriophages use tail fibers to attach to host cell walls.
4. **HIV Entry**: The gp120 glycoproteins bind to CD4 receptors on the host helper T-cell. The viral lipid envelope then fuses with the host cell membrane (or enters via endocytosis), releasing the entire capsid into the host cytoplasm.
5. **Bacteriophage Entry**: The bacteriophage tail fibers attach to specific receptors on the bacterial cell wall. The phage sheath contracts, puncturing the cell wall and membrane, and injects its viral DNA directly into the host cytoplasm, leaving the protein capsid shell outside.

• **Mark 1**: HIV has a lipid envelope (with glycoproteins) whereas bacteriophage is non-enveloped / has a protein head/capsid only.
• **Mark 2**: HIV contains RNA (and reverse transcriptase / integrase) whereas bacteriophage contains DNA (and no reverse transcriptase).
• **Mark 3**: HIV uses gp120 glycoproteins to bind to CD4 receptors on helper T-cells, leading to membrane fusion / endocytosis (entire capsid enters cell).
• **Mark 4**: Bacteriophage uses tail fibers/baseplate to attach to bacterial cell walls.
• **Mark 5**: Bacteriophage injects its DNA directly into the host cell (leaving the protein capsid outside the host).

1. **Mashing/Crushing**: Crush or grind the sunflower seed using a pestle and mortar to break the cell walls and increase the surface area, releasing the lipids.
2. **Extraction**: Add ethanol to the crushed seed sample in a test tube and shake vigorously to dissolve any lipids present.
3. **Decanting**: Allow the mixture to settle or filter it to obtain a clear liquid extract (supernatant containing dissolved lipids in ethanol).
4. **Emulsion formation**: Decant this liquid extract into a separate tube containing distilled water (or add water to the extract) and mix gently.
5. **Principle and Positive Result**: A positive result is indicated by a cloudy white/milky emulsion. Lipids are non-polar hydrophobic molecules that are soluble in organic solvents like ethanol, but highly insoluble in polar water. When the ethanol extract is mixed with water, the lipids precipitate out of solution as microscopic droplets suspended in the water, scattering light to create the cloudy appearance.

• **Mark 1**: Crush/grind the seed sample (using a pestle and mortar).
• **Mark 2**: Add ethanol to the sample and shake/mix thoroughly (to dissolve lipids).
• **Mark 3**: Decant or filter the liquid into a clean tube containing distilled water.
• **Mark 4**: State positive observation: a cloudy-white/milky emulsion/suspension forms.
• **Mark 5**: Explain principle: Lipids are soluble in organic solvents (ethanol) but insoluble in water; adding water causes the dissolved lipids to precipitate out as tiny droplets that scatter light.

The apoplast pathway involves the movement of water through the cellulose cell walls and intercellular spaces of the plant tissues. However, when water reaches the endodermis, its progress is blocked by a band of waterproof suberin called the Casparian strip. This forces the water to cross the cell membrane and enter the cytoplasm, switching to the symplast pathway.

Award 1 mark for the correct option (B).

- Reject A because cytoplasm and plasmodesmata are components of the symplast pathway.
- Reject C because the apoplast pathway is passive and does not directly utilize ATP from active transport.
- Reject D because water does not need to enter the vacuole (tonoplast) before entering the xylem via the apoplast.

During early ventricular systole, the pressure inside the ventricles rises above the pressure inside the atria as the ventricular walls begin to contract. This pressure difference forces the atrioventricular (AV) valves to close. However, because this pressure is not yet high enough to exceed the pressure within the aorta and pulmonary artery, the semilunar valves remain closed. This brief period is also known as isovolumetric ventricular contraction.

Award 1 mark for the correct option (B).

- Reject A: During atrial systole, AV valves are open.
- Reject C: During late ventricular systole, the semilunar valves are open to allow blood to be ejected into the arteries.
- Reject D: During diastole, the semilunar valves are closed, but the AV valves open during the later stage of diastole.

At the venule end of a capillary bed, the blood hydrostatic pressure has dropped significantly due to friction and the previous loss of fluid. However, because large plasma proteins cannot leave the capillary, they maintain a high oncotic (colloid osmotic) pressure within the blood. Since the hydrostatic pressure inside the capillary is now less than the oncotic pressure of the plasma, there is a net movement of water back into the capillary by osmosis.

Award 1 mark for the correct option (B).

- Reject A: If hydrostatic pressure were greater than oncotic pressure, fluid would still be filtered out of the capillary.
- Reject C: Hydrostatic pressure in the tissue fluid is extremely low and is not the main driver compared to blood hydrostatic/oncotic forces.
- Reject D: The oncotic pressure of the plasma is always greater than that of tissue fluid because proteins remain in the blood.

Anatomical adaptations refer to physical structures of the body (e.g., a thick layer of blubber for insulation). Behavioural adaptations are the actions an organism performs to survive (e.g., huddling together in groups to reduce heat loss). Physiological adaptations are internal body processes (e.g., shivering to generate metabolic heat).

Award 1 mark for the correct option (A) which correctly matches all three adaptations to their categories.

Carl Woese introduced the three-domain system of classification based on his work in molecular phylogeny, specifically by sequencing and comparing the genes encoding ribosomal RNA (rRNA). Because ribosomes are highly conserved and found in all self-replicating cells, differences in rRNA sequences provided a reliable measure of evolutionary divergence.

Award 1 mark for identifying ribosomal RNA (rRNA) (B) as the key molecule.

- Reject A (mRNA) as it is transient and not used for deep phylogenies.
- Reject C (Cytochrome c) as it is used for protein-based phylogeny but was not the primary basis of Woese's three-domain model.
- Reject D (membrane phospholipids) as this was discovered later to support but not initiate the system.

To calculate the heterozygosity index:

\[ H = \frac{\text{Number of heterozygotes}}{\text{Total number of individuals in the population}} \]

Given:
- Number of heterozygotes = \(120\)
- Total population size = \(250\)

\[ H = \frac{120}{250} = 0.48 \]

Award 1 mark for the correct calculation and option (C).

- Reject A: This is \(40 / 250\) (homozygous recessive proportion).
- Reject B: This is \(90 / 250\) (homozygous dominant proportion).
- Reject D: This is the proportion of homozygotes in the population \((90 + 40) / 250 = 0.52\).

High specific heat capacity means that water requires a relatively large amount of heat energy to raise its temperature by \(1\text{ }^{\circ}\text{C}\). This property prevents rapid temperature changes in oceans, lakes, and ponds, thereby providing a stable thermal environment for aquatic life to survive without experiencing extreme metabolic stress.

Award 1 mark for the correct option (C).

- Reject A: This is due to water's polar nature and solvent properties.
- Reject B: This is due to the lower density of ice compared to liquid water.
- Reject D: This is due to water's high latent heat of vaporisation.

The Bohr effect refers to the shift of the oxygen-haemoglobin dissociation curve to the right in the presence of elevated carbon dioxide concentrations (or lower pH/higher acidity). This decreases haemoglobin's affinity for oxygen, meaning that haemoglobin releases oxygen more readily to rapidly respiring tissues, such as skeletal muscles during exercise.

Award 1 mark for the correct option (B).

- Reject A: An increase in blood pH represents less carbon dioxide (alkalosis) and would shift the curve to the left, increasing oxygen affinity.
- Reject C: While temperature affects haemoglobin, it is not part of the definition of the Bohr effect (which specifically concerns \(\text{CO}_2\) and \(\text{H}^+\)).
- Reject D: An increase in carbon dioxide concentration shifts the curve to the right, not the left.

To find the net filtration pressure, we calculate the balance between the forces pushing fluid out of the capillary and the forces drawing fluid into the capillary:

* **Forces moving fluid OUT of the capillary (outward forces):**
1. Hydrostatic pressure of blood in the capillary = \(4.7\text{ kPa}\)
2. Oncotic pressure of tissue fluid = \(0.1\text{ kPa}\) (acting to pull fluid out)
* Total outward force = \(4.7 + 0.1 = 4.8\text{ kPa}\)

* **Forces moving fluid INTO the capillary (inward forces):**
1. Hydrostatic pressure of tissue fluid = \(0.2\text{ kPa}\) (acting to push fluid in)
2. Oncotic pressure of blood in the capillary = \(3.3\text{ kPa}\) (acting to pull fluid in)
* Total inward force = \(0.2 + 3.3 = 3.5\text{ kPa}\)

* **Net Filtration Pressure:**
\(\text{Net pressure} = \text{Total outward force} - \text{Total inward force} = 4.8 - 3.5 = +1.3\text{ kPa}\)

Since the net pressure is positive (outward force is greater than inward force), the net movement of fluid is **out of the capillary** into the surrounding tissue space to form tissue fluid.

* **A** is the correct answer (1 mark).
* **B** is incorrect because a net outward pressure of \(+1.3\text{ kPa}\) moves fluid out of the capillary, not into it.
* **C** is incorrect because it subtracts the tissue fluid oncotic pressure from the net hydrostatic difference incorrectly: \((4.7 - 0.2) - (3.3 + 0.1) = 1.1\text{ kPa}\).
* **D** is incorrect because it uses incorrect pressure additions and direction: \((4.7 - 3.3) + (0.2 - 0.1) = 1.5\text{ kPa}\) and incorrect inward direction.

The Casparian strip is a band of suberin (a waterproof, waxy substance) located in the cell walls of the endodermis. It acts as an impermeable barrier that blocks the apoplast pathway (movement through cell walls). This forces water and dissolved mineral ions to cross the selectively permeable cell surface membrane into the cytoplasm, continuing via the symplast pathway. This mechanism allows the plant to regulate which ions enter the xylem and prevents the backflow of water from the vascular cylinder.

1. Reference to the Casparian strip being waterproof / containing suberin which blocks the apoplast pathway / cell wall pathway (1 mark).
2. Explains that water/mineral ions are forced to enter the cytoplasm / symplast pathway across the selectively permeable membrane (1 mark).
3. Mentions that this allows the plant to control/selectively absorb mineral ions OR prevents the backflow of water from the xylem (0.5 marks).

First, calculate the change in mass:
\( \text{Change in mass} = 3.91\text{ g} - 4.25\text{ g} = -0.34\text{ g} \)

Next, calculate the percentage change relative to the initial mass:
\( \text{Percentage change} = \left( \frac{-0.34}{4.25} \right) \times 100 = -8.0\% \)

This represents an 8.0% decrease.

1. Correct calculation of mass loss of \( -0.34\text{ g} \) (1 mark).
2. Correct method for percentage change calculation: \( \frac{-0.34}{4.25} \times 100 \) (1 mark).
3. Final answer correctly rounded to two significant figures: \( -8.0\% \) or \( 8.0\% \) decrease (0.5 marks). [Reject -8% or 8% without the decimal place due to significant figures requirement].

An artery wall is adapted to high pressure in several ways:
1. A thick layer of elastic fibres (elastin) in the tunica media allows the artery wall to stretch to accommodate surges of high pressure blood and recoil to maintain pressure when the heart relaxes.
2. Smooth muscle tissue can contract to constrict the lumen, controlling blood flow and maintaining pressure.
3. A thick outer layer containing tough collagen fibres provides structural strength to prevent the artery from bursting under high pressure.

1. Elastic fibres/elastin stretch under high pressure and recoil to maintain pressure / smooth out blood flow (1 mark).
2. Smooth muscle contracts to constrict the lumen and control/maintain pressure (1 mark).
3. Collagen fibres (in tunica externa) provide structural strength / prevent the wall from bursting (0.5 marks).

A low Simpson's Index of Diversity value indicates that a habitat is dominated by only one or a few species, resulting in low species richness and/or low species evenness. Biologically, this means the ecosystem is unstable and highly vulnerable. If an environmental change occurs, such as a disease outbreak or climate shift, the dominant species may be severely affected, which can lead to the collapse of the entire community due to a lack of alternative ecological niches.

1. Explains that the habitat has low species richness/evenness or is dominated by a few species (1 mark).
2. Identifies that the ecosystem is unstable / fragile to environmental changes (1 mark).
3. Notes that a threat (like disease or climate change) to a dominant species could collapse the entire food web / community (0.5 marks).

At the arterial end of a capillary, the hydrostatic pressure generated by the contracting heart is higher than the oncotic pressure (osmotic pressure) pulling water back in. This pressure difference forces water and small, soluble molecules (such as glucose, amino acids, ions, and oxygen) out of the blood plasma through the small pores in the capillary walls. Large molecules, such as plasma proteins and red blood cells, are too large to pass through and remain inside the capillary.

1. Mentions that hydrostatic pressure at the arterial end is greater than the oncotic / osmotic pressure gradient (1 mark).
2. Explains that water and small dissolved solutes (e.g. glucose, oxygen, ions) are forced out of capillary pores/fenestrations (1 mark).
3. Explains that large plasma proteins or red blood cells remain inside the capillary because they are too large to fit through pores (0.5 marks).

Anatomical features can be misleading due to convergent evolution, where unrelated organisms evolve similar structures to adapt to similar environments. Modern molecular phylogeny compares genetic material (such as DNA base sequences, ribosomal RNA, or amino acid sequences of conserved proteins like cytochrome c). This provides objective, quantitative data reflecting true evolutionary ancestry, which revealed that Archaea and Bacteria are biochemically and genetically distinct enough to warrant separate domains despite looking similar.

1. Identifies that anatomical features can be misleading due to convergent evolution / organisms looking similar but having different evolutionary histories (1 mark).
2. Explains that molecular phylogeny uses objective, quantitative data from DNA, RNA, or protein/amino acid sequences (1 mark).
3. Mentions specific molecular markers such as ribosomal RNA (rRNA) or cytochrome c used to define the three domains (0.5 marks).

First, calculate the total number of cells in mitosis:
\( \text{Cells in mitosis} = 32 \text{ (prophase)} + 12 \text{ (metaphase)} + 6 \text{ (anaphase)} + 10 \text{ (telophase)} = 60 \text{ cells} \).

Calculate the Mitotic Index (MI):
\( \text{MI} = \frac{\text{Number of cells in mitosis}}{\text{Total number of cells}} = \frac{60}{400} = 0.15 \text{ (or } 15\%\text{)} \).

Convert the total cell cycle duration into minutes:
\( 15 \text{ hours} \times 60 \text{ minutes/hour} = 900 \text{ minutes} \).

Estimate the duration of mitosis:
\( \text{Duration} = 0.15 \times 900 \text{ minutes} = 135 \text{ minutes} \).

1. Correct calculation of mitotic index as \( 0.15 \) or \( 15\% \) (1 mark).
2. Correct conversion of 15 hours to 900 minutes (1 mark).
3. Correct calculation of final duration as 135 minutes (0.5 marks).

Anatomical adaptations are physical, structural features of an organism's body. An example in a camel is having wide, padded feet to prevent sinking into sand, or a hump containing fat reserves. Physiological adaptations are internal biochemical, metabolic, or cellular processes. An example in a camel is the ability of its kidneys to produce highly concentrated urine to conserve water, or producing metabolic water from fat oxidation.

1. Defines anatomical adaptations as structural/physical features AND provides a correct example (e.g., wide padded feet, hump for fat storage, long eyelashes) (1 mark).
2. Defines physiological adaptations as chemical/metabolic/internal processes AND provides a correct example (e.g., highly concentrated urine, ability to tolerate high body temperature fluctuation, metabolic water production) (1 mark).
3. Emphasizes the difference between structure and biochemistry/function (0.5 marks).

1. Calculate the total surface area (SA) of the cylindrical cell: \(SA = 2\pi r h + 2\pi r^2 = 2\pi(10)(60) + 2\pi(10)^2 = 1200\pi + 200\pi = 1400\pi \approx 4398.23\ \mu\text{m}^2\). 2. Calculate the volume (V) of the cylinder: \(V = \pi r^2 h = \pi (10)^2 (60) = 6000\pi \approx 18849.56\ \mu\text{m}^3\). 3. Divide SA by V: \(\text{SA:V} = 1400\pi / 6000\pi = 7/30 \approx 0.233\ \mu\text{m}^{-1}\). Rounded to two decimal places, this is 0.23 \(\mu\text{m}^{-1}\).

1 mark for calculating correct surface area (\(1400\pi\) or \(4398.23\ \mu\text{m}^2\)) OR volume (\(6000\pi\) or \(18849.56\ \mu\text{m}^3\)). 1 mark for correct calculation of ratio (0.23 or 0.233). 0.5 marks for correct unit (\(\mu\text{m}^{-1}\) or equivalent like \(\mu\text{m}^2/\mu\text{m}^3\)).

The sinoatrial node (SAN) initiates a wave of depolarization across the atria, causing atrial systole. This electrical impulse reaches the atrioventricular node (AVN). The AVN introduces a short delay (approx. 0.1 seconds) before passing the electrical impulse down the Bundle of His and Purkyne fibres. This delay is vital because it ensures the atria have finished contracting and fully emptied their blood into the ventricles before ventricular systole begins, preventing backflow and ensuring maximum pumping efficiency.

1 mark: AVN receives the excitation wave from the SAN and passes it to the Bundle of His / Purkyne tissue. 1 mark: AVN delays the impulse (by approx. 0.1s) to allow complete atrial contraction/emptying before ventricular contraction begins. 0.5 marks: This prevents simultaneous contraction of atria and ventricles, ensuring efficient blood flow.

Xylem vessels must withstand high negative pressure (tension) generated by transpiration without collapsing. They achieve this via the deposition of lignin in their cell walls, which provides mechanical strength. Additionally, xylem vessels are dead cells with no cytoplasm, organelles, or end walls, forming a hollow, continuous tube. This lack of obstruction allows an uninterrupted, cohesive column of water to be pulled upwards with minimal resistance. Pits in the lateral walls also allow water to move between adjacent vessels if a blockage occurs.

1 mark: Deposition of lignin in cell walls prevents the vessels from collapsing inward under tension/negative pressure. 1 mark: No living contents (empty lumen) and lack of end walls create a continuous, unbroken tube for water transport. 0.5 marks: Pits in walls allow lateral movement of water bypass blockages.

1. Find the proportion of cells in metaphase: \(15 / 250 = 0.06\). 2. Convert the total cycle time from hours to minutes: \(20 \text{ hours} \times 60 \text{ minutes/hour} = 1200 \text{ minutes}\). 3. Calculate the duration of metaphase: \(0.06 \times 1200 \text{ minutes} = 72 \text{ minutes}\).

1 mark for calculating the correct fraction of cells in metaphase (\(15/250\) or 0.06). 1 mark for converting the cycle time to minutes (1200 minutes) and completing the calculation to get 72. 0.5 marks for providing the correct unit (minutes).

During transcription, RNA polymerase matches complementary RNA nucleotides to the DNA template strand. A pairs with U (instead of T in DNA), T pairs with A, C pairs with G, and G pairs with C. This results in the mRNA sequence: 5'- A U G C C G A A U U G A -3'. This sequence consists of four codons: AUG, CCG, AAU, and UGA. UGA is a stop codon, which signals the termination of translation and does not code for an amino acid. Therefore, only the first three codons code for amino acids, resulting in a maximum of 3 amino acids.

1 mark for the correct complementary mRNA sequence (5'- A U G C C G A A U U G A -3' or equivalent base sequence). 1 mark for stating that the maximum number of amino acids is 3. 0.5 marks for explaining that the fourth codon (UGA) is a stop codon which does not code for an amino acid.

Genetic variation exists within a bacterial population due to random mutations, occasionally producing an allele conferring antibiotic resistance. When the population is exposed to the antibiotic (which acts as a strong selection pressure), non-resistant bacteria are killed. The resistant individuals survive and reproduce (differential survival) via binary fission. They pass on the resistant allele to their offspring, increasing the frequency of this allele in the gene pool over successive generations.

1 mark: Selection pressure is applied by the antibiotic, killing non-resistant bacteria. 1 mark: Resistant bacteria survive and reproduce (differential survival/reproduction), passing the resistant allele to offspring. 0.5 marks: Mention of random mutation causing the initial variation or allele frequency increasing over generations.

Triglycerides consist of one glycerol molecule bound to three fatty acids via ester bonds, making them completely hydrophobic (non-polar) and insoluble in water. In contrast, phospholipids consist of one glycerol molecule bound to two fatty acids and one highly polar, hydrophilic phosphate group. This structure makes phospholipids amphipathic, meaning they have a hydrophilic 'head' and hydrophobic 'tails', allowing them to spontaneously form bilayers in aqueous environments.

1 mark: Triglycerides have three fatty acids attached to glycerol, whereas phospholipids have two fatty acids and a phosphate group. 1 mark: Triglycerides are entirely hydrophobic/non-polar, whereas phospholipids are amphipathic (hydrophilic head and hydrophobic tail). 0.5 marks: Contrast in function/behavior (e.g., triglycerides form energy storage droplets, phospholipids form bilayers/cell membranes).

At the arterial end of a capillary, the hydrostatic pressure of the blood is relatively high and exceeds the oncotic pressure (the osmotic pressure exerted by large plasma proteins). This net outward pressure forces water and small dissolved solutes out of the capillary into the surrounding intercellular space, forming tissue fluid. At the venous end, the hydrostatic pressure has dropped significantly due to friction and fluid loss, falling below the oncotic pressure. Consequently, the net movement of water is reversed, drawing water and waste products back into the capillary by osmosis.

1 mark: At the arterial end, hydrostatic pressure is greater than oncotic pressure, causing net filtration/forcing fluid out. 1 mark: At the venous end, hydrostatic pressure falls below oncotic pressure, causing net reabsorption/drawing fluid back in by osmosis. 0.5 marks: Credit mention of plasma proteins causing the oncotic pressure or the role of lymphatic vessels draining excess tissue fluid.

1. Calculate the radius (\(r\)) of the capillary tube:
\(r = \frac{0.8\text{ mm}}{2} = 0.4\text{ mm}\)

2. Calculate the volume (\(V\)) of water lost using the formula for the volume of a cylinder (\(V = \pi r^2 h\), where \(h\) is the distance the bubble moved):
\(V = \pi \times (0.4\text{ mm})^2 \times 45\text{ mm}\)
\(V = \pi \times 0.16 \times 45 = 7.2\pi \approx 22.619\text{ mm}^3\)

3. Calculate the rate of water uptake per minute:
\(\text{Rate} = \frac{22.619\text{ mm}^3}{15\text{ minutes}} = 1.5079\text{ mm}^3\text{ min}^{-1}\)

4. Round to 3 significant figures:
\(1.51\text{ mm}^3\text{ min}^{-1}\)

• One mark for calculating the correct radius (\(0.4\text{ mm}\)) and setting up the volume equation: \(\pi \times 0.4^2 \times 45\) [1 mark]
• One mark for dividing the calculated volume by 15 (minutes) to find the rate: \(\frac{22.62}{15}\) [1 mark]
• One mark for the correct final answer of \(1.51\) (accept answers in the range \(1.50\) to \(1.51\) depending on the rounding of \(\pi\)) to 3 significant figures [1 mark].

Under normal physiological conditions, tissue fluid is formed when the high hydrostatic pressure of blood at the arteriole end of a capillary exceeds the opposing oncotic pressure (generated by plasma proteins), forcing fluid out into the intercellular spaces. Most of this fluid is normally reabsorbed at the venule end because hydrostatic pressure drops below oncotic pressure. In cases of hypertension, the elevated hydrostatic pressure at the arteriole end forces a larger volume of fluid out of the capillary. Additionally, the hydrostatic pressure remains high at the venule end, reducing the net movement of fluid back into the blood vessel. The excess fluid remains in the intercellular spaces, resulting in oedema.

• State that high blood pressure increases the hydrostatic pressure at the arteriole end of the capillaries [1 mark].
• Explain that this greater hydrostatic pressure forces more fluid/water out of the capillaries into the intercellular/tissue spaces [1 mark].
• Explain that because hydrostatic pressure remains higher than normal at the venule end, it exceeds or opposes the oncotic pressure, reducing fluid reabsorption back into the capillary [1 mark].

Molecular phylogeny involves comparing the molecular structures, specifically the sequence of bases in DNA or RNA, or the sequence of amino acids in highly conserved proteins (such as cytochrome c or rubisco) from different organisms. By sequencing the selected gene or protein across the three plant species, scientists can align the sequences and identify mutations or differences. A high degree of sequence similarity indicates fewer mutations have occurred since the species diverged. Therefore, species with the highest similarity are more closely related and share a more recent common ancestor. This sequence data is used to construct phylogenetic trees representing their evolutionary pathways.

• Identify that molecular phylogeny involves analyzing and comparing DNA base sequences, RNA base sequences, or protein amino acid sequences [1 mark].
• Explain that a higher percentage of similarity (or fewer sequence differences) indicates that species are more closely related [1 mark].
• Explain that more similar sequences indicate they shared a more recent common ancestor, which is used to construct a phylogenetic tree [1 mark].

To ensure a valid and reliable experiment using a potometer:
1. Cut the leafy shoot underwater (and at an angle) to prevent air bubbles from entering the xylem vessels, which would break the water column.
2. Assemble the potometer apparatus completely underwater to prevent air locks and ensure a continuous water column.
3. Apply petroleum jelly (vaseline) to all joints to ensure the system is completely airtight and watertight.
4. Dry the leaves thoroughly before beginning the experiment, as wet leaves block stomata and restrict transpiration.
5. Allow the shoot to equilibrate/acclimatise in the set environment for at least 10 minutes before recording any movement of the air bubble.

Award 1 mark for each correct point up to a maximum of 3.8 marks (4 marks equivalent):
- Cut shoot underwater / at an angle to prevent air entering xylem (1)
- Assemble apparatus underwater to ensure no air bubbles are trapped (1)
- Apply grease/petroleum jelly to joints to ensure airtight seals (1)
- Dry leaves before starting the investigation to clear stomata (1)
- Allow acclimatisation time before taking readings (1)

1. Calculate Net Filtration Pressure (NFP) at the arteriole end:
\(\text{NFP} = \text{Hydrostatic pressure} + \text{Oncotic pressure} = 4.3\text{ kPa} - 3.1\text{ kPa} = +1.2\text{ kPa}\) (outward force).
2. At the arteriole end, the hydrostatic pressure (\(4.3\text{ kPa}\)) is greater than the inward oncotic pressure (\(3.1\text{ kPa}\)), resulting in a net outward pressure that forces water and small solutes out of the capillary into the surrounding intercellular space, forming tissue fluid.
3. At the venule end, the hydrostatic pressure drops to \(1.5\text{ kPa}\), which is lower than the inward oncotic pressure of \(3.1\text{ kPa}\). This causes a net inward pressure (\(-1.6\text{ kPa}\)), meaning water is reabsorbed back into the capillary from the tissue fluid down a water potential gradient established by the plasma proteins that remained inside the capillary.

Award marks as follows:
- Correct calculation of net filtration pressure: \(+1.2\text{ kPa}\) (accept 1.2) (1)
- Explanation that hydrostatic pressure is greater than oncotic pressure at the arteriole end, forcing fluid out (1)
- Explanation that oncotic pressure is greater than hydrostatic pressure at the venule end, pulling fluid back in (1)
- Reference to plasma proteins remaining in capillary creating a water potential gradient / osmosis (1)

1. Calculate Simpson's Index of Diversity for Field A:
\(D_A = 1 - 0.38 = 0.62\).
2. Compare with Field B:
Field A (\(D = 0.62\)) has a higher biodiversity index than Field B (\(D = 0.45\)).
3. Ecological Stability:
A higher index of diversity indicates a more stable ecosystem with complex food webs. In Field A, if one species is affected by a threat (such as a disease or environmental change), other species can fill the niche, meaning the ecosystem is less likely to be disrupted. Field B is more dominated by fewer species, making it highly vulnerable to environmental changes.

Award marks as follows:
- Correct calculation of Simpson's Index of Diversity for Field A as 0.62 (1)
- Statement that Field A has higher biodiversity / is more biodiverse than Field B (1)
- Link higher index of diversity to greater ecosystem stability / more complex food webs (1)
- Explanation that in Field A, the loss of one species is less likely to collapse the community because of alternative species / niches (1)

To ensure a valid investigation, the student must control the following variables:
1. **Surface area and volume of the beetroot pieces**: Use a cork borer to obtain cylinders of identical diameter, then use a scalpel and ruler to cut them to equal lengths (e.g., 1 cm).
2. **Prior pigment leakage**: Rinse the cut beetroot discs thoroughly in distilled water and dry them on paper towels before placing them in the ethanol solutions to remove any vacuolar pigment (betalain) released due to mechanical damage during cutting.
3. **Temperature**: Incubate all tubes in a temperature-controlled water bath (e.g., at \(25^\circ\text{C}\)) as temperature itself affects membrane fluidity.
4. **Volume of solution and contact time**: Ensure the exact same volume of ethanol solution is used in each tube (e.g., \(10\text{ cm}^3\)) and that all tubes are incubated for the identical amount of time (e.g., 20 minutes) using a stopwatch.

Award marks as follows (max 3.8 marks / 4 marks equivalent):
- Control beetroot size/surface area using cork borer/ruler/scalpel (1)
- Wash beetroot discs with water and dry to remove pigment released during cutting (1)
- Use a water bath to control temperature across all treatments (1)
- Keep volume of ethanol solution constant AND incubation time constant (1)

1. **Atrioventricular (AV) valve state**:
- At \(0.1\text{ s}\), the AV valve is open because the pressure in the left atrium (\(1.2\text{ kPa}\)) is higher than the pressure in the left ventricle (\(0.8\text{ kPa}\)), causing blood to flow from the atrium to the ventricle.
- By \(0.2\text{ s}\), the AV valve has closed. This is because ventricular systole begins, causing ventricular pressure to rise rapidly to \(6.8\text{ kPa}\), which is greater than the atrial pressure (\(0.5\text{ kPa}\)). The pressure gradient forces the AV valve shut to prevent backflow.
2. **Semi-lunar valve opening**:
- The semi-lunar valve opens when the pressure inside the left ventricle exceeds the pressure in the aorta (\(10.0\text{ kPa}\)).
- Looking at the table, ventricular pressure is \(6.8\text{ kPa}\) at \(0.2\text{ s}\) (below aortic pressure) and rises to \(14.5\text{ kPa}\) at \(0.3\text{ s}\) (above aortic pressure). Therefore, the valve first opens between \(0.2\text{ s}\) and \(0.3\text{ s}\).

Award marks as follows:
- At 0.1 s, the AV valve is open because atrial pressure is higher than ventricular pressure (1)
- At 0.2 s, the AV valve is closed because ventricular pressure is higher than atrial pressure (1)
- Semi-lunar valve opens between 0.2 s and 0.3 s (1)
- Correct reason: Ventricular pressure exceeds aortic diastolic pressure (10.0 kPa) only after 0.2 s / reaches 14.5 kPa at 0.3 s (1)

1. **Genetic Variation**: A random, spontaneous mutation occurs in the DNA of a bacterium in the population, resulting in a new allele that confers resistance to the antibiotic.
2. **Selection Pressure**: The presence of the antibiotic acts as a strong selective pressure.
3. **Differential Survival**: Bacteria without the resistant allele (susceptible strain) are killed by the antibiotic, while those possessing the resistance allele survive.
4. **Reproduction and Inheritance**: The surviving resistant bacteria reproduce by binary fission, passing the advantageous allele for antibiotic resistance to their offspring.
5. **Change in Allele Frequency**: Over multiple generations, the frequency of the resistant allele increases in the population. Consequently, the population becomes increasingly resistant, requiring a much higher concentration of antibiotic (MIC increases from \(0.5\) to \(16.0\text{ }\mu\text{g cm}^{-3}\)) to inhibit growth.

Award marks as follows:
- Mutation occurs resulting in a resistant allele (1)
- Antibiotic acts as a selection pressure (1)
- Resistant bacteria survive while non-resistant bacteria die (differential survival) (1)
- Surviving bacteria reproduce and pass on the resistant allele, increasing its frequency over generations (1)