2022 CCEA A-Level Life and Health Sciences 0008 Practice Paper with Answers

(a) The longest continuous chain contains five carbon atoms (pentane), with a methyl branch on carbon 2 (numbering from the end that gives the lowest locant). The IUPAC name is therefore 2-methylpentane.
(b) Counting atoms in \( \text{CH}_3\text{CH}(\text{CH}_3)\text{CH}_2\text{CH}_2\text{CH}_3 \): 6 carbon atoms and 14 hydrogen atoms, so the molecular formula is \( \text{C}_6\text{H}_{14} \). This is an alkane, so it belongs to the homologous series with general formula \( \text{C}_n\text{H}_{2n+2} \). Final answer: 2-methylpentane, \( \text{C}_6\text{H}_{14} \), \( \text{C}_n\text{H}_{2n+2} \).

(a) 1 mark: parent chain correctly identified as pentane; 1 mark: correct locant and name '2-methylpentane' (reject 4-methylpentane). (b) 1 mark: \( \text{C}_6\text{H}_{14} \); 1 mark: \( \text{C}_n\text{H}_{2n+2} \); [4]

(a) Balancing carbon: 2 propane molecules give 6 carbon atoms, matching 6 CO. Balancing hydrogen: \( 2 \times 8 = 16 \) H atoms requires 8 \( \text{H}_2\text{O} \). Balancing oxygen: right-hand side has \( 6(1) + 8(1) = 14 \) O atoms, requiring 7 \( \text{O}_2 \) on the left. The balanced equation is \( 2\text{C}_3\text{H}_8 + 7\text{O}_2 \rightarrow 6\text{CO} + 8\text{H}_2\text{O} \).
(b) Incomplete combustion in limited air also produces unburned hydrocarbons and carbon particulates (soot), since there is insufficient oxygen to fully oxidise every carbon atom.

(a) 1 mark: correct formulae and states/products (CO, H2O); 1 mark: correctly balanced with coefficients 2, 7, 6, 8. (b) 1 mark: any one of carbon particulates/soot or unburned hydrocarbons; [3]

(a) A primary alcohol has the \( -\text{OH} \) group on a carbon bonded to only one other carbon, e.g. butan-1-ol, \( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} \). A tertiary alcohol has the \( -\text{OH} \) group on a carbon bonded to three other carbons, e.g. 2-methylpropan-2-ol, \( (\text{CH}_3)_3\text{COH} \). Both have molecular formula \( \text{C}_4\text{H}_{10}\text{O} \).
(b) Structural isomers are compounds that share the same molecular formula but differ in the structural arrangement (connectivity) of their atoms.

(a) 1 mark: correct primary alcohol structure with formula \( \text{C}_4\text{H}_{10}\text{O} \); 1 mark: correct tertiary alcohol structure with formula \( \text{C}_4\text{H}_{10}\text{O} \). (b) 1 mark: same molecular formula; 1 mark: different structural/connectivity arrangement of atoms; [4]

(a) But-1-ene reacts with HBr by electrophilic addition, since the electron-rich C=C double bond attacks the electrophile \( \text{H}^{\delta+} \) of the polar H–Br bond.
Step 1: A curly arrow moves from the C=C pi bond to the H atom of HBr, breaking the H–Br bond heterolytically (curly arrow from the bond to Br). Hydrogen bromide is polarised as \( \text{H}^{\delta+}-\text{Br}^{\delta-} \), so H adds to C1 (the terminal carbon), generating the more stable secondary carbocation on C2: \( \text{CH}_3\text{CH}_2\overset{+}{\text{C}}\text{HCH}_3 \), and releasing \( \text{Br}^{-} \).
Step 2: A curly arrow moves from a lone pair on \( \text{Br}^{-} \) to the positively charged carbon, forming a new C–Br bond and giving the major product, 2-bromobutane, \( \text{CH}_3\text{CH}_2\text{CHBrCH}_3 \).

(a) 1 mark: electrophilic addition. (b) 1 mark: curly arrow from C=C to H of HBr with heterolytic fission of H–Br shown; 1 mark: correct secondary carbocation intermediate drawn/named on C2 (Markovnikov addition); 1 mark: curly arrow from Br- lone pair to carbocation forming 2-bromobutane; [4]

Classification depends on the number of carbon atoms directly bonded to the carbon bearing the \( -\text{OH} \) group.
(a) In \( \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \) (propan-1-ol), the \( -\text{OH} \)-bearing carbon is attached to only one other carbon, so it is a primary alcohol.
(b) In \( \text{CH}_3\text{CH}(\text{OH})\text{CH}_3 \) (propan-2-ol), the \( -\text{OH} \)-bearing carbon is attached to two other carbons, so it is a secondary alcohol.
(c) In \( (\text{CH}_3)_3\text{COH} \) (2-methylpropan-2-ol), the \( -\text{OH} \)-bearing carbon is attached to three other carbons, so it is a tertiary alcohol.

1 mark each for (a) primary, (b) secondary, (c) tertiary, correctly identified; [3]

(a) The test tube containing hex-1-ene changes from orange to colourless, as the bromine is used up in an addition reaction. The test tube containing hexane remains orange, as no reaction occurs.
(b) Hex-1-ene contains a C=C double bond, which is an area of high electron density (from the pi bond). This induces a temporary dipole in the approaching \( \text{Br}_2 \) molecule, so the alkene acts as an electrophile-attracting species and reacts by electrophilic addition to form 1,2-dibromohexane, decolourising the bromine water. Hexane is a saturated alkane containing only strong, non-polar C–C and C–H sigma bonds, which are unreactive towards bromine water under these conditions, so no colour change is observed.

(a) 1 mark: hex-1-ene decolourises bromine water; 1 mark: hexane causes no colour change (remains orange). (b) 1 mark: C=C pi bond is a region of high electron density enabling electrophilic addition with bromine; 1 mark: hexane has only unreactive sigma C-C/C-H bonds, no double bond present; [4]

(a) The dichromate(VI) ion, \( \text{Cr}_2\text{O}_7^{2-} \), is reduced from orange \( \text{Cr(VI)} \) to green \( \text{Cr}^{3+} \) as it oxidises the alcohol, so the colour change observed is orange to green.
(b) Propan-1-ol, a primary alcohol, is oxidised (with only mild/controlled oxidation, e.g. using distillation to remove the aldehyde before further oxidation) to the aldehyde propanal: \( \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} + [\text{O}] \rightarrow \text{CH}_3\text{CH}_2\text{CHO} + \text{H}_2\text{O} \). This is balanced: 3 C, 8 H (as 6+2 across OH and [O] balance with product H2O) and 1 O beyond the starting OH are all accounted for on both sides.
(c) Fehling's solution starts as a blue \( \text{Cu}^{2+} \) complex and forms a brick-red precipitate of \( \text{Cu}_2\text{O} \) with an aldehyde (which is easily oxidised further), but gives no colour change with a carboxylic acid.

(a) 1 mark: orange to green. (b) 1 mark: correct formulae for propan-1-ol and propanal; 1 mark: correctly balanced equation including [O] and H2O. (c) 1 mark: Fehling's solution or Tollens' reagent (or Benedict's solution), named correctly; [4]

(a) Addition polymerisation opens the C=C double bond of the monomer to form single C–C bonds linking successive units. The repeat unit is written as \( -[\text{CH}_2-\text{CHCl}]_n- \), with the two open bonds on the left and right of the square bracket representing the continuation of the chain to neighbouring repeat units, and the Cl substituent correctly retained on the second carbon.
(b) Addition polymers such as PVC have strong, saturated, non-polar C–C backbones that are chemically inert and resistant to microbial breakdown, so they do not biodegrade and persist as waste in landfill for a very long time.

(a) 1 mark: correct skeletal repeat unit \( -\text{CH}_2-\text{CHCl}- \) with substituents in the right positions; 1 mark: bonds shown extending outward from the bracket with subscript n. (b) 1 mark: chemically inert/non-biodegradable (or equivalent, e.g. resistant to microbial attack); [3]

(a) An IR spectrum shows absorptions at characteristic wavenumbers corresponding to the vibration of particular bonds, allowing the functional groups present in a molecule (e.g. O–H, C=O, C–H) to be identified.
(b) A mass spectrum shows the mass-to-charge ratio, \( m/z \), of the molecular ion, \( \text{M}^{+} \), giving the relative molecular mass of the compound, and the pattern of fragment ion peaks can reveal parts of the molecular structure.
(c) IR spectroscopy alone can identify which functional groups are present but cannot determine the overall molecular mass or full carbon skeleton, so several isomers with the same functional groups cannot be distinguished. Mass spectrometry alone gives the molecular mass and some structural clues from fragmentation, but cannot unambiguously identify functional groups. Using both together, a chemist can match the molecular mass and fragmentation pattern from MS with the functional groups confirmed by IR, greatly narrowing down the number of possible structures and giving much stronger evidence for the correct identity.

(a) 1 mark: identifies functional groups present, from characteristic absorption wavenumbers. (b) 1 mark: gives relative molecular mass from M+ (accept fragmentation gives structural clues). (c) 1 mark: IR alone cannot give mass/full structure, MS alone cannot confirm functional groups; 1 mark: combining both narrows down possible structures/gives stronger confirmatory evidence; [4]

(a) The C=C double bond is made up of one sigma bond (from head-on orbital overlap along the internuclear axis) and one pi bond (from sideways overlap of adjacent p-orbitals, above and below the plane of the sigma bond). Rotating one carbon relative to the other about the C=C axis would require breaking this pi-bond overlap, which needs a significant amount of energy. This creates an energy barrier to rotation, so the double bond is effectively 'locked' in one geometry at room temperature, allowing distinct cis and trans isomers to exist.
(b) For cis-trans (E-Z) isomerism to occur, each of the two carbon atoms of the C=C double bond must be bonded to two different substituent groups (i.e. no carbon of the double bond may carry two identical groups); in but-2-ene each double-bond carbon carries a methyl group and a hydrogen atom, which are different, satisfying this condition.

(a) 1 mark: C=C consists of a sigma and a pi bond, pi bond from sideways p-orbital overlap; 1 mark: rotation would break pi-overlap, creating an energy barrier to rotation. (b) 1 mark: each carbon of the C=C must bear two different substituents; 1 mark: correctly applied/exemplified for but-2-ene (methyl and H on each carbon, which differ); [4]

(a) If unreacted salicylic acid remains in the sample, the mixture develops a violet/purple colouration when treated with iron(III) chloride solution.
(b) Iron(III) chloride solution gives a characteristic violet/purple colouration specifically with phenols, because the free phenolic –OH group forms a coloured complex with \( \text{Fe}^{3+} \) ions. Salicylic acid retains this free phenolic –OH group, so it gives a positive (coloured) test. In the synthesis of aspirin, this phenolic –OH group is converted (esterified with ethanoic anhydride) into an ester group, \( -\text{OCOCH}_3 \); pure aspirin therefore has no free phenolic –OH group and gives no colour change with iron(III) chloride, so any colouration indicates the sample is impure (contains unreacted salicylic acid).

(a) 1 mark: violet/purple colour forms. (b) 1 mark: salicylic acid has a free phenolic -OH group which complexes with Fe3+; 1 mark: this -OH is esterified in aspirin, so pure aspirin gives no colour change; [3]

(a) A small molecule of water, \( \text{H}_2\text{O} \), is eliminated at each new amide (peptide-type) linkage formed between the diacid and diamine monomers.
(b) Addition polymerisation involves monomers containing a C=C double bond; the pi bond opens and monomer units simply join together to form the polymer chain, with no atoms lost — the polymer has the same empirical composition as the sum of its monomers. Condensation polymerisation, by contrast, involves monomers each carrying two reactive functional groups (for example a dicarboxylic acid, with two \( -\text{COOH} \) groups, reacting with a diamine, with two \( -\text{NH}_2 \) groups); each time two functional groups react to form a new covalent (amide) linkage, a small molecule — here water — is eliminated as a by-product, so the polymer chain contains fewer atoms than the sum of the original monomers.

(a) 1 mark: water. (b) 1 mark: addition polymers form from unsaturated (C=C) monomers with no by-product lost; 1 mark: condensation polymers form from monomers with two reactive functional groups (e.g. diacid + diamine); 1 mark: a small molecule (water) is lost at each new linkage formed; [4]

(a) Assume 100 g of the hydrocarbon: mass of C = 83.7 g, mass of H = 16.3 g.
Moles of C = \( \dfrac{83.7}{12} = 6.98 \) mol. Moles of H = \( \dfrac{16.3}{1} = 16.3 \) mol.
Divide by the smaller value (6.98): C: \( 6.98/6.98 = 1.00 \); H: \( 16.3/6.98 = 2.34 \approx 7/3 \). Multiplying both by 3 to clear the fraction gives the simplest whole-number ratio C : H = 3 : 7, so the empirical formula is \( \text{C}_3\text{H}_7 \).
(b) Empirical formula mass = \( 3(12) + 7(1) = 43 \). Since the molecular mass is 86, \( n = 86/43 = 2 \). Molecular formula = \( (\text{C}_3\text{H}_7)_2 = \text{C}_6\text{H}_{14} \).
Check (second route): for hexane, \( \text{C}_6\text{H}_{14} \), \( M = 6(12)+14(1) = 86 \) ✓, and %C \( = 72/86 \times 100 = 83.7\% \), %H \( = 14/86 \times 100 = 16.3\% \) ✓, confirming the answer is self-consistent. Final answer: empirical formula \( \text{C}_3\text{H}_7 \), molecular formula \( \text{C}_6\text{H}_{14} \).

(a) 1 mark: mol C = 6.98; 1 mark: mol H = 16.3; 1 mark: ratio simplified correctly to 1 : 2.33; 1 mark: whole-number ratio C3H7 stated. (b) 1 mark: empirical formula mass = 43 correctly calculated; 1 mark: n = 2 and molecular formula C6H14 correctly stated; [6]

(a) Checking atom balance: left side has 10 C and 22 H. Right side has \( \text{C}_8\text{H}_{18} \) (8 C, 18 H) plus \( \text{C}_2\text{H}_4 \) (2 C, 4 H), giving 10 C and 22 H — already balanced with all coefficients equal to 1: \( \text{C}_{10}\text{H}_{22} \rightarrow \text{C}_8\text{H}_{18} + \text{C}_2\text{H}_4 \).
(b) \( M(\text{C}_{10}\text{H}_{22}) = 10(12) + 22(1) = 142 \) g mol\(^{-1}\). Moles of decane \( = \dfrac{14.2}{142} = 0.100 \) mol. From the 1:1 stoichiometry, moles of ethene formed = 0.100 mol. \( M(\text{C}_2\text{H}_4) = 2(12)+4(1) = 28 \) g mol\(^{-1}\). Mass of ethene \( = 0.100 \times 28 = 2.80 \) g.
Check (second route, mass balance): mass of octane formed = \( 0.100 \times (8(12)+18(1)) = 0.100 \times 114 = 11.4 \) g; total product mass \( = 11.4 + 2.80 = 14.2 \) g, which equals the mass of decane reacted, confirming conservation of mass. Final answer: 2.80 g.

(a) 1 mark: correct formulae; 1 mark: correctly balanced (1:1:1). (b) 1 mark: M(decane) = 142 correctly calculated; 1 mark: mol decane = 0.100 correctly calculated; 1 mark: 1:1 mole ratio applied to give mol ethene = 0.100; 1 mark: mass of ethene = 2.80 g (using M = 28); [6]

(a) \( \text{C}_3\text{H}_6 + \text{H}_2\text{O} \rightarrow \text{C}_3\text{H}_7\text{OH} \). Checking: left side 3 C, 8 H (6+2), 1 O; right side \( \text{C}_3\text{H}_7\text{OH} = \text{C}_3\text{H}_8\text{O} \), 3 C, 8 H, 1 O — balanced with all coefficients 1.
(b) \( M(\text{C}_3\text{H}_6) = 3(12)+6(1) = 42 \) g mol\(^{-1}\). Moles of propene \( = \dfrac{4200}{42} = 100 \) mol. From 1:1 stoichiometry, moles of propan-2-ol formed = 100 mol. \( M(\text{C}_3\text{H}_7\text{OH}) = 3(12)+8(1)+16 = 60 \) g mol\(^{-1}\). Mass \( = 100 \times 60 = 6000 \) g \( = 6.00 \) kg.
Check (second route, via mass ratio): the mass ratio product/reactant should equal \( M(\text{product})/M(\text{reactant}) = 60/42 = 1.4286 \); \( 4.20 \text{ kg} \times 1.4286 = 6.00 \) kg ✓, confirming the answer. Final answer: 6.00 kg.

(a) 1 mark: correct formulae for propene and propan-2-ol; 1 mark: correctly balanced 1:1:1 equation. (b) 1 mark: M(propene) = 42 correctly calculated; 1 mark: mol propene = 100 correctly calculated; 1 mark: 1:1 ratio applied, mol propan-2-ol = 100; 1 mark: mass = 6.00 kg (using M = 60); [6]

\( M(\text{C}_6\text{H}_{12}\text{O}_6) = 6(12)+12(1)+6(16) = 180 \) g mol\(^{-1}\). Moles of glucose \( = \dfrac{900\,000}{180} = 5000 \) mol. From the equation, 1 mol glucose produces 2 mol ethanol, so moles of ethanol \( = 2 \times 5000 = 10\,000 \) mol. \( M(\text{C}_2\text{H}_5\text{OH}) = 2(12)+6(1)+16 = 46 \) g mol\(^{-1}\). Mass of ethanol \( = 10\,000 \times 46 = 460\,000 \) g \( = 460 \) kg.
Check (second route, mass ratio): mass ratio \( = \dfrac{2 \times 46}{180} = \dfrac{92}{180} = 0.5111 \); \( 900 \times 0.5111 = 460.0 \) kg ✓. Final answer: 460 kg.

1 mark: mol glucose = 5000 correctly calculated (using M = 180); 1 mark: mole ratio 1:2 correctly applied to give mol ethanol = 10 000; 1 mark: M(ethanol) = 46 correctly calculated; 1 mark: mass of ethanol in g = 460 000 g; 1 mark: correctly converted to 460 kg; [5]

(a) The monomer, propene, is \( \text{CH}_2=\text{CHCH}_3 \). Addition polymerisation opens the C=C double bond, joining n monomer units into a chain: \( n\,\text{CH}_2=\text{CHCH}_3 \rightarrow -[\text{CH}_2-\text{CH}(\text{CH}_3)]_n- \), with the repeat unit shown in square brackets, the methyl side-group retained on alternate backbone carbons, and open bonds shown extending from the bracket to link to neighbouring units.
(b) The repeat unit \( -[\text{CH}_2-\text{CH}(\text{CH}_3)]- \) has the same molecular formula as the monomer, \( \text{C}_3\text{H}_6 \), so \( M(\text{repeat unit}) = 3(12)+6(1) = 42 \) g mol\(^{-1}\). \( n = \dfrac{2.10 \times 10^5}{42} = 5000 \).
Check (second route): \( 5000 \times 42 = 210\,000 = 2.10 \times 10^5 \) ✓. Final answer: n = 5000.

(a) 1 mark: correct displayed formula of propene monomer with 'n' coefficient; 1 mark: correct repeat unit structure -[CH2-CH(CH3)]-; 1 mark: bonds shown extending from bracket with subscript n, correct equation format. (b) 1 mark: M(repeat unit) = 42 correctly calculated; 1 mark: correct division set up; 1 mark: n = 5000; [6]

(a) For \( \text{C}_n\text{H}_{2n+1}\text{OH} \), the relative molecular mass is \( M = 12n + (2n+1) + 17 = 14n + 18 \). Setting this equal to 74: \( 14n + 18 = 74 \Rightarrow 14n = 56 \Rightarrow n = 4 \). The molecular formula is therefore \( \text{C}_4\text{H}_{10}\text{O} \) (butan-1-ol, given as straight-chain).
Check (second route): for \( \text{C}_4\text{H}_{10}\text{O} \), \( M = 4(12) + 10(1) + 16 = 48+10+16 = 74 \) ✓, matching the given m/z.
(b) The broad absorption at 3230–3550 \( \text{cm}^{-1} \) is characteristic of an O–H stretch in a hydrogen-bonded alcohol, confirming the alcohol functional group is present. The absence of any absorption near 1715 \( \text{cm}^{-1} \) (the region characteristic of a C=O stretch in aldehydes, ketones and carboxylic acids) confirms that no carbonyl group is present, i.e. the sample is the unoxidised alcohol rather than an oxidation product such as an aldehyde or carboxylic acid.

(a) 1 mark: correct expression 14n+18=74 set up; 1 mark: n = 4 correctly solved; 1 mark: molecular formula C4H10O correctly stated. (b) 1 mark: 3230-3550 cm-1 identified as O-H stretch, confirms alcohol present; 1 mark: no absorption near 1715 cm-1 confirms absence of C=O; 1 mark: correct conclusion that the compound has not been oxidised to an aldehyde/carboxylic acid; [6]

Moles of salicylic acid \( = \dfrac{2.00}{138} = 0.01449 \) mol. Since salicylic acid reacts in a 1:1 mole ratio to form aspirin, the theoretical (maximum) moles of aspirin = 0.01449 mol. Theoretical mass of aspirin \( = 0.01449 \times 180 = 2.609 \) g.
Percentage yield \( = \dfrac{\text{actual mass}}{\text{theoretical mass}} \times 100 = \dfrac{1.95}{2.609} \times 100 = 74.7\% \).
Check (second route): actual moles of aspirin obtained \( = 1.95/180 = 0.01083 \) mol; percentage yield by moles \( = 0.01083/0.01449 \times 100 = 74.7\% \) ✓, consistent with the mass-based calculation. Final answer: 74.7%.

1 mark: mol salicylic acid = 0.0145 correctly calculated; 1 mark: 1:1 mole ratio correctly applied; 1 mark: theoretical mass of aspirin = 2.61 g correctly calculated; 1 mark: percentage yield formula correctly set up (actual/theoretical x 100); 1 mark: correct substitution 1.95/2.61 x 100; 1 mark: final answer 74.7% (accept 74.6-74.8%); [6]

Moles of hexanedioic acid \( = \dfrac{14\,600}{146} = 100 \) mol. Since the diamine is in excess, all 100 mol of diacid is incorporated into repeat units, requiring 100 mol of diamine and releasing \( 2 \times 100 = 200 \) mol of water.
Mass of repeat unit formed per mole = mass of diacid + mass of diamine − mass of 2 water molecules \( = 146 + 116 - 2(18) = 262 - 36 = 226 \) g mol\(^{-1}\) (per repeat unit).
Mass of nylon-6,6 \( = 100 \times 226 = 22\,600 \) g \( = 22.6 \) kg.
Check (second route, mass balance): total mass of monomers reacted \( = 14.6 + (100 \times 116/1000) = 14.6 + 11.6 = 26.2 \) kg; mass of water lost \( = 200 \times 18 = 3600 \) g \( = 3.6 \) kg; mass of polymer \( = 26.2 - 3.6 = 22.6 \) kg ✓, confirming the answer by conservation of mass. Final answer: 22.6 kg (a theoretical maximum, assuming complete reaction).

1 mark: mol hexanedioic acid = 100 correctly calculated; 1 mark: recognises 2 mol water lost per mol repeat unit formed; 1 mark: mass of one repeat unit = 146+116-36 = 226 g mol-1 correctly calculated; 1 mark: mass of nylon = 100 x 226 = 22 600 g; 1 mark: correctly converted to 22.6 kg; 1 mark: correct conclusion identifying this as a theoretical/maximum yield; [6]

A full-mark answer should discuss, in a well-organised and coherent way:

Combustion pollutants: In a limited supply of air, incomplete combustion of alkanes produces toxic carbon monoxide (CO), unburned hydrocarbons, and carbon particulates (soot), alongside water and some CO2. Combustion in plentiful air more fully oxidises carbon to CO2, but the high temperatures involved also cause atmospheric nitrogen and oxygen to combine, forming oxides of nitrogen (NOx); fuels containing sulfur impurities also produce sulfur dioxide (SO2). These pollutants have serious environmental effects: CO is toxic, binding preferentially to haemoglobin and reducing oxygen transport in the blood; particulates contribute to smog and respiratory disease; NOx and SO2 dissolve in atmospheric moisture to form acid rain, which damages aquatic ecosystems, forests and stone buildings; unburned hydrocarbons contribute to photochemical smog.

Catalytic converters: A catalytic converter contains a solid (heterogeneous) catalyst, typically platinum, palladium and rhodium coated onto a honeycomb support, which provides a large surface area. Exhaust gas pollutants are adsorbed onto the catalyst surface (chemisorption), which weakens their bonds and lowers the activation energy for reaction. This allows CO to be oxidised to CO2, and NOx to be reduced to harmless N2 gas, before the exhaust gases leave the vehicle, significantly reducing the toxicity of vehicle emissions (though CO2, a greenhouse gas, is still released).

Polymer waste strategies: Addition polymers such as polythene and PVC are chemically inert (saturated, non-polar backbones) and so do not readily biodegrade, causing them to persist in landfill for a very long time. Strategies to manage this include: (1) Incineration, which recovers useful energy from the polymer waste and reduces landfill volume, but can release toxic gases (e.g. HCl from PVC, or other combustion products) unless the incinerator is fitted with gas-scrubbing equipment to remove them, adding cost; (2) Recycling, which conserves crude oil resources and reduces landfill, but requires plastics to be correctly sorted by type (since different polymers cannot simply be melted together), which is labour-intensive and reduces the quality/uses of the recycled material over repeated cycles; (3) Using waste polymer as a chemical feedstock for cracking, breaking it back down into smaller, useful hydrocarbon molecules, though this requires significant energy input; (4) Ongoing chemical research into biodegradable polymers, which can be broken down by microorganisms, reducing the long-term landfill burden, though these can currently be more expensive to produce and may not have identical properties to conventional polymers.

A good answer weighs these strategies against one another (for example, noting that no single strategy alone solves the disposal problem, and that a combination — reduce, reuse, recycle, recover energy — is generally advocated) using accurate scientific terminology and correct spelling, punctuation and grammar throughout.

Excellent (7-9 marks): Detailed, accurate and well-structured discussion covering combustion pollutants with correct chemistry and environmental effects, the mechanism of the catalytic converter (chemisorption on a solid metal catalyst converting CO to CO2 and NOx to N2), and at least two polymer disposal strategies each evaluated with a clear advantage and a clear limitation; addresses all three bulleted prompts; high standard of spelling, punctuation and grammar (SPaG) with accurate scientific terminology throughout.
Good / Very Good (4-6 marks): Reasonable coverage of most of the three areas, but with some omissions, imbalance, or limited evaluative depth (e.g. strategies listed but not clearly evaluated); mostly accurate chemistry; generally sound SPaG with minor lapses.
Basic (1-3 marks): Basic, list-like or fragmentary answer; only one area covered in any detail, or significant scientific inaccuracies; weak organisation and/or noticeably poor SPaG.
0 marks: No relevant content, or entirely incorrect. [9]

Two thermometers commonly used to measure body temperature are the mercury-in-glass (or alcohol-in-glass) clinical thermometer and the infrared tympanic (ear) thermometer. A mercury-in-glass thermometer requires no power source or electronics, is simple, cheap and gives a directly readable scale, but takes longer to reach a stable reading and must be sterilised between uses to avoid cross-infection. An infrared tympanic thermometer detects infrared radiation from the eardrum and gives a very rapid reading (a few seconds), reducing patient discomfort and the time needed per measurement, and uses disposable probe covers which lowers cross-infection risk, though it can give a less accurate reading if not correctly positioned in the ear canal.

1 mark: first thermometer type named correctly (e.g. mercury-in-glass); 1 mark: valid advantage of this type given; 1 mark: second thermometer type named correctly (e.g. infrared tympanic); 1 mark: valid advantage of this type given; [4]

(a) The systolic value is the maximum pressure exerted by blood on the artery walls, which occurs as the ventricles of the heart contract and eject blood into the arteries.
(b) The diastolic value is the minimum pressure in the arteries, which occurs as the ventricles relax and refill with blood between contractions.
(c) A typical normal blood pressure reading for a healthy adult aged 18-40 is approximately 120/80 mmHg (accept any value in the normal range, e.g. 110-120 over 70-80 mmHg).

1 mark: systolic correctly defined as max pressure during ventricular contraction; 1 mark: diastolic correctly defined as min pressure during ventricular relaxation; 1 mark: normal reading given as approximately 120/80 mmHg (accept range 110-120 / 70-80); [3]

(a) An ECG measures the electrical activity (excitation waves) generated by the heart muscle as it depolarises and repolarises during each heartbeat.
(b) A resting heart rate of 145 beats per minute is well above the normal resting range (60-80 bpm), so this condition is called tachycardia.
(c) A normal ECG trace shows a regular, repeating pattern with clearly identifiable P wave, QRS complex and T wave for each heartbeat. In ventricular fibrillation, the trace instead shows rapid, chaotic, irregular oscillations with no organised or repeating QRS complexes, reflecting uncoordinated contraction of the ventricles.

(a) 1 mark: electrical activity/excitation of the heart. (b) 1 mark: tachycardia. (c) 1 mark: normal trace has regular repeating P-QRS-T pattern; 1 mark: fibrillation trace is rapid, chaotic/irregular with no organised QRS complexes; [4]

(a) An EEG measures the electrical activity produced by neurons in the brain, detected via electrodes placed on the scalp.
(b) An EEG trace can help diagnose epilepsy, by detecting abnormal spikes or sharp waves associated with seizure activity (other valid answers: sleep disorders, some causes of altered consciousness).
(c) EEG can continuously monitor brain electrical activity in real time over an extended period, which is useful for capturing intermittent events such as seizures, whereas MRI produces a static structural image and is not suited to continuous, real-time functional monitoring.

(a) 1 mark: electrical activity of the brain (via scalp electrodes). (b) 1 mark: epilepsy (or other valid condition e.g. sleep disorder). (c) 1 mark: valid advantage, e.g. continuous/real-time monitoring possible with EEG unlike MRI; [3]

X-ray imaging works because X-rays are absorbed to different extents by tissues of different density: the higher the density of the tissue, the greater the absorption of X-rays as they pass through the body. Bone is much denser than soft tissue (due to its high calcium/mineral content), so it absorbs a large proportion of the incident X-ray beam. This means relatively little radiation reaches the photographic film or detector positioned beneath the bone, so that region remains largely unexposed and appears white on the developed image. Soft tissue is much less dense, so it absorbs fewer X-rays; more radiation passes through and reaches the detector, exposing it more and producing a darker grey appearance in the image.

1 mark: X-rays absorbed differently by tissues of different density (higher density = more absorption); 1 mark: bone is dense, absorbs more X-rays; 1 mark: less radiation reaches detector beneath bone, so it appears white/unexposed; 1 mark: soft tissue is less dense, absorbs less, so more radiation reaches detector giving darker image; [4]

(a) In a coherent bundle, each optical fibre occupies the same relative position at both ends of the bundle. In an incoherent bundle, the fibres are arranged randomly, so a fibre's position at one end does not correspond to its position at the other end.
(b) The coherent bundle is used to carry the image from inside the body back to the eyepiece or camera, because keeping each fibre's relative position consistent at both ends means that the light collected from each point of the image is delivered to the corresponding point at the viewing end, preserving the spatial arrangement of the image (an incoherent bundle would scramble the image into a meaningless jumble of light, though it is perfectly adequate for simply carrying illumination to the target).

(a) 1 mark: coherent = same relative fibre position at both ends; 1 mark: incoherent = fibre positions jumbled/random between ends. (b) 1 mark: coherent bundle carries the image, because it preserves the spatial arrangement of light so the image is not scrambled; [3]

(a) Deep structures such as the liver and kidney are typically imaged using ultrasound frequencies of 1-6 MHz. Lower frequencies are chosen for deep structures because they are attenuated (absorbed and scattered) less strongly than higher frequencies as they travel through tissue, so they can penetrate deeply enough to reach and return from the liver; the trade-off is that lower frequencies give poorer resolution of fine structural detail than higher frequencies, but for a large, deep organ this loss of fine resolution is an acceptable compromise for adequate penetration.
(b) A coupling gel is used because there would otherwise be a thin layer of air between the probe and the skin; air has a very different acoustic impedance from soft tissue, so almost all the ultrasound would be reflected at an air gap. The gel excludes air and closely matches the acoustic impedance of skin, allowing ultrasound to be transmitted into and out of the body without a large loss of energy.

(a) 1 mark: 1-6 MHz (or 'low frequency') stated; 1 mark: lower frequency penetrates more deeply/less attenuated; 1 mark: trade-off of poorer resolution of fine structures correctly noted. (b) 1 mark: gel excludes air/matches impedance so ultrasound is transmitted without large energy loss at an air gap; [4]

A practical endoscope needs several channels in addition to the illumination and image-collection fibre bundles, so that it can be used effectively for both diagnosis and minor treatment. These include: an irrigation channel, used to flush water or saline to clean the lens or clear the field of view; a channel to allow the passage of surgical tools, such as biopsy forceps or snares, enabling tissue samples to be taken or small procedures performed; and often a separate channel for air insufflation, used to gently inflate the cavity being examined (e.g. the stomach or bowel) to improve visibility.

1 mark each for any three of: irrigation channel; channel for surgical tools/instruments; air insufflation channel; steering/control mechanism for the tip (any three valid, distinct features); [3]

(a) Magnetic resonance imaging (MRI) uses strong magnetic fields together with radio-frequency pulses to align and then detect the relaxation of hydrogen nuclei in the body's tissues, producing detailed images without exposing the patient to ionising radiation; it is widely used to image soft tissue in fine detail, for example the brain, spinal cord or joints.
(b) Conventional gamma ray imaging (gamma camera imaging) uses a gamma-emitting radiopharmaceutical tracer, injected into or taken up by the patient, and a gamma camera detects the emitted gamma rays to build up an image of where the tracer has concentrated in the body; it is used, for example, in bone scans to detect metastatic cancer or fractures, or in thyroid imaging.

(a) 1 mark: MRI correctly named; 1 mark: valid clinical use given. (b) 1 mark: gamma ray/gamma camera imaging correctly named; 1 mark: valid clinical use given; [4]

(a) Alpha radiation is the most strongly ionising but least penetrating; it is stopped by only a few centimetres of air or a sheet of paper. Gamma radiation is the least ionising but most penetrating, requiring several centimetres of lead (or a similar thickness of dense material) to substantially reduce its intensity.
(b) Although alpha radiation cannot penetrate the outer layer of skin from outside the body, if an alpha-emitting source is ingested or inhaled it is in direct contact with living cells. Because alpha particles are relatively large, heavy and highly charged (compared to beta or gamma), they interact very strongly with matter, causing intense ionisation over a very short range. This means nearly all their energy is deposited in a small volume of tissue, causing severe, concentrated localised damage to cells and DNA, making an internal alpha source more dangerous than an external one of similar activity.

(a) 1 mark: alpha stopped by air/paper; 1 mark: gamma requires several cm of lead. (b) 1 mark: alpha is strongly ionising due to its large mass/charge, depositing energy over a very short range; 1 mark: this causes concentrated/severe damage to living cells if the source is internal (ingested/inhaled); [4]

(a) The activity, A, of a radioactive source is defined as the number of nuclear disintegrations (decay events) occurring per unit time (per second). It is measured in Becquerel (Bq).
(b) One Becquerel represents one nuclear disintegration occurring in one second.

(a) 1 mark: activity defined as number of disintegrations/decays per second; 1 mark: unit correctly stated as Becquerel (Bq). (b) 1 mark: 1 Bq = one disintegration per second; [3]

The physical half-life, \( T_{1/2} \), of a radionuclide is the time taken for the activity of that material to decrease to half its original value, due purely to the process of radioactive (nuclear) decay. The biological half-life is the time taken for half of the radioisotope present to be removed from the body by natural metabolic and excretory processes (e.g. via the kidneys), independent of nuclear decay.
A doctor selecting a radiopharmaceutical needs to consider both, because the isotope must remain in the body, and remain radioactive, for long enough to complete the diagnostic scan and obtain a clear image, but should then be eliminated (either by decaying away or by being excreted from the body) as quickly as possible afterwards, to minimise the total radiation dose received by the patient. An isotope with either half-life too long would deliver an unnecessarily high dose; too short, and the scan could not be completed reliably.

1 mark: physical half-life correctly defined (time for activity to halve, due to nuclear decay); 1 mark: biological half-life correctly defined (time for half the isotope to be removed from the body by metabolic processes); 1 mark: recognises effective retention in the body depends on both; 1 mark: explains the need to balance long enough for the scan against minimising patient radiation dose; [4]

Technetium-99m is a widely used gamma-emitting radioactive tracer in diagnostic nuclear medicine.
(a) It emits gamma radiation.
(b) It is used in a wide variety of medical imaging procedures, for example bone scans, where it is taken up preferentially by areas of high bone turnover (such as fractures or metastatic cancer), allowing these to be identified as regions of increased activity on the resulting image.
(c) Its physical half-life is short (about 6 hours), meaning the isotope decays away quickly once the scan is complete, so it does not continue to expose the patient to radiation for an extended period, keeping the overall radiation dose low while still allowing enough time to perform the scan.
(Accept equally valid alternative answers, for example rubidium-82, a positron emitter used in PET perfusion imaging of the heart because it is rapidly taken up by heart muscle; or thallium-201, used in cardiac imaging and cancer detection.)

(a) 1 mark: radiation type correctly stated for the chosen isotope (e.g. gamma for technetium-99m). (b) 1 mark: valid clinical application named; 1 mark: correctly linked to a relevant property of the isotope. (c) 1 mark: valid additional suitable property given (e.g. short physical half-life, or rapid uptake by target tissue); [4]

(a) A typical resting pulse rate for a healthy adult is 60-80 beats per minute. It can be measured manually by placing two fingers (not the thumb, which has its own pulse) over a pulse point such as the radial artery at the wrist, and counting the number of pulses felt over a fixed period of time, for example counting for 15 seconds and multiplying by 4, or counting for a full 60 seconds; each pulse felt corresponds to one contraction (systole) of the left ventricle forcing a pulse of blood through the artery.
(b) Regular endurance training causes adaptations in the heart, including an increase in the size and strength of the left ventricle (cardiac hypertrophy) and an increased stroke volume (the volume of blood ejected per heartbeat). Because each beat of a trained athlete's heart pumps a larger volume of blood, fewer beats per minute are required to deliver the same resting cardiac output (blood flow) needed to meet the body's oxygen demand, resulting in a lower resting pulse rate than in an untrained person.

(a) 1 mark: 60-80 bpm correctly stated; 1 mark: correct manual method described (pulse point + counting over a fixed time); 1 mark: correctly links each pulse felt to one ventricular contraction. (b) 1 mark: identifies increased stroke volume/cardiac hypertrophy in trained athletes; 1 mark: correctly explains fewer beats needed per minute to meet resting oxygen demand as a result; [5]

(a) \( R = \left( \dfrac{Z_2-Z_1}{Z_2+Z_1} \right)^2 \)
(b) \( Z_2 - Z_1 = 6.30\times10^6 - 1.63\times10^6 = 4.67\times10^6 \). \( Z_2+Z_1 = 6.30\times10^6+1.63\times10^6 = 7.93\times10^6 \). \( R = \left( \dfrac{4.67\times10^6}{7.93\times10^6} \right)^2 = (0.5889)^2 = 0.347 \).
Check (second route, recompute the ratio directly with a calculator to more decimal places): \( 4.67/7.93 = 0.58891 \); squared \( = 0.34681 \), which rounds to 0.347 to 3 s.f., confirming the value.
(c) Since R is the fraction of the incident intensity that is reflected at the boundary, the fraction transmitted into the bone is \( 1 - R = 1 - 0.347 = 0.653 \), i.e. about 65.3% of the incident ultrasound intensity is transmitted into the bone (34.7% is reflected back).

(a) 1 mark: correct equation R = ((Z2-Z1)/(Z2+Z1))^2. (b) 1 mark: Z2-Z1 = 4.67x10^6 correctly calculated; 1 mark: Z2+Z1 = 7.93x10^6 correctly calculated; 1 mark: ratio correctly squared; 1 mark: R = 0.347 to 3 s.f. (c) 1 mark: recognises transmitted fraction = 1 - R; 1 mark: correct value 0.653 (65.3%) stated with correct interpretation; [7]

Convert time to seconds: \( t = 6.00 \times 3600 = 21\,600 \) s.
\( \lambda t = 3.47\times10^{-5} \times 21\,600 = 0.7495 \).
\( A = A_0 e^{-\lambda t} = 4.00\times10^8 \times e^{-0.7495} = 4.00\times10^8 \times 0.4726 = 1.89\times10^8 \) Bq.
Check (second route, via half-life): \( T_{1/2} = 0.693/\lambda = 0.693/(3.47\times10^{-5}) = 19\,971 \) s \( \approx 5.55 \) hours. Number of half-lives elapsed in 6.00 h \( = 6.00/5.55 = 1.081 \). \( A = A_0 \times (0.5)^{1.081} = 4.00\times10^8 \times 0.4722 = 1.89\times10^8 \) Bq, consistent with the direct calculation. Final answer: \( 1.89 \times 10^8 \) Bq.

1 mark: t correctly converted to 21 600 s; 1 mark: lambda*t = 0.750 correctly calculated; 1 mark: correct use of exponential equation A = A0 e^(-lambda t); 1 mark: e^(-0.750) = 0.473 correctly evaluated; 1 mark: correct substitution of A0 = 4.00x10^8; 1 mark: final answer 1.89x10^8 Bq (accept 1.88-1.90x10^8); [6]

(a) \( \ln A = \ln A_0 - \lambda t \Rightarrow \lambda = \dfrac{\ln A_0 - \ln A}{t} = \dfrac{\ln(8.00\times10^6) - \ln(3.17\times10^6)}{t} \).
\( \ln(8.00\times10^6) = 15.895 \); \( \ln(3.17\times10^6) = 14.969 \). Difference \( = 0.926 \).
\( t = 24.0 \) hours \( = 24.0 \times 3600 = 86\,400 \) s.
\( \lambda = \dfrac{0.926}{86\,400} = 1.07\times10^{-5} \) s\(^{-1}\).
Check (second route): \( A/A_0 = 3.17/8.00 = 0.39625 \); \( \ln(0.39625) = -0.9260 \), matching the difference computed above, confirming the subtraction.
(b) \( T_{1/2} = \dfrac{0.693}{\lambda} = \dfrac{0.693}{1.07\times10^{-5}} = 64\,766 \) s \( = \dfrac{64\,766}{3600} = 18.0 \) hours.
Check (second route, direct from data): after 24.0 h the activity has fallen to \( 3.17/8.00 = 0.396 \) of its original value, i.e. between one and two half-lives have elapsed (since \( 0.5^1 = 0.5 \) and \( 0.5^2 = 0.25 \)); \( 24.0/18.0 = 1.33 \) half-lives, and \( 0.5^{1.33} = 0.398 \), close to the given ratio of 0.396, confirming the half-life is consistent.

(a) 1 mark: ln(A0) and ln(A) correctly evaluated; 1 mark: t correctly converted to 86 400 s; 1 mark: correct rearrangement of ln A = ln A0 - lambda t; 1 mark: lambda = 1.07x10^-5 s-1 (accept 1.06-1.08x10^-5). (b) 1 mark: correct use of T(1/2) = 0.693/lambda; 1 mark: T(1/2) = 18.0 hours (accept 17.8-18.2 hours, consistent with candidate's lambda); [6]

(a) Minute ventilation = tidal volume x breathing rate \( = 0.480 \times 15 = 7.20 \) dm\(^3\) per minute.
(b) Percentage \( = \dfrac{0.480}{4.20} \times 100 = 11.4\% \).
Check (second route): \( 11.4\% \) of \( 4.20 = 0.114 \times 4.20 = 0.479 \) dm\(^3\) \( \approx 0.480 \) dm\(^3\) ✓, confirming the percentage.
(c) During quiet, resting breathing the body's oxygen demand and carbon dioxide production are relatively low, so only a small volume of air needs to be exchanged with each breath to maintain normal gas exchange; the remaining, much larger, lung capacity (inspiratory and expiratory reserve volumes) is held in reserve so that tidal volume can increase substantially during exercise or exertion, when oxygen demand rises sharply.

(a) 1 mark: correct method (tidal volume x rate); 1 mark: 7.20 dm3/min. (b) 1 mark: correct percentage method; 1 mark: 11.4%. (c) 1 mark: identifies that resting oxygen demand is low, so only a small volume needs exchanging; 1 mark: correctly links this to reserve capacity available for exercise/exertion; [6]

(a) Corrected count rate = total count rate - background \( = 372 - 24 = 348 \) counts per minute.
(b) After each half-life the source's (corrected) activity halves. After 3 half-lives, the corrected count rate \( = 348 \times (0.5)^3 = 348 \times 0.125 = 43.5 \) counts per minute.
Check (second route, step by step): after 1 half-life: \( 348/2 = 174 \); after 2 half-lives: \( 174/2 = 87.0 \); after 3 half-lives: \( 87.0/2 = 43.5 \) counts per minute ✓, matching the direct calculation.
(c) Background radiation, arising from cosmic rays and naturally occurring radioactive materials in the environment (e.g. in rocks, building materials, and the human body), is always present and is detected by the Geiger-Muller tube alongside radiation from the source under investigation. If this background contribution were not subtracted, the measured (total) count rate would overestimate the true activity of the source, and would not fall to genuinely zero as the source fully decays (it would appear to level off at the background value), leading to an inaccurate decay curve and an incorrect derived half-life or decay constant.

(a) 1 mark: 348 counts per minute. (b) 1 mark: correct method using (0.5)^3; 1 mark: correct working shown; 1 mark: 43.5 counts per minute. (c) 1 mark: background radiation is always present, from cosmic rays/environmental sources; 1 mark: it adds to the true source reading if not subtracted; 1 mark: correctly explains this would distort the derived decay curve/half-life if not corrected for; [7]

(a) The pulse travels down to the boundary and back, a total distance of \( 2 \times 0.0850 = 0.1700 \) m, in time \( 1.13\times10^{-4} \) s.
Speed \( = \dfrac{\text{distance}}{\text{time}} = \dfrac{0.1700}{1.13\times10^{-4}} = 1504 \) m s\(^{-1}\).
Check (second route): time for 0.1700 m at 1504 m/s \( = 0.1700/1504 = 1.130\times10^{-4} \) s ✓, matching the given time.
(b) Using the assumed speed \( c = 1540 \) m s\(^{-1}\), distance travelled \( = c \times t = 1540 \times 1.13\times10^{-4} = 0.174 \) m; this is the total there-and-back distance, so the displayed depth \( = 0.174/2 = 0.0870 \) m.
Comparing to the true depth of 0.0850 m, the machine overestimates the depth by \( 0.0870-0.0850 = 0.0020 \) m, an error of \( 0.0020/0.0850 \times 100 = 2.4\% \). This overestimate occurs because the true speed of sound in this particular tissue (1504 m/s) is slightly lower than the fixed average value of 1540 m/s that the machine assumes for all soft tissue, so the machine calculates a slightly greater depth than the true value for the same travel time.

(a) 1 mark: total distance = 0.1700 m (there and back) correctly identified; 1 mark: correct rearrangement speed = distance/time; 1 mark: correct substitution; 1 mark: speed = 1504 m/s (accept 1500-1510 m/s). (b) 1 mark: displayed depth = 0.0870 m correctly calculated; 1 mark: correct comparison/error calculated (~0.0020 m or ~2.4%); 1 mark: correct physical explanation that true tissue speed is lower than the assumed 1540 m/s; [7]

A full-mark answer should discuss, in a well-organised and coherent way:

Physical principle: Ultrasound B-scan imaging relies on a piezoelectric transducer emitting high-frequency sound pulses (typically 1-18 MHz) into the body. At each boundary between tissues of different specific acoustic impedance, some of the ultrasound is reflected (the fraction given by the intensity reflection coefficient, R) and detected by the same transducer; the time delay between emission and detection of each echo is used to calculate the depth of each boundary, and multiple scan lines are combined by computer to build up a two-dimensional greyscale image. CT scanning, in contrast, relies on ionising X-rays: an X-ray tube and detector rotate around the patient, taking many X-ray absorption measurements (projections) from different angles through a thin slice of the body; because different tissues absorb X-rays to different extents (bone strongly, soft tissue weakly, as in conventional X-ray imaging), a computer uses these many projections to reconstruct a detailed cross-sectional image of the internal anatomy.

Resolution/penetration and dose trade-offs: For ultrasound, choosing a higher frequency improves the resolution of fine structures (as the wavelength is shorter) but reduces the depth of penetration (since higher-frequency sound is attenuated more strongly by tissue), so high frequencies (7-18 MHz) suit superficial structures like the thyroid or breast, while lower frequencies (1-6 MHz) are needed to reach and resolve deeper organs like the liver, at the cost of poorer fine detail. For CT, a higher radiation dose (or a greater number of thinner slices) generally improves image resolution and contrast, but this directly increases the ionising radiation dose delivered to the patient, so clinicians must balance diagnostic image quality against the cumulative radiation exposure and associated risk.

Safety evaluation: Ultrasound uses non-ionising mechanical (sound) energy; at the intensities used diagnostically it causes no known significant tissue damage, so it is considered very safe, including for repeated use and for imaging a foetus during pregnancy, and requires no special radiation shielding or dose monitoring. CT scanning uses ionising X-radiation, which can damage DNA and increases the lifetime risk of cancer with cumulative exposure; a single CT scan typically delivers a much higher effective radiation dose than a conventional X-ray, so CT is used only when clinically justified (when its diagnostic benefit outweighs the radiation risk), doses are kept 'as low as reasonably achievable' (ALARA principle), and it is used more cautiously in children and pregnant patients, for whom ultrasound is often preferred where it can provide adequate diagnostic information.

A good answer draws an explicit overall comparison (e.g. ultrasound for safety-critical, repeatable, real-time soft-tissue imaging; CT for rapid, detailed cross-sectional imaging, particularly of bone and complex trauma, where the diagnostic benefit justifies the radiation dose) using accurate terminology and correct spelling, punctuation and grammar throughout.

Excellent (7-9 marks): Detailed, accurate and well-structured comparison covering the physical principle of both techniques correctly, a clear explanation of the frequency/dose-resolution trade-off for both, and a balanced, well-reasoned safety evaluation referencing the non-ionising nature of ultrasound versus the ionising nature of CT X-rays; addresses all three bulleted prompts with an explicit overall comparison; high standard of SPaG with accurate scientific terminology throughout.
Good / Very Good (4-6 marks): Reasonable coverage of most of the three areas but with some omissions, imbalance between the two techniques, or limited evaluative depth; mostly accurate; generally sound SPaG with minor lapses.
Basic (1-3 marks): Basic or list-like answer, covering only one technique in detail or with significant inaccuracies; weak organisation and/or noticeably poor SPaG.
0 marks: No relevant content, or entirely incorrect. [9]

(a) In a transverse wave, the oscillation of particles (or, for an electromagnetic wave, the oscillation of the electric and magnetic fields) is perpendicular to the direction in which the wave's energy travels. In a longitudinal wave, the oscillation is parallel to the direction of energy transfer, producing alternating regions of compression and rarefaction.
(b) A transverse wave example is light (or any electromagnetic wave, or a wave on a stretched string); a longitudinal wave example is sound.

(a) 1 mark: transverse = oscillation perpendicular to energy transfer direction; 1 mark: longitudinal = oscillation parallel to energy transfer direction. (b) 1 mark: correct example of each given; [3]

(a) A displacement-distance graph (a 'snapshot' of the wave at one instant) allows the wavelength to be read directly, as the distance between two successive identical points (e.g. successive peaks), as well as the amplitude (the maximum displacement from the equilibrium position). A displacement-time graph (showing the motion of one particle over time) allows the time period, T, to be read directly, as the time for one complete oscillation, as well as the amplitude.
(b) Frequency is the reciprocal of the time period: \( f = \dfrac{1}{T} = \dfrac{1}{0.0800} = 12.5 \) Hz.

(a) 1 mark: displacement-distance graph gives wavelength; 1 mark: displacement-time graph gives time period. (b) 1 mark: correct relationship f = 1/T used; 1 mark: f = 12.5 Hz; [4]

(a) The pinna is the visible, external part of the ear; its curved shape collects sound waves from the surrounding air and funnels them into the auditory canal.
(b) The auditory canal is a tube that channels the collected sound waves inward to the tympanic membrane; its length and shape also cause resonance that amplifies certain frequencies (around 3-4 kHz).
(c) The tympanic membrane is a thin, taut membrane that vibrates when sound waves strike it, converting the sound wave into mechanical vibrations that are passed on to the ossicles of the middle ear.

1 mark each for correct function of: (a) pinna - collects/funnels sound; (b) auditory canal - channels sound to eardrum; (c) tympanic membrane - vibrates, transmitting vibration to ossicles; [3]

(a) The three ossicles, in order from the ear drum, are the malleus (hammer), then the incus (anvil), then the stapes (stirrup).
(b) The ossicles act as a chain of small levers, mechanically transmitting the vibrations of the tympanic membrane across the air-filled middle ear to the oval window of the cochlea. Because the oval window has a much smaller area than the tympanic membrane, and the ossicle lever arrangement provides additional mechanical advantage, the ossicles amplify the pressure of the vibration. This amplification is needed to efficiently transfer vibrational energy from air into the denser fluid of the inner ear, which would otherwise reflect most of the sound energy at the boundary (impedance matching).

(a) 1 mark: malleus and incus correctly named and ordered; 1 mark: stapes correctly named and ordered. (b) 1 mark: ossicles mechanically transmit/amplify vibration from eardrum to oval window; 1 mark: correctly explains the need for amplification/impedance matching between air and cochlear fluid; [4]

(a) For a string fixed at both ends, the two end points are always nodes (zero displacement). The third harmonic has three equally-spaced antinodes (points of maximum displacement) between the ends, separated by additional nodes, giving the repeating pattern along the string: node - antinode - node - antinode - node - antinode - node (4 nodes total, 3 antinodes total), each antinode located at the midpoint between two adjacent nodes.
(b) For a string fixed at both ends, the length L is related to the wavelength by \( L = n\dfrac{\lambda}{2} \), where n is the harmonic number. For the third harmonic, \( n = 3 \): \( 0.900 = 3 \times \dfrac{\lambda}{2} \Rightarrow \lambda = \dfrac{2 \times 0.900}{3} = 0.600 \) m.
Check (second route): three half-wavelengths must fit the string length, so one half-wavelength = 0.900/3 = 0.300 m, giving \( \lambda = 2 \times 0.300 = 0.600 \) m ✓, consistent.

(a) 1 mark: correct number and even spacing of nodes/antinodes (4 nodes, 3 antinodes) shown/described; 1 mark: ends of string correctly shown/described as nodes. (b) 1 mark: correct relationship L = n(lambda/2) applied with n = 3; 1 mark: wavelength = 0.600 m; [4]

A vibration generator is used to continuously vibrate one end of a stretched string, sending a transverse wave travelling along it. This wave reflects off the fixed end of the string and travels back in the opposite direction. Because the incident wave and the reflected wave have the same frequency, amplitude and wavelength but travel in opposite directions, they superpose to form a standing (stationary) wave pattern with a fixed shape that does not itself travel along the string. At certain fixed points along the string, the incident and reflected waves are always exactly out of phase (in antiphase) with one another, so by the principle of superposition their displacements always cancel to give zero resultant displacement — these fixed points are called nodes. Midway between adjacent nodes, the two waves are always exactly in phase, reinforcing each other to give a point of maximum resultant displacement — an antinode.

1 mark: vibration generator produces a travelling wave that reflects off the fixed end, generating two waves travelling in opposite directions of the same f, A and lambda; 1 mark: correct application of superposition — at nodes the two waves are always in antiphase, cancelling; 1 mark: at antinodes the two waves are always in phase, reinforcing, giving maximum displacement; [3]

(a) The cornea is the transparent, curved front surface of the eye; being strongly curved, it provides the majority of the eye's total refractive (focusing) power.
(b) The iris is the coloured, muscular ring surrounding the pupil; it controls the diameter of the pupil, and hence the amount of light entering the eye, contracting in bright light and dilating in dim light.
(c) The retina is the light-sensitive layer lining the back of the eye, containing the photoreceptor cells (rod and cone cells) that detect light and convert it into electrical nerve impulses.
(d) The optic nerve carries the electrical nerve impulses generated by the photoreceptor cells in the retina to the visual processing centres of the brain.

1 mark each for correctly stating the function of: (a) cornea - main refractive power; (b) iris - controls pupil size/light entering; (c) retina - contains photoreceptors, converts light to nerve impulses; (d) optic nerve - carries impulses to the brain; [4]

(a) Rod cells are highly sensitive to light and are responsible for vision in dim/low-light conditions, but they cannot distinguish colour, giving only monochrome (black-and-white) vision. Cone cells are less sensitive and require brighter light conditions to function well, but they are responsible for colour vision and high visual acuity (fine detail).
(b) Cone cells are most densely concentrated in the fovea, a small central region of the macula, which is why visual acuity and colour perception are sharpest when an object is viewed directly (foveal vision).

(a) 1 mark: rods sensitive/function in dim light, no colour vision; 1 mark: cones need brighter light, provide colour vision/fine detail. (b) 1 mark: fovea (or macula); [3]

(a) The critical angle is the specific angle of incidence, in the optically denser medium, for which the refracted ray emerges at an angle of refraction of exactly 90°, travelling along the boundary between the two media.
(b) Total internal reflection occurs when light travelling in the denser medium strikes the boundary with the less dense medium at an angle of incidence greater than the critical angle; in this case no light is refracted out of the denser medium, and all of the light is reflected back into it, obeying the law of reflection.

(a) 1 mark: critical angle correctly defined as the angle of incidence giving a 90 degree angle of refraction. (b) 1 mark: condition correctly stated as angle of incidence greater than critical angle; 1 mark: correctly states light travelling from denser to less dense medium/all light reflected, none refracted out; [3]

(a) A single-mode fibre has a much narrower core diameter than a multi-mode fibre.
(b) In a multi-mode fibre, light rays entering the fibre core at different angles undergo total internal reflection and travel along different-length zig-zag paths (different 'modes') to reach the far end. Since these different paths have different lengths, rays that started at the same instant (as part of the same data pulse) arrive at slightly different times, causing the pulse to spread out in time — an effect called modal dispersion. Over long distances, this pulse-spreading becomes severe enough that successive data pulses can overlap and become indistinguishable, limiting the maximum data rate and transmission distance. Because a single-mode fibre has such a narrow core, it only supports essentially one path (mode) for light close to the fibre's axis, so there is no modal dispersion; pulses remain sharp, allowing much higher data rates to be transmitted reliably over much longer distances, which is why single-mode fibre is preferred for long-haul, high-bandwidth links.

(a) 1 mark: single-mode fibre has a much narrower core than multi-mode. (b) 1 mark: multi-mode allows multiple paths/modes of different lengths, causing modal dispersion/pulse spreading; 1 mark: this limits achievable data rate/distance in multi-mode fibre; 1 mark: single-mode's narrow core supports effectively one path, eliminating modal dispersion, allowing higher data rates over longer distances; [4]

(a) A dipole antenna is commonly used both to transmit (generate) and to receive radio signals.
(b) Attenuation refers to the reduction in the strength (power/amplitude) of a radio signal as it travels away from the transmitting antenna, for example due to the signal energy spreading out over an increasing area, or being absorbed/scattered by obstacles. This attenuation can be categorised as path loss (loss due to the signal spreading out and interacting with the environment/obstacles along its path) and free space loss (the unavoidable geometric spreading of signal power over an increasing surface area with distance, even with no obstacles present).

(a) 1 mark: dipole antenna. (b) 1 mark: attenuation correctly defined as reduction in signal strength with distance travelled; 1 mark: correctly names both categories, path loss and free space loss; [3]

(a) The ear shows maximum sensitivity (lowest threshold intensity) at frequencies of approximately 3-4 kHz.
(b) The auditory canal acts similarly to a tube closed at one end (the tympanic membrane). Such a tube has a natural resonant frequency, determined by its length, that falls within the 3-4 kHz range. Sound waves entering the ear at frequencies close to this natural resonant frequency are amplified by resonance within the auditory canal before reaching the eardrum, so a lower external sound intensity is needed for the sound to be perceived, giving the ear its peak sensitivity in this frequency range.
(c) At frequencies well below or well above this range of maximum sensitivity, the threshold intensity increases considerably (sensitivity decreases), meaning a much greater sound intensity is required for the ear to perceive a sound of the same apparent loudness.

(a) 1 mark: 3-4 kHz stated. (b) 1 mark: auditory canal resonates like a closed tube, with resonant frequency in this range; 1 mark: correctly explains this amplifies sound in this range, lowering the threshold intensity/increasing sensitivity. (c) 1 mark: threshold intensity increases (sensitivity decreases) away from this range in both directions; [4]

(a) \( v = f\lambda = 2.50\times10^3 \times 0.136 = 340 \) m s\(^{-1}\).
(b) Since v is unchanged, \( \lambda = \dfrac{v}{f} = \dfrac{340}{4.00\times10^3} = 0.0850 \) m.
Check (second route): \( f\lambda = 4.00\times10^3 \times 0.0850 = 340 \) m/s ✓, matching the speed found in (a).
(c) Since \( v = f\lambda \) and v (the speed of sound in air) remains constant, \( \lambda \) must decrease as f increases, because \( \lambda \) and f are inversely proportional to one another at fixed v; this is confirmed numerically above, as increasing f from 2.50 kHz to 4.00 kHz reduced \( \lambda \) from 0.136 m to 0.0850 m.

(a) 1 mark: correct use of v = f(lambda); 1 mark: v = 340 m/s. (b) 1 mark: correct rearrangement lambda = v/f; 1 mark: lambda = 0.0850 m. (c) 1 mark: wavelength decreases as frequency increases; 1 mark: correctly explains this as an inverse proportionality at constant v (from v = f(lambda)); [6]

(a) \( \dfrac{I}{I_0} = \dfrac{3.16\times10^{-9}}{1\times10^{-12}} = 3160 \). db level \( = 10\log_{10}(3160) = 10 \times 3.500 = 35.0 \) dB.
Check (second route): \( 10^{3.5} = 3162 \approx 3160 \) ✓, confirming the logarithm.
(b) \( I = I_0 \times 10^{65.0/10} = 1\times10^{-12} \times 10^{6.5} \). \( 10^{6.5} = 3.162\times10^6 \). \( I = 1\times10^{-12} \times 3.162\times10^6 = 3.16\times10^{-6} \) W m\(^{-2}\).
Check (second route, via part (a)): substitute this I back into the dB formula: \( 10\log_{10}\left(\dfrac{3.16\times10^{-6}}{1\times10^{-12}}\right) = 10\log_{10}(3.16\times10^6) = 10 \times 6.500 = 65.0 \) dB ✓, confirming the result is self-consistent.

(a) 1 mark: I/I0 = 3160 correctly calculated; 1 mark: correct use of log10 formula; 1 mark: 35.0 dB. (b) 1 mark: correct rearranged exponential formula set up; 1 mark: 10^6.5 = 3.16x10^6 correctly evaluated; 1 mark: I = 3.16x10^-6 W/m2; [6]

(a) \( P = \dfrac{1}{f} = \dfrac{1}{0.400} = 2.50 \) D.
(b) \( f = \dfrac{1}{P} = \dfrac{1}{2.50} = 0.400 \) m \( = 40.0 \) cm.
Check (second route): substituting f = 0.400 m back into P = 1/f gives P = 2.50 D, matching the value given, confirming the calculation is self-consistent with part (a).
(c) A positive value of lens power indicates a converging (convex) lens, since only a converging lens has a positive (real, finite) focal length under the sign convention used; a diverging (concave) lens has a negative power.

(a) 1 mark: correct use of P = 1/f; 1 mark: P = 2.50 D. (b) 1 mark: correct rearrangement f = 1/P; 1 mark: f = 40.0 cm (0.400 m). (c) 1 mark: positive power indicates converging/convex lens; 1 mark: correctly contrasted with diverging/concave lens having negative power; [6]

For the first resonance in a closed tube, \( L = \dfrac{\lambda}{4} \Rightarrow \lambda = 4L = 4 \times 0.168 = 0.672 \) m.
\( v = f\lambda = 512 \times 0.672 = 344 \) m s\(^{-1}\).
Check (second route): \( \lambda = v/f = 344/512 = 0.672 \) m ✓, matching the wavelength found from the tube length, confirming the answer. Final answer: v = 344 m/s, which is close to the accepted value for the speed of sound in air at room temperature (~343 m/s), supporting its plausibility.

1 mark: correct relationship L = lambda/4 identified for a closed-end tube; 1 mark: lambda = 4L correctly rearranged; 1 mark: lambda = 0.672 m correctly calculated; 1 mark: v = f(lambda) correctly applied; 1 mark: v = 344 m/s; [5]

\( v - v_s = 340 - 25.0 = 315 \) m s\(^{-1}\).
\( f' = 700 \times \dfrac{340}{315} = 700 \times 1.0794 = 755.6 \) Hz \( \approx 756 \) Hz (3 s.f.).
Check (second route): \( \dfrac{f'}{f} = \dfrac{756}{700} = 1.080 \), and \( \dfrac{v}{v-v_s} = \dfrac{340}{315} = 1.079 \) ✓, consistent. As expected for a source approaching the observer, \( f' > f \), since the wavefronts are compressed ahead of the moving source.

1 mark: v - vs = 315 m/s correctly calculated; 1 mark: ratio v/(v-vs) = 1.079 correctly calculated; 1 mark: correct substitution f' = 700 x 1.079; 1 mark: f' = 756 Hz (3 s.f., accept 755-756 Hz); 1 mark: correct recognition that f' > f as expected for an approaching source; 1 mark: fully correct method with units/sig figs shown throughout; [6]

The phase difference is found from the path difference as a fraction of one wavelength, since one full wavelength corresponds to a phase difference of \( 2\pi \) radians (360°).
Fraction of a wavelength \( = \dfrac{0.150}{0.400} = 0.375 \).
(a) Phase difference (radians) \( = 0.375 \times 2\pi = 2.356 \) rad \( \approx 2.36 \) rad.
(b) Phase difference (degrees) \( = 0.375 \times 360 = 135^{\circ} \).
Check (second route, converting between units): \( 135^{\circ} \times \dfrac{\pi}{180} = 2.356 \) rad ✓, consistent with the radians answer in (a).

1 mark: path difference expressed as a fraction of wavelength (0.375); (a) 1 mark: correct use of phase = fraction x 2(pi); 1 mark: 2.36 rad. (b) 1 mark: 135 degrees (correct use of fraction x 360); [5]

The object distance is the book's distance from the lens: \( u = +0.250 \) m (real object).
The image formed is virtual (on the same side as the object, at the patient's near point), so \( v = -0.400 \) m.
\( P = \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{-0.400} + \dfrac{1}{0.250} = -2.50 + 4.00 = +1.50 \) D.
Check (second route): a converging (positive power) lens is required to correct long sight (hyperopia), since long-sighted patients need extra converging power to focus on near objects — the positive sign of the answer is therefore physically consistent with the type of correction needed.

1 mark: u = +0.250 m correctly identified as a real object distance; 1 mark: v = -0.400 m correctly identified as a virtual image distance with correct sign; 1 mark: correct substitution into P = 1/v + 1/u; 1 mark: 1/v = -2.50 D and 1/u = 4.00 D each correctly evaluated; 1 mark: P = +1.50 D, with correct positive sign confirming a converging lens is needed to correct long sight; [5]

A full-mark answer should discuss, in a well-organised and coherent way:

Structure and function: The pinna (outer ear) collects sound waves from the environment and funnels them along the auditory canal, whose length and shape cause it to resonate at frequencies around 3-4 kHz, amplifying sounds in this range before they reach the tympanic membrane (ear drum), which vibrates in response. These vibrations pass across the air-filled middle ear via the three ossicles (malleus, incus and stapes), acting as a lever system that mechanically amplifies the vibration and efficiently transfers it from the tympanic membrane to the much smaller oval window of the cochlea, matching the acoustic impedance between air and the denser fluid of the inner ear (without this amplification, most of the sound energy would simply be reflected at the air-fluid boundary). In the inner ear, these vibrations set up pressure waves in the fluid-filled cochlea, which stimulate hair cells lining the cochlea; this mechanical stimulation is converted into electrical nerve impulses, which are carried by the auditory nerve to the brain, where they are interpreted as sound.

Subjectivity of loudness: Because the auditory canal resonates most strongly around 3-4 kHz, the ear is most sensitive (requires the lowest physical intensity to perceive a sound) in this frequency range; sensitivity falls off (a greater intensity is needed) at both lower and higher frequencies. This frequency-dependence is captured in equal loudness curves, which show that sounds of very different physical intensity, at different frequencies, can be perceived as equally loud, and conversely that the same physical intensity is perceived as louder near 3-4 kHz than at the extremes of the audible range. Since loudness is a perceptual response of the brain to this frequency-dependent physiological sensitivity (and also varies somewhat between individuals), it does not correspond directly or linearly to the physical sound intensity (measured objectively in W m-2 or dB); this is why loudness is measured using the subjective unit of phons rather than a purely physical intensity scale.

Hearing protection: Prolonged exposure to high-intensity sound can permanently damage the hair cells of the cochlea, causing noise-induced hearing loss, since (unlike some other cells) damaged cochlear hair cells do not regenerate. Ear defenders or earplugs are commonly used to protect against this, by physically reducing the intensity of sound reaching the tympanic membrane and hence the cochlea. Evaluating this measure: it can be highly effective at reducing cumulative noise exposure and is relatively low-cost and simple to use, but its effectiveness depends heavily on it being worn consistently, for the full duration of exposure, and correctly fitted to form an adequate seal; it also cannot eliminate all risk from extremely loud, sudden (impulsive) noises, which may still cause some damage even with protection in place.

A good answer draws these three areas together into a coherent discussion, using accurate anatomical and physical terminology, with correct spelling, punctuation and grammar throughout.

Excellent (8-10 marks): Detailed, accurate and well-structured discussion covering the roles of all three parts of the ear correctly, a clear and correct explanation of why loudness is subjective (linking auditory canal resonance, frequency response and equal loudness curves), and a genuinely evaluative discussion of one hearing-protection measure (advantage and limitation both given); addresses all three bulleted prompts; high standard of SPaG with accurate technical terminology throughout.
Good / Very Good (5-7 marks): Reasonable coverage of most of the three areas but with some omissions, imbalance, or limited evaluative depth on the hearing-protection measure; mostly accurate; generally sound SPaG with minor lapses.
Basic (1-4 marks): Basic, list-like or fragmentary answer; only one or two areas covered in any detail, or notable scientific inaccuracies; weak organisation and/or noticeably poor SPaG.
0 marks: No relevant content, or entirely incorrect. [10]

A full-mark answer should discuss, in a well-organised and coherent way:

Total internal reflection and fibre structure: An optical fibre consists of a central core, made of glass or plastic with a relatively high refractive index, surrounded by a cladding layer of lower refractive index, with a protective outer coating. Light is launched into the core at a shallow angle; because the core has a higher refractive index than the cladding, light travelling within the core that strikes the core-cladding boundary at an angle of incidence greater than the critical angle undergoes total internal reflection, remaining entirely within the core rather than being partially refracted (and lost) into the cladding. By undergoing repeated total internal reflection along the length of the fibre, light (and the data signal it carries, encoded as pulses) can be guided over very long distances with relatively little loss.

Single-mode vs multi-mode: A multi-mode fibre has a relatively wide core, allowing light to enter and be guided along many different possible paths (modes) at different angles, each undergoing total internal reflection at slightly different points; because these different-angle paths have different lengths, light that entered the fibre at the same instant, as part of one data pulse, arrives at the far end at slightly different times, causing the pulse to broaden in time — this effect is called modal dispersion. Over long distances, this pulse-broadening becomes severe enough that successive pulses overlap and cannot be reliably distinguished, limiting the maximum data rate and the distance the fibre can be used over. A single-mode fibre has a much narrower core, so narrow that it supports essentially only one path (mode) for light, travelling very close to the fibre's axis; because there is effectively only one path length, modal dispersion is eliminated, allowing pulses to remain sharp and enabling much higher data rates over much longer distances than multi-mode fibre, though single-mode fibre requires more precise (and so more expensive) equipment to couple light into its very narrow core, and multi-mode fibre remains cheaper and easier to install for shorter links, such as within a single building.

Advantages over copper cable: Fibre-optic cable can carry a far greater bandwidth (a far higher rate of data transmission) than copper cable of a similar size, because it uses light of very high frequency to carry the signal. Fibre also suffers much lower attenuation (signal loss) over distance than copper cable, meaning fewer signal-boosting repeaters are needed along a long-distance link, reducing cost and complexity. Because the signal in a fibre-optic cable is carried by light rather than by an electric current, fibre-optic cable is immune to electromagnetic interference from nearby electrical equipment or cables, unlike copper, and it is also more secure, since tapping into a fibre-optic signal without detection is far more difficult than tapping a copper cable. These advantages must be weighed against fibre's greater fragility (glass fibres can crack if bent too sharply) and the more specialised, costly equipment and skill needed to install, splice and terminate fibre-optic cable compared with copper.

A good answer draws an explicit overall evaluation (e.g. that despite higher installation cost, fibre-optic cable is now generally preferred for high-bandwidth, long-distance and security-sensitive communication links) using accurate terminology and correct spelling, punctuation and grammar throughout.

Excellent (7-9 marks): Detailed, accurate and well-structured discussion correctly explaining total internal reflection and fibre structure, clearly comparing single-mode and multi-mode fibres with correct reference to modal dispersion, and evaluating at least two advantages of fibre over copper with a balancing limitation; addresses all three bulleted prompts; high standard of SPaG with accurate technical terminology throughout.
Good / Very Good (4-6 marks): Reasonable coverage of most of the three areas but with some omissions, imbalance, or limited evaluative depth (e.g. advantages listed but not clearly evaluated against a limitation); mostly accurate; generally sound SPaG with minor lapses.
Basic (1-3 marks): Basic, list-like or fragmentary answer; only one area covered in any detail, or significant scientific inaccuracies; weak organisation and/or noticeably poor SPaG.
0 marks: No relevant content, or entirely incorrect. [9]

(a) A DNA nucleotide is made up of three components: a deoxyribose sugar, a phosphate group, and one of four nitrogenous bases (adenine, thymine, cytosine or guanine).
(b) Specific base pairing means that, across the two strands of the double helix, adenine (A) always pairs with thymine (T), and cytosine (C) always pairs with guanine (G), held together by hydrogen bonds between the paired bases.

(a) 1 mark: deoxyribose (sugar); 1 mark: phosphate and a nitrogenous base both correctly named. (b) 1 mark: A pairs with T; 1 mark: C pairs with G; [4]

(a) In semi-conservative replication, each of the two new DNA double helices produced is made up of one original (parental) strand, conserved from the original molecule, and one newly synthesised complementary strand; this was demonstrated experimentally by Meselson and Stahl.
(b) DNA helicase is the enzyme responsible for unwinding the DNA double helix and breaking the hydrogen bonds between complementary base pairs on the two strands, separating them to expose single-stranded template DNA to which new, complementary nucleotides can then bind.

(a) 1 mark: each new molecule has one parental and one new strand; 1 mark: correctly attributed to Meselson and Stahl's model/described as semi-conservative. (b) 1 mark: DNA helicase unwinds the double helix; 1 mark: correctly explains it breaks hydrogen bonds between base pairs, exposing template strands; [4]

Meiosis introduces genetic variation through two main mechanisms. First, independent segregation (assortment) of homologous chromosomes occurs during meiosis I: the maternal and paternal chromosomes of each homologous pair are distributed randomly and independently into the daughter cells, so that different gametes end up with different combinations of maternal and paternal chromosomes. Second, genetic recombination by crossing over occurs when homologous chromosomes pair up during meiosis I and exchange corresponding sections of DNA at points called chiasmata; this creates chromosomes carrying new combinations of alleles that were not present on either original parental chromosome, further increasing the genetic diversity of the resulting gametes.

1 mark: independent segregation of homologous chromosomes named; 1 mark: correctly explained as producing different maternal/paternal chromosome combinations; 1 mark: crossing over/genetic recombination named; 1 mark: correctly explained as exchange of DNA sections between homologous chromosomes creating new allele combinations; [4]

(a) The heterozygous parent (Pp) produces gametes P and p in equal proportions; the homozygous recessive parent (pp) produces only gametes p. The Punnett square (P p x p p) gives offspring genotypes: Pp, Pp, pp, pp — i.e. two boxes of Pp and two boxes of pp.
(b) Since P is dominant, Pp offspring are purple-flowered and pp offspring are white-flowered. The genotype ratio 2 Pp : 2 pp simplifies to 1:1, giving a phenotypic ratio of 1 purple : 1 white.

(a) 1 mark: correct gametes identified for each parent (P, p from heterozygote; p, p from homozygous recessive); 1 mark: correctly constructed 2x2 Punnett square; 1 mark: correct offspring genotypes shown (Pp, Pp, pp, pp). (b) 1 mark: genotype ratio 1 Pp : 1 pp correctly stated; 1 mark: phenotypic ratio 1 purple : 1 white correctly deduced; [5]

(a) The heterozygous female parent (\( X^R X^r \)) produces gametes \( X^R \) and \( X^r \) in equal proportions; the male parent (\( X^R Y \)) produces gametes \( X^R \) and Y in equal proportions. Combining these gives four equally likely offspring genotypes: \( X^R X^R \) (red-eyed female), \( X^R X^r \) (red-eyed female, carrier), \( X^R Y \) (red-eyed male), and \( X^r Y \) (white-eyed male).
(b) All female offspring inherit at least one \( X^R \) allele from their father, so all females are red-eyed. Among the male offspring, half are \( X^R Y \) (red-eyed) and half are \( X^r Y \) (white-eyed), giving a 1:1 ratio of red-eyed to white-eyed males; overall, white eye colour is only seen in males, illustrating the characteristic pattern of X-linked inheritance.

(a) 1 mark: correct gametes identified for both parents; 1 mark: correctly constructed genetic diagram/cross; 1 mark: all four offspring genotypes correctly shown. (b) 1 mark: all female offspring correctly identified as red-eyed; 1 mark: male offspring correctly identified as 1:1 red:white eyed, showing the X-linked pattern; [5]

(a) The genotype of an organism is its genetic constitution — the specific combination of alleles it carries at a locus (for example, Tt for stem height in pea plants). The phenotype is the observable characteristic that results from this genotype, in combination with any environmental influence (for example, being a tall plant). A given genotype (e.g. Tt) produces a particular phenotype (tall, since T is dominant).
(b) A homozygous genotype has two identical alleles at a given locus (for example, TT or tt), while a heterozygous genotype has two different alleles at that locus (for example, Tt).

(a) 1 mark: genotype correctly defined as the genetic constitution/allele combination; 1 mark: phenotype correctly defined as the observable/expressed characteristic (interaction with environment). (b) 1 mark: homozygous correctly defined as two identical alleles at a locus; 1 mark: heterozygous correctly defined as two different alleles at a locus; [4]

(a) The father (\( I^A I^B \)) produces gametes \( I^A \) and \( I^B \) in equal proportions; the mother (\( I^O I^O \)) produces only \( I^O \) gametes. Combining these gives offspring genotypes \( I^A I^O \) and \( I^B I^O \), in equal (1:1) proportions.
(b) Because \( I^A \) and \( I^B \) are both dominant to \( I^O \), an \( I^A I^O \) genotype gives blood group A, and an \( I^B I^O \) genotype gives blood group B. The children are therefore expected to be blood group A or blood group B, in a 1:1 ratio (no child can be group AB or group O from this particular cross).

(a) 1 mark: correct gametes identified for each parent; 1 mark: correctly constructed genetic diagram; 1 mark: offspring genotypes IAIO and IBIO correctly shown. (b) 1 mark: phenotypes correctly identified as group A and group B; 1 mark: correct 1:1 ratio stated; [5]

(a) Insulin extracted from cattle or pigs has a slightly different amino acid sequence from human insulin, because it is produced from a different species' insulin gene, and this difference can cause adverse immune reactions in some patients.
(b) To produce Humulin, the human gene coding for insulin is isolated and inserted into a bacterium, such as E. coli, using genetic engineering (recombinant DNA) techniques. The genetically modified bacteria are then cultured on a large scale in fermenters; as they grow and divide, they express the inserted human gene and synthesise human insulin, identical in sequence to that made naturally in the human pancreas, which is then extracted, purified and formulated for clinical use.

(a) 1 mark: correctly identifies animal insulin has a different amino acid sequence from human insulin. (b) 1 mark: human insulin gene inserted into a bacterium (e.g. E. coli) using genetic engineering; 1 mark: modified bacteria cultured on a large scale (fermenter) and express the gene; 1 mark: insulin produced is extracted/purified, identical to natural human insulin; [4]

One advantage is that insulin produced by genetically modified bacteria is identical in structure to naturally occurring human insulin, since it is made directly from the human insulin gene; this results in fewer adverse (immune) reactions in patients compared with animal-derived insulin, which differs slightly in amino acid sequence. A second advantage is that genetically modified bacteria can be grown rapidly, in very large quantities, in fermenters under controlled conditions, allowing much larger amounts of insulin to be produced at a lower production cost than the difficult and limited process of extracting insulin from animal pancreases; producing insulin this way also avoids ethical or religious objections some patients may have to using an animal-derived product.

1 mark: fewer adverse/immune reactions identified as an advantage; 1 mark: correctly explained by chemical identity to human insulin; 1 mark: larger quantities/lower production cost identified as a second advantage; 1 mark: correctly explained by ease of large-scale bacterial culture (accept fewer ethical/religious objections as an alternative second point); [4]

(a) Gene therapy is the technique of introducing genetic material (a functioning copy of a gene) into a patient's cells, in order to treat or prevent a disease that is caused by a defective or missing gene.
(b) In cystic fibrosis, a defective gene (CFTR) results in abnormal ion transport across cell membranes, causing unusually thick and sticky mucus to build up, particularly in the lungs; this obstructs the airways, impairs normal lung function, and creates conditions that favour recurrent, chronic lung infections.

(a) 1 mark: gene therapy correctly defined as introducing genetic material into cells; 1 mark: correctly links this to treating/preventing disease caused by a defective gene. (b) 1 mark: identifies abnormally thick/sticky mucus builds up in the lungs; 1 mark: correctly links this to airway obstruction/susceptibility to chronic lung infection; [4]

Gene therapy can act in three distinct ways. First, by repairing the defective gene itself, correcting the specific fault within the gene's DNA sequence. Second, by replacing the faulty gene entirely with a normal, functioning copy, so the defective version is removed and substituted. Third, by adding (supplementing with) a normal copy of the gene into the patient's cells, while leaving the original defective gene still in position; the added normal gene then compensates for the defective one by providing a working copy for the cell to use.

1 mark each for correctly identifying and briefly describing each of: repairing the defective gene; replacing the faulty gene with a normal copy; adding a normal gene while leaving the defective gene in place; plus 1 mark for all three being clearly and distinctly presented; [4]

(a) The human genome (and that of other organisms) contains many regions of repetitive, non-coding DNA, in which a short base sequence is repeated a variable number of times. The exact number of repeats at each of these regions varies considerably between individuals. Because there are many such variable regions across the genome, the probability that two unrelated individuals share the identical combination of repeat numbers at all of the regions analysed is extremely low; this means the overall pattern obtained (the genetic fingerprint) is, for practical purposes, unique to each individual (other than identical twins), allowing individuals to be distinguished or compared using this pattern.
(b) Besides forensic science (identifying suspects or victims from DNA evidence), genetic fingerprinting is also used in medical diagnosis (for example, identifying genetic disease markers) and in determining genetic relationships, such as paternity testing, as well as in assessing genetic variability within populations of animals or plants for breeding programmes.

(a) 1 mark: genome contains repetitive, non-coding sequences that vary in repeat number between individuals; 1 mark: correctly explains the probability of two individuals sharing the same pattern at multiple regions is very low; 1 mark: correctly concludes the resulting fingerprint pattern is unique to an individual. (b) 1 mark each for two valid fields other than forensic science, e.g. medical diagnosis, paternity/relationship testing, animal or plant breeding; [5]

(a) A similarity is that both embryonic and adult stem cells are unspecialised cells that are capable of self-renewal (dividing repeatedly to produce more stem cells) and of differentiating into other, more specialised cell types. A difference is that embryonic stem cells are pluripotent — able to differentiate into almost any of the cell types found in the body — whereas adult stem cells are generally multipotent, restricted to differentiating into a narrower range of cell types related to the tissue in which they are found (for example, bone marrow stem cells mainly differentiate into blood cell types).
(b) Because embryonic stem cells are pluripotent, they have the potential to be directed to differentiate into almost any required cell or tissue type, giving them a much broader range of potential medical applications (for example, generating replacement cells for many different damaged tissues or organs) than adult stem cells, whose more limited (multipotent) differentiation capability restricts their use mainly to regenerating the specific tissue types related to their tissue of origin.

(a) 1 mark: valid similarity given (self-renewal/differentiation capability, both unspecialised); 1 mark: valid difference given (pluripotent vs multipotent). (b) 1 mark: correctly explains embryonic stem cells can differentiate into almost any cell type (pluripotent); 1 mark: correctly links this to a broader range of potential medical applications than the more restricted adult stem cells; [4]

One ethical concern is uncertainty about the long-term effects of genetically modified organisms on human health and on the environment; for example, there is concern that genes engineered into a crop plant could transfer to wild relative species (gene flow), with unpredictable ecological consequences, or that long-term human health effects of consuming GM products are not yet fully understood. A second, distinct ethical concern relates to the commercial patenting and ownership of genetically modified organisms and the genes used to create them; opponents argue this can concentrate control over agriculture and the food supply in the hands of a small number of corporations, potentially restricting the ability of farmers, particularly in developing countries, to freely save and replant seed, raising questions of fairness and access.

1 mark: first distinct ethical concern identified (e.g. unknown long-term health/environmental effects, gene flow to wild populations); 1 mark: correctly explained/exemplified; 1 mark: second distinct ethical concern identified (e.g. patenting/commercial ownership, corporate control, access for developing-world farmers); 1 mark: correctly explained/exemplified; [4]

(a) A Tt x Tt cross gives an expected phenotypic ratio of 3 tall : 1 short. Expected tall \( = \dfrac{3}{4} \times 160 = 120 \); expected short \( = \dfrac{1}{4} \times 160 = 40 \).
(b) \( \chi^2 = \dfrac{(O-E)^2}{E}(\text{tall}) + \dfrac{(O-E)^2}{E}(\text{short}) = \dfrac{(132-120)^2}{120} + \dfrac{(28-40)^2}{40} = \dfrac{144}{120} + \dfrac{144}{40} = 1.20 + 3.60 = 4.80 \).
Check (second route): total observed 132+28 = 160 = total expected 120+40 ✓, confirming no arithmetic slip in setting up O and E; recomputing each term independently, \( 144/120 = 1.2 \) and \( 144/40 = 3.6 \), sum 4.8, consistent.
(c) With 2 phenotypic classes, degrees of freedom = 2 − 1 = 1, matching the given critical value of 3.84. Since the calculated \( \chi^2 \) value of 4.80 is greater than the critical value of 3.84, the null hypothesis (that the observed data fit the expected 3:1 ratio) is rejected: there is a statistically significant difference between the observed and expected results at the 5% significance level, meaning the observed ratio deviates significantly from a true 3:1 ratio.

(a) 1 mark: 3:1 ratio stated; 1 mark: expected values 120 and 40 correctly calculated. (b) 1 mark: (O-E)^2/E calculated correctly for tall (1.20); 1 mark: (O-E)^2/E calculated correctly for short (3.60); 1 mark: correct summation method; 1 mark: chi-squared = 4.80. (c) 1 mark: correctly compares 4.80 to 3.84 (4.80 > 3.84); 1 mark: correct conclusion that the null hypothesis is rejected/significant deviation from 3:1 at 5% level; [8]

(a) Total offspring = 320. Expected round-yellow \( = \dfrac{9}{16}\times320 = 180 \); round-green \( = \dfrac{3}{16}\times320 = 60 \); wrinkled-yellow \( = \dfrac{3}{16}\times320 = 60 \); wrinkled-green \( = \dfrac{1}{16}\times320 = 20 \).
(b) \( \chi^2 = \dfrac{(189-180)^2}{180} + \dfrac{(61-60)^2}{60} + \dfrac{(59-60)^2}{60} + \dfrac{(11-20)^2}{20} \)
\( = \dfrac{81}{180} + \dfrac{1}{60} + \dfrac{1}{60} + \dfrac{81}{20} = 0.450 + 0.0167 + 0.0167 + 4.050 = 4.53 \).
Check (second route): total observed \( = 189+61+59+11 = 320 \), equal to total expected \( 180+60+60+20 = 320 \) ✓, confirming the categories were set up correctly before computing chi-squared.
(c) Degrees of freedom = number of classes − 1 = 4 − 1 = 3, matching the given critical value of 7.81. Since the calculated \( \chi^2 \) value of 4.53 is less than the critical value of 7.81, the null hypothesis is accepted: there is no statistically significant difference between the observed and expected results at the 5% significance level, so the data are consistent with the expected 9:3:3:1 dihybrid ratio.

(a) 1 mark: all four expected values (180, 60, 60, 20) correctly calculated. (b) 1 mark: (O-E)^2/E correctly calculated for round-yellow and round-green; 1 mark: (O-E)^2/E correctly calculated for wrinkled-yellow and wrinkled-green; 1 mark: correct summation method shown; 1 mark: chi-squared = 4.53 (accept 4.5-4.54). (c) 1 mark: degrees of freedom = 3 correctly stated; 1 mark: correct conclusion that null hypothesis is accepted, as 4.53 < 7.81; [8]

(a) Since the unknown fragment's migration distance (24 mm) lies between the migration distances of the 500 bp marker (30 mm) and the 1000 bp marker (20 mm), its size can be estimated by comparing where it falls between these two known reference points: larger fragments migrate a shorter distance (they move more slowly through the gel matrix), so as migration distance decreases from 30 mm to 20 mm, fragment size increases from 500 bp to 1000 bp; the unknown fragment's size can be estimated by proportionally interpolating between these two calibration values according to where its migration distance (24 mm) falls within this range.
(b) Using linear interpolation between (30 mm, 500 bp) and (20 mm, 1000 bp):
Fraction of the way from 30 mm to 20 mm \( = \dfrac{24-30}{20-30} = \dfrac{-6}{-10} = 0.6 \).
Estimated size \( = 500 + 0.6 \times (1000-500) = 500 + 0.6\times500 = 500+300 = 800 \) bp.
Check (second route, using the rate of change directly): rate \( = \dfrac{1000-500}{20-30} = \dfrac{500}{-10} = -50 \) bp per mm; size at 24 mm \( = 500 + (-50)\times(24-30) = 500 + (-50)\times(-6) = 500+300 = 800 \) bp, consistent with the fraction method. Final answer: approximately 800 base pairs.

(a) 1 mark: correctly identifies that larger fragments migrate a shorter distance (inverse relationship); 1 mark: recognises the unknown's migration distance lies between the two given calibration points; 1 mark: describes using proportional/interpolated comparison between the calibration points to estimate size. (b) 1 mark: correct fraction (0.6) calculated; 1 mark: correct application of interpolation formula; 1 mark: correct working shown; 1 mark: 800 bp; [7]

(a) Each Ff parent produces gametes F and f in equal proportions. Combining these in a Punnett square gives four equally likely offspring genotypes: FF, Ff, Ff, ff — i.e. a 1 FF : 2 Ff : 1 ff genotype ratio.
(b)(i) Only the ff genotype results in cystic fibrosis: probability \( = \dfrac{1}{4} = 25\% \).
(ii) The Ff genotype is an unaffected carrier: probability \( = \dfrac{2}{4} = \dfrac{1}{2} = 50\% \).
(iii) The FF genotype is neither affected nor a carrier: probability \( = \dfrac{1}{4} = 25\% \).
Check (second route): the three probabilities must sum to 1 (100%) since they cover all possible outcomes: \( 25\% + 50\% + 25\% = 100\% \) ✓, confirming the probabilities are correctly assigned and complete.
(c) This probability information (a 1 in 4, or 25%, chance that any given child will have cystic fibrosis) can be given to the couple during genetic counselling, allowing them to make informed decisions about family planning, such as considering prenatal genetic testing during a pregnancy, or being aware of the risk before deciding to conceive.

(a) 1 mark: correct gametes F, f identified for each parent; 1 mark: correctly constructed 2x2 Punnett square; 1 mark: correct offspring genotypes (FF, Ff, Ff, ff) shown. (b) 1 mark: (i) 25% (1 in 4) correct; 1 mark: (ii) correct method (2/4 identified); 1 mark: (ii) 50% (1 in 2) correct; 1 mark: (iii) 25% (1 in 4) correct. (c) 1 mark: correctly explains use in genetic counselling/informed reproductive decision-making (e.g. prenatal testing); [8]

A full-mark answer should discuss, in a well-organised and coherent way:

What gene therapy is: Gene therapy is the introduction of genetic material into a patient's cells to treat or prevent a disease caused by a defective gene. It can act in one of three ways: by repairing the defective gene itself; by replacing the faulty gene entirely with a normal, functioning copy; or by adding (supplementing with) a normal copy of the gene while leaving the original defective gene in place, so that the added gene compensates for the faulty one.

Application to cystic fibrosis: Cystic fibrosis is caused by a defective gene (CFTR), which disrupts normal ion transport across cell membranes; this causes abnormally thick, sticky mucus to accumulate, particularly in the lungs, obstructing the airways, impairing lung function and predisposing patients to recurrent chronic lung infections, as well as affecting the digestive system. Initial attempts at gene therapy for cystic fibrosis focused on delivering a normal, functioning copy of the CFTR gene directly to the epithelial cells lining the lungs — for example, by inhaling the gene packaged inside a modified virus or a liposome (fat droplet) as a delivery vector — aiming to supplement the defective gene and restore normal ion transport. Early clinical trials achieved only limited and short-lived success, in part because it proved difficult to deliver the gene efficiently to enough cells and to sustain long-term expression, and because the body's immune response to repeated use of viral vectors reduced their effectiveness over time; research into more effective delivery methods continues.

Ethical evaluation — somatic vs germ cell therapy: Somatic cell therapy modifies the genes of ordinary (non-reproductive) body cells, such as lung epithelial cells in the cystic fibrosis example above; because these changes affect only the cells of the treated individual and are not passed on to their children, somatic cell therapy is generally considered less ethically contentious, though questions of safety, long-term effectiveness and equitable access to an expensive treatment remain. Germ cell (sperm, egg, or early embryo) therapy, by contrast, would introduce heritable changes that are passed on to all future generations descended from the treated individual; this raises much greater ethical concern, because those future generations cannot consent to the genetic change made on their behalf, the change would permanently alter the human gene pool in ways that are difficult to reverse or fully predict, and there is concern that the same techniques could be misused for non-therapeutic genetic 'enhancement' rather than treating disease (a 'slippery slope' argument). For these reasons, germ cell gene therapy is currently prohibited or very tightly restricted in clinical practice in many countries, including the UK, while somatic cell gene therapy for serious genetic diseases is more widely accepted and has progressed to licensed clinical treatments for some conditions.

A good answer weighs these ethical considerations against the genuine medical benefit gene therapy could offer patients with serious inherited conditions, using accurate terminology and correct spelling, punctuation and grammar throughout.

Excellent (7-9 marks): Detailed, accurate and well-structured discussion correctly explaining all three mechanisms of gene therapy, the cause of cystic fibrosis and how gene therapy has been applied/attempted to treat it, and a genuinely evaluative comparison of the ethical issues raised by somatic versus germ cell therapy; addresses all three bulleted prompts; high standard of SPaG with accurate scientific terminology throughout.
Good / Very Good (4-6 marks): Reasonable coverage of most of the three areas but with some omissions, imbalance, or limited evaluative depth on the ethical comparison; mostly accurate; generally sound SPaG with minor lapses.
Basic (1-3 marks): Basic, list-like or fragmentary answer; only one area covered in any detail, or significant scientific inaccuracies; weak organisation and/or noticeably poor SPaG.
0 marks: No relevant content, or entirely incorrect. [9]