2025 CCEA A-Level Life and Health Sciences 0008 Practice Paper with Answers

The longest continuous carbon chain is five carbons (pentane); a methyl branch is attached at carbon 3, numbered to give the lowest locant. Name: 3-methylpentane.

1 mark: correct parent chain identified as pentane; 1 mark: correct name '3-methylpentane' with locant 3. [2]

Structural isomers share the same molecular formula but differ in how the atoms are connected. For C4H10 there are exactly two possible arrangements: the straight chain butane, and the branched 2-methylpropane.

1 mark: correct definition (same molecular formula, different structural arrangement/connectivity of atoms); 1 mark: correct number (two) with both isomers named (butane, 2-methylpropane). [2]

Elimination of HBr from 2-bromobutane can remove a hydrogen from C1 or C3. By Zaitsev's rule the major product is the more substituted (more stable) alkene, formed by loss of H from C3: but-2-ene.

1 mark: correct application of Zaitsev's rule (more substituted alkene is the major product); 1 mark: correct name but-2-ene. [2]

Rotation about a C=C double bond is restricted, so if each carbon of the double bond is attached to two different groups, two distinct spatial arrangements are possible. For but-2-ene, the two methyl groups (the higher-priority substituent on each carbon) can lie on the same side of the double bond (Z-but-2-ene) or on opposite sides (E-but-2-ene).

1 mark: correct explanation of restricted rotation about C=C with two different groups on each carbon; 1 mark: correctly identifies E and Z forms of but-2-ene by relative positions of the methyl groups. [2]

Alcohols are classified by the number of carbon atoms directly bonded to the carbon bearing the -OH group: one carbon group = primary (propan-1-ol), two carbon groups = secondary (propan-2-ol), three carbon groups = tertiary (2-methylpropan-2-ol).

1 mark: all three correctly classified; 1 mark: correct classification rule stated (number of carbon groups attached to the C-OH carbon). [2]

Propene, CH2=CH-CH3, undergoes addition polymerisation: the C=C double bond opens and monomers join end to end without loss of any atoms, giving the repeat unit -[CH2-CH(CH3)]-.

1 mark: correct repeat unit showing the -CH2-CH(CH3)- backbone in square brackets; 1 mark: 'addition polymerisation' stated. [2]

The broad absorption at 2500-3300 cm-1 is caused by the O-H bond stretching within the strongly hydrogen-bonded carboxylic acid dimer, which broadens the peak compared with an alcohol O-H. Its presence, together with a carbonyl absorption near 1700 cm-1, confirms a -COOH group.

1 mark: correctly identifies O-H stretch (broad, hydrogen-bonded) in the 2500-3300 cm-1 range; 1 mark: correctly identifies the carboxylic acid (-COOH) functional group. [2]

Aspirin retains the carboxylic acid group of salicylic acid and gains an ester group where the phenolic -OH has been acylated by ethanoic anhydride, so it contains both an ester and a carboxylic acid group. The reaction converting the phenol to the ester is an esterification (acylation).

1 mark: both functional groups named (ester and carboxylic acid); 1 mark: correct reaction type (esterification/acylation) named. [2]

The three isomers arise from different ways of arranging the same five carbon atoms into a continuous or branched chain: the unbranched chain (pentane), a single branch (2-methylbutane), and a doubly-branched chain (2,2-dimethylpropane). Because only the carbon skeleton differs, not the functional groups, this is chain (skeletal) isomerism.

1 mark each for two of the three correct isomer formula+name pairs (up to 2 marks); 1 mark: correct identification as chain/skeletal isomerism. Accept any valid ordering. [3]

Both compounds share the molecular formula C4H10O but place the oxygen atom in a different functional group: as a hydroxyl group (alcohol) or as an ether linkage between two alkyl groups. Because the functional group itself differs, this is functional group isomerism.

1 mark: correct alcohol structure and name; 1 mark: correct ether structure and name; 1 mark: 'functional group isomerism' correctly identified. [3]

Step 1: the electron-rich pi bond of ethene acts as a nucleophile and attacks the electrophilic, slightly positive hydrogen atom of the polar H-Br bond; the H-Br sigma bond breaks heterolytically, with both electrons going to bromine, forming Br- and a carbocation intermediate, CH3CH2+. Step 2: the bromide ion generated in step 1 acts as a nucleophile, using a lone pair to bond to the positively charged carbon of the carbocation, forming bromoethane, CH3CH2Br. Overall this is a two-step electrophilic addition mechanism proceeding via a carbocation intermediate.

1 mark: identifies pi bond as nucleophile attacking delta-positive H of HBr; 1 mark: correct heterolytic fission of H-Br bond shown/described; 1 mark: carbocation intermediate correctly identified; 1 mark: Br- identified as the nucleophile in step 2; 1 mark: correct final product bromoethane named. [5]

Addition of HBr to propene can proceed via a secondary carbocation (leading to 2-bromopropane) or a primary carbocation (leading to 1-bromopropane). Alkyl groups are electron-donating by the positive inductive effect, so the secondary carbocation, with two alkyl groups adjacent to the positive charge, is stabilised more than the primary carbocation, which has only one. The more stable carbocation forms faster and predominates, so 2-bromopropane is the major product (Markovnikov addition) and 1-bromopropane the minor product.

1 mark: correct major product identified (2-bromopropane); 1 mark: correct minor product identified (1-bromopropane); 1 mark: secondary carbocation intermediate for the major product correctly identified; 1 mark: reference to positive inductive effect of alkyl groups stabilising the secondary carbocation; 1 mark: explicit comparison showing secondary is more stable than primary, hence forms preferentially. [5]

Initiation: UV light supplies enough energy to break the Cl-Cl bond homolytically, producing two chlorine radicals, Cl2 -> 2Cl•. Propagation (a repeating cycle that consumes reactants and produces the product): Cl• abstracts a hydrogen atom from methane, Cl• + CH4 -> •CH3 + HCl; the methyl radical then reacts with another chlorine molecule, •CH3 + Cl2 -> CH3Cl + Cl•, regenerating a chlorine radical to continue the chain. Termination: the reaction stops when two radicals collide and combine, removing them from the system, e.g. Cl• + Cl• -> Cl2, or •CH3 + Cl• -> CH3Cl, or •CH3 + •CH3 -> C2H6.

1 mark: initiation step with UV-induced homolytic fission of Cl2 to 2Cl•; 1 mark: first propagation step Cl• + CH4 -> •CH3 + HCl; 1 mark: second propagation step •CH3 + Cl2 -> CH3Cl + Cl•; 1 mark: at least one valid termination step (radical-radical combination); 1 mark: correct use of dot notation for radicals throughout. [5]

Partial oxidation to the aldehyde occurs first: CH3CH2OH + [O] -> CH3CHO + H2O. If the aldehyde is distilled off as it forms, it is removed from the oxidising mixture before further reaction can occur, giving ethanal as the isolated product. If instead the mixture is heated under reflux, the ethanal remains in contact with the oxidising agent and is further oxidised: CH3CHO + [O] -> CH3COOH, giving ethanoic acid as the isolated product. The colour of the dichromate(VI) ion changes from orange to green as Cr(VI) is reduced to Cr(III), providing visible evidence that oxidation is occurring.

1 mark: correct equation ethanol to ethanal; 1 mark: correct equation ethanal to ethanoic acid; 1 mark: distillation identified as needed to isolate ethanal (prevents further oxidation); 1 mark: reflux identified as needed for complete oxidation to ethanoic acid; 1 mark: colour change of dichromate(VI) (orange to green) correctly stated as evidence of oxidation. [5]

A carboxylic acid group on hexanedioic acid reacts with an amine group on 1,6-diaminohexane: -COOH + H2N- -> -CONH- + H2O, eliminating a molecule of water as the two monomers join through an amide linkage; this repeats at both ends of each monomer to build up the polymer chain. Unlike addition polymers, which retain every atom of the monomer in a pure carbon backbone, condensation polymers such as nylon lose a small molecule (here water) at each linkage and contain hetero-atom (C-N or C-O) links within the backbone.

1 mark: correct equation showing -COOH + H2N- reacting; 1 mark: correct amide linkage -CONH- shown as product; 1 mark: water correctly named as the eliminated small molecule; 1 mark: valid structural difference stated (e.g. hetero-atoms in backbone vs pure C backbone, or loss of small molecule vs no loss); 1 mark: reference to the two monomer functional groups (diamine + diacid) required for condensation. [5]

The phenolic -OH group of 2-hydroxybenzoic acid is acylated by ethanoic anhydride, forming aspirin (2-acetoxybenzoic acid) and ethanoic acid as a by-product: 2-hydroxybenzoic acid + (CH3CO)2O -> aspirin + CH3COOH. The phosphoric(V) acid is not consumed in the reaction and so acts as a catalyst: it protonates the carbonyl oxygen of the anhydride, increasing the electrophilicity of the carbonyl carbon and making it more susceptible to nucleophilic attack by the phenolic oxygen, which speeds up the acylation.

1 mark: correct organic reactants named/formulae given (2-hydroxybenzoic acid + ethanoic anhydride); 1 mark: correct organic products named (aspirin + ethanoic acid); 1 mark: correct overall equation; 1 mark: phosphoric(V) acid correctly identified as a catalyst (not consumed, lowers activation energy); 1 mark: mechanism-level detail - protonates the anhydride carbonyl, increasing electrophilicity for nucleophilic attack by the phenolic -OH. [5]

Moles of 2-hydroxybenzoic acid = 2.50 / 138 = 0.0181 mol. Since the reaction is 1:1, theoretical moles of aspirin = 0.0181 mol, so theoretical mass = 0.0181 x 180 = 3.26 g. Percentage yield = (actual / theoretical) x 100 = (2.61 / 3.26) x 100 = 80.1%.

1 mark: correct moles of 2-hydroxybenzoic acid (0.0181 mol); 1 mark: correct theoretical mass of aspirin (3.26 g); 1 mark: correct percentage yield calculation set up (actual/theoretical x 100); 1 mark: correct final answer 80.1% (accept 79-81%, ecf from earlier steps). [4]

Combustion equation: \( \text{C}_2\text{H}_5\text{OH} + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O} \). Moles of ethanol = 5.00 / 46 = 0.109 mol. From the 1:2 mole ratio, moles of CO2 = 2 x 0.109 = 0.217 mol. Volume of CO2 at RTP = 0.217 x 24 = 5.22 dm³.

1 mark: correct balanced combustion equation for ethanol; 1 mark: correct moles of ethanol (0.109 mol); 1 mark: correct mole ratio applied (2 mol CO2 per mol ethanol, giving 0.217 mol CO2); 1 mark: correct final volume 5.22 dm³ (ecf). [4]

(a) A melting point that is lower than, and over a broader range than, the sharp literature value indicates the product still contains impurities, which lower and broaden the melting point. (b) Moles of 2-hydroxybenzoic acid = 3.00 / 138 = 0.0217 mol. Theoretical mass of aspirin = 0.0217 x 180 = 3.91 g. Percentage yield = (3.40 / 3.91) x 100 = 86.9%.

(a) 1 mark: correctly states the melting point range being lower/broader than the literature value indicates the product is impure. (b) 1 mark: correct moles of 2-hydroxybenzoic acid (0.0217 mol); 1 mark: correct theoretical mass of aspirin (3.91 g); 1 mark: correct percentage yield method (actual/theoretical x100); 1 mark: correct final answer 86.9% (ecf, accept 86-88%). [5]

Moles of but-1-ene = 11.2 / 24 = 0.467 mol. The reaction is 1:1, so theoretical moles of butan-2-ol = 0.467 mol, giving a theoretical mass = 0.467 x 74 = 34.5 g. Applying the 65% yield: actual mass = 34.5 x 0.65 = 22.4 g.

1 mark: correct moles of but-1-ene (0.467 mol); 1 mark: correct 1:1 mole ratio applied; 1 mark: correct theoretical mass of butan-2-ol (34.5 g); 1 mark: correct application of 65% yield; 1 mark: correct final answer 22.4 g (ecf). [5]

(a) A relative molecular mass of 74 with one oxygen atom is consistent with a saturated alcohol of formula C4H10O, such as 2-methylpropan-2-ol. (b) The mass lost between the molecular ion and the base peak is 74 - 59 = 15, corresponding to loss of a methyl radical, CH3•. For 2-methylpropan-2-ol this loss gives a stabilised tertiary carbocation fragment, [(CH3)2COH]+, at m/z = 59, which is why this fragment forms the base peak.

1 mark: correct molecular formula C4H10O deduced from Mr = 74; 1 mark: correct mass loss calculated (74 - 59 = 15); 1 mark: correctly identifies the fragment lost as CH3• (methyl radical); 1 mark: correct fragment ion formula/structure at m/z 59 given; 1 mark: reference to the stability of the resulting (tertiary) carbocation fragment explaining why this is the base peak. [5]

The absorption at 1740 cm-1 is characteristic of a C=O stretch, and its high wavenumber together with the absence of any 2500-3300 cm-1 O-H absorption rules out a carboxylic acid, pointing instead to an ester. A molecular ion of 74 with an ester group fits methyl ethanoate, CH3COOCH3. The peak at m/z 59 corresponds to loss of a methyl radical (M - 15) from the molecular ion, and the peak at m/z 43 corresponds to the CH3CO+ acylium ion, formed by loss of OCH3 (M - 31).

1 mark: correct compound identified (methyl ethanoate); 1 mark: 1740 cm-1 correctly assigned to C=O (ester) stretch; 1 mark: absence of 2500-3300 cm-1 band correctly used to rule out a carboxylic acid/alcohol O-H; 1 mark: m/z 59 correctly explained as loss of CH3• (M-15) from the molecular ion; 1 mark: m/z 43 correctly explained as the CH3CO+ acylium ion (loss of OCH3, M-31). [5]

The middle CH2 of propan-1-ol (CH3-CH2-CH2-OH) is not directly bonded to oxygen, so it falls in the typical alkyl CH2 region, delta 1.2-1.5 ppm. By the n+1 rule, its splitting is determined by all non-equivalent protons on adjacent carbons: 3 protons from the neighbouring CH3 plus 2 protons from the neighbouring CH2OH gives 5 neighbouring protons, so the signal is split into a sextet (5+1 = 6 peaks). Its relative integration is 2H, part of the overall 3:2:2:1 ratio for CH3:CH2:CH2OH:OH across the full spectrum.

1 mark: correct chemical shift range (delta 1.2-1.5, accept 1.0-1.7) justified as an alkyl CH2 not directly bonded to O; 1 mark: correct splitting pattern (sextet) identified; 1 mark: correct application of the n+1 rule showing 5 neighbouring protons (3+2); 1 mark: correct relative integration value of 2H stated; 1 mark: correct overall ratio 3:2:2:1 given for the full spectrum. [5]

The 1715 cm-1 absorption confirms a C=O group is present, consistent with a ketone or aldehyde. The absence of any absorption in the 2500-3300 cm-1 region rules out a carboxylic acid isomer. The single NMR signal shows that all the protons in the molecule are chemically equivalent; in propanone, CH3COCH3, both methyl groups are in identical environments by symmetry, so all six protons give one signal. Propanal, CH3CH2CHO, is ruled out because its CH3, CH2 and CHO protons are all in different (non-equivalent) environments and would give three separate signals, not one.

1 mark: X correctly identified as propanone; 1 mark: 1715 cm-1 correctly assigned to the C=O stretch, consistent with a ketone; 1 mark: absence of 2500-3300 cm-1 correctly used to rule out a carboxylic acid isomer/functional group; 1 mark: single NMR signal correctly linked to the six equivalent protons in propanone (by symmetry); 1 mark: propanal correctly ruled out with valid reasoning (three non-equivalent proton environments, so three signals expected); 1 mark: clear, logical structure linking all three pieces of evidence to the final identification. [6]

React 2-hydroxybenzoic acid with excess ethanoic anhydride and a few drops of concentrated phosphoric(V) acid catalyst, warming gently in a water bath at about 50-60°C for around 15 minutes. Cool the mixture and add cold water, which decomposes the excess unreacted anhydride and induces crystallisation of the crude aspirin. Filter the crude solid under reduced pressure using a Buchner funnel and flask, washing with a little cold water. Recrystallise the crude product by dissolving it in a minimum volume of hot solvent, allowing the solution to cool slowly so that pure crystals form while impurities remain in solution, then filter again and wash with a little cold solvent. Dry the crystals in a desiccator or low-temperature oven. Confirm purity by determining the melting point (a pure sample gives a sharp melting point close to the literature value of 135°C, whereas an impure sample melts lower and over a broader range) and/or by thin-layer chromatography, where a pure product gives a single spot with a single Rf value.

Level 3 (7-9 marks): a full, logically ordered method covering reaction conditions (warming with phosphoric(V) acid catalyst), quenching/crystallisation with cold water, vacuum filtration of the crude product, recrystallisation to purify, drying, and a valid purity check (melting point range and/or TLC) with a clear expected result (sharp m.p. near 135°C / single TLC spot); fluent use of specialist terms (e.g. 'recrystallisation', 'vacuum/Buchner filtration', 'Rf value') throughout. Level 2 (4-6 marks): most stages of the method present but with gaps in detail, ordering errors, or an underdeveloped purity check; reasonable use of specialist terms. Level 1 (1-3 marks): a fragmented or largely incomplete method with few correct stages; limited use of specialist terminology. 0 marks: no creditable response. [9]

Technetium-99m emits gamma radiation only, at an energy suitable for external detection without causing excessive tissue damage, and its short half-life of about 6 hours means it decays away quickly after imaging, minimising the radiation dose received by the patient.

1 mark each for any two valid properties: pure/predominantly gamma emitter (no particulate radiation); short half-life (~6 hours); suitable gamma energy for detection without excessive tissue damage; readily forms compounds that target specific organs. [2]

Half-life is defined as the average time required for the activity, or equivalently the number of undecayed radioactive nuclei in a sample, to decrease to half of its initial value.

1 mark: reference to time taken; 1 mark: for activity/number of nuclei to halve. [2]

Ultrasound uses high-frequency sound waves rather than ionising radiation, so there is no radiation dose or associated cancer risk to the developing fetus, and because images are produced continuously the sonographer can view real-time movement, such as the fetal heartbeat.

1 mark each for any two: non-ionising/no radiation dose to fetus; real-time/moving image; relatively low cost/portable equipment; no known harmful side effects at diagnostic intensities. [2]

Acoustic impedance, Z, is a measure of how much a tissue resists the passage of an ultrasound wave, and is calculated as the product of the tissue's density and the speed of sound through it. Reflection of ultrasound occurs at boundaries between tissues of different acoustic impedance.

1 mark: reference to density x speed of sound (Z = density x speed); 1 mark: correctly described as a measure of a tissue's resistance to ultrasound / determines reflection at a boundary. [2]

An ECG records the small electrical potential differences generated on the skin as the heart's electrical impulses spread through the cardiac muscle. On a normal trace, the QRS complex is the sharp deflection that corresponds to depolarisation and contraction of the ventricles.

1 mark: correct description of what ECG measures (electrical activity of the heart); 1 mark: QRS complex correctly named. [2]

A pulse oximeter shines red and infrared light through a finger or earlobe. Oxygenated and deoxygenated haemoglobin absorb these two wavelengths of light to different extents, so by measuring the ratio of light absorbed at each wavelength the device calculates the percentage of haemoglobin that is oxygen-saturated (SpO2).

1 mark: reference to differing absorption of red/infrared light by oxygenated vs deoxygenated haemoglobin; 1 mark: ratio of absorbances used to calculate % saturation (SpO2). [2]

In diagnostic use, a low-activity gamma-emitting tracer with a short half-life is used purely to produce an image, keeping the patient's radiation dose as low as reasonably possible. In therapeutic use, a much higher activity source, often emitting more strongly ionising beta or alpha radiation, is used deliberately to destroy diseased tissue such as tumour cells, since the aim is to deposit a large, damaging dose in a targeted region.

1 mark: correct description of diagnostic use (low-dose gamma tracer for imaging); 1 mark: correct description of therapeutic use (higher dose/alpha or beta emitter to destroy tissue). [2]

Barium has a much higher atomic number than the elements found in soft tissue, so it absorbs X-rays far more strongly. When it coats or fills the digestive tract, this creates a strong contrast on the radiograph between the barium-filled tract and the surrounding soft tissue, making the outline of the digestive tract clearly visible.

1 mark: high atomic number/density of barium causing strong X-ray absorption; 1 mark: creates visible contrast against soft tissue on the image. [2]

Radiographers reduce their exposure by standing behind a lead-lined screen or leaving the room while the X-ray is taken, and by wearing a personal dosimeter badge that records their cumulative radiation dose over time so that exposure can be monitored and controlled.

1 mark each for any two: lead/protective shielding or leaving the room; dosimeter badges; maximising distance from source; minimising time of exposure; lead aprons for patients/staff where applicable. [2]

A CT scanner rotates an X-ray source and detector around the patient, taking many projections from different angles which a computer reconstructs into cross-sectional (slice) images, whereas a conventional X-ray produces only a single flat 2D projection image. Because CT requires many exposures, it delivers a substantially higher total radiation dose to the patient than a single conventional X-ray.

1 mark each for any two valid differences: cross-sectional/3D image vs single 2D projection; multiple exposures from different angles vs single exposure; higher radiation dose for CT vs conventional X-ray; computer reconstruction required for CT. [2]

Three examples: the heart's electrical activity can be recorded using an ECG; blood oxygen saturation can be measured using a pulse oximeter; and blood pressure can be measured using a sphygmomanometer. Other valid pairs include lung volume with a spirometer, or body temperature with a thermometer.

1 mark each for any three valid quantity + correct device pairs (max 3): heart electrical activity/ECG; oxygen saturation/pulse oximeter; blood pressure/sphygmomanometer; lung volume/spirometer; body temperature/thermometer. [3]

Absorbed dose is defined as the energy absorbed per unit mass of the irradiated tissue, and its SI unit is the gray (Gy), where 1 Gy = 1 joule per kilogram. Absorbed dose alone does not fully describe biological risk because different types of radiation (e.g. alpha, beta, gamma) cause different amounts of biological damage for the same absorbed dose; a radiation weighting factor must be applied to obtain the equivalent dose, measured in sieverts, which better reflects biological risk.

1 mark: correct definition of absorbed dose (energy per unit mass); 1 mark: correct unit gray (Gy); 1 mark: correct explanation referencing radiation weighting factor/equivalent dose (sieverts) needed because different radiations cause different biological damage per unit absorbed dose. [3]

A strong magnetic field aligns the spin (magnetic moment) of hydrogen nuclei in the body. A radiofrequency pulse is then applied, which knocks the nuclei out of alignment. As the nuclei relax back into alignment with the field they emit radiofrequency signals, which are detected by the scanner and computer-processed to build a detailed image; because different tissues relax at different rates, this produces the contrast seen between tissue types in the final image.

1 mark: strong magnetic field aligns hydrogen nuclei/proton spins; 1 mark: RF pulse applied, knocking nuclei out of alignment; 1 mark: nuclei relax and emit detectable RF signal; 1 mark: different relaxation rates in different tissues give rise to image contrast. [4]

Fast-moving electrons decelerate rapidly on striking the metal target; the loss of kinetic energy is converted into electromagnetic radiation (X-ray photons), a process known as bremsstrahlung ('braking radiation'). Characteristic X-rays are also produced when inner-shell electrons are ejected from target atoms and outer electrons fall to fill the vacancy, releasing a photon of a specific energy. However, only a small fraction (around 1%) of the electrons' kinetic energy is converted into X-rays; the vast majority is converted into heat, which is why the anode must be cooled or rotated.

1 mark: electrons decelerate/collide with the target, converting kinetic energy to EM radiation (bremsstrahlung); 1 mark: reference to characteristic X-ray production via electron shell transitions (or general acceptance of continuous + characteristic spectrum); 1 mark: correct statement that most of the energy converts to heat, not X-rays; 1 mark: correct reference to the need for the anode to be cooled/rotated as a consequence. [4]

The collimator is a thick sheet of lead containing many narrow parallel holes, placed in front of the detector crystal. It only allows gamma photons travelling perpendicular to the detector (i.e. in a straight line from the source) to pass through and reach the scintillation crystal, absorbing photons travelling at other angles. This ensures that each point on the resulting image corresponds accurately to the position of the tracer within the body, rather than the image being blurred by photons arriving at an angle.

1 mark: lead sheet with narrow parallel holes/channels described; 1 mark: only allows gamma rays travelling perpendicular/in a straight line through to reach the detector; 1 mark: absorbs photons at other angles; 1 mark: this ensures spatial accuracy/correct mapping of the tracer's location in the resulting image. [4]

The patient takes a maximal inspiration and then exhales as forcefully and completely as possible into the spirometer, which records the volume of air displaced (for example via a bell and chamber connected to a rotating drum, or a digital flow sensor), producing a trace from which the volume can be read directly. Vital capacity is defined as the maximum volume of air that can be forcibly exhaled from the lungs after a maximal inspiration; it does not include the residual volume of air that always remains in the lungs and cannot be exhaled, which simple spirometry cannot measure.

1 mark: patient takes a maximal inspiration then maximal expiration into the device; 1 mark: correct description of how the spirometer records the volume (e.g. bell/drum trace or digital flow-volume sensor); 1 mark: correct definition of vital capacity (max volume exhaled after max inspiration); 1 mark: reference to residual volume remaining in the lungs and not being measured by simple spirometry. [4]

A short half-life means the isotope decays away quickly after the scan is complete, minimising the total radiation dose delivered to the patient's tissues, whereas an isotope with a much longer half-life would continue irradiating the patient for a far longer time after imaging has finished, for no diagnostic benefit. A practical consequence of technetium-99m's short (about 6 hour) half-life is that it cannot be produced far in advance or stockpiled; hospitals instead use a molybdenum-99/technetium-99m generator, which produces a fresh supply of technetium-99m on site shortly before it is needed.

1 mark: short half-life minimises total radiation dose to the patient after the scan; 1 mark: valid comparison to a long-half-life isotope continuing to irradiate tissue long after imaging; 1 mark: correct practical consequence identified (needs on-site generator/cannot be stockpiled, must be used soon after production); 1 mark: reference to the molybdenum-99/technetium-99m generator system. [4]

A fibre-optic endoscope contains two bundles of thin glass or plastic optical fibres: an incoherent bundle carries light from an external source into the body to illuminate the area being examined, and a coherent bundle, in which the fibres occupy the same relative position at both ends, carries the reflected light back out to form a recognisable image. Light is kept within each individual fibre by total internal reflection at the boundary between the higher-refractive-index core and the lower-refractive-index cladding surrounding it, as the light repeatedly strikes this boundary at an angle of incidence greater than the critical angle.

1 mark: two fibre bundles described (illumination and imaging); 1 mark: coherent bundle correctly identified as required for forming the image (fibres in matching positions at each end); 1 mark: core has higher refractive index than the surrounding cladding; 1 mark: total internal reflection correctly identified as the principle keeping light in the fibre; 1 mark: reference to angle of incidence exceeding the critical angle at the core-cladding boundary. [5]

In transmission mode, an alternating voltage applied across the crystal causes it to mechanically vibrate at high frequency, generating an ultrasound wave (the converse piezoelectric effect). In receiving mode, returning reflected ultrasound waves mechanically deform the crystal, which generates a small alternating voltage that is amplified and processed to build the image (the direct piezoelectric effect). Coupling gel is applied because air has a very different acoustic impedance from skin, so without the gel almost all of the ultrasound would be reflected at the skin surface rather than entering the body. The gel has an acoustic impedance close to that of skin/soft tissue, eliminating the air gap between the transducer and the skin and allowing efficient transmission of ultrasound into the body.

1 mark: converse piezoelectric effect correctly described (voltage -> mechanical vibration, generating ultrasound); 1 mark: direct piezoelectric effect correctly described (returning wave -> mechanical deformation -> voltage signal); 1 mark: coupling gel correctly explained as removing the air gap between probe and skin; 1 mark: reference to acoustic impedance mismatch between air and skin causing (near-)total reflection without the gel; 1 mark: gel's acoustic impedance described as closely matching that of skin/soft tissue for efficient transmission. [5]

In this context a 'lead' refers to a specific combination or pairing of electrodes on the body used to record the potential difference between two points, giving a particular 'view' of the heart's electrical activity from a different angle; using 12 leads gives 12 such views, allowing a more complete picture of the heart's electrical activity to be built up. Electrodes are conductive pads placed on the skin, often with conductive gel to improve electrical contact, that detect the tiny potential differences generated as electrical impulses (waves of depolarisation and repolarisation) spread through the heart muscle and conduct outward through body tissue to the skin surface; these signals are then amplified and processed to produce the displayed ECG trace.

1 mark: 'lead' correctly defined as a specific electrode pairing/combination giving a particular view of the heart's electrical activity; 1 mark: reference to 12 leads giving multiple views/angles of the heart; 1 mark: electrodes correctly described as detecting small potential differences via skin contact (often gel-assisted); 1 mark: reference to depolarisation/repolarisation of heart muscle conducting through body tissue to the skin; 1 mark: reference to amplification/processing of the signal to produce the displayed trace. [5]

Number of half-lives elapsed = 24 / 6.0 = 4. Activity halves each half-life: 800 -> 400 -> 200 -> 100 -> 50 MBq after 4 half-lives.

1 mark: correct number of half-lives calculated (24/6 = 4); 1 mark: correct method (800 x (1/2)^n or repeated halving) set up; 1 mark: correct intermediate halvings shown (800->400->200->100->50); 1 mark: correct final answer 50 MBq; 1 mark: correct unit MBq retained throughout. [5]

The activity falls from 640 Bq to 40 Bq, a reduction by a factor of 16. Since 16 = 2^4, this represents 4 half-lives. The half-life is therefore 30 minutes / 4 = 7.5 minutes.

1 mark: correct recognition that 640 to 40 represents a reduction by a factor of 16; 1 mark: correct number of half-lives calculated (16 = 2^4, so 4 half-lives); 1 mark: correct division of 30 minutes by 4; 1 mark: correct final answer 7.5 minutes; 1 mark: working shown clearly with units. [5]

The pulse travels to the boundary and back, so total distance = speed x time = 1540 x 26x10⁻⁶ = 0.0400 m. The depth of the boundary is half this round-trip distance: 0.0400 / 2 = 0.0200 m = 2.0 cm.

1 mark: correct recognition that the pulse travels to the boundary and back (distance = speed x time for round trip); 1 mark: correct total distance calculated (1540 x 26x10⁻⁶ = 0.0400 m); 1 mark: correct halving to find depth (0.0200 m); 1 mark: correct final answer 2.0 cm (or 0.020 m, or 20 mm); 1 mark: correct unit conversion/consistency shown throughout. [5]

Equivalent dose = absorbed dose x radiation weighting factor = 0.050 x 20 = 1.0 Sv. The weighting factor is applied because alpha radiation causes far more biological damage per unit absorbed dose than the reference radiation used to define the gray.

1 mark: correct formula stated/used (equivalent dose = absorbed dose x radiation weighting factor); 1 mark: correct substitution (0.050 x 20); 1 mark: correct final answer 1.0 Sv; 1 mark: correct unit sievert (Sv) given; 1 mark: reasoning shown as to why weighting factor is applied (accounts for greater biological damage of alpha radiation). [5]

The R-R interval is the time between successive heartbeats. Heart rate = 60 / R-R interval = 60 / 0.80 = 75 beats per minute.

1 mark: correct recognition that the R-R interval represents the time for one heartbeat; 1 mark: correct formula (heart rate = 60 / R-R interval); 1 mark: correct substitution (60/0.80); 1 mark: correct final answer 75 bpm; 1 mark: correct unit (beats per minute) stated. [5]

(a) Using \( I = I_0 e^{-\mu x} \): the exponent is \( \mu x = 0.30 \times 4.0 = 1.2 \), so \( I = I_0 e^{-1.2} \approx 0.30 I_0 \), i.e. about 30% of the initial intensity is transmitted. (b) Bone has a higher density and a higher linear attenuation coefficient than soft tissue, so it absorbs and scatters a greater proportion of the X-ray photons passing through it, meaning less intensity would be transmitted. It is precisely this difference in attenuation between bone and soft tissue that produces the contrast seen on an X-ray image.

(a) 1 mark: correct exponential attenuation formula stated (I = I0 e^(-mu x)); 1 mark: correct exponent calculated (mu x = 0.30 x 4.0 = 1.2); 1 mark: correct final percentage ~30% (accept 29-31%). (b) 1 mark: correctly states transmitted intensity would be lower/less; 1 mark: correctly explains bone has a higher attenuation coefficient/density than soft tissue; 1 mark: correctly links this difference in attenuation to the formation of image contrast. [6]

X-ray imaging relies on the differential absorption of ionising X-ray photons by different tissues (denser tissue such as bone absorbs more), producing a single 2D projection image; it is quick and cheap but gives limited soft-tissue contrast and involves ionising radiation. CT extends this principle by taking many X-ray projections from different angles and computer-reconstructing them into detailed cross-sectional images, giving much better spatial and soft-tissue detail than plain X-ray, but at a considerably higher radiation dose. Ultrasound relies on the reflection of high-frequency sound waves at boundaries between tissues of differing acoustic impedance; it is non-ionising and safe for repeated use (e.g. in pregnancy), relatively cheap and portable, and gives real-time moving images, but has lower resolution than CT/MRI and cannot image through bone or gas-filled structures well. MRI relies on the resonance behaviour of hydrogen nuclei placed in a strong magnetic field and disturbed by radiofrequency pulses, with the emitted signal used to build an image; it gives excellent soft-tissue contrast without any ionising radiation, but scans are slow, the equipment is very expensive, and it is unsuitable for patients with certain metal implants or pacemakers due to the strong magnetic field.

Level 3 (9-12 marks): comparison of all four techniques (X-ray, CT, ultrasound, MRI) is complete and accurate, with correct physical principle, a valid comment on image quality/resolution, and a valid risk/limitation identified for each; the answer is explicitly comparative (not four separate descriptions) and uses specialist terminology fluently and accurately throughout (e.g. 'attenuation', 'acoustic impedance', 'resonance', 'ionising'). Level 2 (5-8 marks): most techniques covered with reasonable accuracy but one technique may be missing, underdeveloped, or contain an error; some comparative language used; adequate use of specialist terms. Level 1 (1-4 marks): only one or two techniques addressed with limited accuracy, largely descriptive rather than comparative, or very limited/inaccurate use of specialist terminology. 0 marks: no creditable response. [12]

The three ossicles form a small lever system that transmits vibrations from the eardrum across the air-filled middle ear to the oval window of the cochlea, amplifying the pressure in the process so that vibrations are efficiently passed into the fluid-filled inner ear.

1 mark: correctly identifies transmission of vibrations from eardrum to oval window/cochlea; 1 mark: reference to amplification (via lever action/pressure concentration over a smaller area). [2]

The cochlea is the coiled, fluid-filled structure of the inner ear; within it, specialised hair cells in the organ of Corti are displaced by the fluid vibrations, generating nerve impulses that travel along the auditory nerve.

1 mark: cochlea correctly named; 1 mark: hair cells (organ of Corti) correctly named. [2]

The cornea, at the front of the eye, provides most of the eye's total refractive power and is fixed in shape, doing the bulk of the initial bending of incoming light. The lens provides a smaller, but variable and adjustable, amount of refractive power through the process of accommodation, fine-tuning the focus for objects at different distances.

1 mark: cornea correctly described as providing most/fixed refractive power; 1 mark: lens correctly described as providing variable/adjustable focusing (accommodation). [2]

The retina contains rods and cones. Rods are highly sensitive to low light levels and function well in dim conditions but do not distinguish colour, whereas cones require brighter light to function and enable both colour vision and finer visual detail (higher acuity).

1 mark: rods and cones both correctly named; 1 mark: valid functional difference stated (e.g. rods for dim light/no colour, cones for bright light/colour and detail). [2]

In a transverse wave, the particles or fields oscillate at right angles (perpendicular) to the direction in which the wave transfers energy; light and other electromagnetic waves are transverse. In a longitudinal wave, the oscillations are parallel to the direction of energy transfer, with regions of compression and rarefaction; sound waves are longitudinal.

1 mark: correct distinction (perpendicular vs parallel oscillation relative to wave travel) with correct example for at least one type; 1 mark: both correct examples given (light = transverse, sound = longitudinal). [2]

The critical angle is the angle of incidence, measured in the optically denser core, at which the refracted ray travels exactly along the boundary between the core and cladding, i.e. the angle of refraction is 90 degrees. At angles of incidence greater than this, total internal reflection occurs instead of refraction.

1 mark: correct reference to angle of incidence at which refraction angle = 90 degrees; 1 mark: correctly framed in terms of light travelling from the denser (core) to less dense (cladding) medium. [2]

A standing wave forms when two waves of the same frequency and wavelength, travelling in opposite directions, superpose; for the resulting pattern of nodes and antinodes to be clear and stable, the two waves must also have similar (comparable) amplitude.

1 mark: two waves of the same frequency/wavelength travelling in opposite directions superposing; 1 mark: similar/comparable amplitude of the two waves. [2]

Radio waves span an extremely wide range of the electromagnetic spectrum, with wavelengths from around 1 mm up to several kilometres, corresponding to frequencies from about 3 kHz up to 300 GHz. Common household applications include broadcast radio and television signals, Wi-Fi, and mobile phone communication.

1 mark: correct approximate range given (wavelength or frequency); 1 mark: valid household application named. [2]

The typical range of human hearing is approximately 20 Hz to 20,000 Hz (20 kHz). With increasing age, sensitivity to higher frequencies typically declines, a condition known as presbycusis, which reduces the effective upper limit of the audible range.

1 mark: correct frequency range (20 Hz-20 kHz, accept close values); 1 mark: correct description of age-related decline in high-frequency hearing (presbycusis). [2]

The Eustachian tube connects the middle ear cavity to the pharynx at the back of the throat. Its function is to equalise air pressure on either side of the eardrum, allowing the eardrum to vibrate freely and respond correctly to incoming sound.

1 mark: correctly identifies it connects the middle ear to the pharynx/throat; 1 mark: correctly explains its function is to equalise pressure either side of the eardrum. [2]

Myopia occurs when the eyeball is too long, or the refractive power of the cornea/lens is too great, so that light from a distant object is brought to a focus in front of the retina rather than on it. It is corrected by placing a diverging (concave) lens in front of the eye, which diverges the incoming light slightly before it enters the eye, moving the focal point back onto the retina.

1 mark: correct cause (eyeball too long / refractive power too great, image focuses in front of retina); 1 mark: correct correction (diverging/concave lens). [2]

Digital signals consist of discrete on/off pulses that can be regenerated exactly at intervals along the fibre using repeaters, removing accumulated noise rather than just amplifying it, which gives much better noise immunity over long distances than an analogue signal. Digital signals can also be more easily processed, compressed, encrypted, and combined with other signals using multiplexing.

1 mark each for any two valid advantages: easier/exact regeneration free of accumulated noise; better noise immunity generally; easier signal processing/compression/encryption/multiplexing; more efficient use of bandwidth via multiplexing. [2]

In the fundamental mode of a string fixed at both ends, exactly half a wavelength fits along the string's length, so the wavelength is 2L. Since the string is fixed at both ends, a node (a point of zero displacement) occurs at each end; a single antinode, the point of maximum displacement, occurs at the midpoint of the string.

1 mark: correct wavelength expressed as 2L; 1 mark: nodes correctly located at both fixed ends; 1 mark: single antinode correctly located at the midpoint. [3]

Sound waves are first collected by the pinna and travel along the ear canal to the eardrum (tympanic membrane), which vibrates. These vibrations are transmitted and amplified by the three ossicles (malleus, incus, stapes) across the middle ear to the oval window of the cochlea. Vibrations in the fluid of the cochlea displace hair cells in the organ of Corti, generating nerve impulses that travel along the auditory nerve to the brain.

1 mark: pinna, ear canal and eardrum correctly sequenced; 1 mark: ossicles and oval window correctly sequenced; 1 mark: cochlea/hair cells and auditory nerve correctly named as the final conversion/transmission stage. [3]

(a) The third harmonic of a string fixed at both ends has 3 antinodes and 4 nodes (n antinodes and n+1 nodes for the nth harmonic). (b) Wavelength = 2L/n = (2 x 0.80)/3 = 0.53 m. (c) Frequency = speed/wavelength = 120/0.533 = 225 Hz.

(a) 1 mark: correct nodes (4) and antinodes (3) both stated. (b) 1 mark: correct formula wavelength = 2L/n; 1 mark: correct substitution and answer (0.53 m). (c) 1 mark: correct formula f = v/wavelength; 1 mark: correct final answer 225 Hz (ecf from (b)). [5]

Using wavelength = speed/frequency: in air, wavelength_1 = 340/500 = 0.68 m. In the denser medium, wavelength_2 = 1500/500 = 3.0 m. The frequency remains constant at 500 Hz because it is determined by the source producing the wave, not by the medium through which it travels; only the wave's speed and wavelength change when it crosses into a new medium.

1 mark: correct formula wavelength=v/f used; 1 mark: correct wavelength_1 = 0.68 m; 1 mark: correct wavelength_2 = 3.0 m; 1 mark: correctly states frequency remains constant (determined by the source); 1 mark: correct explanation that only speed and wavelength change at a boundary between media. [5]

For a tube closed at one end, the fundamental mode has a quarter wavelength fitting the tube length: L = wavelength/4, so wavelength = 4L = 4 x 0.25 = 1.0 m. Frequency = speed/wavelength = 340/1.0 = 340 Hz.

1 mark: correct relationship for a closed-end tube fundamental (L = wavelength/4); 1 mark: correct wavelength = 1.0 m; 1 mark: correct formula f = v/wavelength; 1 mark: correct substitution; 1 mark: correct final answer 340 Hz. [5]

The path difference expressed in wavelengths is 1.0 / 0.40 = 2.5 wavelengths. Since this is a half-integer (odd multiple of half a wavelength), the two waves arrive out of phase and destructive interference occurs; constructive interference would instead require the path difference to be a whole number of wavelengths.

1 mark: correct calculation of path difference in terms of wavelength (1.0/0.40 = 2.5 wavelengths); 1 mark: correct identification that 2.5 wavelengths is a half-integer number of wavelengths; 1 mark: correct conclusion - destructive interference; 1 mark: correct general rule stated (constructive at whole-number wavelength path differences, destructive at half-integer/odd-half-wavelength path differences); 1 mark: coherent, logical justification linking calculation to conclusion. [5]

For a tube open at both ends, the nth harmonic has wavelength = 2L/n. For the second harmonic (n=2), wavelength = (2 x 0.60)/2 = 0.60 m. The second harmonic of an open-open tube has 2 nodes (interior points of zero displacement) and 3 antinodes (including one at each open end).

1 mark: correct formula wavelength = 2L/n for an open-open tube; 1 mark: correct substitution (n=2, L=0.60); 1 mark: correct wavelength 0.60 m; 1 mark: correct number of nodes (2); 1 mark: correct number of antinodes stated for support (3) or correct description of the node/antinode pattern. [5]

(a) In the fundamental mode, wavelength = 2L = 2 x 0.65 = 1.30 m. Speed = f x wavelength = 196 x 1.30 = 255 m/s. (b) The new wavelength = 2 x 0.487 = 0.974 m. New frequency = speed/wavelength = 255/0.974 = 262 Hz. (c) Shortening the vibrating length L reduces the wavelength of the fundamental mode, since wavelength = 2L. Because the wave speed on the string is unchanged, and f = v/wavelength, a smaller wavelength produces a larger frequency, which is heard as a higher pitch.

(a) 1 mark: correct wavelength=2L=1.30 m; 1 mark: correct v=f x wavelength=255 m/s (accept 254-256). (b) 1 mark: correct new wavelength=2x0.487=0.974 m; 1 mark: correct new f=v/wavelength approx 262 Hz (ecf). (c) 1 mark: correct explanation that shorter L gives smaller wavelength, and constant v with f=v/wavelength means smaller wavelength gives higher f, raising pitch. [6]

The corrective lens must form a virtual image of a very distant object (object at infinity) exactly at the patient's far point, 2.0 m in front of the eye, on the same side as the object. For an object at infinity, the image distance equals the focal length, so f = -2.0 m (negative, by sign convention, since the lens is diverging and the image is virtual). Power = 1/f = 1/(-2.0) = -0.50 D.

1 mark: correct principle stated - lens must create a virtual image at the far point (2.0 m) for an object at infinity; 1 mark: correct sign convention applied (diverging lens, negative focal length); 1 mark: correct focal length f = -2.0 m; 1 mark: correct formula P = 1/f used; 1 mark: correct final power -0.50 D. [5]

The critical angle satisfies sin(critical angle) = n_cladding / n_core = 1.45 / 1.50 = 0.967. Taking the inverse sine gives critical angle = sin⁻¹(0.967) ≈ 75.2 degrees.

1 mark: correct formula sin(critical angle) = n2/n1 (cladding/core); 1 mark: correct substitution (1.45/1.50 = 0.967); 1 mark: correct method to find inverse sine; 1 mark: correct final answer approx 75.2 degrees (accept 75.0-75.5); 1 mark: units/degrees correctly stated. [5]

Total attenuation = 0.20 x 15 = 3.0 dB. Using dB = 10 log10(P1/P2): 3.0 = 10 log10(2.0/P2), so log10(2.0/P2) = 0.30, giving 2.0/P2 = 10^0.30 = 2.0. Therefore P2 = 2.0/2.0 = 1.0 mW.

1 mark: correct total attenuation calculated (0.20 x 15 = 3.0 dB); 1 mark: correct dB formula used (dB = 10 log10(P1/P2)); 1 mark: correct rearrangement to find P1/P2 (=10^0.3 approx 2.0); 1 mark: correct substitution to find P2; 1 mark: correct final answer approx 1.0 mW. [5]

Rearranging the lens formula: \( \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{15} - \frac{1}{10} = \frac{2-3}{30} = -\frac{1}{30} \), so v = -30 cm; the negative sign shows the image is virtual, formed on the same side of the lens as the object. Magnification = v/u = 30/10 = 3, so the image is upright, virtual and magnified three times.

1 mark: correct lens formula rearranged (1/v = 1/f - 1/u); 1 mark: correct substitution (1/15 - 1/10); 1 mark: correct value of v = -30 cm with correct sign interpretation (virtual image); 1 mark: correct magnification formula (m = v/u) applied; 1 mark: correct final magnification = 3 (with correct description as upright/virtual/magnified). [5]

Sound intensity level (dB) = \( 10 \log_{10}(I / I_0) \). Here \( I/I_0 = (1.0\times10^{-4}) / (1.0\times10^{-12}) = 1.0\times10^{8} \). Since \( \log_{10}(10^8) = 8 \), the sound intensity level = 10 x 8 = 80 dB.

1 mark: correct formula stated (dB = 10 log10(I/I0)); 1 mark: correct ratio calculated (I/I0 = 10^8); 1 mark: correct log evaluated (log10(10^8) = 8); 1 mark: correct final answer 80 dB; 1 mark: unit (dB) correctly stated with working clearly shown. [5]

(a) Rearranging \( \text{dB} = 10\log_{10}(I/I_0) \): \( 110 = 10\log_{10}(I/I_0) \), so \( \log_{10}(I/I_0) = 11 \), giving \( I/I_0 = 10^{11} \), so \( I = 10^{11} \times 1.0\times10^{-12} = 0.10 \) W m⁻². (b) 110 - 85 = 25 dB above the guideline. (c) Prolonged exposure to loud sound causes excessive mechanical vibration that damages or destroys the delicate hair cells (stereocilia) in the cochlea's organ of Corti; because these hair cells do not regenerate once destroyed, this results in permanent, irreversible (sensorineural) hearing loss, particularly affecting high-frequency sensitivity.

(a) 1 mark: correct formula rearranged (I = I0 x 10^(dB/10)); 1 mark: correct final answer I = 0.10 W m⁻². (b) 1 mark: correct subtraction 110 - 85 = 25 dB. (c) 1 mark: correct identification of hair cell/cochlea damage; 1 mark: correct link to permanent hearing loss because hair cells do not regenerate. [6]

Information, such as sound or data, is first converted into a digital (binary) electrical signal, then encoded as pulses of light using an LED or laser at the transmitter. This light travels along the core of the optical fibre, remaining confined within the fibre by total internal reflection at the core-cladding boundary: because the core has a higher refractive index than the surrounding cladding, light striking the boundary at an angle of incidence greater than the critical angle is totally internally reflected back into the core rather than escaping. At the receiving end, a photodiode (detector) converts the light pulses back into an electrical digital signal, which is then decoded. Over long distances, two factors degrade the signal: attenuation, the progressive loss of signal power largely due to absorption and scattering of light within the fibre, and dispersion, the broadening and overlapping of pulses (for example because different light paths take slightly different times in a multimode fibre), both of which can cause errors in the received signal if left uncorrected. To counter this, the signal must be periodically regenerated using repeaters, which detect the degraded pulses and retransmit a clean new signal.

Level 3 (9-12 marks): a complete, logically sequenced account covering encoding of information as a digital signal, conversion to light pulses (LED/laser), total internal reflection correctly explained with reference to refractive index and critical angle, detection/decoding at the receiver, AND two correctly explained limiting factors (attenuation and dispersion); fluent, accurate use of specialist terms throughout (e.g. 'total internal reflection', 'critical angle', 'attenuation', 'dispersion', 'refractive index'). Level 2 (5-8 marks): most stages present with reasonable accuracy but with gaps (e.g. only one limiting factor explained, or TIR explanation incomplete); adequate specialist terminology. Level 1 (1-4 marks): fragmented or largely incomplete account, limited accurate content, little or inaccurate specialist terminology. 0 marks: no creditable response. [12]

In DNA, adenine always pairs with thymine (A-T) and cytosine always pairs with guanine (C-G), a rule known as complementary base pairing. Each pair of bases is held together by hydrogen bonds between the two strands of the double helix.

1 mark: correct base pairing (A-T and C-G); 1 mark: hydrogen bonds correctly named. [2]

The genetic code is described as degenerate because most of the 20 amino acids are specified by more than one three-base codon; for example, several different codons can all code for the same amino acid, giving the code a degree of built-in redundancy.

1 mark: reference to most amino acids having more than one codon; 1 mark: correct use of the term 'degenerate'/redundancy in context. [2]

Meiosis introduces variation in at least two ways: crossing over, in which homologous chromosomes exchange sections of DNA during prophase I, creating new combinations of alleles on a chromosome; and independent assortment, in which each pair of homologous chromosomes lines up and separates independently of every other pair during metaphase I, creating many possible combinations of maternal and paternal chromosomes in the resulting gametes.

1 mark each for any two: crossing over/recombination; independent assortment; (random fertilisation, if the question scope is extended). [2]

A human gamete contains 23 chromosomes and is described as haploid, while a somatic (body) cell contains 46 chromosomes and is described as diploid. This difference is significant because it means fertilisation, the fusion of a haploid sperm and a haploid egg, restores the full diploid number of 46 chromosomes in the resulting zygote, with one complete set inherited from each parent.

1 mark: correct chromosome numbers (23 gamete, 46 somatic); 1 mark: correct explanation that fertilisation restores the diploid number, combining genetic material from both parents. [2]

Totipotent stem cells, found in the very earliest embryo, have the potential to differentiate into any cell type of the body, and also into extra-embryonic tissue such as the placenta. Pluripotent stem cells, such as those of the later embryo, can differentiate into almost any cell type found in the body, but have lost the ability to form extra-embryonic tissue.

1 mark: correct definition of totipotent (any cell type including extra-embryonic tissue); 1 mark: correct definition of pluripotent (almost any body cell type, but not extra-embryonic tissue). [2]

Harvesting embryonic stem cells typically requires the destruction of the early embryo from which they are taken; some people consider the embryo to have a moral status or a right to life from the point of fertilisation, and object to its destruction for research or therapeutic purposes on ethical grounds.

1 mark: valid ethical concern identified (e.g. destruction of the embryo, moral status of the embryo, consent issues); 1 mark: brief explanation/justification of why this is a concern. [2]

Escherichia coli (E. coli) is the bacterium most commonly used as a host for producing recombinant human insulin, and the human insulin gene is typically inserted into the bacterium using a plasmid, a small circular piece of DNA, as the vector.

1 mark: E. coli correctly named; 1 mark: plasmid correctly named as the vector. [2]

Restriction enzymes cut DNA at specific base sequences known as recognition sites, often leaving short single-stranded overhangs called 'sticky ends'. Using the same restriction enzyme to cut both the human insulin gene and the plasmid produces complementary sticky ends on each, allowing the two pieces of DNA to base-pair together and subsequently be joined (ligated) into a single recombinant plasmid.

1 mark: correct function described (cuts DNA at a specific recognition sequence); 1 mark: correct explanation that using the same enzyme produces complementary/matching sticky ends for ligation. [2]

Somatic cell gene therapy targets and alters genes only in the body (non-reproductive) cells of a patient, so any genetic change is confined to that individual and is not passed on to their offspring. Germline gene therapy instead alters genes in reproductive cells or very early embryos, meaning any genetic change would be inherited by future generations.

1 mark: correct definition of somatic gene therapy (body cells, not inherited); 1 mark: correct definition of germline gene therapy (reproductive cells/embryo, inherited by offspring). [2]

A commonly used vector in gene therapy is a modified, disabled virus, such as an adenovirus, engineered to deliver the functional gene into a patient's cells without causing disease. A risk of using a viral vector is that it may trigger an unwanted immune response against the virus itself, or that it may insert the new gene at a random location in the patient's genome, potentially disrupting another important gene (insertional mutagenesis).

1 mark: valid viral vector named (e.g. adenovirus, retrovirus); 1 mark: valid risk correctly stated (immune response or insertional mutagenesis/disruption of another gene). [2]

The DNA double helix first unwinds and the two strands separate as the hydrogen bonds between complementary bases are broken, a process catalysed by the enzyme helicase. DNA polymerase then uses each original (template) strand to synthesise a new, complementary strand by adding free nucleotides according to the base-pairing rule. Because each resulting daughter DNA molecule consists of one original (parental) strand and one newly synthesised strand, the process is described as semi-conservative.

1 mark: strands separate/unwind (helicase, hydrogen bonds broken); 1 mark: DNA polymerase synthesises a new complementary strand from each template strand using base pairing; 1 mark: correct explanation that each daughter molecule has one original and one new strand (semi-conservative). [3]

A null hypothesis states that there is no real, significant difference between the observed results of an investigation and the results that were expected, and that any difference seen is due to chance alone. A suitable null hypothesis for a genetic cross investigation would be: 'there is no significant difference between the observed and expected ratio of offspring phenotypes in this genetic cross.' This hypothesis can then be tested statistically, for example using a chi-squared test.

1 mark: correct general definition of a null hypothesis (no significant difference/deviation is due to chance); 1 mark: valid genetics-specific example null hypothesis given; 1 mark: correct link to statistical testing (e.g. reference to being tested using a chi-squared test). [3]

Each Ff parent produces gametes F and f in equal proportion. Combining these gametes in a Punnett square gives offspring genotypes in the ratio 1 FF : 2 Ff : 1 ff. Therefore the probability of an affected child (ff) is 1/4, and the probability of an unaffected carrier (Ff) is 2/4 = 1/2.

1 mark: correct parental genotypes/gametes shown (F, f from each parent); 1 mark: correct genotypic ratio derived (1 FF : 2 Ff : 1 ff); 1 mark: correct probability affected (1/4); 1 mark: correct probability unaffected carrier (1/2). [4]

The cross Hh x hh produces offspring in the ratio 1 Hh : 1 hh, so half of the children are expected to inherit the H allele and develop Huntington's disease. Because H is dominant, any child inheriting even a single copy of H from the affected parent will develop the disease regardless of the other, unaffected (hh) parent's genotype; an unaffected partner cannot 'dilute' or prevent the inheritance of a dominant allele.

1 mark: correct cross set up (Hh x hh); 1 mark: correct ratio derived (1 Hh : 1 hh); 1 mark: correct final probability 1/2 (50%); 1 mark: correct explanation that dominance means one copy of H is sufficient to cause the disease irrespective of the second, unaffected parent's genotype. [4]

The cross X^H X^h x X^H Y gives daughters X^H X^H and X^H X^h (both unaffected, though one type is a carrier), and sons X^H Y (unaffected) and X^h Y (affected). Sons therefore have a 1/2 probability of being affected. Daughters always receive the father's X chromosome, which carries X^H, so no daughter can be affected (probability 0), although half of the daughters will be carriers.

1 mark: correct genetic diagram/cross with sex-linked notation (X^H X^h x X^H Y); 1 mark: correct son genotypes/ratio derived (1 X^H Y : 1 X^h Y); 1 mark: correct probability sons affected (1/2); 1 mark: correct probability daughters affected (0), with correct reasoning that daughters always receive their father's X chromosome (X^H). [4]

The heterozygous parent RrYy produces four gamete types in equal proportion: RY, Ry, rY and ry. The homozygous recessive parent rryy produces only ry gametes. Combining these gives four offspring genotypes, RrYy (round yellow), Rryy (round green), rrYy (wrinkled yellow) and rryy (wrinkled green), each in equal proportion, giving a phenotypic ratio of 1:1:1:1. Because one parent is homozygous recessive for both genes, this is described as a (dihybrid) test cross, used to determine the genotype of the other parent.

1 mark: correct gametes derived from the heterozygous parent (RY, Ry, rY, ry); 1 mark: correct single gamete type from the homozygous recessive parent (ry); 1 mark: correct genetic diagram/Punnett square showing all four offspring genotypes; 1 mark: correct 1:1:1:1 phenotypic ratio stated; 1 mark: correctly named as a test cross. [5]

A standard AaBb x AaBb dihybrid cross gives offspring in the genotypic ratio 9 A_B_ : 3 A_bb : 3 aaB_ : 1 aabb. Applying the phenotype rules given: A_B_ (9 parts) is purple; aaB_ (3 parts) is red; both A_bb and aabb (3+1 = 4 parts) are white, since bb masks the effect of gene A. This gives a final phenotypic ratio of 9 purple : 3 red : 4 white, a modified ratio caused by epistasis, the masking of one gene's phenotypic expression by another gene at a different locus.

1 mark: correct standard dihybrid ratio derived (9:3:3:1 for A_B_:A_bb:aaB_:aabb); 1 mark: correct phenotype assigned to each genotype class per the question (purple/red/white); 1 mark: correct combination of the two white classes (3+1=4); 1 mark: correct final ratio 9:3:4; 1 mark: correctly named as epistasis. [5]

(a) Expected round = 3/4 x 160 = 120; expected wrinkled = 1/4 x 160 = 40. (b) For round: (O-E) = 132-120 = 12, so \( (O-E)^2/E = 144/120 = 1.2 \). For wrinkled: (O-E) = 28-40 = -12, so \( (O-E)^2/E = 144/40 = 3.6 \). Summing: \( \chi^2 = 1.2 + 3.6 = 4.8 \).

(a) 1 mark: correct expected values (120 round, 40 wrinkled). (b) 1 mark: correct (O-E) values calculated for both categories (+12 and -12); 1 mark: correct (O-E)^2/E values calculated (1.2 and 3.6); 1 mark: correct summation giving chi-squared = 4.8; 1 mark: working shown clearly using the given formula. [5]

Since the calculated chi-squared value (4.8) is greater than the critical value (3.84) at the 5% significance level, the null hypothesis is rejected. This means there is a statistically significant difference between the observed and expected results, i.e. it is unlikely that this difference arose by chance alone; this suggests the observed data does not fit the expected 3:1 ratio, and that some other factor may be influencing the results.

1 mark: correct comparison made (4.8 > 3.84); 1 mark: correct conclusion that the null hypothesis is rejected; 1 mark: correct statement that there is a significant difference between observed and expected results (unlikely to be due to chance alone); 1 mark: correctly frames the significance level used (5%/p=0.05) in the conclusion; 1 mark: valid comment on the implication (e.g. the assumed genetic model/ratio may not be correct, or another factor is influencing the results). [5]

Degrees of freedom is calculated as the number of categories minus 1. For a dihybrid cross with four phenotypic categories, degrees of freedom = 4 - 1 = 3. From the table given, the critical value at the 5% (p=0.05) significance level for 3 degrees of freedom is 7.81.

1 mark: correct formula stated (degrees of freedom = number of categories - 1); 1 mark: correct calculation (degrees of freedom = 3); 1 mark: correct critical value identified from the table (7.81); 1 mark: correct significance level referenced (5%/p=0.05); 1 mark: clear working/reasoning shown throughout. [5]

Expected values: 180, 60, 60, 20. For each category, \( (O-E)^2/E \): (190-180)^2/180 = 100/180 = 0.56; (55-60)^2/60 = 25/60 = 0.42; (58-60)^2/60 = 4/60 = 0.07; (17-20)^2/20 = 9/20 = 0.45. Summing these gives \( \chi^2 = 0.56+0.42+0.07+0.45 \approx 1.49 \).

1 mark: correct expected values used (180, 60, 60, 20); 1 mark: correct (O-E)^2/E calculated for at least two categories; 1 mark: correct (O-E)^2/E calculated for all four categories (0.56, 0.42, 0.07, 0.45); 1 mark: correct summation method shown; 1 mark: correct final chi-squared approx 1.49 (accept 1.4-1.5). [5]

(a) The data shows a steady, broadly consistent increase in the global production of recombinant insulin over the 40-year period, rising from 4 tonnes in 1985 to 62 tonnes in 2025. (b) Two valid reasons: recombinant insulin is structurally identical to human insulin, so it causes fewer immune or allergic reactions than insulin extracted from animals; and it can be mass-produced cheaply and in essentially unlimited supply by culturing genetically engineered bacteria, without depending on a limited animal-derived supply.

(a) 1 mark: correct description of trend (steady/consistent increase over time). (b) 1 mark each for any two valid reasons: identical structure to human insulin, fewer immune/allergic reactions; unlimited/scalable, cost-effective bacterial production not dependent on animal supply; purer product with less contamination risk; ethical concerns about animal-sourced insulin reduced. [4]

(a) Percentage increase = (change/original) x 100 = ((1610-850)/850) x 100 = (760/850) x 100 = 89.4%. (b) A valid reason includes improved techniques for isolating and culturing stem cells, wider clinical acceptance and expanded donor registries, or an increasing range of approved clinical applications, all of which would increase the number of eligible patients and successful procedures over time.

(a) 1 mark: correct method (change/original x100); 1 mark: correct final answer approx 89.4% (accept 89-90%). (b) 1 mark: valid, relevant reason suggested; 1 mark: reason clearly linked/explained in relation to the increasing trend. [4]

(a) The gene therapy group maintained a consistently high survival rate, staying above 90% throughout the study, whereas the control group's survival rate declined steadily and substantially, from 100% to 55% by year 5, indicating gene therapy was associated with markedly better survival outcomes. (b) The 5-year follow-up period may be too short to detect rare or delayed long-term side effects, such as insertional mutagenesis leading to cancer years later, and/or the sample size or trial conditions may limit how confidently the results generalise to the wider patient population.

(a) 1 mark: correct comparative description of both trends (gene therapy stable/high, control declining); 1 mark: valid quantitative/qualitative detail cited from the data (e.g. 90% vs 55%). (b) 1 mark: valid limitation identified (e.g. short follow-up period, small sample size, risk of undetected long-term/rare effects); 1 mark: clear explanation of why this limits confidence in long-term safety conclusions. [4]

(a) The difference is 78 - 52 = 26 percentage points. (b) Insulin analogues can be engineered to have modified absorption rates, for example a faster onset or a longer, steadier action, allowing them to more closely mimic the body's natural pattern of insulin release than standard human insulin, which can improve overall blood glucose control.

(a) 1 mark: correct subtraction (78-52=26 percentage points). (b) 1 mark: valid reason suggested (e.g. modified absorption/action profile better mimicking natural insulin release); 1 mark: reason clearly explained/linked to improved glucose control. [3]

A functional copy of the gene is inserted into a suitable vector, commonly a disabled virus such as an adenovirus, which delivers it into the patient's target cells (somatic gene therapy); once inside, the gene is expressed to produce the missing or non-functional protein, restoring normal cell function. Several issues must be considered before such treatment is used widely. Technically, it must be ensured that the gene is delivered specifically and efficiently to the correct cells or tissue, that it integrates safely without disrupting other genes (insertional mutagenesis) or triggering a harmful immune response against the vector, and that expression is sustained long enough to be clinically useful. Ethically, a clear distinction must be drawn between somatic gene therapy, which is not inherited, and germline gene therapy, which would be passed on to future generations and raises far greater ethical concern. Socially, the high cost of gene therapies may limit access and create inequality between patients who can and cannot afford treatment, and informed consent is essential given the uncertain long-term risks involved. Finally, extensive, long-term clinical trials are needed before widespread use, given historical cases of adverse immune reactions and vector-related cancers seen in early gene therapy trials.

Level 3 (7-9 marks): a clear, logically sequenced description of the gene therapy mechanism (vector, delivery, gene expression) AND a well-developed discussion covering technical, ethical and social/safety issues, using specialist terminology accurately and fluently throughout (e.g. 'vector', 'somatic', 'germline', 'insertional mutagenesis', 'informed consent'). Level 2 (4-6 marks): a reasonable description of the mechanism with some relevant issues discussed, but coverage is narrower (e.g. only one or two categories of issue) or contains minor inaccuracies; adequate use of specialist terms. Level 1 (1-3 marks): a fragmented or largely inaccurate account, with few relevant issues identified and limited use of specialist terminology. 0 marks: no creditable response. [9]