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According to Chargaff's rules, in a double-stranded DNA molecule, the percentage of adenine (A) equals thymine (T), and the percentage of guanine (G) equals cytosine (C). If A = 28%, then T = 28%. The remaining 44% of the bases are G and C, meaning C = 22% and G = 22%. In a 100 base-pair fragment, there are 28 A-T base pairs and 22 G-C base pairs. Each A-T pair has 2 hydrogen bonds, and each G-C pair has 3 hydrogen bonds. Therefore, the total number of hydrogen bonds is: \((28 \times 2) + (22 \times 3) = 56 + 66 = 122\).

1 mark for correct calculations showing 22% cytosine and 122 hydrogen bonds (Option A).

In differential centrifugation, organelles are separated based on their size and density. The largest and densest organelles sediment first at low gravitational forces (low speed), while smaller and lighter organelles require much higher speeds to sediment. Nuclei are the largest/densest organelles and sediment first, followed by mitochondria (and lysosomes), and finally ribosomes (the smallest/least dense organelles) sediment at very high speeds.

1 mark for identifying the correct sedimentation order of nuclei, then mitochondria, then ribosomes (Option C).

A competitive inhibitor binds reversibly to the active site of the enzyme, competing directly with the substrate. At very high substrate concentrations, the substrate molecules outcompete the inhibitor molecules, allowing the enzyme to still reach its maximum rate of reaction. Thus, \(V_{\text{max}}\) remains unchanged. However, because the inhibitor reduces the affinity of the enzyme for the substrate at lower concentrations, a higher concentration of substrate is required to reach half of \(V_{\text{max}}\), which means the Michaelis constant (\(K_{\text{m}}\)) increases.

1 mark for identifying that competitive inhibition leaves \(V_{\text{max}}\) unchanged and increases \(K_{\text{m}}\) (Option B).

Vital capacity (VC) is the maximum volume of air that can be expired after a maximum inspiration. It is calculated by summing the tidal volume (TV), the inspiratory reserve volume (IRV), and the expiratory reserve volume (ERV): \(\text{VC} = \text{TV} + \text{IRV} + \text{ERV} = 500\text{ cm}^3 + 2500\text{ cm}^3 + 1200\text{ cm}^3 = 4200\text{ cm}^3\). The residual volume is not included in the vital capacity (it is included in the total lung capacity).

1 mark for the correct calculation of vital capacity as \(4200\text{ cm}^3\) (Option C).

The active loading of sucrose into companion cells relies on a proton gradient. Hydrogen ions (protons, \(\text{H}^+\)) are actively pumped out of the companion cell cytoplasm into the apoplast (cell wall space) using proton pumps. This process directly utilizes ATP. Sucrose is then co-transported back into the companion cell down the proton electrochemical gradient via facilitated diffusion through sucrose-proton cotransporter proteins, which does not directly hydrolyze ATP.

1 mark for identifying the active transport of hydrogen ions out of the companion cell as the ATP-consuming step (Option A).

During early ventricular systole, the ventricles begin to contract, causing ventricular pressure to rise rapidly. As soon as the ventricular pressure exceeds the atrial pressure, the atrioventricular (AV) valves snap shut to prevent backflow. However, the ventricular pressure is not yet high enough to exceed the pressure in the aorta and pulmonary artery, so the semilunar valves remain closed. This brief period of contraction with all valves closed is known as isovolumetric contraction.

1 mark for selecting the correct phase of the cardiac cycle where all valves are closed (Option B).

Proto-oncogenes code for proteins that stimulate normal cell cycle progression. A gain-of-function mutation in just one allele converts it into an oncogene, causing over-activity and potential cancer development (it is dominant). In contrast, tumor suppressor genes code for proteins that inhibit cell division or repair DNA; these typically require both alleles to be inactivated by loss-of-function mutations before regulation of the cell cycle is lost (it is recessive).

1 mark for identifying that a single mutated allele of a proto-oncogene vs. both mutated alleles of a tumor suppressor gene are required for oncogenesis (Option B).

Penicillin is a beta-lactam antibiotic. Its mechanism of action involves inhibiting the enzyme transpeptidase (also known as penicillin-binding protein). This enzyme is responsible for catalyzing the cross-linking of peptidoglycan chains, which forms the strong, rigid bacterial cell wall. Inhibiting this enzyme results in a weak cell wall, leading to osmotic lysis and cell death in actively growing bacteria.

1 mark for identifying the inhibition of transpeptidase and peptidoglycan cross-linking as penicillin's mechanism of action (Option C).

Goblet cells are specialized epithelial cells that synthesize and secrete mucus (primarily consisting of mucins, which are heavily glycosylated proteins). This requires an extensive network of rough endoplasmic reticulum (for protein synthesis), Golgi apparatus (for glycosylation and packaging), and secretory vesicles (for exocytosis). Mature red blood cells lack organelles, palisade cells are plant cells, and squamous epithelial cells are extremely thin and not specialized for high levels of protein secretion.

1 mark for correct identification of option A.

At the arteriole end, the net filtration pressure is calculated as: Hydrostatic pressure - Oncotic pressure = \(4.3 - 3.1 = 1.2\text{ kPa}\) outwards. At the venule end, the net pressure is: Hydrostatic pressure - Oncotic pressure = \(1.6 - 3.1 = -1.5\text{ kPa}\) (which corresponds to a net reabsorption pressure of \(1.5\text{ kPa}\) inwards).

1 mark for correct calculation and selection of option A.

A competitive inhibitor binds reversibly to the active site of the enzyme, competing directly with the substrate. Because it increases the concentration of substrate needed to achieve half of the maximum velocity, the apparent Michaelis-Menten constant (\(K_m\)) increases. However, at infinitely high substrate concentrations, the substrate outcompetes the inhibitor, and the maximum rate of reaction (\(V_{max}\)) remains unchanged.

1 mark for identifying that Km increases and Vmax remains unchanged (Option A).

According to Chargaff's rules, if Cytosine (C) = 34%, then Guanine (G) must also = 34%. This accounts for 68% of the total base composition. The remaining bases (Adenine + Thymine) make up \(100\% - 68\% = 32\%\). Since \(A = T\), Adenine = \(32\% / 2 = 16\%\). Cytosine and guanine base pairs are held together by 3 hydrogen bonds.

1 mark for the correct calculation of adenine (16%) and correct number of hydrogen bonds (3) in option B.

The QRS complex on an ECG represents the rapid depolarisation of the ventricles. This electrical depolarisation triggers the subsequent mechanical contraction of the ventricles, known as ventricular systole. Atrial depolarisation is represented by the P wave, and ventricular repolarisation is represented by the T wave.

1 mark for correctly matching the QRS complex to ventricular depolarisation/systole (Option B).

Stomatal opening is initiated by the active transport of potassium ions (\(K^+\)) into the guard cells. This solute accumulation decreases (makes more negative) the water potential inside the guard cells. Consequently, water enters the guard cells from surrounding epidermal cells by osmosis down a water potential gradient, causing the guard cells to become turgid and swell, pulling the stomatal pore open.

1 mark for identifying active transport of potassium ions into the cell, leading to decreased water potential, osmotic water entry, and turgidity (Option B).

Penicillin functions by inhibiting the enzyme transpeptidase (also known as a penicillin-binding protein). This enzyme is responsible for catalyzing the formation of peptide cross-links between the peptidoglycan chains in the bacterial cell wall. Inhibition of transpeptidase weakens the cell wall, causing the bacterial cell to burst due to osmotic pressure.

1 mark for correctly identifying penicillin's mode of action on transpeptidase and peptidoglycan cross-linking (Option C).

Multipotent stem cells (such as haematopoietic stem cells in bone marrow) can differentiate into a limited range of cell types, typically within a specific tissue. Totipotent cells can differentiate into any embryonic body cell as well as extra-embryonic cells like the placenta. Pluripotent cells can differentiate into any tissue cell in the body but NOT extra-embryonic tissues. Unipotent cells can only differentiate into a single cell type.

1 mark for correctly matching the definition of multipotent stem cells in Option C.

Total nucleotides = 1200, representing 600 base pairs. Since 30% of nucleotides are cytosine (360 bases), there must be an equal amount of guanine (360 bases), forming 360 C-G base pairs. Each C-G pair has 3 hydrogen bonds, giving 360 * 3 = 1080 hydrogen bonds. The remaining 40% of nucleotides are adenine and thymine (480 bases in total), which forms 240 A-T base pairs. Each A-T pair has 2 hydrogen bonds, giving 240 * 2 = 480 hydrogen bonds. The total number of hydrogen bonds is 1080 + 480 = 1560.

1 mark for the correct answer B (1560).

Inhibitor X increases Km from 2.5 to 5.0 mmol dm^-3 but leaves Vmax unchanged, which is characteristic of a competitive inhibitor that binds to the active site and decreases the apparent affinity for the substrate. Inhibitor Y decreases Vmax but leaves Km unchanged, showing it is a non-competitive inhibitor that does not bind to the active site.

1 mark for the correct option A.

At 100x magnification, 1 epu represents 10 micrometers. When switching to 400x magnification, the magnification increases by a factor of 4. This means each epu represents an actual distance 4 times smaller, so 1 epu = 10 / 4 = 2.5 micrometers. Since the cell spans 12 epu at 400x, its actual length is 12 * 2.5 = 30 micrometers.

1 mark for the correct option B (30 micrometers).

An antibody consists of heavy and light chains. The variable regions are located at the N-terminal of both the heavy and light chains, forming the specific antigen-binding sites. Constant regions are made of both heavy and light chains. Variable regions differ between different antibodies to allow specificity. Disulfide bonds are present throughout the constant regions to stabilize the structure.

1 mark for the correct option A.

First, convert the image length from millimetres to micrometres:
\(45\text{ mm} \times 1,000 = 45,000\text{ }\mu\text{m}\).

Next, use the magnification formula:
\(\text{Actual size} = \frac{\text{Image size}}{\text{Magnification}}\)

Substitute the values:
\(\text{Actual size} = \frac{45,000\text{ }\mu\text{m}}{15,000} = 3\text{ }\mu\text{m}\).

Award [1 mark] for correct conversion of units and setting up the division (e.g., \(45,000 / 15,000\) or showing \(0.003\text{ mm}\)).
Award [0.5 marks] for the final correct value of 3 (with or without units).

When companion cells actively load sucrose into the phloem sieve tube element, the concentration of dissolved solutes inside the sieve tube increases. This increase in solute concentration significantly lowers the water potential inside the sieve tube element. As a result, water moves from the surrounding tissues (such as adjacent xylem vessels or companion cells), where the water potential is higher, into the sieve tube by osmosis down a water potential gradient.

Award [1 mark] for stating that active loading of sucrose increases solute concentration, which lowers/decreases the water potential inside the sieve tube element.
Award [0.5 marks] for explaining that water moves in by osmosis down a water potential gradient.

In double-stranded DNA, complementary base pairing rules state that cytosine (C) always pairs with guanine (G), meaning their percentages are equal:
\(\%C = \%G = 22\%\)

Therefore, the combined percentage of cytosine and guanine is:
\(22\% + 22\% = 44\%\)

The remaining percentage must be divided equally between adenine (A) and thymine (T):
\(100\% - 44\% = 56\%\) (for A and T combined)

Thus, the percentage of adenine is:
\(\%A = \frac{56\%}{2} = 28\%\).

Award [1 mark] for identifying that cytosine and guanine total \(44\%\) or that adenine and thymine total \(56\%\).
Award [0.5 marks] for the final correct answer of \(28\%\).

The semi-lunar valves open during the stage of ventricular systole (ventricular contraction). As the muscular walls of the ventricles contract, they reduce the volume of the ventricular chambers, rapidly increasing the internal hydrostatic pressure of the blood. Once this ventricular pressure exceeds the pressure in the major elastic arteries (the aorta and pulmonary artery), the pressure gradient forces the semi-lunar valves open.

Award [1 mark] for correctly identifying 'ventricular systole' (or ventricular contraction).
Award [0.5 marks] for explaining that the pressure inside the ventricles exceeds the pressure in the aorta / pulmonary artery.

A competitive inhibitor has a structure similar to the substrate and competes for the same active site. However, at extremely high substrate concentrations, substrate molecules vastly outnumber the competitive inhibitor molecules. The probability of an inhibitor molecule collision with an active site becomes negligible compared to a substrate molecule. Consequently, the active sites are almost entirely occupied by substrate, allowing the reaction to reach its maximum rate (\(V_{\max}\)).

Award [1 mark] for stating that the inhibitor has negligible/no effect at high substrate concentration, or that the rate of reaction reaches its maximum (\(V_{\max}\)).
Award [0.5 marks] for explaining that the excessive substrate molecules outcompete the inhibitor for the active sites.

Herd immunity refers to the resistance to the spread of a contagious disease within a population that results if a sufficiently high proportion of individuals are immune (often achieved via vaccination). Because most of the population is immune, the pathogen struggle to find susceptible hosts to replicate in. This chain of transmission is broken, making it highly unlikely that an unvaccinated, susceptible individual will come into contact with an infectious person, thereby protecting them indirectly.

Award [1 mark] for defining herd immunity as a large/high proportion of the population being immune.
Award [0.5 marks] for explaining that this limits/stops the spread of the pathogen, reducing the risk of exposure for unvaccinated individuals.

Under normal conditions, proto-oncogenes code for proteins (like growth factor receptors or transcription factors) that stimulate cell division and promote progression through cell cycle checkpoints. In contrast, tumor suppressor genes code for proteins that halt the cell cycle when errors are detected, repair damaged DNA, or trigger apoptosis (programmed cell death) if the damage is irreparable, preventing uncontrolled growth.

Award [1 mark] for stating that proto-oncogenes stimulate / promote cell division / cell cycle progression.
Award [0.5 marks] for stating that tumor suppressor genes inhibit / slow cell division, repair DNA, or trigger apoptosis.

The inner surface of alveoli is lined with a thin layer of moisture, which creates substantial cohesive forces due to hydrogen bonding between water molecules (surface tension). Surfactant (a phospholipid-protein mixture) disrupts these cohesive forces, lowering surface tension. This prevents the alveoli from completely collapsing (atelectasis) as their volume decreases during expiration, and significantly reduces the elastic recoil, making it much easier to reinflate and expand them during inspiration.

Award [1 mark] for identifying that surfactant lowers/reduces surface tension inside the alveoli.
Award [0.5 marks] for explaining that this prevents alveolar collapse (during expiration) OR decreases the physical work/force needed to inflate/expand the lungs (during inspiration).

To calculate actual size: \(\text{Actual size} = \frac{\text{Image size}}{\text{Magnification}}\). First, convert the image size from millimetres (mm) to micrometres (\(\mu\text{m}\)): \(48\text{ mm} \times 1000 = 48\ 000\ \mu\text{m}\). Next, divide by the magnification: \(\frac{48\ 000\ \mu\text{m}}{40\ 000} = 1.2\ \mu\text{m}\).

1 mark for correct conversion of units (48 mm to 48,000 \(\mu\text{m}\)) OR correct rearrangement of the formula. 0.5 marks for correct final answer of 1.2 \(\mu\text{m}\) (accept 1.2).

The QRS complex represents the electrical impulse spreading through the ventricles, causing ventricular depolarisation. This depolarisation is what initiates the contraction of the ventricular muscle (ventricular systole).

1 mark for identifying 'ventricular depolarisation' or 'depolarisation of the ventricles'. 0.5 marks for linking this electrical activity to ventricular contraction/systole.

Herd immunity is achieved when a high percentage of the population becomes immune to an infectious disease (usually through vaccination). This disrupts the chains of transmission, making it highly unlikely for any non-immune or unvaccinated individual to encounter an infectious carrier.

1 mark for stating that a high/sufficient proportion of the population is immune/vaccinated. 0.5 marks for explaining that this significantly reduces transmission pathways, meaning unvaccinated individuals are highly unlikely to encounter an infected person.

DNA nucleotides contain a deoxyribose pentose sugar, whereas RNA nucleotides contain a ribose pentose sugar. Additionally, DNA nucleotides can have the nitrogenous base thymine (but never uracil), while RNA nucleotides can have uracil (but never thymine).

1 mark for identifying the difference in pentose sugar (deoxyribose vs ribose). 0.5 marks for identifying the difference in nitrogenous bases (thymine in DNA vs uracil in RNA).

The moist lining of the alveoli creates surface tension. Surfactant, composed of phospholipids and proteins, lowers this surface tension. This prevents the alveoli from completely collapsing (atelectasis) when air is exhaled, ensuring that a large surface area remains available for efficient gas exchange and making re-inflation easier.

1 mark for stating that surfactant lowers/reduces surface tension to prevent alveolar collapse (during exhalation). 0.5 marks for explaining that this maintains a large surface area for diffusion or reduces the physical work needed to inflate the lungs.

The lock-and-key hypothesis suggests that the enzyme's active site is completely complementary to the substrate shape before binding. In contrast, the induced-fit model states that the active site is flexible; as the substrate binds, the active site changes shape slightly to mould itself around the substrate, forming a more precise fit and putting strain on the substrate bonds to lower activation energy.

1 mark for explaining that in induced-fit, the active site changes shape/moulds around the substrate upon binding. 0.5 marks for contrasting this with the lock-and-key model where the active site is rigid/pre-formed and complementary.

Water molecules are polar and form hydrogen bonds with one another, causing them to stick together (cohesion). This cohesion ensures that water in the xylem vessels forms an uninterrupted, continuous column. When transpiration occurs at the leaves, water is pulled upwards, and because of cohesion, the entire column moves up under tension.

1 mark for identifying that hydrogen bonding causes cohesion, which keeps water molecules together to form a continuous/unbroken column. 0.5 marks for stating that this allows the entire column to be pulled up under tension (transpiration pull).

Pluripotent stem cells (such as embryonic stem cells) have the potential to differentiate into all the cell types that make up the adult body, but not extraembryonic tissues (like the placenta). Multipotent stem cells (such as adult stem cells) are more restricted, only having the capacity to differentiate into a limited range of specialized cell types belonging to a specific lineage or tissue (e.g., haematopoietic stem cells can only form different types of blood cells).

1 mark for defining pluripotent as having the ability to differentiate into any/all body cell types (except extraembryonic). 0.5 marks for defining multipotent as only being able to differentiate into a limited range or a specific lineage of cell types.

First, convert the image length from millimetres to micrometres: \(48\text{ mm} \times 1000 = 48,000\text{ }\mu\text{m}\). Then, use the magnification formula: \(\text{Actual size} = \text{Image size} / \text{Magnification}\). Therefore, \(\text{Actual size} = 48,000 / 15,000 = 3.2\text{ }\mu\text{m}\).

1 mark for correct conversion of units (48,000 micrometres) or setting up the correct division \(48 / 15,000\). 0.5 marks for the correct final answer of 3.2.

To find the heart rate, use the formula: \(\text{Cardiac Output} = \text{Stroke Volume} \times \text{Heart Rate}\). First, convert cardiac output to \(\text{cm}^3\text{ min}^{-1}\): \(5.4\text{ dm}^3 = 5400\text{ cm}^3\). Then, calculate the heart rate: \(5400 / 75 = 72\text{ bpm}\).

1 mark for converting the cardiac output to \(5400\text{ cm}^3\) (or stroke volume to \(0.075\text{ dm}^3\)) and rearranging the formula. 0.5 marks for the correct final heart rate of 72.

A competitive inhibitor competes with the substrate for the active site, which increases the substrate concentration required to reach half of \(V_{\text{max}}\), thereby increasing the Michaelis constant (\(K_m\)). However, because high substrate concentrations can fully overcome competitive inhibition, the maximum velocity (\(V_{\text{max}}\)) remains unchanged.

1 mark for identifying that the Michaelis constant (\(K_m\)) increases. 0.5 marks for stating that the maximum velocity (\(V_{\text{max}}\)) remains unchanged.

In double-stranded DNA, cytosine (C) pairs with guanine (G), so G is also 34%. This accounts for 68% of the total bases. The remaining bases must be adenine (A) and thymine (T), which sum to \(100\% - 68\% = 32\%\). Since A pairs with T, the percentage of thymine is equal to the percentage of adenine: \(32\% / 2 = 16\%\).

1 mark for showing that the remaining adenine and thymine bases constitute 32% of the total DNA. 0.5 marks for the correct final percentage of 16%.

The variable region has a unique amino acid sequence (primary structure) which determines its specific three-dimensional folding (tertiary structure). This shape is complementary to a specific antigen, allowing the antibody to bind to the antigen and form an antigen-antibody complex.

1 mark for explaining that its unique amino acid sequence / tertiary structure is complementary to a specific antigen. 0.5 marks for stating that this allows binding / the formation of an antigen-antibody complex.

First, hydrogen ions are actively transported (pumped) out of the companion cell into the surrounding cell wall using ATP, creating a concentration gradient. The hydrogen ions then diffuse back down their concentration gradient into the companion cell via co-transporter proteins, carrying sucrose molecules with them against their concentration gradient.

1 mark for describing the active transport of hydrogen ions out of the companion cell to establish a concentration gradient. 0.5 marks for describing the passive co-transport of hydrogen ions back in alongside sucrose.

1. Carbon dioxide reacts with water in erythrocytes to form carbonic acid (\(\text{H}_2\text{CO}_3\)), which is catalysed by the enzyme carbonic anhydrase.
2. Carbonic acid rapidly dissociates into hydrogen ions (\(\text{H}^+\)) and hydrogencarbonate ions (\(\text{HCO}_3^-\)).
3. Hydrogencarbonate ions diffuse out of the cell into the plasma down a concentration gradient.
4. To maintain electrical neutrality, chloride ions (\(\text{Cl}^-\)) diffuse into the erythrocyte from the plasma. This process is known as the chloride shift.

Marking points (max 3 marks):
- MP1: Carbon dioxide reacts with water to form carbonic acid, catalysed by carbonic anhydrase. [1]
- MP2: Carbonic acid dissociates into hydrogen ions and hydrogencarbonate ions (\(\text{HCO}_3^-\)). [1]
- MP3: Hydrogencarbonate ions diffuse out of the erythrocyte and chloride ions (\(\text{Cl}^-\)) diffuse in to maintain electrical neutrality (the chloride shift). [1]

1. Ethanol acts as a competitive inhibitor of alcohol dehydrogenase because its molecular shape is similar to the substrate, ethylene glycol, and is complementary to the enzyme's active site.
2. Ethanol binds temporarily to the active site, preventing ethylene glycol from binding and thus reducing the rate of formation of enzyme-substrate complexes.
3. This competitive inhibition can be overcome by increasing the concentration of the substrate (ethylene glycol).
4. At high substrate concentrations, the probability of substrate molecules binding to the active site is much greater than that of the inhibitor, allowing the maximum rate of reaction (\(V_{\max}\)) to be reached.

Marking points (max 4 marks):
- MP1: Inhibitor (ethanol) has a shape similar / complementary to the active site of the enzyme (alcohol dehydrogenase). [1]
- MP2: Inhibitor binds to/blocks the active site, preventing the formation of enzyme-substrate complexes. [1]
- MP3: The effect of the competitive inhibitor is overcome by increasing the substrate (ethylene glycol) concentration. [1]
- MP4: At high substrate concentrations, the maximum rate of reaction (\(V_{\max}\)) is still achieved. [1]

1. Spindle fibres are made of microtubules, which must assemble and disassemble to coordinate the alignment and separation of chromosomes during mitosis.
2. Preventing the disassembly of microtubules means that spindle fibres cannot shorten during anaphase.
3. As a result, sister chromatids cannot be pulled to opposite poles of the cell, arresting the cell cycle during mitosis.
4. This halt in cell division triggers cell death pathways (apoptosis), preventing the survival and uncontrolled division of the cancer cells.

Marking points (max 3 marks):
- MP1: Spindle fibres (made of microtubules) are required for separating sister chromatids/chromosomes during mitosis. [1]
- MP2: Preventing microtubule disassembly stops spindle fibres from shortening, so chromatids cannot be pulled to opposite poles during anaphase. [1]
- MP3: This arrests the cell cycle (in mitosis) OR prevents cytokinesis, which triggers programmed cell death / apoptosis. [1]

1. Surfactant is a mixture of phospholipids and proteins that coats the inner surface of alveoli.
2. Its primary role is to reduce the surface tension of the moisture layer lining the alveoli, preventing the cohesive forces of water from pulling the alveolar walls inward.
3. This prevents the alveoli from collapsing completely when the infant exhales.
4. In IRDS, the lack of surfactant leads to high surface tension, causing widespread alveolar collapse (atelectasis).
5. Consequently, the infant must expend a huge amount of muscular effort to generate enough pressure to force the collapsed alveoli open during each inhalation, leading to respiratory fatigue and insufficient gas exchange.

Marking points (max 4 marks):
- MP1: Surfactant reduces the surface tension of the liquid lining the alveoli. [1]
- MP2: This prevents the walls of the alveoli from sticking together / collapsing during exhalation. [1]
- MP3: In IRDS (surfactant deficiency), high surface tension causes the alveoli to collapse completely (atelectasis) upon exhalation. [1]
- MP4: Extreme muscular effort is required to re-inflate the collapsed alveoli during inhalation, resulting in breathing difficulty / fatigue / poor oxygenation. [1]

To calculate the actual size, use the formula:
\(\text{Actual size} = \frac{\text{Image size}}{\text{Magnification}}\)

First, convert the image measurement from millimetres (mm) to micrometres (\(\mu\text{m}\)):
\(15 \text{ mm} \times 1000 = 15,000 \mu\text{m}\)

Next, divide by the magnification:
\(\text{Actual size} = \frac{15,000 \mu\text{m}}{30,000} = 0.5 \mu\text{m}\)

1 mark: Correct conversion of 15 mm to 15,000 \(\mu\text{m}\) OR correct rearrangement of the formula showing image size divided by magnification.
1 mark: Correct calculation of the actual size (0.5 \(\mu\text{m}\)).

Both nucleotides share a similar structural blueprint of a pentose sugar, a phosphate group, and a nitrogenous base. However, they differ in the nature of their pentose sugars: RNA contains ribose, which has a hydroxyl group (\(-\text{OH}\)) at carbon 2, while DNA contains deoxyribose, which lacks this oxygen atom at carbon 2, having only a hydrogen (\(-\text{H}\)) atom.

1 mark: Identifies ribose in RNA as having a hydroxyl group (\(-\text{OH}\)) on carbon-2 / C2.
1 mark: Identifies deoxyribose in DNA as having a hydrogen atom (\(-\text{H}\)) / lacking oxygen on carbon-2 / C2.

According to the induced-fit model, the enzyme's active site is not completely rigid. When the substrate collides with the active site, it induces a conformational change. This shape change aligns catalytic groups and puts mechanical strain on the substrate bonds, facilitating the formation of the transition state and lowering the activation energy.

1 mark: Explains that the active site undergoes a conformational change/shapes around the substrate upon binding.
1 mark: Explains that this change puts strain on the chemical bonds of the substrate, reducing the activation energy.

When activated, T-helper cells release cell-signalling molecules called cytokines (including interleukins). These cytokines diffuse to nearby lymphocytes, binding to specific receptors. This signals B-lymphocytes to undergo clonal expansion (mitosis) and differentiate into antibody-secreting plasma cells, and also enhances the phagocytic activity of macrophages and stimulates T-killer cells.

1 mark: Stimulate B-lymphocytes to divide/undergo clonal expansion or differentiate into antibody-secreting plasma cells.
1 mark: Activate cytotoxic T-cells (T-killer cells) or stimulate phagocytic activity in macrophages.

The internal surfaces of mammalian alveoli are lined with a thin layer of moisture. Surfactant, composed of lipids and proteins, reduces the cohesive forces of water molecules at this air-liquid interface. This significantly lowers surface tension, preventing the small alveoli from collapsing during exhalation and reducing the work needed to expand them upon inspiration.

1 mark: Lowers/reduces the surface tension of the fluid layer lining the inner walls of the alveoli.
1 mark: Prevents the alveoli from collapsing/sticking together during exhalation OR decreases the physical effort needed to reinflate them during inhalation.

Active transport is used to pump hydrogen ions (\(\text{H}^+\)) out of the companion cell cytoplasm, across the plasma membrane, and into the surrounding apoplast (cell wall). This requires ATP. This creates an electrochemical gradient of \(\text{H}^+\). The hydrogen ions then diffuse back into the companion cell down this gradient through a co-transporter protein, which simultaneously transports sucrose against its concentration gradient into the cell.

1 mark: Hydrogen ions (\(\text{H}^+\)) are actively pumped (using ATP) out of the companion cell into the apoplast/cell wall.
1 mark: This creates an electrochemical/concentration gradient, allowing hydrogen ions to diffuse back down their gradient through a co-transporter protein, bringing sucrose along with them.

Proto-oncogenes code for proteins that promote normal cell growth and division. A mutation can transform a proto-oncogene into an oncogene, which is overactive. This causes continuous, unregulated cell division (mitosis) that is independent of normal regulatory growth factors, culminating in the development of a tumour mass.

1 mark: Mutation transforms the proto-oncogene into an oncogene.
1 mark: This leads to the continuous stimulation of cell division / uncontrolled mitosis (resulting in a tumour).

An ECG records the electrical events occurring in the cardiac muscle over time. The P wave corresponds to atrial depolarisation, the QRS complex corresponds to ventricular depolarisation, and the T wave represents ventricular repolarisation (recovery/relaxation phase of the ventricles).

1 mark: Refers to repolarisation / electrical recovery.
1 mark: Correctly identifies this event as occurring in the ventricles/ventricular muscle.

8 divisions of the stage micrometer equal 800 \(\mu\text{m}\) (since 8 multiplied by 100 is 800). Because 50 eyepiece graticule divisions align with these 8 stage divisions, 50 eyepiece divisions represent 800 \(\mu\text{m}\). Therefore, one eyepiece graticule unit is equal to 800 divided by 50, which equals 16 \(\mu\text{m}\).

1 mark: Correct calculation of the total length of the stage micrometer divisions (800 micrometres). 1 mark: Correct final answer of 16 micrometres.

Prosthetic groups are non-protein organic molecules or metal ions that are permanently integrated into the structure of a protein (for example, the iron-containing haem group in haemoglobin). In contrast, coenzymes only form temporary, weak bonds with the enzyme during the catalytic cycle, assisting in chemical group transfer before dissociating.

1 mark: Identifies that a prosthetic group is permanently or covalently attached to the protein. 1 mark: Identifies that a temporary coenzyme binds reversibly or transiently during catalysis.

During DNA replication, DNA polymerase links nucleotides together. A condensation reaction occurs between the 5-carbon phosphate group of an incoming activated nucleotide and the 3-carbon hydroxyl group of the pentose sugar of the existing strand. This forms a covalent phosphodiester bond and eliminates one water molecule.

1 mark: Identifies the reaction as a condensation reaction with the release of a water molecule. 1 mark: States the bond forms between the 5-carbon phosphate and the 3-carbon hydroxyl group of the adjacent pentose sugar.

The apoplast pathway involves water moving through the non-living cell walls. When it reaches the endodermis, it encounters the Casparian strip, which is a band of waterproof suberin embedded in the cell walls. This impenetrable barrier prevents water from moving further through the walls, forcing it to enter the living cytoplasm via the selectively permeable cell surface membrane of the endodermal cells.

1 mark: Identifies the presence of the waterproof Casparian strip made of suberin in the endodermal cell walls. 1 mark: Explains that this blocks apoplastic flow, forcing water across the cell membrane into the cytoplasm/symplast.

During a primary immune response, an antigen enters the body. Out of a huge population of lymphocytes, only those few cells carrying antigen receptors complementary to the specific antigen will bind to it. This binding selects these specific cells for subsequent clonal expansion via mitotic cell division.

1 mark: Explains that a specific lymphocyte (B or T cell) is selected/bound by the antigen. 1 mark: Identifies that this selection depends on complementary fit between the cell receptors and the specific antigen.

The P wave on an ECG corresponds to atrial depolarisation. This is the spread of electrical excitation from the sinoatrial node across both atria, which directly triggers atrial contraction (systole) to complete the filling of the ventricles with blood.

1 mark: Identifies the P wave as atrial depolarisation. 1 mark: Mentions that this triggers atrial contraction or atrial systole.

Proto-oncogenes are normal genes that code for proteins regulating cell growth and division. A mutation can convert a proto-oncogene into an oncogene. This causes the gene to be permanently active or overexpressed, leading to constant stimulation of cell division, bypassing of key cell cycle checkpoints, and resulting in the uncontrolled mitosis that produces a tumour.

1 mark: Explains that the mutation converts the proto-oncogene into an active oncogene. 1 mark: Explains that this leads to uncontrolled cell division/mitosis or failure of cell cycle control.

Type II alveolar cells secrete a phospholipid-protein mixture called surfactant. This layer reduces the cohesive forces of water molecules lining the alveoli, thereby reducing surface tension. Without surfactant, the high surface tension would cause alveoli to collapse completely during exhalation and require extreme muscular effort to re-inflate during inhalation.

1 mark: Identifies that surfactant reduces the surface tension of the fluid layer lining the alveoli. 1 mark: Explains that this prevents alveolar collapse (during expiration) or makes expansion/inflation easier (during inspiration).

The 2 divisions of the stage micrometer measure \(2 \times 0.1\text{ mm} = 0.2\text{ mm}\). In micrometres, this is \(0.2 \times 1000 = 200\mu\text{m}\). To find the value of 1 eyepiece division, divide this distance by 40: \(200\mu\text{m} / 40 = 5\mu\text{m}\).

Award 1 mark for converting the stage micrometer distance to \(200\mu\text{m}\) (or \(0.2\text{ mm}\)). Award 1 mark for the correct final answer of \(5\mu\text{m}\).

According to the induced-fit model, the enzyme's active site is not a perfect match for the substrate initially. As the substrate binds, the active site moulds around it. This conformational change puts strain on the bonds of the substrate, destabilising them and reducing the activation energy needed for the reaction.

Award 1 mark for stating that substrate binding causes a conformational change/change in shape of the active site. Award 1 mark for stating that this puts strain on or weakens the substrate bonds, making them easier to break.

DNA polymerase moves along the template strand and catalyses the formation of phosphodiester bonds between the phosphate group of an incoming free activated nucleotide and the 3' carbon of the pentose sugar on the growing strand. This forms the sugar-phosphate backbone.

Award 1 mark for mentioning that it catalyses condensation reactions. Award 1 mark for specifying that it forms phosphodiester bonds (between adjacent nucleotides or to form the sugar-phosphate backbone).

During ventricular systole, blood is forced into the arteries under high pressure, causing the elastic fibres in the tunica media to stretch. During ventricular diastole, when heart pressure drops, the elastic fibres recoil. This push sustains blood pressure and smooths out the flow, turning a pulsatile flow into a continuous one.

Award 1 mark for describing that elastic fibres stretch during ventricular systole / under high pressure. Award 1 mark for describing that they recoil during ventricular diastole / under low pressure to maintain blood pressure/push blood.

The QRS complex corresponds to the depolarisation of the ventricles, while the P wave corresponds to the depolarisation of the atria. Because the ventricular walls are much thicker and contain significantly more cardiac muscle tissue than the atrial walls, the electrical activity generated during ventricular depolarisation is much stronger, resulting in a larger peak on the ECG.

Award 1 mark for identifying the QRS complex as ventricular depolarisation. Award 1 mark for explaining that the ventricles have a larger muscle mass/thicker walls than the atria, generating a stronger electrical signal.

When potassium ions (\(\text{K}^+\)) are actively transported into the guard cells, it lowers the water potential inside the cells. Water then moves down a water potential gradient from surrounding epidermal cells into the guard cells by osmosis. This causes the guard cells to become turgid and swell. Because the inner cell walls of the guard cells are thicker and less elastic than the outer walls, they bend outwards, opening the stomatal pore.

Award 1 mark for explaining that \(\text{K}^+\)\ entry lowers the water potential, causing water to enter by osmosis. Award 1 mark for explaining that the guard cells become turgid and bend outwards to open the pore (due to asymmetric wall thickness).

Penicillin is a beta-lactam antibiotic that inhibits transpeptidase, the enzyme responsible for forming cross-links in peptidoglycan, a major component of bacterial cell walls. This weakens the cell wall, causing the bacterium to burst due to osmotic pressure (osmotic lysis). Since human cells are eukaryotic and do not have cell walls or peptidoglycan, they are completely unaffected by penicillin.

Award 1 mark for identifying that penicillin inhibits peptidoglycan synthesis/cross-linking in bacterial cell walls (causing osmotic lysis). Award 1 mark for stating that human cells lack cell walls/peptidoglycan and are therefore unaffected.

Following phagocytosis, the pathogen is enclosed in a phagosome which fuses with a lysosome to form a phagolysosome. Hydrolytic enzymes digest the pathogen. The resulting antigenic fragments are bound to Major Histocompatibility Complex (MHC) proteins and transported to the macrophage's outer cell surface membrane, where they are displayed to activate T lymphocytes.

Award 1 mark for stating that the macrophage digests/breaks down the pathogen (using lysosomal enzymes). Award 1 mark for stating that it displays/presents the pathogen's antigens on its outer cell surface membrane using MHC proteins.

During DNA replication, the lagging strand is synthesised in short, discontinuous sections called Okazaki fragments because DNA polymerase can only synthesise DNA in the 5' to 3' direction. DNA ligase is the enzyme responsible for joining these fragments together. It catalyses the formation of covalent phosphodiester bonds between the phosphate group of one nucleotide and the deoxyribose sugar of the adjacent nucleotide. This process seals the nicks in the sugar-phosphate backbone, producing a continuous strand of DNA.

Award up to 2 marks (scaled to 1.76):

1. Joins Okazaki fragments together / seals nicks in the sugar-phosphate backbone; [1 mark]
2. By catalysing the formation of phosphodiester bonds (between adjacent nucleotides / deoxyribose and phosphate); [1 mark]

DO NOT ACCEPT: formation of hydrogen bonds.

During DNA replication, the lagging strand is synthesised in short, discontinuous sections called Okazaki fragments because DNA polymerase can only synthesise DNA in the 5' to 3' direction. DNA ligase is the enzyme responsible for joining these fragments together. It catalyses the formation of covalent phosphodiester bonds between the phosphate group of one nucleotide and the deoxyribose sugar of the adjacent nucleotide. This process seals the nicks in the sugar-phosphate backbone, producing a continuous strand of DNA.

Award up to 2 marks (scaled to 1.76):

1. Joins Okazaki fragments together / seals nicks in the sugar-phosphate backbone; [1 mark]
2. By catalysing the formation of phosphodiester bonds (between adjacent nucleotides / deoxyribose and phosphate); [1 mark]

DO NOT ACCEPT: formation of hydrogen bonds.

The proton pump uses ATP to actively transport hydrogen ions (\(H^+\)) out of the companion cell cytoplasm into the cell wall/apoplast, creating an electrochemical gradient. The sucrose-\(H^+\) cotransporter then allows hydrogen ions to diffuse back down their gradient into the cell, carrying a sucrose molecule against its concentration gradient simultaneously.

1. Active transport of hydrogen ions (\(H^+\)) out of the companion cell into the cell wall/apoplast using ATP [1]
2. Development of a higher concentration/electrochemical gradient of \(H^+\) outside the companion cell [1]
3. Facilitated diffusion of \(H^+\) back into the companion cell through the cotransporter protein, which co-transports sucrose against its concentration gradient [1]

Higher carbon dioxide concentration results in a shift of the oxygen dissociation curve to the right. This is known as the Bohr effect. Carbon dioxide reacts with water to form carbonic acid, which dissociates into hydrogen ions and hydrogencarbonate ions. The hydrogen ions bind to haemoglobin, causing a conformational change that reduces its affinity for oxygen. Consequently, haemoglobin releases oxygen more readily to tissues that need it for aerobic respiration.

1. Shift of the oxygen dissociation curve to the right (Bohr effect) [1]
2. Carbon dioxide reacts to form hydrogen ions (\(H^+\)) which lower pH [1]
3. \(H^+\) ions bind to haemoglobin, lowering its affinity for oxygen [1]
4. This allows haemoglobin to release/unload more oxygen to the actively respiring tissues [1]

An IgG antibody is bivalent, meaning it possesses two identical antigen-binding sites within its variable regions. The flexible hinge region allows these binding sites to adjust their distance and angle. When each binding site attaches to an antigen on a different pathogen, the antibody links the pathogens together. Repeated binding events by multiple antibodies result in the clumping (agglutination) of the pathogens.

1. IgG possesses two antigen-binding sites / variable regions [1]
2. Hinge region provides flexibility to bind to antigens at different distances/angles [1]
3. Allows the antibody to bind to two different pathogens/antigens simultaneously [1]
4. Clumps pathogens together, restricting their movement or making them easier for phagocytes to engulf [1] (max 3)

DNA polymerase synthesises DNA exclusively in the 5' to 3' direction. Because the two template strands of the DNA double helix are antiparallel, one template runs 3' to 5' while the other runs 5' to 3'. On the 3' to 5' template, replication can proceed continuously in the same direction that the replication fork is opening (leading strand). On the 5' to 3' template, replication must proceed in the opposite direction, moving away from the fork. This requires synthesis to occur in short, discontinuous sections (Okazaki fragments) as new areas of template are exposed.

1. DNA polymerase can only synthesise DNA / add nucleotides in the 5' to 3' direction [1]
2. The template strands are antiparallel (one runs 3' to 5', the other runs 5' to 3') [1]
3. Leading strand is synthesised continuously because polymerase moves in the same direction as the unwinding replication fork [1]
4. Lagging strand is synthesised discontinuously (in Okazaki fragments) because polymerase must move away from the fork [1]

### Indicative Content

**Structural adaptations of the alveolus and capillaries:**
* **Very short diffusion pathway:** The alveolar wall consists of a single layer of thin, flattened squamous epithelial cells. The capillary wall consists of a single layer of squamous endothelial cells. This minimizes the distance over which oxygen and carbon dioxide must diffuse.
* **Narrow capillary lumen:** Capillaries are so narrow that red blood cells are compressed against the capillary wall, shortening the diffusion distance and slowing blood flow to allow maximum time for gas exchange.
* **Large surface area:** The presence of millions of alveoli provides an extremely large surface-area-to-volume ratio for diffusion.
* **Moist lining:** A thin layer of moisture lines the alveoli, allowing respiratory gases to dissolve before diffusing across the epithelial membrane.
* **Maintenance of steep concentration gradients:** A dense, continuous network of pulmonary capillaries carries oxygenated blood away and brings deoxygenated blood to the alveoli, while ventilation continuously replenishes oxygen and removes carbon dioxide.

**Role of pulmonary surfactant:**
* **Reducing surface tension:** The moist lining of the alveolus creates a cohesive force between water molecules (surface tension) that tends to pull the alveolar walls inward, causing them to collapse.
* **Preventing collapse:** Surfactant, secreted by Type II alveolar cells, contains phospholipids and proteins that insert between water molecules, significantly reducing surface tension.
* **Maintaining surface area:** By preventing the collapse of alveoli during expiration (atelectasis), surfactant ensures that the total surface area available for gas exchange remains high.
* **Increasing lung compliance:** It reduces the pressure and muscular effort required to expand the lungs during inspiration, ensuring efficient ventilation.

**Level 3 (5–6 marks):**
* Detailed description of at least three structural adaptations of the alveolus (e.g., thin squamous epithelium, capillary proximity, moist surface, large surface area) and an explanation of how they enhance diffusion.
* Comprehensive explanation of the role of surfactant, linking surface tension reduction to the prevention of alveolar collapse and maintenance of surface area.
* Information is organized clearly, using appropriate biological terms accurately.

**Level 2 (3–4 marks):**
* Description of at least two structural adaptations of the alveolus with some linkage to diffusion.
* Basic explanation of the role of surfactant (e.g., reduces surface tension / stops lungs sticking together) but may lack detail on how this prevents collapse or maintains surface area.
* Structure of the response is generally logical but may lack precise terminology in places.

**Level 1 (1–2 marks):**
* Outlines one or two basic features of alveoli (e.g., thin walls, large surface area) with little explanation of their significance.
* Mention of surfactant without a clear description of its role or mechanism.
* The response is poorly structured and relies on basic terms.

**Level 0 (0 marks):**
* No response or no creditworthy content.

### Indicative Content

**Structure of the IgG antibody molecule:**
* **Polypeptide chains:** It is a Y-shaped glycoprotein made of four polypeptide chains: two identical long heavy chains and two identical short light chains.
* **Disulfide bridges:** Covalent disulfide bonds link the heavy chains to each other and to the light chains, maintaining the tertiary and quaternary structure of the protein.
* **Variable regions:** Located at the tips of the 'Y' arms. They have a highly specific primary structure/amino acid sequence, resulting in a unique 3D shape that is complementary to a specific antigen, allowing high-affinity binding.
* **Constant region:** The main body of the 'Y' shape (the stem of the heavy chains). It has a highly conserved amino acid sequence within a class of antibodies, allowing recognition by host immune cells.
* **Hinge region:** A flexible region in the middle of the heavy chains that allows the distance between the two antigen-binding sites to vary.

**Relation of structure to function:**
* **Bivalency (two antigen-binding sites):** Because it has two identical variable regions, a single antibody can bind to two antigens on different pathogen cells simultaneously. This leads to **agglutination** (clumping), which prevents the spread of pathogens and makes them easier for phagocytes to engulf.
* **Neutralisation:** Binding of the variable regions to antigens (such as viral attachment proteins or bacterial toxins) physically blocks them, preventing pathogens from entering host cells or toxins from binding to cell receptors.
* **Opsonisation:** The constant region acts as a ligand for specific receptors (Fc receptors) on phagocytes (such as macrophages and neutrophils). Once the variable regions are bound to the pathogen, the constant region 'flags' the pathogen, facilitating engulfment and destruction during phagocytosis.
* **Flexibility via the hinge region:** Enables the antibody to bind to antigens that are varying distances apart on the surface of a pathogen or on separate pathogens, increasing binding efficiency.

**Level 3 (5–6 marks):**
* Detailed description of antibody structure including variable and constant regions, heavy and light chains, disulfide bonds, and the hinge region.
* Precise functional links established for at least three distinct mechanisms (agglutination, opsonisation, neutralisation, and/or structural flexibility).
* High level of biological accuracy and clear, logical structuring of the response.

**Level 2 (3–4 marks):**
* Description of several structural features (e.g., variable/constant regions, light/heavy chains).
* Links made to at least two functions (e.g., antigen binding and phagocytosis/agglutination), though the explanations of the mechanisms may be less detailed.
* Clear presentation with minor omissions or technical errors.

**Level 1 (1–2 marks):**
* Simple description of antibody structure (e.g., Y-shape, variable region determines specificity).
* Basic statement of function (e.g., binds to pathogens or helps phagocytes) without detailed linking of structure to the mechanism.
* Fragmented or unstructured response.

**Level 0 (0 marks):**
* No response or no creditworthy content.