2023 CCEA A-Level Life and Health Sciences 0008 Practice Paper with Answers

Structural isomers are compounds that share the same molecular formula but have different structural (arrangement of atoms) formulae. For example, \( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3 \) (butane) and \( (\text{CH}_3)_3\text{CH} \) (2-methylpropane) both have molecular formula \( \text{C}_4\text{H}_{10} \) but different carbon skeletons.

1 mark: same molecular formula but different structural arrangement/formula stated; 1 mark: valid example given with correct formulae (e.g. butane and 2-methylpropane, both \( \text{C}_4\text{H}_{10} \)); [2]

The Cl–Cl bond in \( \text{Cl}_2 \) must be broken homolytically to generate chlorine radicals. This requires ultraviolet light (or strong sunlight) to provide sufficient energy. The subsequent reaction of methane with chlorine radicals, replacing an H atom with a Cl atom, proceeds by a free-radical substitution mechanism.

1 mark: UV light/sunlight correctly stated as the essential condition; 1 mark: mechanism correctly named as free-radical substitution; [2]

Add a few drops of orange bromine water to separate samples of each compound and shake. Hex-1-ene contains a C=C double bond, which undergoes electrophilic addition with bromine, so the orange bromine water is decolourised (turns colourless). Hexane is saturated and contains no C=C bond, so no reaction occurs and the bromine water remains orange.

1 mark: bromine water (aqueous bromine) stated as reagent/test; 1 mark: correct observation given — decolourised with hex-1-ene, no change with hexane; [2]

The carbon atom bonded to the –OH group is itself bonded to three other carbon atoms (two ethyl groups and one methyl group) and no hydrogen atom. An alcohol is classified as tertiary when the carbinol carbon is attached to three alkyl/aryl groups, so 3-methylpentan-3-ol is a tertiary alcohol.

1 mark: tertiary correctly stated; 1 mark: correct reason — carbon bearing OH is bonded to three other carbon atoms; [2]

An addition polymer's repeat unit is derived by opening the C=C double bond of its monomer. Reversing this for \( (-\text{CH}_2-\text{CHCl}-)_n \) gives the monomer \( \text{CH}_2=\text{CHCl} \), which is chloroethene (also called vinyl chloride).

1 mark: correct monomer structure \( \text{CH}_2=\text{CHCl} \) deduced; 1 mark: correct name chloroethene (accept vinyl chloride); [2]

The broad absorption at 3300 \( \text{cm}^{-1} \) is characteristic of an O–H bond, broadened by hydrogen bonding, consistent with an alcohol group. The sharp absorption at 1715 \( \text{cm}^{-1} \) is characteristic of a C=O (carbonyl) stretch, indicating a carbonyl-containing group such as an aldehyde, ketone or ester.

1 mark: 3300 \( \text{cm}^{-1} \) correctly assigned to O–H (alcohol, hydrogen-bonded); 1 mark: 1715 \( \text{cm}^{-1} \) correctly assigned to C=O (carbonyl); [2]

The phenolic –OH group of salicylic acid is converted into an ester (ethanoyloxy) group by reaction with ethanoic anhydride (acetic anhydride), \( (\text{CH}_3\text{CO})_2\text{O} \), usually with a few drops of concentrated acid catalyst and gentle heating.

1 mark: ethanoic anhydride (accept acetic anhydride) correctly named; 1 mark: correct formula \( (\text{CH}_3\text{CO})_2\text{O} \); [2]

Nylon-6,6 is formed by condensation polymerisation between hexanedioic acid and hexane-1,6-diamine. At each amide linkage formed, an –OH from the acid and an –H from the amine are eliminated together as a molecule of water, \( \text{H}_2\text{O} \).

1 mark: condensation polymerisation correctly named; 1 mark: water correctly identified as the molecule eliminated; [2]

Numbering the longest chain from the carbon bearing the –OH group (to give it the lowest locant) gives a four-carbon chain (butan-1-ol) with a methyl branch on carbon 3, so the compound is 3-methylbutan-1-ol.

1 mark: correct parent chain identified as butan-1-ol; 1 mark: correct full name with locant, 3-methylbutan-1-ol; [2]

Complete combustion of a hydrocarbon in excess oxygen produces carbon dioxide and water only. Balancing carbon (3), hydrogen (8, giving 4 \( \text{H}_2\text{O} \)) and then oxygen (3×2 + 4×1 = 10, requiring 5 \( \text{O}_2 \)) gives \( \text{C}_3\text{H}_8(g) + 5\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(l) \).

1 mark: correct products \( 3\text{CO}_2 \) and \( 4\text{H}_2\text{O} \) with carbon and hydrogen balanced; 1 mark: \( 5\text{O}_2 \) correctly balanced and correct state symbols included; [2]

But-2-ene reacts with hydrogen gas, \( \text{H}_2 \), in the presence of a nickel catalyst (typically around 150–200°C), which adds across the C=C double bond to give butane. This is an addition reaction, specifically catalytic hydrogenation.

1 mark: \( \text{H}_2 \) with Ni catalyst (and heat) correctly stated; 1 mark: reaction correctly named as addition/hydrogenation; [2]

E/Z isomerism arises because rotation about the C=C double bond is restricted by the π bond, and each carbon of the double bond in but-2-ene is bonded to two different groups (H and \( \text{CH}_3 \)); this allows two distinct, non-interconvertible spatial arrangements. In Z-but-2-ene the two methyl groups lie on the same side of the double bond, giving the molecule a small net dipole moment and therefore additional dipole–dipole forces between molecules on top of London forces. E-but-2-ene is more symmetrical and effectively non-polar, relying only on (weaker) London forces. Consequently Z-but-2-ene has the higher boiling point.

1 mark: restricted rotation about the C=C π bond correctly explained; 1 mark: each double-bond carbon bears two different groups, allowing two distinct arrangements; 1 mark: Z-isomer correctly identified as having the higher boiling point with correct reason (net dipole/extra dipole–dipole forces vs symmetrical, non-polar E-isomer); [3]

The colour changes from orange to green. Acidified dichromate(VI) ions, \( \text{Cr}_2\text{O}_7^{2-} \) (orange, chromium in the +6 state), act as the oxidising agent and are themselves reduced to \( \text{Cr}^{3+} \) ions (green) as they oxidise the alcohol. Butan-2-ol is a secondary alcohol, so it is oxidised to a ketone, butan-2-one, \( \text{CH}_3\text{COCH}_2\text{CH}_3 \), which has no hydrogen on the carbonyl carbon and therefore cannot be oxidised further.

1 mark: colour change orange to green correctly stated; 1 mark: correct explanation — \( \text{Cr}_2\text{O}_7^{2-} \) (Cr(VI), orange) reduced to \( \text{Cr}^{3+} \) (green) as the alcohol is oxidised; 1 mark: product correctly identified as a ketone (butan-2-one), which resists further oxidation; [3]

LDPE chains are highly branched, so the chains cannot pack closely together; this reduces the area of contact between neighbouring chains, weakening the London (van der Waals) forces between them, giving a flexible polymer with lower density and a lower melting point. HDPE chains are largely linear (unbranched) and can pack closely together in a more ordered, more crystalline arrangement, increasing the area of contact and therefore the strength of the London forces between chains, giving a denser, more rigid polymer with a higher melting point.

1 mark: LDPE chains correctly described as highly branched, unable to pack closely; 1 mark: HDPE chains correctly described as linear/unbranched, packing closely in a more ordered structure; 1 mark: correct link to intermolecular (London/van der Waals) forces — weaker in LDPE (flexible, lower m.p.), stronger in HDPE (rigid, higher m.p.); [3]

The molecular ion peak, formed when a molecule loses one electron in the mass spectrometer without fragmenting, appears at a mass-to-charge ratio equal to the relative molecular mass of the compound, so \( M_r = 46 \), consistent with ethanol, \( \text{CH}_3\text{CH}_2\text{OH} \). The peak at m/z = 31 corresponds to loss of a \( \text{CH}_3 \) radical (mass 15) from the molecular ion (46 − 15 = 31), giving the fragment ion \( \text{CH}_2=\overset{+}{\text{O}}\text{H} \) (protonated formaldehyde / \( \text{CH}_2\text{OH}^+ \)).

1 mark: molecular ion correctly explained as giving \( M_r \) of the compound (= 46); 1 mark: loss of \( \text{CH}_3 \) (mass 15) correctly identified from 46 − 31; 1 mark: fragment at m/z = 31 correctly suggested as \( \text{CH}_2\text{OH}^+ \) (accept equivalent formula); accept ECF from candidate's identified compound; [3]

Crude aspirin contains impurities such as unreacted salicylic acid and ethanoic acid by-product. Recrystallisation removes these: the crude solid is dissolved in a minimum volume of hot solvent, the hot solution is filtered to remove any insoluble impurities, then allowed to cool slowly so that pure aspirin crystallises out (soluble impurities remain dissolved in the mother liquor), and the crystals are collected by vacuum filtration. Pure aspirin has a known, sharp melting point of approximately 135°C. An impure sample melts over a wider temperature range and at a lower temperature than the pure compound, so comparing the observed melting point (and the sharpness of its range) with the literature value indicates the purity of the product.

1 mark: recrystallisation correctly explained — dissolve in minimum hot solvent, cool slowly so pure crystals form, impurities remain in solution; 1 mark: pure aspirin correctly stated to have a sharp, known melting point (accept ~135°C); 1 mark: correct explanation that an impure sample melts over a wider range/at a lower temperature, allowing purity to be assessed by comparison with the literature value; [3]

Nylon-6,6 chains are linked by amide, \( -\text{CONH}- \), groups. The N–H of one chain can hydrogen bond to the C=O of an adjacent chain, so extensive hydrogen bonding occurs between neighbouring polymer chains. Poly(ethene) chains are non-polar hydrocarbon chains with no such polar groups, so the only intermolecular forces between chains are (much weaker) London/van der Waals forces. Because hydrogen bonds are considerably stronger than London forces, nylon-6,6 has much greater tensile strength than poly(ethene).

1 mark: nylon chains correctly described as linked by amide (–CONH–) groups capable of hydrogen bonding between chains; 1 mark: poly(ethene) chains correctly described as non-polar, held together only by (weaker) London/van der Waals forces; 1 mark: correct conclusion linking stronger hydrogen bonding in nylon to its greater tensile strength compared with poly(ethene); [3]

A and B show functional group isomerism, since they have the same molecular formula but belong to different homologous series (different functional groups). Propanal, \( \text{CH}_3\text{CH}_2\text{CHO} \), has its carbonyl carbon at the end of the carbon chain, bonded to one hydrogen atom and one alkyl group, making it an aldehyde. Propanone, \( \text{CH}_3\text{COCH}_3 \), has its carbonyl carbon within the chain, bonded to two alkyl groups and no hydrogen, making it a ketone.

1 mark: functional group isomerism correctly named; 1 mark: propanal correctly described as an aldehyde — carbonyl carbon at chain end, bonded to one H and one alkyl group; 1 mark: propanone correctly described as a ketone — carbonyl carbon within the chain, bonded to two alkyl groups; [3]

Initiation: \( \text{Cl}_2 \xrightarrow{\text{UV light}} 2\text{Cl}^{\bullet} \). Ultraviolet light provides enough energy to break the Cl–Cl bond homolytically, each chlorine atom keeping one electron of the shared pair, generating two chlorine radicals.

Propagation (chain-carrying steps, each regenerating a radical):
\( \text{Cl}^{\bullet} + \text{C}_2\text{H}_6 \rightarrow \text{C}_2\text{H}_5^{\bullet} + \text{HCl} \)
\( \text{C}_2\text{H}_5^{\bullet} + \text{Cl}_2 \rightarrow \text{C}_2\text{H}_5\text{Cl} + \text{Cl}^{\bullet} \)
In the first step a chlorine radical abstracts a hydrogen atom from ethane, forming HCl and an ethyl radical; in the second, the ethyl radical reacts with a chlorine molecule to form chloroethane and regenerate a chlorine radical, which can continue the chain.

Termination: two radicals combine to form a stable molecule, ending the chain, e.g. \( 2\text{Cl}^{\bullet} \rightarrow \text{Cl}_2 \), or \( \text{C}_2\text{H}_5^{\bullet} + \text{Cl}^{\bullet} \rightarrow \text{C}_2\text{H}_5\text{Cl} \), or \( 2\text{C}_2\text{H}_5^{\bullet} \rightarrow \text{C}_4\text{H}_{10} \).

1 mark: correct initiation equation \( \text{Cl}_2 \rightarrow 2\text{Cl}^{\bullet} \) with homolytic fission under UV light explained; 1 mark: correct first propagation step \( \text{Cl}^{\bullet} + \text{C}_2\text{H}_6 \rightarrow \text{C}_2\text{H}_5^{\bullet} + \text{HCl} \); 1 mark: correct second propagation step \( \text{C}_2\text{H}_5^{\bullet} + \text{Cl}_2 \rightarrow \text{C}_2\text{H}_5\text{Cl} + \text{Cl}^{\bullet} \); 1 mark: valid termination equation given (any two radicals combining to a stable molecule); 1 mark: correct explanation that propagation regenerates a radical and continues the chain while termination consumes two radicals; [5]

The C=C double bond of but-1-ene is a region of high electron density and acts as a nucleophile, attacking the slightly positive hydrogen atom of the polar H–Br molecule; the H–Br bond breaks heterolytically, with both electrons going to the bromine, releasing \( \text{Br}^- \) and generating a carbocation intermediate on the double-bond carbon that did not bond to H.

Protonation at C1 gives a secondary carbocation at C2: \( \text{CH}_3\text{CH}_2\overset{+}{\text{C}}\text{HCH}_3 \). Protonation at C2 gives a primary carbocation at C1: \( \overset{+}{\text{C}}\text{H}_2\text{CH}_2\text{CH}_2\text{CH}_3 \).

The secondary carbocation is more stable than the primary carbocation because the two alkyl (electron-donating) groups attached to the positively charged carbon disperse the positive charge by their positive inductive effect, whereas the primary carbocation has only one such group. The reaction therefore proceeds preferentially via the more stable secondary carbocation, which is then attacked by the \( \text{Br}^- \) nucleophile (a lone pair on Br⁻ forms the new C–Br bond) to give 2-bromobutane as the major product; only a minor amount of 1-bromobutane forms via the less-favoured primary carbocation route.

1 mark: C=C correctly described as electron-rich, attacking the \( \delta+ \) hydrogen of HBr; 1 mark: heterolytic fission of H–Br correctly shown, giving \( \text{Br}^- \) and a carbocation intermediate; 1 mark: both possible carbocations correctly identified (secondary at C2 and primary at C1); 1 mark: secondary carbocation correctly explained as more stable due to the electron-donating inductive effect of two alkyl groups; 1 mark: correct conclusion that the more stable carbocation route dominates, giving 2-bromobutane as the major product (Markovnikov addition); [5]

Equation: \( \text{C}_6\text{H}_4(\text{OH})\text{COOH} + (\text{CH}_3\text{CO})_2\text{O} \rightarrow \text{C}_6\text{H}_4(\text{OCOCH}_3)\text{COOH} + \text{CH}_3\text{COOH} \)

This is an esterification reaction, proceeding by nucleophilic acyl substitution (a condensation reaction, since a small molecule, ethanoic acid, is eliminated). Mechanistically, a lone pair on the oxygen of the phenolic –OH group of salicylic acid attacks the electrophilic carbonyl carbon of one of the C=O groups in ethanoic anhydride (the carbonyl carbon is electrophilic due to the electronegative oxygen atoms withdrawing electron density). This forms a tetrahedral intermediate, which collapses, breaking a C–O bond within the anhydride and expelling an ethanoate ion as the leaving group; this ethanoate ion is then protonated (loses to/gains H⁺ in solution) to form ethanoic acid. A new ester (ethanoyloxy) linkage is left at the former phenolic position of the ring, while the –COOH group of salicylic acid takes no part in the reaction and remains unchanged in the aspirin product. The second organic product formed is therefore ethanoic acid, \( \text{CH}_3\text{COOH} \).

1 mark: correct balanced equation/formulae for salicylic acid + ethanoic anhydride → aspirin + ethanoic acid; 1 mark: reaction correctly named as esterification/nucleophilic acyl substitution (condensation); 1 mark: mechanism correctly described — nucleophilic O of the phenolic OH attacks the electrophilic carbonyl carbon of the anhydride; 1 mark: ethanoate leaving group correctly linked to formation of ethanoic acid as the second product; 1 mark: correct recognition that only the phenolic OH reacts, the –COOH of salicylic acid is unchanged in aspirin; [5]

Repeat unit: \( (-\text{HN}(\text{CH}_2)_6\text{NHCO}(\text{CH}_2)_8\text{CO}-)_n \).

The new bond formed at each linkage is an amide linkage (analogous to a peptide bond). Mechanistically, the lone pair on the nitrogen atom of an –NH2 group of the diamine acts as a nucleophile and attacks the electrophilic carbonyl carbon of an –COCl group of the diacyl dichloride (the carbonyl carbon is strongly electrophilic due to the electronegative chlorine and oxygen atoms). This forms a tetrahedral intermediate that collapses, expelling chloride ion, \( \text{Cl}^- \); the resulting protonated amide nitrogen then loses a proton (to \( \text{Cl}^- \), forming HCl) to give the neutral amide linkage, \( -\text{CONH}- \). This condensation occurs at both ends of every monomer unit, building up the long polyamide chain and releasing one molecule of hydrogen chloride, HCl, for every linkage formed.

1 mark: correct repeat unit showing amide linkages with correct chain lengths, \( (\text{CH}_2)_6 \) and \( (\text{CH}_2)_8 \); 1 mark: linkage correctly named as amide; 1 mark: HCl correctly identified as the by-product/small molecule eliminated; 1 mark: correct description of the amine nitrogen lone pair attacking the electrophilic carbonyl carbon of the acyl chloride; 1 mark: correct description of chain build-up via repeated condensation at both ends of each monomer; [5]

Propan-1-ol is a primary alcohol; with a limited/controlled amount of oxidising agent it is oxidised in a single step to the aldehyde, propanal: \( \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} + [\text{O}] \rightarrow \text{CH}_3\text{CH}_2\text{CHO} + \text{H}_2\text{O} \).

To isolate propanal, the reaction mixture is gently heated with the apparatus set up for distillation (not reflux). Propanal has a lower boiling point (~49°C) than propan-1-ol and a much lower boiling point than propanoic acid, so as soon as it forms it evaporates and is distilled off, removing it from further contact with the oxidising agent before it can be oxidised further.

If the reaction is instead heated under reflux (using a condenser to return all vapours to the flask) with excess oxidising agent, the aldehyde remains in contact with the dichromate(VI) and is oxidised further to the carboxylic acid: \( \text{CH}_3\text{CH}_2\text{CHO} + [\text{O}] \rightarrow \text{CH}_3\text{CH}_2\text{COOH} \). The final product structural formula is therefore \( \text{CH}_3\text{CH}_2\text{COOH} \) (propanoic acid).

1 mark: correct first oxidation equation, propan-1-ol → propanal using [O], with water as by-product; 1 mark: distillation apparatus correctly linked to the aldehyde's lower boiling point, distilling off as formed and removed from the oxidising agent; 1 mark: correct explanation that reflux (vapours condensed and returned) would instead allow full oxidation; 1 mark: correct second oxidation equation, propanal → propanoic acid; 1 mark: correct final structural formula \( \text{CH}_3\text{CH}_2\text{COOH} \) given; [5]

Repeat unit: \( (-\text{CH}_2-\text{CH}(\text{CH}_3)-)_n \), i.e. a two-carbon backbone unit bearing one hydrogen and one methyl substituent on alternating carbons, with a single bond extending from each end of the unit to join the neighbouring repeat units.

No by-product is formed because addition polymerisation involves only the breaking of the π component of each monomer's C=C double bond; the electrons from the broken π bonds are used to form new C–C σ bonds directly between adjacent monomer units. No atoms are substituted or lost from the monomer in this process, so every atom present in the propene monomers is retained in the polymer chain — the polymer's empirical formula is identical to that of the monomer, giving 100% atom economy and no by-product.

1 mark: correct repeat unit \( -[\text{CH}_2-\text{CH}(\text{CH}_3)]- \) shown with bonds extending from both ends; 1 mark: methyl group correctly placed on alternate carbon, consistent with propene's structure; 1 mark: correct explanation that only the π bond of C=C opens to form new σ bonds between monomers; 1 mark: correct conclusion that all monomer atoms are retained in the polymer (100% atom economy), hence no by-product; [4]

Moles of salicylic acid \( = \dfrac{2.50}{138} = 0.01812 \text{ mol} \).

Salicylic acid and aspirin react in a 1:1 mole ratio, so theoretical (maximum possible) moles of aspirin \( = 0.01812 \text{ mol} \).

Theoretical mass of aspirin \( = 0.01812 \times 180 = 3.26 \text{ g} \).

Percentage yield \( = \dfrac{\text{actual mass}}{\text{theoretical mass}} \times 100 = \dfrac{2.35}{3.26} \times 100 = 72.1\% \).

1 mark: moles of salicylic acid correctly calculated (2.50/138 = 0.0181 mol); 1 mark: correct 1:1 mole ratio between salicylic acid and aspirin identified; 1 mark: theoretical mass of aspirin correctly calculated (0.0181 × 180 = 3.26 g); 1 mark: percentage yield formula correctly applied (actual/theoretical × 100); 1 mark: correct final answer, 72.1% (accept 71.9–72.3% for rounding; ECF from candidate's theoretical mass); [5]

Total mass of reactants \( = 146 + 116 = 262 \).

Two condensation (amide-forming) linkages occur per repeat unit (one at each end of the monomer chain), each releasing one molecule of water, so total mass of by-product \( = 2 \times 18 = 36 \), which is consistent with the given repeat unit mass: \( 262 - 36 = 226 \).

Atom economy \( = \dfrac{\text{mass of desired product}}{\text{total mass of reactants}} \times 100 = \dfrac{226}{262} \times 100 = 86.3\% \).

1 mark: total mass of reactants correctly calculated (146 + 116 = 262); 1 mark: two molecules of water (36 in total) correctly identified as eliminated per repeat unit; 1 mark: mass of desired product correctly confirmed/used (226); 1 mark: atom economy formula correctly applied ((226/262) × 100); 1 mark: correct final answer, 86.3% (accept 86–87%; ECF from candidate's masses); [5]

Moles of ethanol \( = \dfrac{4.60}{46} = 0.100 \text{ mol} \).

Ethanol and ethyl ethanoate are in a 1:1 mole ratio, so theoretical mass of ethyl ethanoate \( = 0.100 \times 88 = 8.80 \text{ g} \).

Percentage yield \( = \dfrac{6.16}{8.80} \times 100 = 70.0\% \).

1 mark: moles of ethanol correctly calculated (4.60/46 = 0.100 mol); 1 mark: theoretical mass of ethyl ethanoate correctly calculated (0.100 × 88 = 8.80 g); 1 mark: percentage yield formula correctly applied; 1 mark: correct final answer, 70.0% (ECF from candidate's theoretical mass); [4]

A complete answer should address the following three areas in a well-organised, scientifically accurate account.

Synthesis: salicylic acid (2-hydroxybenzoic acid) is reacted with ethanoic anhydride (or, less commonly, ethanoyl chloride), with a few drops of a concentrated acid catalyst such as phosphoric(V) or sulfuric(VI) acid, and heated under reflux for a set time. This acetylates the phenolic –OH group of salicylic acid via nucleophilic acyl substitution, converting it into the ester (ethanoyloxy) group of aspirin, while the –COOH group of salicylic acid is unaffected. Ethanoic acid is formed as a by-product.

Purification: the reaction mixture is cooled (often in an ice bath) to encourage crystallisation of the crude aspirin, which is collected by vacuum (Büchner) filtration. The crude solid is then recrystallised: it is dissolved in a minimum volume of a suitable hot solvent, the hot solution is filtered if necessary to remove insoluble impurities, and allowed to cool slowly so that pure aspirin crystals form (soluble impurities remain in the mother liquor). The purified crystals are collected by vacuum filtration, washed with a small volume of cold solvent to remove surface impurities, and dried.

Analysis: the melting point of the dried product is measured and compared with the literature value for aspirin (approximately 135°C); a sharp melting point over a narrow range close to this literature value indicates high purity, whereas a lower and/or broader melting range indicates impurity. Infrared spectroscopy can confirm the structural change: the broad phenolic O–H absorption (roughly 3200–3550 cm⁻¹) present in salicylic acid is absent in pure aspirin, while an additional ester C=O absorption (around 1750 cm⁻¹) appears alongside the retained carboxylic acid C=O absorption (around 1710 cm⁻¹). \( ^1\text{H} \) NMR spectroscopy can confirm the presence of the new acetyl \( \text{CH}_3 \) singlet (around \( \delta \) 2.3 ppm, integrating for 3H) that is absent from salicylic acid. Mass spectrometry can confirm the molecular ion peak at m/z = 180, matching the relative molecular mass of aspirin. Finally, a percentage yield calculation (actual mass obtained relative to the theoretical mass from the limiting reagent) is used to assess the efficiency of the synthesis.

Level of Response marking (10 marks), integrating scientific content with quality of written communication (spelling, punctuation, grammar and use of specialist terminology).

Excellent (8–10 marks): well-structured response using correct scientific terminology throughout, free from significant errors in spelling, punctuation and grammar. Must cover all three areas accurately: (i) synthesis — salicylic acid reacted with ethanoic anhydride, acid catalyst, reflux/heating, acetylation of the phenolic OH; (ii) purification — crystallise/cool, vacuum filtration, recrystallisation (dissolve in minimum hot solvent, cool slowly, filter, wash, dry), impurities remain in mother liquor; (iii) analysis — melting point compared with literature value (~135°C) to assess purity; IR used to show loss of phenolic O–H and presence of ester C=O alongside retained acid C=O; NMR used to show the new acetyl CH3 singlet (~2.3 ppm); mass spectrometry used to confirm molecular ion at m/z = 180; percentage yield calculated.

Good (5–7 marks): covers most of the three areas with mostly correct terminology, but treatment may be incomplete, may omit specific reagents/values, or contain minor scientific inaccuracies; communication generally clear with occasional lapses in spelling/grammar.

Basic (1–4 marks): some relevant knowledge shown but largely descriptive/list-like; addresses only one or two of the three areas superficially; may contain significant scientific errors or omissions; poorly organised with noticeable spelling, punctuation and grammar errors that hinder communication.

0 marks: no relevant, creditable content; not attempted or entirely irrelevant.

[10]

The molecular formula \( \text{C}_4\text{H}_{10}\text{O} \) has zero degrees of unsaturation, consistent with a saturated alcohol (or ether). The signal at \( \delta \) 1.2 ppm integrates for 9H and is a singlet, meaning nine equivalent protons with no neighbouring non-equivalent protons (by the n+1 rule, a singlet has zero adjacent H); this is consistent with three chemically equivalent \( \text{CH}_3 \) groups attached to a central carbon bearing no hydrogen. The 1H signal at \( \delta \) 1.6 ppm, which exchanges with \( \text{D}_2\text{O} \) (a diagnostic test for O–H/N–H protons), is the hydroxyl proton. This pattern is only consistent with the highly symmetric tertiary alcohol \( (\text{CH}_3)_3\text{COH} \) (2-methylpropan-2-ol), in which the three methyl groups are equivalent by symmetry and are adjacent only to the quaternary carbon (which bears no H).

1 mark: molecular formula/zero degrees of unsaturation correctly used to identify a saturated alcohol; 1 mark: 9H singlet correctly interpreted as three equivalent \( \text{CH}_3 \) groups with no adjacent H (n+1 rule); 1 mark: 1H exchangeable (with \( \text{D}_2\text{O} \)) signal correctly identified as the O–H proton; 1 mark: correct structure deduced — 2-methylpropan-2-ol (tert-butanol), \( (\text{CH}_3)_3\text{COH} \); [4]

The molecular ion at m/z = 74 gives \( M_r = 74 \), which with molecular formula \( \text{C}_3\text{H}_6\text{O}_2 \) fits the saturated carboxylic acid propanoic acid, \( \text{CH}_3\text{CH}_2\text{COOH} \) (\( \text{C}_n\text{H}_{2n}\text{O}_2 \) with n = 3). The very broad IR absorption at 2500–3300 \( \text{cm}^{-1} \) is characteristic of the O–H bond of a carboxylic acid, broadened by strong intermolecular hydrogen bonding (dimer formation), distinguishing it from the narrower O–H absorption of an alcohol. The sharp absorption at 1710 \( \text{cm}^{-1} \) is the C=O stretch of the carboxylic acid carbonyl. In the mass spectrum, loss of the \( -\text{COOH} \) group (mass 45) from the molecular ion (\( 74 - 45 = 29 \)) gives the ethyl cation, \( \text{C}_2\text{H}_5^+ \), at m/z = 29.

1 mark: \( M_r = 74 \) with formula \( \text{C}_3\text{H}_6\text{O}_2 \) correctly identified as propanoic acid; 1 mark: broad 2500–3300 \( \text{cm}^{-1} \) absorption correctly assigned to carboxylic acid O–H (hydrogen-bonded); 1 mark: 1710 \( \text{cm}^{-1} \) absorption correctly assigned to C=O of the carboxylic acid; 1 mark: m/z = 29 fragment correctly assigned as \( \text{C}_2\text{H}_5^+ \) from loss of COOH; [4]

Ethyl ethanoate, \( \text{CH}_3\text{COOCH}_2\text{CH}_3 \), has three distinct proton environments.

The acetyl \( \text{CH}_3 \) (attached to the carbonyl carbon) appears at \( \delta \) approximately 2.0 ppm as a singlet, integrating for 3H; it shows no splitting because the carbonyl carbon it is attached to has no hydrogen atoms (n+1 rule with n = 0 neighbouring H gives a singlet).

The \( \text{OCH}_2 \) group appears at \( \delta \) approximately 4.1 ppm (deshielded by the adjacent oxygen) as a quartet, integrating for 2H; it is split into a quartet by the three equivalent protons of the neighbouring \( \text{CH}_3 \) group (n+1 = 3+1 = 4).

The ethyl \( \text{CH}_3 \) group appears at \( \delta \) approximately 1.3 ppm as a triplet, integrating for 3H; it is split into a triplet by the two protons of the neighbouring \( \text{OCH}_2 \) group (n+1 = 2+1 = 3).

1 mark: three distinct proton environments correctly identified; 1 mark: acetyl \( \text{CH}_3 \) correctly described — \( \delta \) ~2.0 ppm, singlet, 3H (no adjacent H); 1 mark: \( \text{OCH}_2 \) correctly described — \( \delta \) ~4.1 ppm, quartet, 2H (coupled to adjacent \( \text{CH}_3 \), n+1 rule); 1 mark: ethyl \( \text{CH}_3 \) correctly described — \( \delta \) ~1.3 ppm, triplet, 3H (coupled to adjacent \( \text{CH}_2 \)); [4]

Systolic pressure is the peak pressure exerted by blood against the walls of the arteries at the moment the ventricles of the heart contract and eject blood into the circulation. It is the higher of the two values recorded in a blood pressure reading, e.g. the '120' in a reading of 120/80 mmHg.

1 mark: identifies maximum/peak arterial pressure; 1 mark: correctly links this to ventricular contraction/systole; [2]

For a healthy resting adult, the heart rate normally lies between 60 and 100 beats per minute. Values consistently below 60 bpm indicate bradycardia and values consistently above 100 bpm indicate tachycardia.

1 mark: range 60-100 stated; 1 mark: correct unit, beats per minute (bpm); [2]

Attenuation is the progressive decrease in the intensity of an X-ray beam as it travels through tissue. It occurs because photons are removed from the beam by processes such as the photoelectric effect and Compton scattering, with denser tissues (e.g. bone) attenuating the beam more strongly than soft tissue.

1 mark: reduction/decrease in intensity/number of photons; 1 mark: as beam passes through tissue/matter (due to absorption and/or scattering); [2]

Barium sulfate has a much higher atomic number than the surrounding soft tissue, so it strongly absorbs X-rays. Coating the lining of the gastrointestinal tract with barium sulfate increases the attenuation contrast between the gut and neighbouring soft tissue, allowing the shape and any abnormalities of the gut to be clearly outlined on the radiograph.

1 mark: increases attenuation/absorption of X-rays by the gut; 1 mark: improves contrast between gut and surrounding soft tissue on the image; [2]

The activity, A, of a radioactive sample is the rate at which unstable nuclei in the sample decay, i.e. the number of disintegrations per second. One becquerel is defined as one disintegration per second (1 Bq = 1 s⁻¹).

1 mark: number of decays/disintegrations per second/unit time; 1 mark: unit becquerel (Bq); [2]

Half-life, \( t_{1/2} \), is the time required for the activity of a radioactive source, or the number of radioactive nuclei present, to decrease to one half of its initial value. It is a constant for a given isotope, independent of the amount of the isotope present.

1 mark: time for activity/number of nuclei to halve; 1 mark: reference to 'original/initial value'; [2]

A pulse oximeter shines red light (about 660 nm) and infrared light (about 940 nm) through, e.g., a fingertip. Oxyhaemoglobin absorbs more infrared light and less red light, whereas deoxyhaemoglobin absorbs more red light and less infrared light. By detecting the transmitted light at both wavelengths and comparing the pulsatile (arterial) component of absorption, the device calculates the percentage oxygen saturation, SpO₂.

1 mark: oxy- and deoxyhaemoglobin absorb red/infrared light differently; 1 mark: ratio of absorption at the two wavelengths used to determine oxygen saturation; [2]

Acoustic impedance is defined by \( Z = \rho c \), where \( \rho \) is the density of the medium and \( c \) is the speed of the ultrasound wave in that medium. It determines how much of an ultrasound wave is reflected at a boundary between two different tissues.

1 mark: Z = density × speed of sound (in that medium); 1 mark: correct unit kg m⁻² s⁻¹; [2]

On an electrocardiogram, the P wave is the small deflection that occurs first in each cardiac cycle. It corresponds to the spread of electrical depolarisation across the two atria, which triggers atrial contraction and forces blood into the ventricles.

1 mark: depolarisation of the atria; 1 mark: correctly linked to (impending) atrial contraction; [2]

Because unsealed radioactive sources can contaminate skin, surfaces or be ingested/inhaled as well as irradiate the body, staff apply the ALARA (As Low As Reasonably Achievable) principle. Acceptable precautions include wearing gloves and protective clothing to avoid contamination, handling sources with remote tools or behind shielding to increase distance, keeping handling time as short as possible, storing sources in lead-lined containers, and wearing personal dosimeters to monitor cumulative exposure.

1 mark each for any two valid precautions from: protective clothing/gloves; remote handling/shielding to increase distance; minimising time near source; lead-lined storage; personal dosimeter/film badge monitoring; (max 2); [2]

Air has a very low acoustic impedance compared with skin and soft tissue. If air were present between the probe and the skin, the large mismatch in acoustic impedance at the air-skin boundary would cause almost all of the ultrasound to be reflected, so almost no signal would enter the body and no useful image could be formed. The gel has an acoustic impedance close to that of skin, so it excludes air pockets and provides good acoustic coupling, allowing most of the ultrasound energy to be transmitted into the tissue rather than reflected at the surface.

1 mark: gel excludes/removes air between probe and skin; 1 mark: air has a very different acoustic impedance from tissue/skin, causing (near-total) reflection at an air gap; 1 mark: gel provides good acoustic coupling/matching, allowing (more of) the beam to be transmitted into the body; [3]

Technetium-99m is favoured for diagnostic scans for several reasons. Its half-life of about 6 hours is long enough to prepare the radiopharmaceutical, inject the patient and complete the scan, but short enough that the patient's radiation dose falls away quickly, minimising long-term risk. It is a pure gamma emitter, so no alpha or beta particles are emitted, which reduces the (highly ionising) dose absorbed by the patient's tissues; gamma rays are also penetrating enough to escape the body and be detected by a gamma camera outside the patient. It can also be readily attached to a range of chemical tracers so that its distribution reflects the process (e.g. bone metabolism, blood flow) being investigated.

1 mark: relatively short half-life (~6 hours) limits patient dose while allowing time for the procedure; 1 mark: pure gamma emitter, so no (highly ionising) alpha/beta dose to the patient; 1 mark: gamma radiation is penetrating enough to be detected outside the body by a gamma camera; [3]

The cuff is inflated above the expected systolic pressure, which occludes the brachial artery completely so no sound is heard. As the cuff pressure is slowly released, at the point where cuff pressure falls just below the systolic pressure, blood is forced through the still partially-compressed artery in turbulent bursts, producing the first tapping Korotkoff sound; this pressure is recorded as the systolic pressure. As the cuff continues to deflate, the sounds continue while the artery remains partially compressed. The sounds disappear (or become muffled) when the cuff pressure falls below the diastolic pressure, because the artery is no longer compressed and blood flow returns to smooth, silent laminar flow; this pressure is recorded as the diastolic pressure.

1 mark: sounds first appear as turbulent flow begins when cuff pressure falls just below systolic pressure, giving the systolic reading; 1 mark: sounds continue while artery remains partially occluded; 1 mark: sounds disappear/change when normal (laminar, silent) flow resumes as cuff pressure falls below diastolic pressure, giving the diastolic reading; [3]

One advantage of MRI over CT is that it does not use ionising radiation; instead it uses a strong magnetic field and radio waves, so there is no cumulative radiation dose risk to the patient, which is particularly important for repeated scans or paediatric patients. One disadvantage of MRI is that a scan typically takes much longer than a CT scan (many minutes rather than seconds), during which the patient must remain still in a narrow, noisy bore; MRI is also unsuitable for patients with certain ferromagnetic implants, cochlear implants or pacemakers, and is generally more expensive and less widely available than CT.

1 mark: valid advantage identified (e.g. no ionising radiation used); 1 mark: correct explanation of advantage; 1 mark: valid disadvantage with explanation (e.g. longer scan time / contraindicated with certain metal implants); accept any one credible advantage and one credible disadvantage, each explained; [3]

For diagnostic imaging, the radioisotope only needs to remain active for long enough to be administered, distributed within the body and imaged by the detector; a short half-life means the activity - and therefore the dose to the patient's healthy tissue - decays away quickly once the scan is complete, minimising unnecessary long-term radiation risk. For radiotherapy, the aim is the opposite: a large, sustained absorbed dose is required within the tumour over the course of treatment in order to kill malignant cells, so sources and treatment schedules are chosen to deliver a high dose over the required treatment period, even though this means healthy tissue near the tumour also receives a larger dose, which is managed by careful beam shaping and shielding rather than by choosing an extremely short half-life.

1 mark: diagnosis requires only brief activity, so short half-life minimises patient dose/risk after the scan; 1 mark: therapy requires sustained/higher dose delivered to tumour tissue to destroy cancer cells; 1 mark: valid comparison drawn between the two competing requirements (dose minimisation vs cell-killing dose); [3]

A tympanic thermometer contains an infrared sensor that detects the infrared radiation emitted by the tympanic membrane (eardrum), which shares its blood supply with the hypothalamus (the body's temperature-regulating centre) and so closely reflects core temperature. Because the sensor measures emitted radiation rather than waiting for heat to conduct into a probe, the reading is obtained within one or two seconds. However, the reading can be unreliable if the probe is not correctly aligned with the eardrum, if there is a build-up of earwax blocking the infrared path, or if the ear canal shape varies (e.g. in young children), all of which can cause an inaccurately low or high reading compared with core body temperature.

1 mark: detects infrared radiation emitted by the eardrum, giving a near-instant reading; 1 mark: eardrum temperature closely reflects core/hypothalamic temperature; 1 mark: valid source of error stated (e.g. misalignment in ear canal, earwax blocking sensor); [3]

(a) \( \text{Heart rate} = \dfrac{60}{0.75} = 80 \) beats per minute.
(b) \( \text{Cardiac output} = \text{HR} \times \text{SV} = 80 \times 68 = 5440 \text{ cm}^3\text{min}^{-1} = 5.44 \text{ dm}^3\text{min}^{-1} \).
(c) \( \text{MAP} = \text{DBP} + \frac{1}{3}(\text{SBP} - \text{DBP}) = 76 + \frac{1}{3}(128 - 76) = 76 + \frac{1}{3}(52) = 76 + 17.3 = 93.3 \text{ mmHg} \).
(d) 93.3 mmHg lies between 70 mmHg and 100 mmHg, so the patient's MAP indicates adequate perfusion of the major organs.

(a) 1 mark: correct method, HR = 60/T; 1 mark: 80 bpm; (b) 1 mark: CO = HR × SV (allow ECF from (a)); 1 mark: correct answer 5.44 dm³ min⁻¹ (or equivalent, e.g. 5440 cm³ min⁻¹) with unit; (c) 1 mark: correct substitution into MAP formula; 1 mark: 93.3 mmHg (accept 93-94 mmHg); allow ECF throughout; (d) 1 mark: yes/within normal range, consistent with candidate's value from (c); [7]

(a) \( I = I_0 e^{-\mu x} = 5.0 \times 10^4 \times e^{-(0.150)(8.0)} = 5.0 \times 10^4 \times e^{-1.20} = 5.0 \times 10^4 \times 0.301 = 1.5 \times 10^4 \) counts s⁻¹.
(b) Total attenuation exponent = \( \mu_{\text{tissue}}x_{\text{tissue}} + \mu_{\text{bone}}x_{\text{bone}} = 1.20 + (0.480 \times 2.0) = 1.20 + 0.96 = 2.16 \). \( I = 5.0 \times 10^4 \times e^{-2.16} = 5.0 \times 10^4 \times 0.1153 = 5.8 \times 10^3 \) counts s⁻¹ (5770 counts s⁻¹).
(c) Bone has a much higher linear attenuation coefficient than soft tissue (denser, higher effective atomic number), so it absorbs and scatters a much larger fraction of the incident X-ray photons. Fewer photons reach the detector/film beneath the bone, so that region receives less exposure and appears bright/white on the developed image, while soft tissue, which transmits more photons, appears darker.

(a) 1 mark: correct substitution \( \mu x = 1.20 \); 1 mark: 1.5×10⁴ counts s⁻¹; (b) 1 mark: total exponent = 2.16 (or equivalent two-stage calculation); 1 mark: correct method e⁻²·¹⁶; 1 mark: 5.8×10³ counts s⁻¹ (accept 5700-5800); allow ECF from (a); (c) 1 mark: bone has higher μ/absorbs more X-rays; 1 mark: fewer photons reach detector beneath bone → less exposure → appears white/bright; [7]

(a) \( \lambda = \dfrac{\ln 2}{t_{1/2}} = \dfrac{0.693}{6.0} = 0.116 \text{ h}^{-1} \).
(b) \( A = A_0 e^{-\lambda t} = 800 \times e^{-(0.116)(6.0)} = 800 \times e^{-0.693} = 800 \times 0.500 = 400 \text{ MBq} \) (as expected, since 6.0 hours is very close to one half-life).
(c) The activity must fall to \( \dfrac{50}{800} = \dfrac{1}{16} = \left(\dfrac{1}{2}\right)^4 \) of its initial value, i.e. exactly 4 half-lives must elapse. Time \( = 4 \times t_{1/2} = 4 \times 6.0 = 24.0 \) hours (or by the exponential equation: \( 50 = 800e^{-0.116t} \Rightarrow t = \dfrac{\ln(800/50)}{0.116} = \dfrac{\ln 16}{0.116} = \dfrac{2.77}{0.116} \approx 23.9 \) hours). So the waste may be disposed of approximately 24 hours after 08:00, i.e. at approximately 08:00 the following day.

(a) 1 mark: correct substitution ln2/6.0; 1 mark: 0.116 h⁻¹ (accept 0.115-0.116); (b) 1 mark: correct substitution into A = A₀e⁻λt (allow ECF from (a)); 1 mark: 400 MBq (accept 395-405); (c) 1 mark: recognises 50/800 = 1/16 = (½)⁴ or sets up exponential equation correctly; 1 mark: correct method (4 half-lives, or logarithmic solution) with ECF from (a)/(b); 1 mark: answer of 24 hours (accept 23.5-24.5 hours) with correct conclusion (approx. 08:00 next day); [7]

(a) \( D = \dfrac{E}{m} = \dfrac{6.3}{0.45} = 14.0 \text{ Gy} \).
(b) \( H = D \times w_R = 14.0 \times 1 = 14.0 \text{ Sv} \).
(c) \( H = D \times w_R = 14.0 \times 20 = 280 \text{ Sv} \).
(d) The alpha particles would cause far greater biological damage for the same absorbed dose (14.0 Gy in both cases), because alpha particles have a much larger radiation weighting factor (20, compared with 1 for gamma rays). This reflects the fact that alpha particles are densely ionising over a very short range (high linear energy transfer), depositing their energy intensely along a short track within the tissue, causing more severe localised cellular/DNA damage than the same absorbed dose of sparsely-ionising gamma radiation.

(a) 1 mark: correct substitution E/m; 1 mark: 14.0 Gy; (b) 1 mark: H = D × wR (allow ECF from (a)); 1 mark: 14.0 Sv; (c) 1 mark: correct substitution wR = 20 (allow ECF from (a)); 1 mark: 280 Sv; (d) 1 mark: alpha particles identified (consistent with candidate's larger value from (b)/(c)); 1 mark: valid reason, e.g. higher wR / more densely ionising / higher LET causes greater biological damage per unit absorbed dose; [8]

Apparatus: a calibrated fingertip pulse oximeter, a light source of variable and measurable intensity (e.g. a dimmable lamp with a light meter/lux meter), and a stable test subject (volunteer) seated at rest.
Procedure: with the volunteer seated and at rest, attach the pulse oximeter to the same finger throughout. Take and record repeated SpO₂ and pulse rate readings at a series of measured ambient light intensities (independent variable), e.g. 0, 200, 500, 1000 lux, allowing the reading to stabilise each time before recording. Repeat each light level several times and calculate a mean SpO₂ and pulse rate for each.
Variables to control: use the same finger and the same volunteer throughout; keep the volunteer still and at rest (to avoid motion artefact and heart-rate changes affecting the reading); keep the ambient temperature constant; use the same oximeter/sensor and the same light source spectrum for every reading, changing only its intensity.
Precaution for validity: shield the sensor/finger from any additional stray light other than the controlled source (e.g. with a light-excluding sleeve) so that only the intended light intensity affects the reading, and compare readings against a baseline reading taken in darkness to identify any deviation caused by the light.

1 mark: suitable apparatus identified (pulse oximeter + variable/measurable light source); 1 mark: independent variable clearly identified as ambient light intensity, varied systematically; 1 mark: dependent variable(s) identified (SpO₂ and/or pulse rate) with repeats/mean taken to improve reliability; 1 mark: at least two relevant controlled variables stated (e.g. same finger/subject, subject at rest, same sensor); 1 mark: valid precaution to ensure accurate readings (e.g. shielding from stray light, comparison with baseline/reference reading); 1 mark: overall coherent, logically ordered method; [6]

Apparatus: an ultrasound transducer/probe connected to a pulse generator and an oscilloscope (or the scanner's own timing display), coupling gel, and a tissue-mimicking phantom containing a flat reflecting target set at a precisely known, adjustable depth, d.
Procedure: apply coupling gel to exclude air, and place the transducer on the surface of the phantom directly above the target. Send a short ultrasound pulse into the phantom and use the oscilloscope/scanner to measure the pulse-echo time, t, i.e. the time between the transmitted pulse and the received echo from the reflector. Since the pulse travels to the reflector and back, the speed is calculated from \( v = \dfrac{2d}{t} \).
Improving accuracy: repeat the measurement of t at each depth several times and take a mean to reduce random timing error; better still, repeat the whole procedure for several different known depths, d, and plot a graph of 2d (or d) against t - the speed is then obtained from the gradient of the (straight-line) graph, which averages out random error across all the readings rather than relying on a single measurement.

1 mark: correct apparatus (transducer/oscilloscope/scanner, phantom with reflector at known depth, coupling gel); 1 mark: gel used to ensure good coupling/exclude air; 1 mark: pulse-echo time, t, correctly identified as the measured quantity; 1 mark: correct relationship v = 2d/t (factor of 2 for the return path) used to calculate speed; 1 mark: repeats taken and mean calculated to reduce random error; 1 mark: use of multiple known depths and a distance-time graph (speed from gradient) to improve accuracy; [6]

Apparatus: a GM tube connected to a counter/ratemeter, a stopwatch/timer, the radioactive source (held at a fixed distance from the tube using a clamp/source holder), and appropriate shielding/tongs for handling the source.
Accounting for background radiation: before introducing the source, record the count over a set time period (e.g. 5 minutes) with no source present, and calculate the background count rate. This background count rate must be subtracted from every subsequent reading taken with the source present, since background radiation is always contributing to the detected counts.
Data collection and processing: with the source placed at a fixed distance from the (unmoved) GM tube, start the timer and record the number of counts detected in short, regular time intervals (e.g. every 30 s) as the source decays. Convert each reading into a corrected count rate by subtracting the background rate. Plot a graph of corrected count rate against time; because the source decays exponentially, the half-life can be read directly from the curve as the time taken for the (corrected) count rate to fall to half of any chosen value. Alternatively, plot the natural logarithm of the corrected count rate against time, which gives a straight line of gradient \( -\lambda \), from which \( t_{1/2} = \dfrac{\ln 2}{\lambda} \) can be calculated.
Precaution: keep the source at a fixed, unchanged distance and geometry relative to the GM tube throughout (e.g. using a clamp) so that only the decreasing activity of the source, not a changing geometry, affects the count rate; also use tongs and keep exposure time to a minimum to limit dose to the experimenter, and repeat the whole procedure (or take several readings at each time) to improve precision.

1 mark: correct apparatus (GM tube + counter/ratemeter, timer, source at fixed distance); 1 mark: background count rate measured (with source absent) before the main experiment; 1 mark: background subtracted from every subsequent reading to obtain corrected count rate; 1 mark: readings taken at regular, sufficiently short time intervals appropriate to the half-life; 1 mark: correct graphical method described (count rate vs time and half-life read from curve, or ln(count rate) vs time giving gradient -λ); 1 mark: source/tube geometry kept fixed throughout (distance not changed); 1 mark: valid safety or precision precaution (e.g. tongs/minimising exposure time, repeating readings); [7]

A full-mark response would address all four bullet points with accurate physics and a reasoned final judgement, for example:

Physical principles: CT uses a narrow, rotating X-ray beam and an array of detectors that record the transmitted intensity from many angles around the patient; a computer reconstructs a cross-sectional (axial) image from these attenuation measurements, using the same principle as conventional X-ray (differential absorption of X-rays by different tissues) but processed mathematically into 2D/3D slices. MRI instead places the patient in a strong static magnetic field, which aligns the magnetic moments of hydrogen nuclei (mainly in water and fat) in the body; radiofrequency pulses are then applied to knock these nuclei out of alignment, and as they relax back into alignment they emit a radio signal that is detected and used to construct an image, with the relaxation behaviour depending on the local tissue type.

Image quality: CT gives excellent contrast and spatial resolution for dense structures such as bone, since bone strongly attenuates X-rays compared with surrounding tissue, but different soft tissues (e.g. tendon, cartilage, muscle) attenuate X-rays only slightly differently from one another, giving relatively poor soft-tissue contrast. MRI, by contrast, is particularly sensitive to differences in hydrogen (water/fat) content and relaxation behaviour between soft tissues, giving excellent contrast between ligaments, cartilage, tendons and fluid - making it the superior technique for identifying soft-tissue injuries such as a torn anterior cruciate ligament or meniscal damage in the knee.

Safety: CT uses ionising X-radiation, so the patient receives a radiation dose that carries a small increased long-term cancer risk, particularly relevant if repeated scans are needed or for younger patients; use is therefore justified against clinical benefit (ALARA principle). MRI does not use ionising radiation at all, so it carries no radiation dose risk, but the strong magnetic field means it is contraindicated for patients with certain ferromagnetic implants, cochlear implants, or older pacemakers, and the enclosed bore can cause difficulty for claustrophobic patients.

Practical factors: a CT scan is very fast (often completed within seconds to a minute), making it more suitable for patients who cannot keep still, for emergency situations, or where bone injury is also suspected; it is also more widely available and generally cheaper than MRI. An MRI scan takes considerably longer (typically 20-45 minutes) and requires the patient to remain still throughout, is more expensive and less widely available, but gives the diagnostic detail needed for a soft-tissue knee injury.

Conclusion: for a suspected soft-tissue knee injury specifically, MRI would generally be the technique of choice because its superior soft-tissue contrast and lack of ionising radiation outweigh its longer scan time and higher cost, provided the patient has no contraindications to the magnetic field; CT might still be used first, or in combination, if a bony injury (e.g. fracture) also needs to be excluded quickly.

Assessed using a Level of Response mark scheme integrating scientific content with quality of written communication (QWC).

Level 3 (Excellent): 10-13 marks. The candidate gives a full, accurate and well-balanced comparison addressing all four bullet points (principles, image quality, safety, practical factors) with correct physics throughout, reaches a reasoned conclusion as to which technique is preferable for this injury, and communicates clearly using appropriate scientific terminology with few or no errors of spelling, punctuation or grammar.

Level 2 (Good): 6-9 marks. The candidate addresses most of the four bullet points with generally accurate physics, though coverage may be uneven (e.g. one area covered only briefly) or contain minor inaccuracies; the answer is reasonably well organised and mostly clearly expressed, with occasional lapses in spelling, punctuation or grammar.

Level 1 (Basic): 1-5 marks. The candidate shows some relevant knowledge of CT and/or MRI but the response is limited, addressing only one or two bullet points, or containing significant inaccuracies/omissions; the answer may be poorly organised or list-like, with noticeable errors in spelling, punctuation or grammar that occasionally hinder meaning.

0 marks: no creditable content, or the response does not relate to the question.

Indicative content for the four bullet points: CT - rotating X-ray tube/detector array, computer reconstruction from attenuation data, good for bone, ionising radiation dose, fast scan; MRI - magnetic field aligns hydrogen nuclei, RF pulses and relaxation signal detected, excellent soft-tissue contrast, no ionising radiation, contraindicated with certain metal implants/pacemakers, longer scan time and higher cost; overall judgement that MRI is generally preferred for soft-tissue knee injury while CT may still be used to exclude bony injury quickly. [13]

For an object at infinity, incoming rays are parallel, so \( \dfrac{1}{u} = 0 \). The corrective lens must produce a virtual image at the eye's far point, 4.0 m in front of the lens, so \( v = -4.0 \text{ m} \) (virtual image, real-is-positive convention). Using the lens equation \( \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} \): \( \dfrac{1}{f} = \dfrac{1}{-4.0} - 0 = -0.25 \text{ m}^{-1} \), so \( f = -4.0 \text{ m} \). The power of the lens is \( P = \dfrac{1}{f} = \dfrac{1}{-4.0} = -0.25 \text{ D} \). The negative sign shows the lens must be diverging (concave).

recognise image must form at the far point, v = -4.0 m ; use \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \) with \( \frac{1}{u} = 0 \) for a distant object ; correct substitution giving f = -4.0 m ; power \( P = \frac{1}{f} \) calculated as -0.25 D ; correct statement that the lens is diverging/concave (negative sign) [5]

The critical angle \( \theta_c \) satisfies \( \sin\theta_c = \dfrac{n_2}{n_1} \), where \( n_1 = 1.48 \) (core) and \( n_2 = 1.44 \) (cladding). \( \sin\theta_c = \dfrac{1.44}{1.48} = 0.973 \), so \( \theta_c = \sin^{-1}(0.973) = 76.7^{\circ} \). For total internal reflection to occur at the core-cladding boundary, the angle of incidence at that boundary must be greater than \( 76.7^{\circ} \).

use \( \sin\theta_c = \frac{n_2}{n_1} \) ; correct substitution \( \frac{1.44}{1.48} = 0.973 \) ; \( \theta_c = 76.7^{\circ} \) (accept 76-77°) ; correct statement that angle of incidence must be greater than \( \theta_c \) for TIR to occur [5]

(a) Sound spreads uniformly over a sphere of radius r, so \( I = \dfrac{P}{4\pi r^2} \). Substituting \( P = 0.80 \text{ W} \) and \( r = 5.0 \text{ m} \): \( I = \dfrac{0.80}{4\pi (5.0)^2} = \dfrac{0.80}{314.2} = 2.5 \times 10^{-3} \text{ W m}^{-2} \). (b) The sound intensity level is \( L = 10\log_{10}\left(\dfrac{I}{I_0}\right) = 10\log_{10}\left(\dfrac{2.5\times10^{-3}}{1.0\times10^{-12}}\right) = 10\log_{10}(2.5\times10^{9}) = 10(9.40) = 94.0 \text{ dB} \), approximately.

use \( I = \frac{P}{4\pi r^2} \) ; correct substitution giving \( I \approx 2.5 \times 10^{-3} \text{ W m}^{-2} \) (accept 2.5-2.6 × 10⁻³) ; use \( L = 10\log_{10}\left(\frac{I}{I_0}\right) \) ; correct substitution ; \( L \approx 94 \text{ dB} \) (accept 93-95 dB), error carried forward from (a) [5]

(a) For a pipe open at both ends, the fundamental has an antinode at each end and one node in the middle, so the pipe length equals half a wavelength: \( L = \dfrac{\lambda_1}{2} \), giving \( \lambda_1 = 2L = 2(0.85) = 1.70 \text{ m} \). The frequency is \( f_1 = \dfrac{v}{\lambda_1} = \dfrac{340}{1.70} = 200 \text{ Hz} \). (b) An open pipe supports all integer harmonics, so \( f_n = n f_1 \). The third harmonic frequency is \( f_3 = 3 \times 200 = 600 \text{ Hz} \).

\( \lambda_1 = 2L = 1.70 \text{ m} \) ; use \( f_1 = \frac{v}{\lambda_1} \) ; \( f_1 = 200 \text{ Hz} \) ; recognise open pipe supports all harmonics, \( f_3 = 3f_1 \) ; \( f_3 = 600 \text{ Hz} \) [5]

(a) Using \( c = f\lambda \), rearranged: \( \lambda = \dfrac{c}{f} = \dfrac{3.00\times10^{8}}{97.4\times10^{6}} = 3.08 \text{ m} \). (b) Rearranging again for frequency: \( f = \dfrac{c}{\lambda} = \dfrac{3.00\times10^{8}}{300} = 1.00\times10^{6} \text{ Hz} = 1000 \text{ kHz} \).

use \( \lambda = \frac{c}{f} \) ; correct substitution \( \frac{3.00\times10^8}{97.4\times10^6} \) ; \( \lambda = 3.08 \text{ m} \) ; rearrange to \( f = \frac{c}{\lambda} \) with \( \lambda = 300 \text{ m} \) ; \( f = 1.00\times10^6 \text{ Hz} \) (1000 kHz) [5]

Using Snell's law, \( n_1 \sin\theta_1 = n_2 \sin\theta_2 \), with \( n_1 = 1.00 \) (air), \( \theta_1 = 40^{\circ} \), \( n_2 = 1.52 \): \( \sin\theta_2 = \dfrac{n_1 \sin\theta_1}{n_2} = \dfrac{1.00 \times \sin 40^{\circ}}{1.52} = \dfrac{0.643}{1.52} = 0.423 \). Therefore \( \theta_2 = \sin^{-1}(0.423) = 25.0^{\circ} \).

use \( n_1\sin\theta_1 = n_2\sin\theta_2 \) ; correct value \( \sin40^{\circ} = 0.643 \) ; \( \sin\theta_2 = \frac{0.643}{1.52} = 0.423 \) ; \( \theta_2 = 25.0^{\circ} \) (accept 24.9-25.1°) [4]

Pressure is related to force by \( P = \dfrac{F}{A} \). Since the same force \( F \) is transmitted from the eardrum to the oval window, \( \dfrac{P_{\text{oval}}}{P_{\text{eardrum}}} = \dfrac{F/A_{\text{oval}}}{F/A_{\text{eardrum}}} = \dfrac{A_{\text{eardrum}}}{A_{\text{oval}}} \). Substituting the areas: \( \dfrac{P_{\text{oval}}}{P_{\text{eardrum}}} = \dfrac{55}{3.2} = 17.2 \). The pressure is therefore increased by a factor of approximately 17.

recognise \( P = \frac{F}{A} \) with force constant across the ossicles ; pressure ratio = \( \frac{A_{\text{eardrum}}}{A_{\text{oval window}}} \) ; correct substitution \( \frac{55}{3.2} \) ; answer ≈ 17 (accept 17.0-17.2) [4]

With four nodes (including the two fixed ends) and three antinodes, the string is divided into three equal loops, each loop being half a wavelength long. This is the third harmonic. Since three half-wavelengths fit the length L: \( L = \dfrac{3\lambda}{2} \), so \( \lambda = \dfrac{2L}{3} = \dfrac{2(0.90)}{3} = 0.60 \text{ m} \). The frequency is \( f = \dfrac{v}{\lambda} = \dfrac{60}{0.60} = 100 \text{ Hz} \).

correctly identify third harmonic (three loops/antinodes between the fixed ends) ; use \( \lambda = \frac{2L}{n} \) with n = 3 ; \( \lambda = 0.60 \text{ m} \) ; \( f = \frac{v}{\lambda} = 100 \text{ Hz} \) [4]

The pattern of displacement values (0, 4, 0, -4, 0) repeats every 20 ms, so the period is \( T = 20 \text{ ms} = 0.020 \text{ s} \). The frequency is \( f = \dfrac{1}{T} = \dfrac{1}{0.020} = 50 \text{ Hz} \). The maximum displacement shown in the table is 4 mm, so the amplitude is 4 mm.

correct identification of repeating pattern giving T = 20 ms (0.020 s) ; \( f = \frac{1}{T} = 50 \text{ Hz} \) ; amplitude = 4 mm (maximum displacement) [3]

A closed pipe vibrating at its fundamental has a node at the closed end and an antinode at the open end with nothing in between, so the pipe length corresponds to one-quarter of a wavelength: \( L = \dfrac{\lambda}{4} \). The wavelength is \( \lambda = \dfrac{v}{f} = \dfrac{330}{220} = 1.5 \text{ m} \). The pipe length is therefore \( L = \dfrac{1.5}{4} = 0.375 \text{ m} \).

state the pipe length equals one-quarter of a wavelength (node at closed end, antinode at open end) ; \( \lambda = \frac{v}{f} = 1.5 \text{ m} \) ; \( L = \frac{\lambda}{4} = 0.375 \text{ m} \) [3]

The displacement pattern (0, 3, 0, -3, 0) repeats every 40 cm along x, so the wavelength is \( \lambda = 40 \text{ cm} = 0.40 \text{ m} \). The frequency is \( f = \dfrac{v}{\lambda} = \dfrac{4.0}{0.40} = 10 \text{ Hz} \).

correct identification of repeat distance from the table, \( \lambda = 40 \text{ cm} = 0.40 \text{ m} \) ; use \( f = \frac{v}{\lambda} \) ; \( f = 10 \text{ Hz} \) [3]

From the table, the threshold level stays at or below 20 dB for frequencies from 125 Hz to 2000 Hz, so hearing is close to normal over this range. Between 2000 Hz and 8000 Hz the threshold rises sharply, to 55 dB at 4000 Hz and 60 dB at 8000 Hz, showing a significant loss of sensitivity in this high-frequency region. This pattern, with loss concentrated around 4000 Hz, is characteristic of noise-induced hearing loss, caused by damage to the hair cells at the region of the basilar membrane that responds to these frequencies.

identify 125-2000 Hz as within/at the normal threshold (≤20 dB) ; identify the marked rise in threshold (hearing loss) at 4000-8000 Hz ; suggest link to noise-induced damage of hair cells at the corresponding region of the basilar membrane [3]

The inverse square law states \( I \propto \dfrac{1}{r^2} \), so doubling the distance reduces the intensity to one quarter. The corresponding change in sound level is \( \Delta L = 10\log_{10}\left(\dfrac{1}{4}\right) = -6.0 \text{ dB} \). In the table, the level falls by exactly 6 dB every time the distance doubles (90 → 84 → 78 → 72 dB for 1.0 → 2.0 → 4.0 → 8.0 m), so the data are consistent with the inverse square law. Extending the same pattern, doubling the distance again from 8.0 m to 16 m gives a further fall of 6 dB, so the predicted level at 16 m is \( 72 - 6 = 66 \text{ dB} \).

recognise sound level falls by 6 dB each time the distance doubles in the data ; link this to the inverse square law, \( I \propto \frac{1}{r^2} \) giving \( \Delta L = 10\log_{10}\left(\frac{1}{4}\right) = -6 \text{ dB} \), matching the data ; extrapolate the pattern to give 66 dB at 16 m [3]

The data show that as frequency decreases, the position of maximum response moves further from the base of the cochlea (from 3 mm at 8000 Hz to 31 mm at 250 Hz): high frequencies produce a maximum response near the base (oval window end, where the basilar membrane is narrow and stiff), while low frequencies produce a maximum response near the apex (where the membrane is wider and more flexible). The relationship is not linear - equal successive halvings of frequency (8000 → 4000 → 2000 ...) do not correspond to equal increases in distance, showing a roughly logarithmic mapping. This frequency-to-place mapping is called tonotopic organisation (tonotopy) of the cochlea.

correctly describe inverse relationship: high frequency corresponds to response near the base (small distance), low frequency corresponds to response near the apex (large distance) ; note the relationship is non-linear (unequal distance changes for equal frequency ratios) ; correct term, tonotopic organisation/mapping [3]

The envelope pattern (5, 8, 5, 2, 5) repeats every 1.00 ms, so the period of the modulating signal is \( T = 1.00 \text{ ms} = 1.0\times10^{-3} \text{ s} \). The frequency is \( f = \dfrac{1}{T} = \dfrac{1}{1.0\times10^{-3}} = 1000 \text{ Hz} \) (1 kHz), a typical audio frequency. The envelope traces out how the amplitude of the much higher-frequency carrier wave (typically several MHz) varies with time; the carrier's own frequency is unaffected and remains constant, but its amplitude is made to rise and fall in step with the audio (modulating) signal, so that the audio information is encoded in the shape of the envelope.

correct identification of repeat pattern, T = 1.00 ms ; \( f = \frac{1}{T} = 1000 \text{ Hz} \) (1 kHz) ; correct explanation that the carrier (much higher, e.g. MHz, frequency) has its amplitude varied in step with the modulating/audio signal while its frequency remains constant [3]

Sound is a mechanical (longitudinal) wave: it is transmitted by particles of a medium being displaced and colliding with neighbouring particles, producing regions of compression and rarefaction. In a vacuum there are no particles present to be displaced or to collide, so sound cannot be transmitted. Light is an electromagnetic wave, consisting of oscillating electric and magnetic fields; it does not require particles of a medium to propagate, and so can travel through the vacuum of space.

sound is a mechanical wave that requires particles of a medium to be transmitted (via compressions and rarefactions) ; no particles are present in a vacuum, so sound cannot travel through it ; light is an electromagnetic wave (oscillating electric and magnetic fields) that does not require a medium and so can travel through a vacuum [3]

The three ossicles form a jointed lever system, which by itself gives a small mechanical advantage. More importantly, the ossicles transmit the force collected over the relatively large area of the eardrum onto the much smaller area of the oval window. Since pressure is force divided by area, concentrating the same force onto a smaller area produces a substantial increase in pressure. This pressure amplification is needed to overcome the impedance mismatch between air and the fluid of the cochlea - without it, most of the sound energy would simply be reflected at an air-to-fluid boundary rather than being transmitted into the cochlea.

ossicles act as a lever system, providing some mechanical advantage ; force collected over the large area of the eardrum is concentrated onto the much smaller area of the oval window, amplifying pressure ; this overcomes the impedance mismatch between air and cochlear fluid, reducing the energy that would otherwise be reflected at the boundary [3]

Loud, prolonged sound causes mechanical damage to the hair cells of the organ of Corti on the basilar membrane. Because of the tonotopic organisation of the cochlea, each frequency stimulates hair cells at a particular position along the basilar membrane. The region of the basilar membrane that responds maximally to frequencies around 4000 Hz appears to be particularly susceptible to mechanical stress from typical occupational noise exposure. Hair cells in this region are therefore damaged preferentially, producing a characteristic dip, or 'noise notch', in hearing sensitivity around 4000 Hz on an audiogram, rather than an even loss across the whole frequency range.

noise causes mechanical damage to hair cells of the organ of Corti/basilar membrane ; damage is frequency-specific because of the tonotopic organisation (place coding) of the cochlea ; the region responding to ~4000 Hz is particularly vulnerable, producing a characteristic noise-induced 'notch' at that frequency rather than a uniform loss [3]

Accommodation is the adjustment of the power (focal length) of the eye lens, brought about by the ciliary muscles. To focus on a near object, the ciliary muscles contract, which reduces the tension in the suspensory ligaments; the elastic lens then relaxes into a more curved, rounder shape, increasing its power (decreasing its focal length) so that light from the near object is refracted more strongly to focus on the retina. To focus on a distant object, the ciliary muscles relax, the suspensory ligaments become taut and pull the lens into a thinner, flatter shape, decreasing its power (increasing its focal length) so that the less strongly converging light from a distant object still focuses on the retina.

near object: ciliary muscles contract, suspensory ligaments slacken, lens becomes rounder/thicker, power increases ; distant object: ciliary muscles relax, suspensory ligaments become taut, lens becomes flatter/thinner, power decreases ; correct overall statement that accommodation changes lens power/focal length so the image remains focused on the retina [3]

Total internal reflection occurs at a boundary where light travels from a medium of higher refractive index into one of lower refractive index, provided the angle of incidence exceeds the critical angle. Surrounding the core with a cladding of lower (but similar) refractive index establishes this condition reliably and uniformly along the whole length of the fibre. Using a solid cladding rather than leaving the core bare in air also protects the reflecting surface: it prevents scratches, dust and moisture reaching the core surface, and stops light leaking or scattering into neighbouring fibres where fibres are bundled together (cross-talk). This keeps the critical angle condition consistent and reduces energy losses along the fibre.

TIR requires light to travel from a higher to a lower refractive-index medium at the boundary, so the cladding must have a lower refractive index than the core ; the cladding protects/insulates the core surface from damage, dirt and contact with other fibres, preventing surface defects and light leakage/cross-talk between adjacent fibres ; this keeps the critical-angle condition consistent and reduces signal loss along the fibre's length [3]

The Eustachian tube connects the middle ear to the back of the throat (pharynx). It equalises the air pressure on either side of the eardrum with the outside atmosphere, allowing the eardrum to vibrate freely, and it also allows drainage of fluid or mucus from the middle ear, helping to prevent the build-up of fluid and reduce the risk of middle-ear infection.

any two of: equalises air pressure either side of the eardrum/between the middle ear and the atmosphere ; allows drainage of fluid or mucus from the middle ear ; helps prevent build-up of fluid/reduce risk of infection - 1 mark each, max [2]

Sound intensity level, in decibels, is defined as \( L = 10\log_{10}\left(\dfrac{I}{I_0}\right) \), where \( I \) is the sound intensity and \( I_0 = 1.0\times10^{-12} \text{ W m}^{-2} \) is the threshold-of-hearing reference intensity. A logarithmic scale is used because the human ear can detect an enormous range of intensities, spanning many orders of magnitude (roughly \( 10^{12} \) : 1 from the threshold of hearing to the threshold of pain), which would be impractical to represent using a linear scale.

correct definition/formula, \( L = 10\log_{10}\left(\frac{I}{I_0}\right) \), with \( I_0 \) as the threshold-of-hearing reference intensity ; logarithmic scale needed because the ear responds to a very large range of intensities (many orders of magnitude) that a linear scale could not conveniently represent [2]

A standing wave is produced when two waves of the same frequency (and therefore the same wavelength) travelling in opposite directions along the same line meet and superpose. In an air column, this occurs when a wave travelling from the source is reflected at the closed or open end, so that the incident and reflected waves - of similar amplitude - continuously interfere, producing fixed points of no displacement (nodes) and maximum displacement (antinodes).

two waves of the same frequency/wavelength travelling in opposite directions superpose (interfere) ; produced by a wave and its reflection at a boundary, with comparable amplitude, giving fixed nodes and antinodes [2]

The normal range of human hearing is approximately 20 Hz to 20 000 Hz (20 kHz). With increasing age, sensitivity to the higher frequencies within this range declines, so the upper frequency limit that can be heard typically decreases; this age-related loss of high-frequency hearing is known as presbycusis.

states range as approximately 20 Hz to 20 000 Hz (accept 20 Hz-20 kHz) ; states that sensitivity/upper limit for high frequencies decreases with age (presbycusis) [2]

A converging (convex) lens is used to correct hyperopia. In a hyperopic eye, the eyeball is too short (or the cornea/lens does not converge light strongly enough), so light from near objects would come to a focus behind the retina. A converging lens adds extra convergence to the incoming light before it enters the eye, bringing the point of focus forward so that it falls exactly on the retina.

correctly states converging/convex lens ; correct reason - hyperopic eye under-converges light from near objects (image would form behind the retina) and the converging lens adds power to bring the focus onto the retina [2]

Optical fibres suffer much lower attenuation (signal loss) than copper cables over a given distance, meaning fewer repeaters/amplifiers are needed along a transmission route. Optical fibres also have a much greater bandwidth, allowing them to carry far more information (a much higher data rate) than a copper cable of similar size. Optical fibres are additionally immune to electromagnetic interference and are lighter and more secure than copper cables.

any two of: lower attenuation/signal loss over distance ; much greater bandwidth/information-carrying capacity ; immune to electromagnetic interference ; lighter/thinner or more secure (harder to tap) - 1 mark each, max [2]

Radio waves have a very much longer wavelength than visible light - radio wavelengths range from around a metre to several kilometres, compared with roughly \( 4\times10^{-7} \) to \( 7\times10^{-7} \text{ m} \) for visible light. Because their wavelength is comparable to the size of everyday obstacles such as hills and buildings, radio waves diffract noticeably around such obstacles, allowing reception even when there is no direct line of sight to the transmitter; visible light, with its much shorter wavelength, is not significantly diffracted by objects of this size.

states radio waves have a much (many orders of magnitude) longer wavelength than visible light ; states consequence - radio waves diffract noticeably around obstacles (hills/buildings) of everyday size, enabling reception without direct line of sight, unlike light [2]

Sound waves are first collected by the pinna (outer ear), which funnels them along the ear canal towards the tympanic membrane (eardrum). The eardrum is a thin, flexible membrane that is made to vibrate at the same frequency as the incoming sound wave. These vibrations are passed across the air-filled middle ear by the three ossicles - the malleus, incus and stapes - which are linked in a chain from the eardrum to the oval window of the cochlea. The ossicles act partly as a lever system, but more significantly they transmit the force collected over the relatively large area of the eardrum onto the much smaller area of the oval window; since pressure is force per unit area, this concentration of force produces a substantial increase in pressure. This pressure amplification is essential because it overcomes the impedance mismatch between air and the fluid that fills the cochlea, which would otherwise reflect most of the sound energy at the boundary rather than transmitting it inward.

Vibration of the oval window sets up a pressure wave that travels through the fluid of the cochlea, causing the basilar membrane, which runs along the length of the coiled cochlea, to be displaced up and down. Sitting on the basilar membrane is the organ of Corti, which contains rows of hair cells. As the basilar membrane moves, the stereocilia (hair-like projections) on top of the hair cells are bent, relative to the fixed tectorial membrane above them. This bending opens mechanically-gated ion channels in the hair cell membrane, allowing an influx of ions that depolarises the hair cell. The depolarisation causes the hair cell to release neurotransmitter onto the endings of neurons of the auditory nerve, generating action potentials that travel along the auditory nerve to the brain. In this way mechanical vibration is transduced into an electrical (neural) signal.

The pitch of a sound is coded, in large part, by which region of the basilar membrane is displaced most strongly - a phenomenon known as tonotopic organisation. The basilar membrane is narrow and stiff near the base (close to the oval window) and becomes progressively wider and more flexible towards the apex. As a result, the base of the membrane resonates most strongly in response to high-frequency sounds, while the apex resonates most strongly in response to low-frequency sounds; sounds of intermediate frequency produce a maximum response at intermediate positions along the membrane. Because hair cells at a given position are connected to specific fibres of the auditory nerve, the position of maximum stimulation is preserved as a 'place code' that is carried through to the brain, allowing different frequencies (and hence different pitches) to be distinguished. Loudness is separately coded by the amplitude of basilar membrane displacement and the corresponding rate of nerve impulses generated.

Overall, therefore, sound energy is collected and amplified by the outer and middle ear, converted into a neural signal by the hair cells of the cochlea, and its pitch is encoded by the position of maximal stimulation along the tonotopically organised basilar membrane.

This question is marked using a Level of Response scheme; Quality of Written Communication (spelling, punctuation, grammar and clarity of scientific expression) is assessed alongside scientific content.

Level 3 (Excellent - 10-13 marks): A comprehensive, logically sequenced account that correctly links most or all of the following: pinna collects and funnels sound into the ear canal; eardrum vibrates in response to the sound wave; ossicles (malleus, incus, stapes) transmit and amplify the vibration via lever action and the area difference between eardrum and oval window; vibration of the oval window sets up a pressure wave in the cochlear fluid, displacing the basilar membrane; movement of the basilar membrane bends the stereocilia of hair cells in the organ of Corti, opening ion channels and triggering neurotransmitter release/generation of action potentials in the auditory nerve; tonotopic organisation of the basilar membrane (narrow/stiff base resonates to high frequency, wide/flexible apex resonates to low frequency) explains how pitch is coded by the position of maximum stimulation (place theory). Answer is well organised, uses correct scientific terminology throughout, and is largely free of errors in spelling, punctuation and grammar.

Level 2 (Good - 6-9 marks): Most stages of the pathway are described with reasonable accuracy (e.g. eardrum vibration, ossicle transmission/amplification, cochlear fluid disturbance, hair cell involvement, generation of a nerve impulse), but the account may be incomplete, lack full logical sequencing, or contain minor inaccuracies; reference to tonotopic organisation/pitch coding is present but may be partial or imprecise. Expression is generally clear, with some errors in spelling, punctuation or grammar, or with organisation that is not fully logical.

Level 1 (Basic - 1-5 marks): Answer includes a small number of relevant, largely isolated points (e.g. eardrum vibrates, ossicles are involved, cochlea/hair cells are mentioned) without a coherent overall sequence or sufficient physiological detail; little or no correct reference to tonotopic organisation or how pitch is specifically coded. Scientific terminology is used sparsely or inaccurately; expression may be poor, with a number of errors in spelling, punctuation and grammar that impede communication.

0 marks: No relevant content, or a response that is entirely incorrect or irrelevant to the question. [13]

(a) Mother (\( TtEe \)) is heterozygous at both loci, which assort independently, so she produces four gamete types in equal proportions: \( TE \), \( Te \), \( tE \), \( te \). Father (\( ttee \)) is homozygous recessive at both loci, so he produces only one gamete type: \( te \).

(b) Punnett square (father's single gamete \( te \) across the top; mother's four gametes down the side):
Mother TE x Father te -> TtEe (taster, free earlobes)
Mother Te x Father te -> Ttee (taster, attached earlobes)
Mother tE x Father te -> ttEe (non-taster, free earlobes)
Mother te x Father te -> ttee (non-taster, attached earlobes)
Each genotype occurs with probability 1/4.

(c) Phenotypic ratio = 1 taster/free : 1 taster/attached : 1 non-taster/free : 1 non-taster/attached.

(d) Non-taster with free earlobes = genotype \( ttEe \) = 1/4 of offspring = 25%.

(a) 1 mark: mother produces TE, Te, tE, te; father produces te only (both needed for the mark); (b) 1 mark: correct Punnett square set up with father's single gamete type and mother's four gamete types; 1 mark: all four offspring genotypes correctly derived (TtEe, Ttee, ttEe, ttee); 1 mark: correct phenotypes assigned to each genotype; (c) 1 mark: ratio stated as 1:1:1:1; (d) 1 mark: probability = 1/4 or 25% (ECF from candidate's own diagram); [6]

(a) Mother is an unaffected carrier: \( X^C X^c \). Father has normal vision: \( X^C Y \).

(b) Mother's gametes: \( X^C \), \( X^c \). Father's gametes: \( X^C \), \( Y \).
Cross gives: \( X^C X^C \) (daughter, normal vision), \( X^C X^c \) (daughter, carrier, normal vision), \( X^C Y \) (son, normal vision), \( X^c Y \) (son, colour blind). Each occurs with probability 1/4.

(c) Of the two possible daughter genotypes (\( X^C X^C \) and \( X^C X^c \), each 1/4 of all offspring, i.e. 1/2 of daughters each), half of all daughters produced are expected to be carriers.

(d) A son inherits his single X chromosome from his mother and his Y chromosome from his father; because the father never contributes an X chromosome to a son, the father's X-linked alleles can never be passed to his sons, so an unaffected (\( X^C Y \)) father cannot be the source of the \( X^c \) allele in a colour-blind son.

(a) 1 mark: both parental genotypes correct (X^C X^c and X^C Y); (b) 1 mark: correct gametes identified for each parent; 1 mark: all four offspring genotypes correctly derived; 1 mark: correct phenotypes/sex assigned to each genotype; (c) 1 mark: 1/2 (50%) of daughters are carriers; (d) 1 mark: sons receive X only from mother / father's X never passed to sons; [6]

(a) Father (\( L^M L^N \)) produces gametes \( L^M \) and \( L^N \) in equal proportions. Mother (\( L^M L^M \)) produces only \( L^M \) gametes.
Cross: \( L^M \) (father) x \( L^M \) (mother) -> \( L^M L^M \) (type M); \( L^N \) (father) x \( L^M \) (mother) -> \( L^M L^N \) (type MN).
Expected ratio: 1 type M : 1 type MN.

(b) Type N (genotype \( L^N L^N \)) cannot appear. Because the mother is homozygous \( L^M L^M \), every offspring must inherit at least one \( L^M \) allele from her, so a child cannot be homozygous \( L^N L^N \).

(a) 1 mark: correct gametes for each parent identified; 1 mark: both offspring genotypes correctly derived; 1 mark: correct blood types (phenotypes) assigned; 1 mark: ratio 1:1 stated; (b) 1 mark: type N identified with correct explanation that mother can only donate L^M; [5]

(a) Because \( H \) is dominant, any individual carrying \( H \) would show symptoms of Huntington's disease. As both named parents are unaffected, both must be homozygous recessive, genotype \( hh \). Both therefore produce only \( h \) gametes. Cross \( hh \) x \( hh \): all offspring genotypes are \( hh \) (unaffected), with probability 1. It is therefore not genetically possible, under normal Mendelian inheritance, for two parents who are both genotype \( hh \) to produce a child of genotype \( Hh \); since the affected child must have inherited an \( H \) allele from a biological parent, this raises doubt that the named man (if genuinely unaffected/\( hh \)) is the biological father.

(b) Genetic fingerprinting (DNA profiling): DNA is extracted from the child, mother, and alleged father and specific highly variable regions (short tandem repeats/STRs, or VNTRs) are amplified using PCR. The fragments are separated by size using gel electrophoresis to produce a pattern of bands unique to each individual. Bands in the child's profile that do not match the mother's profile must have come from the biological father; if these bands are absent from the alleged father's profile, paternity is excluded, whereas a match supports paternity.

(a) 1 mark: both unaffected parents must be genotype hh; 1 mark: cross hh x hh shown correctly, gametes both h; 1 mark: valid conclusion that all offspring would be hh, so a Hh child is not possible from these two named parents; (b) 1 mark: names genetic fingerprinting/DNA profiling (PCR amplification of STR/VNTR loci, gel electrophoresis); 1 mark: explains bands not matching the mother must come from the true father, compared against alleged father's profile; [5]

For each class, calculate \( \frac{(O-E)^2}{E} \):
Rough, black: \( \frac{(172-180)^2}{180} = \frac{64}{180} = 0.36 \)
Rough, white: \( \frac{(68-60)^2}{60} = \frac{64}{60} = 1.07 \)
Wavy, black: \( \frac{(54-60)^2}{60} = \frac{36}{60} = 0.60 \)
Wavy, white: \( \frac{(26-20)^2}{20} = \frac{36}{20} = 1.80 \)

\( \chi^2 = 0.36 + 1.07 + 0.60 + 1.80 = 3.82 \) (2 d.p.)

Degrees of freedom = number of classes − 1 = 4 − 1 = 3. The critical value at \( p = 0.05 \) for 3 degrees of freedom is 7.82. Since the calculated \( \chi^2 \) value (3.82) is less than the critical value (7.82), the difference between the observed and expected results is not statistically significant. The null hypothesis (that the offspring fit a 9:3:3:1 ratio) is accepted; any deviation from the expected ratio is likely to be due to chance.

1 mark: correct method shown, (O-E)^2/E calculated for each class; 1 mark: correct chi-squared total = 3.82 (accept 3.8-3.83, ECF from candidate's own (O-E)^2/E values); 1 mark: degrees of freedom = 3 identified and compared correctly with critical value 7.82; 1 mark: valid conclusion — calculated value less than critical value, not significant, accept null hypothesis/data fit 9:3:3:1 ratio; [4]

Expected numbers (3:1 ratio of 200 offspring): tall = \( \frac{3}{4} \times 200 = 150 \); short = \( \frac{1}{4} \times 200 = 50 \).

\( \chi^2 = \frac{(142-150)^2}{150} + \frac{(58-50)^2}{50} = \frac{64}{150} + \frac{64}{50} = 0.43 + 1.28 = 1.71 \)

Degrees of freedom = 2 classes − 1 = 1. The critical value at \( p = 0.05 \) for 1 degree of freedom is 3.84. Since the calculated \( \chi^2 \) (1.71) is less than the critical value (3.84), the difference between observed and expected results is not significant; the data support the hypothesis that stem height follows simple monohybrid dominant/recessive inheritance in a 3:1 ratio.

1 mark: correct expected values (150 and 50) and chi-squared calculated correctly = 1.71 (ECF for arithmetic slips if method correct); 1 mark: degrees of freedom = 1 correctly identified and compared with critical value 3.84; 1 mark: valid conclusion — not significant, data consistent with 3:1 ratio; [3]

\( \chi^2 = \frac{(330-320)^2}{320} + \frac{(150-160)^2}{160} + \frac{(20-20)^2}{20} = \frac{100}{320} + \frac{100}{160} + 0 = 0.31 + 0.63 + 0 = 0.94 \)

The critical value at \( p = 0.05 \) for 1 degree of freedom is 3.84. As the calculated \( \chi^2 \) (0.94) is less than the critical value (3.84), the difference between the observed and Hardy-Weinberg expected genotype numbers is not statistically significant. The null hypothesis is accepted: the observed genotype frequencies do not differ significantly from those predicted by the Hardy-Weinberg equation, so the population can be considered to be in Hardy-Weinberg equilibrium at this locus (consistent with no selection, mutation, migration or non-random mating affecting this allele).

1 mark: correct method and chi-squared value calculated = 0.94 (ECF for arithmetic slips if method correct); 1 mark: degrees of freedom = 1 correctly compared with critical value 3.84; 1 mark: valid conclusion — not significant, population in Hardy-Weinberg equilibrium; [3]

Each set of three consecutive bases along the gene (a triplet, or codon on the mRNA) codes for one specific amino acid. The linear sequence of triplets along the gene therefore specifies the linear sequence in which amino acids are joined together by peptide bonds during translation, and this amino acid sequence is the primary structure of the polypeptide. The genetic code is described as degenerate because most amino acids can be specified by more than one different triplet/codon (there are 64 possible triplets but only 20 amino acids), so a change to one base, particularly the third base of a triplet, does not always result in a different amino acid being incorporated.

1 mark: triplet/codon of three bases codes for one amino acid; 1 mark: sequence of triplets/codons determines the sequence of amino acids (primary structure); 1 mark: degenerate = more than one triplet/codon can code for the same amino acid; [3]

DNA helicase breaks the hydrogen bonds between complementary base pairs, unwinding and separating (unzipping) the two strands of the double helix to form a replication fork. Each of the two original (parental) strands then acts as a template. Free DNA nucleotides in the nucleus align opposite their complementary bases on each template strand by complementary base pairing (A with T, C with G). DNA polymerase then catalyses the formation of phosphodiester bonds between adjacent nucleotides, joining them together into a new strand. Because each of the two resulting DNA molecules is made up of one original (parental) strand and one newly synthesised strand, the process is described as semi-conservative.

1 mark: DNA helicase breaks hydrogen bonds/unwinds and separates the two strands; 1 mark: each parental strand acts as template, free nucleotides align by complementary base pairing; 1 mark: DNA polymerase joins nucleotides (forms phosphodiester bonds) into new strand; each daughter molecule has one old and one new strand (semi-conservative); [3]

DNA polymerase can only add new nucleotides to a growing strand in the 5' to 3' direction. On the leading-strand template, this direction is the same as the direction in which the replication fork opens, so synthesis is continuous. On the lagging-strand template, the fork opens in the direction opposite to that in which polymerase can synthesise, so the new strand must be made discontinuously, as a series of short fragments (Okazaki fragments), each begun separately as more of the template is exposed. DNA ligase then catalyses the formation of phosphodiester bonds between the sugar-phosphate backbones of adjacent Okazaki fragments, joining/sealing them together into one continuous lagging strand.

1 mark: DNA polymerase synthesises only in the 5' to 3' direction; 1 mark: leading strand synthesised continuously, lagging strand synthesised discontinuously as Okazaki fragments because fork opens in opposite direction to synthesis; 1 mark: DNA ligase joins/seals adjacent Okazaki fragments together; [3]

During prophase I of meiosis, homologous chromosomes pair up closely along their length to form a bivalent. Non-sister chromatids of the two homologous chromosomes come into contact and may break and rejoin at points called chiasmata, exchanging equivalent segments of DNA (and therefore alleles) between the maternal and paternal chromatids. This produces chromatids carrying new combinations of alleles that were not present together on either original parental chromosome. Because these recombinant chromosomes are then distributed into different gametes at the two meiotic divisions, crossing over increases the genetic variation among the gametes (and therefore offspring) produced.

1 mark: homologous chromosomes pair to form a bivalent in prophase I; 1 mark: chiasmata form, non-sister chromatids exchange segments/alleles; 1 mark: new/recombinant allele combinations produced, not present on original parental chromosomes, increasing variation in gametes; [3]

A restriction enzyme is used to cut out the gene for human insulin (or its complementary DNA copy), cutting at a specific recognition sequence and leaving fragments with 'sticky ends'. The same restriction enzyme is used to cut open a bacterial plasmid, producing complementary sticky ends. The insulin gene is inserted into the cut plasmid and DNA ligase is used to seal the sugar-phosphate backbones together, forming a recombinant plasmid. This recombinant plasmid is inserted into a host bacterium, such as Escherichia coli, by transformation. Bacteria that have successfully taken up the plasmid are identified and selected, usually using a marker gene, and are then cultured on a large scale in a fermenter, where they express the inserted gene and synthesise human insulin, which is subsequently extracted and purified.

1 mark: restriction enzyme cuts insulin gene and plasmid producing (complementary) sticky ends; 1 mark: DNA ligase inserts gene into plasmid, forming recombinant plasmid; 1 mark: plasmid transformed into host bacterium, selected and cultured (fermenter) to express and produce insulin; [3]

Genetically engineered human insulin has an amino acid sequence identical to naturally occurring human insulin, so it is less likely to be recognised as foreign and trigger an immune or allergic reaction in patients, unlike animal-derived insulin, which differs slightly in structure. In addition, production using genetically modified bacteria grown in large fermenters allows very large, reliable and continuous quantities of insulin to be produced relatively cheaply, without depending on a limited and variable supply of animal pancreases, and it avoids ethical or religious objections that some patients have to using an animal-derived product.

Any two of: 1 mark: identical to human insulin, reduced risk of immune/allergic rejection; 1 mark: can be mass-produced reliably/cheaply in fermenters, not limited by animal pancreas supply; 1 mark: avoids ethical/religious objections to animal-derived products; [3]

GM crops engineered for pest or herbicide resistance can increase crop yield by reducing losses from pests and disease, and can reduce spending on pesticides, potentially increasing the farmer's overall income. However, GM seed is often considerably more expensive to buy than conventional seed, and much GM seed is patented by the company that developed it, so farmers may be legally required to buy new seed each growing season rather than saving seed from their harvest, increasing their long-term costs and dependence on seed companies. Smaller or subsistence farmers, particularly in developing countries, may be unable to afford the initial cost of GM seed and associated inputs, which could widen the economic gap between them and larger, wealthier commercial farms.

Any three of: 1 mark: increased yield/reduced pesticide costs can increase farmer profit; 1 mark: GM seed is expensive/patented, cannot be saved and re-sown, increasing dependence/cost; 1 mark: smaller/subsistence farmers may be unable to afford GM technology, economic disadvantage; 1 mark: valid alternative economic point credited (e.g. export market access/consumer resistance affecting price); [3]

Because changes made to the genome of an embryo affect germline (reproductive) cells, they will be passed on to all future generations descended from that individual. Any unforeseen harmful effects, such as off-target mutations, could therefore be inherited by descendants who never consented to the procedure. There is also concern that germline gene editing could be used not only to correct disease-causing alleles but also to select for or enhance non-medical traits ('designer babies'), which raises issues of equality of access, since only those who could afford the technology would benefit, potentially widening social inequality and affecting society's acceptance of natural human variation and disability.

Any two of: 1 mark: changes are heritable, passed to future generations without their consent/unknown long-term or off-target risks; 1 mark: potential for use beyond disease correction/enhancement of non-medical traits ('designer babies'); 1 mark: access/equality concerns, only the wealthy could afford the technology, widening inequality; [3]

Somatic cell gene therapy involves inserting a functioning allele into the body (somatic) cells of a patient, for example cells of the lungs or bone marrow; the genetic change affects only the treated individual and is not passed on to their offspring, since it does not affect the gametes. Germline gene therapy involves altering the genes present in gametes or in the cells of a very early embryo, so that the change becomes present in every cell of the resulting individual, including their own reproductive cells. As a result, unlike somatic gene therapy, germline changes are heritable and can be passed on to future generations.

1 mark: somatic therapy alters body cells of patient only, effect not passed to offspring; 1 mark: germline therapy alters gametes/early embryo cells, present in all resulting cells including reproductive cells; 1 mark: key difference stated — germline changes are heritable/passed to future generations, somatic changes are not; [3]

A virus, such as an adenovirus, is first genetically modified/disabled so that it cannot cause disease or replicate uncontrollably within the patient, and is used as a vector. A functioning copy of the required human gene is inserted into the viral genome. The modified virus is then introduced into the patient, for example by inhalation into the airway epithelium in the treatment of cystic fibrosis, where it infects the target cells in the same way as a normal virus, injecting its genetic material, including the functioning gene, into the host cell. Once inside the target cell, the functioning gene can be transcribed and translated, so the cell produces the protein it was previously unable to make.

1 mark: virus is disabled/modified so it is harmless, used as a vector; 1 mark: functioning gene inserted into the viral genome; 1 mark: virus infects target host cells, delivering the gene, which is expressed (transcribed/translated) to produce the required protein; [3]

Patients with ADA-SCID lack a functioning allele for the enzyme adenosine deaminase (ADA), so their lymphocytes (T- and B-cells) cannot develop or function normally, leaving the patient highly vulnerable to infection. In gene therapy, bone marrow (haematopoietic) stem cells are removed from the patient, and a functioning copy of the ADA gene is inserted into these cells in vitro, usually using a disabled viral vector. The genetically modified stem cells are then returned to the patient's bone marrow, where they divide and differentiate into lymphocytes that now carry and express the functioning ADA gene, restoring enzyme activity and allowing the immune system to develop and function normally.

1 mark: patient's cells lack a functioning ADA gene, so immune (lymphocyte) function/development is impaired, prone to infection; 1 mark: bone marrow/haematopoietic stem cells removed and a functioning ADA gene inserted (via a vector) in vitro; 1 mark: modified cells returned to patient, express the functioning gene as they differentiate, restoring immune function; [3]

The DNA sample is first heated to around 95 degrees Celsius (denaturation); this breaks the hydrogen bonds between the two strands, separating the double helix into two single strands. The mixture is then cooled to around 50-65 degrees Celsius (annealing), allowing short, single-stranded DNA primers, which are complementary to sequences flanking the target region, to bind to each single strand. The temperature is then raised to around 72 degrees Celsius (extension), at which a heat-stable DNA polymerase (such as Taq polymerase) adds free nucleotides to the primers, synthesising new complementary strands and doubling the amount of the target DNA sequence. This three-step cycle is then repeated many times, amplifying the target DNA exponentially.

1 mark: denaturation (approx. 95C) — DNA heated, hydrogen bonds broken, strands separate; 1 mark: annealing (approx. 50-65C) — primers bind to complementary flanking sequences; 1 mark: extension (approx. 72C) — (Taq) DNA polymerase adds nucleotides, synthesises new strand; cycle repeated to amplify DNA; [3]

Possible limitations/risks of gene therapy include: the effect of somatic gene therapy is often temporary, because the treated cells eventually die and are replaced by cells lacking the functioning gene, meaning treatment may need to be repeated; the viral vector used to deliver the gene may trigger an immune response in the patient; and there is a risk that the inserted gene may integrate into the host genome at a random location, potentially disrupting another important gene, such as a tumour suppressor gene, and causing further health problems such as cancer.

Any two of: 1 mark: effect often temporary, treated cells die and are replaced, treatment must be repeated; 1 mark: immune response triggered by the vector; 1 mark: risk of insertional mutagenesis disrupting another (e.g. tumour suppressor) gene; [2]

Restriction enzymes cut DNA at specific recognition/base sequences, producing fragments with complementary 'sticky ends'; this is used both to cut open the plasmid and to cut out the gene of interest from a larger DNA sample. DNA ligase then catalyses the joining/sealing of the sugar-phosphate backbones of the gene of interest and the cut plasmid together, forming a single, continuous recombinant DNA molecule.

1 mark: restriction enzyme cuts DNA at a specific (recognition) sequence, producing (complementary) sticky ends; 1 mark: DNA ligase joins/seals the gene and plasmid together to form the recombinant plasmid; [2]

DNA fragments (produced, for example, by restriction enzyme digestion or PCR amplification of variable regions of DNA) are loaded into wells at one end of an agarose gel, and an electric current is passed across the gel. Because DNA is negatively charged, due to its phosphate groups, the fragments migrate through the gel towards the positive electrode. Smaller fragments move further/faster through the pores of the gel in a given time than larger fragments, because they experience less resistance passing through the gel matrix; this separates the fragments by size, producing a banding pattern.

1 mark: DNA (negatively charged) fragments loaded in wells, migrate through the gel towards the positive electrode under an electric current; 1 mark: smaller fragments move further/faster because they experience less resistance through the gel matrix, so fragments are separated by size; [2]

A totipotent stem cell can differentiate into any cell type found in the organism, including extra-embryonic/placental tissue, meaning it has the potential to form an entire new organism. A pluripotent stem cell can differentiate into almost any cell type of the body, derived from the three embryonic germ layers, but it cannot form extra-embryonic tissue and so cannot, on its own, develop into a whole new organism.

1 mark: totipotent cells can form any cell type, including extra-embryonic tissue/a whole organism; 1 mark: pluripotent cells can form almost any body cell type but not extra-embryonic tissue/a whole organism; [2]

One example is the treatment of leukaemia using a bone marrow (haematopoietic stem cell) transplant. The patient's own diseased bone marrow is destroyed, typically using chemotherapy or radiotherapy, and is then replaced with healthy multipotent haematopoietic stem cells from a donor. These stem cells divide and differentiate to produce new, healthy blood cells (red blood cells, white blood cells and platelets), restoring normal blood cell production. Other valid examples include using stem cells to repair damaged heart muscle after a heart attack by differentiating into cardiac muscle cells, or treating type 1 diabetes using stem cells differentiated into insulin-producing pancreatic beta cells.

1 mark: valid therapeutic application correctly named (e.g. leukaemia/bone marrow transplant, cardiac repair, type 1 diabetes); 1 mark: correct explanation of how the relevant stem cells differentiate to replace/repair the specific damaged or diseased cell type; [2]

A gene is a length/sequence of DNA, found at a specific locus on a chromosome, that codes for a specific polypeptide or protein (and so influences a particular characteristic). An allele is one of two or more alternative forms of a gene, which differ from other alleles of the same gene by one or a small number of bases, and which can give rise to different versions of the characteristic or protein produced.

1 mark: gene = a length of DNA/sequence of bases, at a specific locus, that codes for a polypeptide/protein or characteristic; 1 mark: allele = an alternative form of a gene, differing in base sequence from other alleles of that gene; [2]

A high-quality answer would address the following areas in a structured, linked discussion:

Medical/economic benefits: Genetic engineering and stem cell technologies offer treatments for conditions that were previously untreatable. For example, gene therapy has been used to treat severe combined immunodeficiency (SCID) by inserting a functioning ADA gene into a patient's stem cells, and cystic fibrosis by delivering a functioning CFTR gene to airway cells using a viral vector. Stem cell transplants are already used to treat leukaemia by replacing diseased bone marrow, and are being researched for conditions such as Parkinson's disease and spinal cord injury. Effective cures could improve patients' quality of life and, over the long term, could reduce the economic burden on health services of managing chronic disease, potentially offsetting the high initial costs of research and treatment.

Ethical concerns: Deriving embryonic stem cells typically involves the destruction of a human embryo, which some people believe has moral status from conception, making this ethically unacceptable to them. Germline gene editing alters the genome of gametes or early embryos; because this change is heritable, it is passed on to all future generations without their consent, and any unforeseen harmful effects (such as off-target mutations) could also be inherited. There is also concern that such technology could be used not only to treat disease but to enhance non-medical traits ('designer babies'), raising concerns about the extent to which humans should be able to alter their own genetic inheritance.

Social concerns: These treatments are often expensive and may not be equally accessible to all patients, which could widen existing health inequalities between wealthier and poorer patients, or between richer and poorer countries. There is also public concern and mistrust around the safety of new genetic technologies and their long-term, unknown effects, and some religious or cultural groups object to genetic modification or embryo research on principle, which can affect public acceptance and uptake of these treatments.

Economic implications: Research, development, clinical trials and regulation of genetic engineering and stem cell technologies are extremely costly, and these costs are often reflected in high treatment prices. Companies that develop genetic technologies frequently patent their techniques or products, allowing them to control pricing and profit from their use, which raises questions about who owns and benefits economically from genetic technologies. Set against this, successfully treating or curing a genetic condition may ultimately save health services money by removing the need for lifelong management of a chronic illness.

A strong response would draw these strands together into a balanced, evidence-based conclusion, for example noting that while genetic engineering and stem cell technologies offer significant medical and long-term economic benefits, their ethical acceptability and the fairness of access to them remain unresolved issues that will shape how widely they are adopted.

Assessed using a Level of Response mark scheme, integrating scientific content with Quality of Written Communication (QWC).

Indicative content: benefits of genetic engineering/stem cell treatments (named examples, e.g. gene therapy for SCID/cystic fibrosis, stem cells for leukaemia/Parkinson's); ethical concerns (embryo destruction/moral status, heritable germline editing, 'designer babies'); social concerns (unequal access/health inequality, public trust and safety concerns, religious/cultural objections); economic implications (high R&D and treatment costs, patenting/ownership and profit, potential long-term savings to health services).

Level 3 (Excellent, 10-13 marks): Detailed, well-balanced discussion that addresses ethical, social AND economic implications, supported by specific, accurate scientific examples and named technologies/conditions. The answer has a clear, logical structure with a coherent line of reasoning, uses appropriate scientific and specialist terminology accurately throughout, and contains few, if any, errors in spelling, punctuation and grammar.

Level 2 (Good, 6-9 marks): A reasonable discussion covering at least two of ethical, social or economic implications, with some relevant scientific detail or examples, though coverage may be uneven or one area is underdeveloped. The answer shows some clear structure; terminology is mostly used appropriately; spelling, punctuation and grammar are mostly accurate, with minor errors that do not significantly hinder meaning.

Level 1 (Basic, 1-5 marks): A limited, largely generic or one-sided discussion (for example, addressing only ethical or only economic implications), with little scientific detail or few relevant examples, and/or presented as a list of unlinked points rather than continuous, structured writing. Errors in spelling, punctuation and grammar are noticeable and may hinder the clear communication of meaning.

0 marks: No relevant content, or a response that does not address the question. [13]