2024 CCEA A-Level Life and Health Sciences 0008 模拟试题及答案详解

(a) The straight-chain aldehyde with molecular formula C4H8O is butanal: \( \text{CH}_3\text{CH}_2\text{CH}_2\text{CHO} \), IUPAC name butanal. (b) A ketone isomer must have the carbonyl group within the chain rather than at the end: butan-2-one, \( \text{CH}_3\text{COCH}_2\text{CH}_3 \). (c) Butanal and butan-2-one have the same molecular formula but the carbonyl functional group is positioned differently (aldehyde vs ketone), so they show structural isomerism, specifically functional group isomerism. (d) A cyclic ether isomer of C4H8O could be, for example, tetrahydrofuran (a five-membered ring containing one oxygen atom, C4H8O). This also shows functional group isomerism relative to butanal, since the oxygen-containing functional group (ether within a ring, no C=O) is entirely different from the aldehyde's C=O group, though the molecular formula is identical. (e) (i) The aldehyde's carbonyl (C=O) bond gives a strong, sharp absorption at approximately 1720 cm-1 (typical range 1680-1750 cm-1 for aldehydes/ketones). (ii) The aldehydic C-H bond (the hydrogen directly attached to the carbonyl carbon) gives a characteristic, relatively weak absorption in the range 2750-2900 cm-1, which is diagnostic for aldehydes (distinguishing them from ketones, which lack this C-H). Final answer: (a) butanal, CH3CH2CH2CHO; (b) butan-2-one, CH3COCH2CH3; (c) structural (functional group) isomerism; (d) tetrahydrofuran (cyclic ether), functional group isomerism; (e) (i) ~1720 cm-1, C=O stretch; (ii) 2750-2900 cm-1, aldehydic C-H stretch.

[1] correct structural formula for butanal; [1] correct IUPAC name 'butanal'; [1] correct structural formula for a valid ketone isomer (e.g. butan-2-one); [1] correct IUPAC name for the ketone; [1] correctly identifies structural/functional group isomerism between (a) and (b); [1] a valid cyclic ether structure drawn for Y (e.g. tetrahydrofuran) with correct molecular formula C4H8O; [1] correctly names the isomerism between Y and the aldehyde as structural/functional group isomerism; [1] correct wavenumber (approx. 1680-1750 cm-1) and bond (C=O) for (e)(i); [1] correct wavenumber range (2750-2900 cm-1) for (e)(ii); [1] correctly identifies the bond in (e)(ii) as C-H (aldehydic C-H). Maximum 10 marks.

(a) As the electron-rich C=C double bond of but-1-ene approaches the H-Br molecule, the high electron density of the pi bond repels and polarises the H-Br bond further, inducing a dipole (delta-positive on H, delta-negative on Br) even though HBr is already polar. The pi electrons of the double bond then act as a nucleophile, attacking the delta-positive hydrogen atom of HBr; this forms a new C-H bond and causes heterolytic fission of the H-Br bond, with both electrons from that bond going to the bromine atom, releasing a bromide ion, Br-. This leaves a positively charged carbon (a carbocation) on the carbon that did not bond to hydrogen. The bromide ion, being nucleophilic, is then attracted to and bonds with this positively charged carbon, forming the final addition product. (b) Addition of H+ can occur at either carbon of the C=C double bond, giving two possible carbocation intermediates: a secondary carbocation (if H adds to the terminal CH2, giving a positive charge on the second carbon, which bears two alkyl groups) or a primary carbocation (if H adds to the second carbon, giving a positive charge on the terminal carbon, which bears only one alkyl group). Alkyl groups are electron-donating (positive inductive effect), which stabilises a carbocation by reducing the concentration of positive charge; a secondary carbocation, with two electron-donating alkyl groups attached to the positively charged carbon, is therefore more stable than a primary carbocation, which has only one. The reaction proceeds preferentially via the more stable secondary carbocation intermediate, so the major product is 2-bromobutane (CH3CHBrCH2CH3), formed when Br- bonds to the secondary carbocation; the minor product, 1-bromobutane, forms via the less stable primary carbocation pathway and is produced in a much smaller amount. (c) Addition polymerisation requires monomers containing a C=C double bond, which opens up to form single bonds linking monomer units together. The repeat unit of poly(but-1-ene), formed from but-1-ene monomers, is -[CH(CH2CH3)-CH2]-, reflecting the two carbons of the original double bond now joined by single bonds to neighbouring repeat units, with the ethyl side chain (CH2CH3) retained from the original but-1-ene structure. (d) Because addition polymers such as this are chemically inert (resistant to chemical breakdown), they do not readily biodegrade, creating a waste management challenge. Two strategies used to manage this waste are: incineration, in which the polymer waste is burned under controlled conditions to release its stored energy (which can be used, for example, to generate electricity), and recycling, in which waste polymer is collected, sorted, melted and reprocessed into new plastic products, reducing the need for new (virgin) polymer to be manufactured from crude oil feedstock. Using the waste polymer as feedstock for cracking (breaking it back down into smaller, useful hydrocarbon molecules) is also an accepted strategy. Final answer: (a) the pi bond acts as a nucleophile, polarising and heterolytically breaking H-Br to form a carbocation intermediate, which is then bonded to by Br-; (b) 2-bromobutane is the major product, formed via the more stable secondary carbocation (stabilised by two electron-donating alkyl groups) rather than the less stable primary carbocation; (c) a C=C double bond is required, giving the repeat unit -[CH(CH2CH3)-CH2]-; (d) incineration to release energy, and recycling/use as cracking feedstock.

(a) [1] the C=C pi bond acts as a nucleophile/electron-rich centre; [1] this induces/increases polarisation of the H-Br bond; [1] the pi electrons attack the delta-positive hydrogen, forming a new C-H bond; [1] heterolytic fission of the H-Br bond occurs, releasing Br-; [1] correctly identifies the resulting intermediate as a carbocation (positively charged carbon). (b) [1] correctly identifies 2-bromobutane as the major product; [1] correctly identifies that the reaction proceeds via the more stable carbocation intermediate; [1] correctly identifies this as the secondary carbocation (compared with a primary carbocation for the minor product); [1] correct explanation that alkyl groups are electron-donating, stabilising the positive charge, so a secondary carbocation (two alkyl groups) is more stable than a primary carbocation (one alkyl group). (c) [1] correctly states a C=C double bond is required; [1] correct repeat unit skeleton showing two linked carbons in square brackets; [1] correct ethyl side chain shown on the repeat unit. (d) [1] mark for each valid, correctly described waste management strategy, to a maximum of 3 (e.g. incineration to release energy [1-2 depending on detail]; recycling [1-2]; use as feedstock for cracking [1-2]), up to the section maximum. Maximum 15 marks overall.

(a)(i) The industrial process used to break long-chain alkanes into shorter, more useful (and more valuable) molecules is called cracking. (ii) Two sets of conditions are used industrially: thermal cracking, which uses a high temperature (typically around 700-900C) and high pressure without a catalyst, tending to produce a higher proportion of alkenes; and catalytic cracking, which uses a lower temperature (typically around 450C), a slight pressure, and a zeolite catalyst, tending to produce a higher proportion of branched and aromatic hydrocarbons useful in petrol. (iii) Cracking decane (C10H22) into octane (C8H18) must also produce a molecule containing the remaining 2 carbons and 4 hydrogens; since cracking produces an alkane and an alkene, the balanced equation is: C10H22 -> C8H18 + C2H4 (ethene). Checking atoms: carbon 10 = 8 + 2; hydrogen 22 = 18 + 4; the equation is balanced. (b)(i) Complete combustion of propane in a plentiful supply of oxygen produces carbon dioxide and water: C3H8 + 5O2 -> 3CO2 + 4H2O. Checking atoms: carbon 3 = 3; hydrogen 8 = 8; oxygen 10 = 10; balanced. (ii) Incomplete combustion, occurring in a limited oxygen supply, can produce carbon monoxide instead of carbon dioxide: 2C3H8 + 7O2 -> 6CO + 8H2O (doubled to avoid a fractional oxygen coefficient). Checking atoms: carbon 6 = 6; hydrogen 16 = 16; oxygen 14 = 14; balanced. (iii) Carbon monoxide is toxic because it binds to haemoglobin in red blood cells more strongly than oxygen does, forming carboxyhaemoglobin; this reduces the blood's capacity to transport oxygen around the body, which can cause headaches, unconsciousness, or death in cases of severe exposure. (iv) Nitrogen oxides (NOx) are also produced in vehicle engines, formed when the nitrogen and oxygen naturally present in air react together under the high temperature and pressure inside the engine; nitrogen oxides contribute to acid rain (by forming nitric acid in the atmosphere) and to photochemical smog. (c) Level of response guidance (indicative content, to be marked using the 4-band rubric below): Advantages - bioethanol is produced from renewable crops (such as sugar cane or maize) that can be regrown, unlike finite fossil fuel reserves of crude oil; because the plants used to make bioethanol absorb carbon dioxide from the atmosphere during photosynthesis as they grow, burning bioethanol is often considered to be closer to carbon neutral than burning petrol, potentially reducing its net contribution to the enhanced greenhouse effect and climate change; using biofuels also reduces a country's dependency on imported crude oil. Disadvantages - growing crops for biofuel production competes for agricultural land that could otherwise be used to grow food crops, which can contribute to higher food prices or encourage deforestation to clear additional land for fuel crops, which itself releases stored carbon and reduces future carbon absorption; the farming, fermentation, distillation and transport processes involved in producing bioethanol themselves require energy (often from fossil fuels) and may involve fertilisers and pesticides with their own environmental impacts, meaning the true net carbon and environmental benefit is smaller than a simple 'carbon neutral' claim suggests; bioethanol also has a lower energy density than petrol, meaning a greater volume must be burned to release the same amount of energy, and higher-ethanol blends can require vehicle engines to be modified. Evaluation - overall, biofuels such as bioethanol can offer a genuine, if partial, reduction in net carbon emissions and reduced dependence on finite fossil fuels compared with petrol, but this benefit is reduced by the land-use, food-supply and production-energy costs involved, meaning biofuels are best regarded as one part of a wider strategy rather than a complete, unproblematic replacement for fossil fuels. Final answer: (a) cracking; thermal (high temp/pressure, no catalyst) and catalytic (lower temp, zeolite catalyst); C10H22 -> C8H18 + C2H4. (b) C3H8 + 5O2 -> 3CO2 + 4H2O; 2C3H8 + 7O2 -> 6CO + 8H2O; CO poisoning (reduces oxygen transport); NOx causing acid rain/smog. (c) biofuels are renewable and reduce net CO2 emissions compared with fossil fuels, but land use, food competition and production energy costs mean the environmental benefit, while real, is only partial.

(a)(i) [1] cracking. (ii) [1] thermal cracking named with a valid condition (e.g. high temperature/pressure, no catalyst); [1] catalytic cracking named with a valid condition (e.g. lower temperature, zeolite catalyst); [1] for a further correct distinguishing detail (e.g. product tendency, or correct approximate temperature) between the two. (iii) [1] correct products (C8H18 and C2H4); [1] correctly balanced equation. (b)(i) [1] correct products (CO2 and H2O); [1] correctly balanced equation. (ii) [1] correct products (CO and H2O); [1] correctly balanced equation (integer coefficients). (iii) [1] valid health hazard correctly explained (e.g. CO binds to haemoglobin, reducing oxygen transport, causing headaches/unconsciousness/death). (iv) [1] a valid pollutant named (e.g. NOx, particulates, SO2) with a correct associated environmental effect. Maximum 12 marks for (a)+(b). (c) Level of response mark scheme (12 marks, 4 bands): Excellent (9-12 marks): discusses at least two distinct, accurate advantages and two distinct, accurate disadvantages of biofuels compared with fossil fuels, reaching a balanced evaluative conclusion; accurate specialist vocabulary throughout; clear, coherent, well-structured writing. Good (5-8 marks): discusses some relevant advantages and disadvantages with generally correct reasoning, though may be less complete or balanced; mostly accurate vocabulary; generally clear writing with minor errors. Basic (1-4 marks): identifies one or two relevant points with limited explanation or one-sided coverage; writing may be list-like or contain errors that hinder meaning. 0 marks: no creditable response. Total maximum for the question: 24 marks.

(a)(i) The oxidation of a primary alcohol to a carboxylic acid uses acidified potassium dichromate(VI) solution (potassium dichromate(VI) in dilute sulfuric acid), heated under reflux (to ensure the alcohol is fully oxidised to the carboxylic acid rather than being lost as a volatile aldehyde). As the oxidising agent (dichromate(VI) ions, containing Cr in the +6 oxidation state) is reduced to Cr3+ ions during the reaction, the colour of the solution changes from orange to green. (ii) The equation, using [O] to represent the oxidising agent (a simplified way of representing the oxidation without balancing the full dichromate redox equation), is: CH3CH2CH2OH + 2[O] -> CH3CH2COOH + H2O. Checking atoms: carbon 3 = 3; hydrogen 8 = 6 + 2 = 8; oxygen 1 + 2 = 2 + 1 = 3; the equation is balanced. (iii) Butan-2-ol is a secondary alcohol (the OH-bearing carbon is attached to two other carbon atoms), so oxidation under these conditions produces a ketone, butan-2-one. Unlike aldehydes, ketones cannot be oxidised further under these conditions because the carbonyl carbon in a ketone is bonded to two carbon-containing groups and has no hydrogen atom attached to it; further oxidation to a carboxylic acid would require breaking a carbon-carbon bond, which acidified potassium dichromate(VI) cannot achieve. (iv) A simple test to distinguish an aldehyde from a ketone is to add Fehling's solution (a blue solution containing Cu2+ ions) to a sample of each and warm gently. An aldehyde is able to reduce the Cu2+ ions in Fehling's solution to Cu+ ions, forming a brick-red precipitate of copper(I) oxide; a ketone cannot reduce Fehling's solution (for the same reason it cannot be oxidised further, given above) and so the solution remains blue with no precipitate formed. (An equivalent test using Tollens' reagent, giving a silver mirror with an aldehyde and no change with a ketone, is equally acceptable.) (b)(i) Geometric (E/Z) isomerism requires two conditions to be met: restricted rotation around the C=C double bond (because rotation would require temporarily breaking the pi bond, which does not happen under normal conditions), and each carbon of the double bond must be attached to two different groups. In but-2-ene (CH3-CH=CH-CH3), each double-bond carbon is attached to one methyl group (CH3) and one hydrogen atom, which are different, so geometric isomerism is possible. In but-1-ene (CH2=CH-CH2-CH3), one of the double-bond carbons (the terminal =CH2 carbon) is attached to two identical hydrogen atoms, so this carbon does not have two different groups attached, meaning but-1-ene cannot show geometric isomerism. (ii) In Z-but-2-ene, the two methyl (CH3) groups are positioned on the same side of the C=C double bond (with the two hydrogen atoms also on the same side as each other, opposite the methyl groups). In E-but-2-ene, the two methyl groups are positioned on opposite sides of the double bond (with each methyl group diagonally opposite a hydrogen atom on the other carbon). (iii) Cahn-Ingold-Prelog (CIP) priority rules assign priority to the two groups attached to each carbon of the double bond based on atomic number, with the group containing the atom of higher atomic number directly attached to the double-bond carbon given higher priority. On the left-hand carbon of CHCl=CHBr, the attached groups are Cl (atomic number 17) and H (atomic number 1), so Cl has higher priority than H. On the right-hand carbon, the attached groups are Br (atomic number 35) and H, so Br has higher priority than H. The isomer is labelled Z (from the German 'zusammen', meaning together) if the two higher-priority groups (here, Cl and Br) are on the same side of the double bond, and E (from 'entgegen', meaning opposite) if they are on opposite sides. Since the question asks which isomer has the two higher-priority groups on the same side, this is, by definition, the Z isomer. (iv) E and Z isomers, despite having the same molecular formula and the same atoms bonded to each other, have a different overall shape/geometry and often a different net dipole moment (for example, a Z isomer with two polar groups on the same side often has a larger overall dipole moment than the corresponding E isomer, where the individual bond dipoles may partially or fully cancel). This difference in shape and polarity affects the strength of intermolecular forces between molecules, which is why E and Z isomers can have different physical properties, such as different melting or boiling points. Final answer: (a) acidified potassium dichromate(VI), reflux, orange to green; CH3CH2CH2OH + 2[O] -> CH3CH2COOH + H2O; butan-2-one, cannot be oxidised further (no H on carbonyl carbon); Fehling's/Tollens' test, brick-red precipitate/silver mirror for aldehyde only. (b) restricted C=C rotation plus two different groups per carbon (but-1-ene fails this at the terminal =CH2); Z-but-2-ene has methyls on the same side, E-but-2-ene on opposite sides; Cl and Br are higher priority than H, so the Z isomer has them on the same side; differing shape/dipole moment changes intermolecular forces and physical properties.

(a)(i) [1] acidified potassium dichromate(VI); [1] heated under reflux; [1] colour change orange to green. (ii) [1] correct products (propanoic acid and water); [1] correctly balanced equation with 2[O]. (iii) [1] correctly names butan-2-one/ketone; [1] correct explanation (no hydrogen on the carbonyl carbon/would require breaking a C-C bond). (iv) [1] correct reagent named (Fehling's, Benedict's or Tollens'); [1] correct positive-result observation for an aldehyde (brick-red precipitate or silver mirror); [1] correctly states a ketone gives no reaction/no colour change under this test. (b)(i) [1] restricted rotation around the C=C double bond; [1] each carbon of the double bond must have two different attached groups; [1] correctly explains that but-1-ene's terminal carbon has two identical H atoms, so does not meet this condition. (ii) [1] correct description of Z-but-2-ene (methyl groups on the same side); [1] correct description of E-but-2-ene (methyl groups on opposite sides). (iii) [1] correctly states priority is assigned by atomic number of the attached atom; [1] correctly identifies Cl and Br as the higher-priority groups (over H) on their respective carbons; [1] correctly identifies the isomer with both higher-priority groups on the same side as Z. (iv) [1] correctly identifies a difference in shape/dipole moment between E and Z isomers; [1] correctly links this to a difference in intermolecular forces/physical properties (e.g. melting or boiling point). Maximum 20 marks.

(a)(i) Propan-2-ol is manufactured by reacting propene with steam (H2O in the gas phase) in the presence of a phosphoric acid catalyst (concentrated sulfuric acid is used similarly in some processes), at a high temperature and pressure. (ii) The addition of water across the C=C double bond of propene proceeds via a carbocation intermediate, in the same way as electrophilic addition of HBr does. Protonation of the double bond can occur at either carbon, giving either a secondary carbocation (positive charge on the central carbon, stabilised by two electron-donating alkyl/methyl groups) or a primary carbocation (positive charge on the terminal carbon, stabilised by only one alkyl group). Since alkyl groups are electron-donating and help stabilise a positive charge, the secondary carbocation is more stable than the primary carbocation, so the reaction proceeds preferentially through this more stable intermediate. Water then bonds to this secondary carbocation, and loss of a proton gives propan-2-ol (with the OH group on the central carbon) as the major product, rather than propan-1-ol (which would form via the less stable primary carbocation). (b)(i) 1-bromopropane reacts with aqueous sodium hydroxide in a simple 1:1 substitution: CH3CH2CH2Br + NaOH -> CH3CH2CH2OH + NaBr. (ii) This is a nucleophilic substitution reaction. The hydroxide ion, OH-, acts as the nucleophile: it has lone pairs of electrons on the oxygen atom and is attracted to the slightly positive carbon atom that is bonded to the electronegative bromine atom. The hydroxide ion attacks this carbon atom, and as the new C-OH bond begins to form, the C-Br bond simultaneously breaks heterolytically, with both electrons from the bond going to the bromine atom; bromine leaves the molecule as a bromide ion, Br-, which is the leaving group. Because 1-bromopropane is a primary haloalkane, this substitution proceeds as a single-step mechanism in which the nucleophile attacks from the side of the carbon atom opposite to the departing bromine atom. Final answer: (a) steam with a phosphoric acid catalyst, high temperature/pressure; propan-2-ol is the major product because it forms via the more stable secondary carbocation. (b) CH3CH2CH2Br + NaOH -> CH3CH2CH2OH + NaBr; nucleophilic substitution in which OH- (nucleophile) attacks the carbon bonded to Br, displacing Br- (leaving group) as the C-Br bond breaks.

(a)(i) [1] steam/water named as the reagent; [1] correct catalyst named (phosphoric acid or concentrated sulfuric acid) with high temperature/pressure. (ii) [1] correctly identifies that the reaction proceeds via a carbocation intermediate; [1] correctly identifies the secondary carbocation as more stable than the primary carbocation; [1] correct explanation in terms of alkyl groups being electron-donating/stabilising the positive charge. (b)(i) [1] correct products (propan-1-ol and NaBr); [1] correctly balanced equation. (ii) [1] correctly identifies OH- as the nucleophile, attacking the carbon bonded to bromine; [1] correctly identifies Br- as the leaving group, with heterolytic fission of the C-Br bond; [1] correctly describes this as a nucleophilic substitution (single-step, back-side attack for a primary haloalkane). Maximum 10 marks.

(a)(i) Salicylic acid reacts with acetic anhydride in a 1:1 ratio, with the OH group of salicylic acid being acetylated (esterified), producing aspirin and acetic acid as a by-product: C7H6O3 + C4H6O3 -> C9H8O4 + CH3COOH. (ii) Moles of salicylic acid = mass / Mr = 2.76 / 138 = 0.0200 mol. (iii) Since the reaction is 1:1 between salicylic acid and aspirin, moles of aspirin (theoretical) = moles of salicylic acid = 0.0200 mol. Theoretical mass of aspirin = moles x Mr = 0.0200 x 180 = 3.60 g. (iv) Percentage yield = (actual yield / theoretical yield) x 100 = (3.15 / 3.60) x 100 = 87.5%. (b)(i) The technique used to purify the crude solid aspirin is recrystallisation. (ii) The crude solid is dissolved in the minimum volume of a hot suitable solvent (enough to just dissolve it), the hot solution is filtered if needed to remove insoluble impurities, then allowed to cool slowly, during which the desired product (aspirin) crystallises out of solution as it becomes less soluble at lower temperature, while soluble impurities remain dissolved in the small volume of solvent; the crystals are then filtered from the solvent, washed with a small amount of cold solvent to remove surface impurities, and dried. (iii) Aqueous iron(III) chloride (FeCl3) reacts with phenols to give a distinctive purple/violet colouration. Salicylic acid contains a phenol (-OH directly attached to the benzene ring), so any unreacted salicylic acid remaining as an impurity in the aspirin sample would give a purple/violet colour with FeCl3 solution. Aspirin itself does not contain a free phenol group (the phenolic OH has been converted to an ester group during the reaction), so pure aspirin gives no colour change (or only a very faint/pale colour) with FeCl3; a purple/violet colour therefore indicates the sample is impure, containing unreacted salicylic acid. (c)(i) The molecular ion peak, M+, is formed when a whole aspirin molecule loses one electron (is ionised) in the mass spectrometer without fragmenting; its mass-to-charge ratio (m/z = 180, given the ion has a single positive charge) is therefore equal to the relative molecular mass of aspirin. (ii) The mass of the fragment lost is found by subtracting the fragment ion's m/z value from the molecular ion's m/z value: 180 - 138 = 42. A neutral fragment of mass 42 lost from aspirin (which contains an -O-CO-CH3 ester/acetyl group) corresponds to the loss of ketene, CH2=C=O (molecular formula C2H2O, Mr = (2x12) + (2x1) + 16 = 24+2+16 = 42), formed by loss of the acetyl portion of the ester group; this regenerates a fragment ion corresponding to the mass of salicylic acid itself (138), consistent with the given Mr of salicylic acid. (d) In combustion analysis, all of the carbon in the original sample ends up in the carbon dioxide produced, and all of the hydrogen ends up in the water produced. Moles of CO2 = mass / Mr = 1.98 / 44 = 0.0450 mol, so moles of C (in the original sample) = 0.0450 mol (since each CO2 contains one C atom), giving a mass of carbon = 0.0450 x 12 = 0.540 g. Moles of H2O = 0.36 / 18 = 0.0200 mol, so moles of H (in the original sample) = 0.0200 x 2 = 0.0400 mol (since each H2O contains two H atoms), giving a mass of hydrogen = 0.0400 x 1 = 0.0400 g. The mass of oxygen in the original 0.900 g sample is found by difference: mass of O = 0.900 - 0.540 - 0.0400 = 0.320 g, so moles of O = 0.320 / 16 = 0.0200 mol. The mole ratio C : H : O = 0.0450 : 0.0400 : 0.0200. Dividing through by the smallest value (0.0200): C : H : O = 2.25 : 2.00 : 1.00. Multiplying through by 4 to obtain whole numbers gives C : H : O = 9 : 8 : 4. The empirical formula is therefore C9H8O4 (which, in this case, is also the molecular formula of aspirin, since Mr = (9x12)+(8x1)+(4x16) = 108+8+64 = 180, matching the molecular ion peak found in part (c)). Final answer: (a) C7H6O3 + C4H6O3 -> C9H8O4 + CH3COOH; 0.0200 mol; 3.60 g; 87.5%. (b) recrystallisation (dissolve in minimum hot solvent, cool to crystallise, filter, wash, dry); FeCl3 gives a purple/violet colour with a phenol (unreacted salicylic acid), indicating impurity, while pure aspirin gives no colour change. (c) M+ represents the whole ionised molecule, mass = Mr; loss of 42 = ketene, CH2=C=O (C2H2O). (d) empirical formula C9H8O4.

(a)(i) [1] correct products; [1] correctly balanced 1:1:1:1 equation. (ii) [1] correct value, 0.0200 mol (accept 2.00 x 10^-2 mol), with working shown. (iii) [1] correct method (using 1:1 mole ratio); [1] correct final mass, 3.60 g. (iv) [1] correct method (actual/theoretical x 100); [1] correct final answer, 87.5%. (b)(i) [1] recrystallisation. (ii) [1] dissolve in minimum hot solvent (and cool to crystallise); [1] filter, wash and dry the crystals. (iii) [1] correctly describes adding FeCl3 and the purple/violet colour with a phenol/salicylic acid; [1] correctly links this colour to an impure sample, with pure aspirin giving no such colour change. (c)(i) [1] correctly describes the molecular ion as the whole molecule having lost one electron, with m/z equal to the relative molecular mass. (ii) [1] correct subtraction showing mass lost = 42; [1] correctly identifies the fragment as ketene/C2H2O; [1] correct supporting reasoning (e.g. linking to loss of the acetyl/ester portion, or correctly calculating Mr of C2H2O = 42). (d) [1] correct moles of CO2 and hence moles/mass of C; [1] correct moles of H2O and hence moles/mass of H; [1] correct mass of O found by difference and correct moles of O; [1] correct mole ratio simplified by dividing by the smallest value; [1] correct final empirical formula, C9H8O4, obtained by scaling to whole numbers. Maximum 21 marks.

(a) The normal range of human core body temperature is approximately 36.5C to 37.5C, with 37C often quoted as the average value. (b) A traditional mercury-in-glass thermometer relies on the thermal expansion of mercury, which takes several minutes to reach thermal equilibrium with the body, giving a slow response time; it also poses a safety risk, since mercury is toxic and the glass thermometer can break, potentially causing injury and mercury contamination (which is why mercury thermometers have been widely phased out of clinical use). A digital electronic thermometer uses a thermistor, a component whose electrical resistance changes in a well-defined way with temperature (typically decreasing as temperature increases); because the electronic circuit can respond almost instantly to this resistance change, a digital thermometer has a much faster response time, typically producing a reading within seconds, and it avoids the safety hazards of mercury and glass, making it safer for clinical use, particularly with children. (c) An infrared tympanic thermometer is a non-contact (or near-contact) thermometer that detects the infrared radiation naturally emitted by the tympanic membrane (eardrum) and surrounding ear canal tissue; because all objects above absolute zero emit infrared radiation, and the intensity of this radiation increases with temperature, the sensor can convert the detected infrared intensity into a temperature reading. The tympanic membrane is a useful measurement site because it is close to, and shares blood supply with, structures near the brain's temperature-regulating centre, so its temperature closely tracks core body temperature. An advantage of this method over a contact thermometer is that it gives a very fast reading (often around one to two seconds) and, since it does not need to be held in prolonged contact with the patient (often using a disposable probe cover), it reduces the risk of cross-infection between patients. (d) If core body temperature falls significantly below the normal range (hypothermia), the rate of the body's metabolic (enzyme-catalysed) reactions slows down, since enzyme activity is temperature-dependent; this can also increase the risk of cardiac arrhythmias (irregular heart rhythms), as the electrical activity controlling the heartbeat is disrupted by the lower temperature. Final answer: (a) approximately 36.5-37.5C; (b) mercury thermometer is slow (minutes) and carries a mercury/glass safety risk, while a digital thermistor thermometer is fast (seconds) and safer; (c) detects infrared radiation from the tympanic membrane, giving a fast, hygienic non-contact reading; (d) slowed metabolic rate/enzyme activity, or increased risk of cardiac arrhythmia.

[1] correct normal range given (approximately 36.5-37.5C, accept values within this general region e.g. 36-37.5C). (b) [1] mercury thermometer correctly described as slow response (minutes); [1] correctly identifies a mercury/glass safety hazard; [1] digital thermistor thermometer correctly described as fast response (seconds) and/or safer. (c) [1] correctly identifies detection of infrared radiation emitted by the tympanic membrane/ear; [1] correct explanation that radiation intensity relates to temperature (or that the site closely reflects core temperature); [1] valid advantage stated (e.g. speed, or reduced cross-infection risk due to non-contact/probe cover use). (d) [1] any valid physiological consequence of hypothermia (e.g. slowed metabolism/enzyme activity, cardiac arrhythmia, reduced consciousness). Maximum 8 marks.

(a) Level of response guidance (indicative content, to be marked using the 4-band rubric below): Physical principles - the electroencephalogram (EEG) works by placing a number of electrodes at standard positions on the scalp, which detect tiny electrical voltage fluctuations, of the order of microvolts, generated by the brain. These voltage fluctuations arise from the synchronised electrical activity of large populations of neurons in the cerebral cortex firing together, producing postsynaptic potentials that are large enough, when summed across many neurons, to be detected at the scalp surface (even though an individual neuron's signal would be far too small to detect this way). The tiny detected signals are amplified electronically and displayed as a continuous trace of voltage against time for each electrode. Diagnostic use - because different states of brain activity (such as being awake and alert, relaxed with eyes closed, or in different stages of sleep) produce characteristically different EEG trace patterns, a patient's EEG trace can be compared with known normal patterns to help identify abnormal brain activity. For example, epilepsy is associated with abnormal, highly synchronised bursts of electrical activity appearing as characteristic spike-and-wave patterns on the EEG trace, which can help confirm a diagnosis and identify the type and origin of seizures; EEG can also contribute to diagnosing sleep disorders (by analysing the pattern of brain waves through different sleep stages) and, in some cases, to assessing the extent of brain injury or confirming brain death (shown by a flat/isoelectric trace with no detectable electrical activity). Final answer: EEG detects small scalp voltage signals produced by synchronised cortical neuron activity, amplifies and displays them as a voltage-time trace, and comparing abnormal trace patterns (e.g. spike-and-wave activity in epilepsy) against normal patterns helps diagnose brain conditions. (b) EEG traces are conventionally divided into four main frequency bands. Delta waves have the lowest frequency, below about 4 Hz, and are associated with deep sleep. Theta waves have a frequency of about 4-8 Hz, associated with drowsiness or light sleep. Alpha waves have a frequency of about 8-13 Hz, associated with a relaxed, wakeful state, typically with the eyes closed. Beta waves have the highest frequency of the four, about 13-30 Hz, associated with an alert, actively concentrating state. (c) Epilepsy is a specific brain condition that can be diagnosed with the aid of a characteristic abnormal EEG pattern, shown as spike-and-wave discharges (sudden, sharp spikes in voltage followed by a slower wave, repeated rhythmically), which are not present in a normal EEG trace. Final answer: (b) delta <4 Hz, theta 4-8 Hz, alpha 8-13 Hz, beta 13-30 Hz; (c) epilepsy, shown by spike-and-wave discharges.

(a) Level of response mark scheme (8 marks, 4 bands). Excellent (7-8 marks): accurately describes the origin of the EEG signal (synchronised neuronal/cortical electrical activity detected via scalp electrodes, amplified and displayed as a voltage-time trace) and discusses at least one specific, correctly named diagnostic application with a described characteristic pattern; accurate specialist vocabulary; clear, coherent, well-structured writing. Good (4-6 marks): describes the general principle of EEG detection with reasonable accuracy and gives at least one relevant diagnostic use, though may lack some detail or precision; mostly accurate vocabulary; generally clear writing with minor errors. Basic (1-3 marks): gives a limited or partially correct description of EEG and/or its use, with little development; writing may be list-like or contain errors that hinder meaning. 0 marks: no creditable response. (b) [1] mark for each correctly named band with a correct approximate frequency range, to a maximum of 4: delta (<4 Hz), theta (4-8 Hz), alpha (8-13 Hz), beta (13-30 Hz). (c) [1] a valid brain condition correctly linked to a named characteristic EEG pattern (e.g. epilepsy - spike-and-wave discharges). Maximum 13 marks.

(a) The activity, A, of a radioactive source is the number of nuclear disintegrations (decays) occurring per second. It is measured in becquerel (Bq), where 1 Bq represents one disintegration per second. (b)(i) Converting the half-life to seconds: T1/2 = 6.0 hours x 3600 s/hour = 21 600 s. Rearranging T1/2 = 0.693/lambda gives lambda = 0.693/T1/2 = 0.693/21600 = 3.21x10^-5 s^-1 (to 3 significant figures). (ii) Converting 18 hours to seconds: t = 18 x 3600 = 64 800 s. Using A = A0 e^(-lambda t): A = 400 x e^(-(3.208x10^-5)(64800)) = 400 x e^(-2.079) = 400 x 0.1250 = 50.0 MBq (to 3 s.f.). This can be confirmed using the half-life relationship directly: 18 hours corresponds to exactly 18/6.0 = 3 half-lives, so the activity is reduced by a factor of (1/2)^3 = 1/8: A = 400/8 = 50.0 MBq, which agrees with the exponential calculation. (iii) Substituting the physical half-life (6.0 hours) and biological half-life (24 hours) into 1/Teff = 1/Tphysical + 1/Tbiological: 1/Teff = 1/6.0 + 1/24 = 4/24 + 1/24 = 5/24. Therefore Teff = 24/5 = 4.8 hours. This effective half-life (4.8 hours) is shorter than either the physical half-life (6.0 hours) or the biological half-life (24 hours) alone, because both physical decay and biological elimination act together to remove the radioactive tracer from the body. (c) A short physical half-life is desirable for a diagnostic radioisotope for two main reasons: first, it reduces the total radiation dose received by the patient, since the isotope remains significantly radioactive for only a relatively short time rather than continuing to irradiate the patient's tissues over a long period; second, it means the isotope's activity falls to a low, safe level relatively quickly after the scan is completed, meaning the patient poses a reduced radiation risk to others (such as family members or hospital staff) sooner, and any radioactive waste (e.g. from the patient's excretion) also becomes safe more quickly. (d) Background radiation is the radiation detected from sources other than the one being investigated, arising from naturally occurring sources such as cosmic rays, radioactive elements in rocks and soil, and radon gas, as well as any other artificial sources present in the environment (e.g. medical or industrial sources). A radiation detector cannot distinguish between counts caused by the source under investigation and counts caused by this background radiation; they are simply added together in the total measured count rate. If background radiation is not accounted for, the measured count rate (and any activity or half-life calculated from it) would be systematically too high, and inaccurate, particularly for sources with low activity or towards the end of a long decay experiment where the source's own count rate has fallen close to the background level. To correct for this, the background count rate is measured separately (with no source present) before or after the experiment, and this background count rate is subtracted from every measured count rate obtained with the source present, giving the true, corrected count rate due to the source alone. (e) Two valid safety precautions when handling and administering radioactive tracers are: using appropriate shielding, such as lead aprons, lead screens or lead-lined containers, to absorb radiation and reduce staff exposure, and minimising the time spent near the radioactive source while maximising the distance from it (since intensity decreases rapidly with distance, following the inverse square law); staff should also wear personal dosimeter badges to monitor and record their cumulative radiation exposure over time, to ensure it remains within safe legal limits. Final answer: (a) disintegrations per second, measured in becquerel (Bq); (b)(i) lambda = 3.21x10^-5 s^-1; (ii) A = 50.0 MBq (confirmed by the 3-half-life shortcut); (iii) Teff = 4.8 hours; (c) reduces patient dose, and allows activity to fall to a safe level (for the patient and others) sooner; (d) background radiation adds to every measured count and must be subtracted to isolate the count rate due to the source alone, otherwise measured activity/half-life would be inaccurately high; (e) shielding and minimising time/maximising distance from the source, plus personal dosimeters to monitor exposure.

(a) [1] correctly defines activity as the number of disintegrations per second; [1] correctly states the unit, becquerel (Bq). (b)(i) [1] correct conversion of half-life to seconds (21 600 s); [1] correct rearrangement of the equation; [1] correct final value, 3.21x10^-5 s^-1 (accept answers rounding to this from correct working; ECF from an incorrect but consistent time conversion). (ii) [1] correct conversion of 18 hours to seconds; [1] correct substitution into A = A0e^(-lambda t); [1] correct final value from the exponential method, approximately 50 MBq; [1] correct alternative confirmation using the half-life/fraction method (3 half-lives, factor of 1/8); [1] correct final confirmed value, 50.0 MBq, with the two methods shown to agree. (iii) [1] correct substitution of both half-lives into the given equation; [1] correct combination of fractions (5/24); [1] correct rearrangement to find Teff; [1] correct final value, 4.8 hours. (c) [1] mark for each valid, distinct reason, to a maximum of 2 (e.g. reduced patient dose; activity falls to a safe level quickly, reducing risk to others/allowing earlier discharge). (d) [1] correctly identifies that background radiation is present from natural/other sources regardless of the source under study; [1] correctly explains that the detector cannot distinguish source counts from background counts, so both are included in a raw measurement; [1] correctly explains the correction method (measuring and subtracting a separately measured background count rate) and/or the consequence of not doing so (systematically high/inaccurate results). (e) [1] mark for each valid, distinct safety precaution, to a maximum of 2 (e.g. shielding; minimising time/maximising distance; personal dosimeters; protective clothing/gloves). Maximum 21 marks.

(a) An X-ray tube contains several key components. A heated filament (cathode) is heated by an electric current, causing thermionic emission, in which electrons gain enough thermal energy to be released from the metal surface. A high voltage supply is connected across the tube, creating a strong electric field that accelerates these electrons from the cathode towards the anode, giving them high kinetic energy. The electrons strike a metal target (anode), typically made of a high-melting-point, high-density metal such as tungsten; as the fast-moving electrons rapidly decelerate within the target, X-ray photons are produced. The whole tube is evacuated (a vacuum), which prevents the accelerating electrons from colliding with air molecules and losing energy before reaching the target. Because only a small fraction of the electrons' kinetic energy is converted into X-ray photons, with the vast majority converted into heat, a cooling system (such as a rotating anode, or cooling fins/oil) is needed to remove this heat and prevent the target from melting. (b) X-ray photons are produced by two distinct mechanisms. The first is bremsstrahlung ('braking radiation'), in which an incoming electron is decelerated as it passes close to the strong electric field of a target atom's nucleus; the kinetic energy lost by the electron as it decelerates is emitted as an X-ray photon, and because electrons can be decelerated by different amounts depending on how closely they pass the nucleus, this produces a continuous range (spectrum) of X-ray photon energies. The second is characteristic radiation, in which an incoming electron collides with and ejects an inner-shell electron from a target atom; an electron from a higher energy level then falls down to fill this inner-shell vacancy, releasing a photon with an energy exactly equal to the difference between the two energy levels, which is characteristic of (and depends only on) the target element used, producing sharp, specific energy peaks. (c) The maximum possible energy of an X-ray photon is produced when an electron loses all of its kinetic energy in a single bremsstrahlung event; this maximum kinetic energy is equal to the energy gained by the electron as it is accelerated through the full tube voltage, given by E = eV, where e is the charge on the electron and V is the accelerating voltage. Substituting e = 1.6x10^-19 C and V = 80 000 V: E = 1.6x10^-19 x 80 000 = 1.28x10^-14 J. Converting to keV: since 1 eV is defined as the energy gained by an electron accelerated through a potential difference of 1 V, an electron accelerated through 80 000 V (80 kV) gains exactly 80 000 eV = 80 keV of energy; so the maximum photon energy is 80 keV, consistent with the joule value calculated (1.28x10^-14 J / 1.6x10^-19 J per eV = 80 000 eV = 80 keV). (d) Denser tissues, such as tooth enamel and bone, contain a greater concentration of atoms (and often atoms of higher atomic number, such as calcium) per unit volume than low-density soft tissue; this means an X-ray beam passing through dense tissue is more likely to interact with and be absorbed (attenuated) by these atoms, so a greater fraction of the incident X-ray intensity is absorbed by dense tissue than by an equivalent thickness of soft tissue. In dental X-ray imaging, X-rays are directed through the teeth and jaw onto a photographic film or digital detector on the other side. Because dense structures such as enamel, dentine and bone absorb more of the X-ray beam, less radiation reaches the film/detector directly behind them, producing a lighter (less exposed) area on the image; soft tissue and any decayed/less dense areas of a tooth absorb less radiation, allowing more X-rays through and producing a darker area, allowing the dentist to distinguish healthy dense tooth structure from areas such as decay, or to see the position of a filling, based on these differences in image density. (e) A precaution used to minimise a dental patient's radiation exposure is the use of a lead apron and/or thyroid collar to shield parts of the body not being imaged, and/or keeping the exposure time as short as possible and collimating (narrowing) the X-ray beam to expose only the specific area of interest. Final answer: (a) heated filament/cathode (thermionic emission of electrons), high voltage supply (accelerates electrons), metal target/anode e.g. tungsten (produces X-rays as electrons decelerate), evacuated tube (prevents energy loss to air), cooling system (removes heat). (b) bremsstrahlung (continuous spectrum from electron deceleration) and characteristic radiation (discrete energies from electron transitions in target atoms). (c) 1.28x10^-14 J = 80 keV. (d) denser tissue absorbs more X-rays (fewer reach the film, lighter image), less dense tissue absorbs fewer (more reach the film, darker image), allowing dense structures like enamel/bone to be distinguished from soft tissue/decay. (e) lead apron/thyroid collar and/or beam collimation/minimal exposure time.

(a) [1] mark for each correctly named component with a correct function, to a maximum of 5: heated filament/cathode (thermionic emission); high voltage supply (accelerates electrons); metal target/anode, e.g. tungsten (X-ray production on electron deceleration); evacuated tube (prevents electron energy loss to air); cooling system (removes heat from target). (b) [1] bremsstrahlung, correctly described (deceleration of electrons in the target's field, giving a continuous spectrum); [1] characteristic radiation, correctly described (electron transitions between energy levels in target atoms, giving discrete/specific energies). (c) [1] correct equation/approach (E = eV); [1] correct substitution; [1] correct value in joules (1.28x10^-14 J); [1] correct value in keV (80 keV), with correct reasoning/conversion shown. (d) [1] correctly explains that denser tissue contains more/denser atoms and absorbs more X-rays (greater attenuation); [1] correctly links this to less radiation reaching the film/detector behind dense tissue, producing a lighter image area; [1] correctly applies this to distinguishing dense structures (enamel/bone/fillings) from soft tissue/decay on a dental X-ray. (e) [1] a valid precaution (e.g. lead apron/thyroid collar, beam collimation, minimal exposure time). Maximum 15 marks.

(a) In MRI, the patient is placed inside a strong external magnetic field. Hydrogen nuclei (protons), which are abundant throughout the body (particularly in water and fat), have a property called spin and behave like tiny magnets; in the strong magnetic field, they align with the field direction and precess (wobble) around this direction at a specific frequency that depends on the strength of the magnetic field. A radiofrequency (RF) pulse, tuned to exactly match this precession frequency, is then applied; this causes resonance, in which the protons absorb energy from the RF pulse and are tipped out of their aligned position. When the RF pulse is switched off, the protons gradually relax back into alignment with the magnetic field, releasing the energy they absorbed as a detectable radiofrequency signal. Because the rate at which this relaxation occurs differs depending on the local tissue environment (for example, differing between fat, water and other soft tissues), these differences in the emitted signal are detected and processed by a computer to build a detailed image that can distinguish between different types of soft tissue. (b) An advantage of MRI compared with an X-ray or CT scan is that it does not use ionising radiation (unlike X-rays, which are ionising), so it does not carry the associated radiation risk to the patient, making it generally safer, particularly for repeated scans or for imaging patient groups where minimising radiation exposure is especially important (e.g. children, or during pregnancy where clinically appropriate). (c) In conventional gamma camera imaging, a patient is administered a gamma-emitting radiopharmaceutical (such as technetium-99m), which becomes concentrated in a particular organ or tissue depending on its chemical properties. Gamma photons are emitted from within the patient's body in all directions. A collimator, consisting of a thick lead block containing many narrow, parallel channels, is placed between the patient and the detector; only gamma photons travelling in the direction of these parallel channels (essentially perpendicular to the detector face) are able to pass through, while photons travelling at other angles are absorbed by the lead between channels. This is essential because it means each point on the detector only receives photons that originated from directly in front of it, allowing the spatial distribution of radioactivity within the body to be accurately mapped, giving the image spatial resolution. The photons that pass through the collimator strike a scintillation crystal (commonly sodium iodide), which absorbs the gamma photon energy and re-emits it as a flash of visible light (a scintillation); an array of photomultiplier tubes behind the crystal detects and amplifies these light flashes, converting them into electrical signals. A computer processes the pattern and intensity of signals across the whole detector array to construct a two-dimensional image showing the distribution and concentration of radioactivity within the body, which is useful for functional imaging, for example showing areas of abnormal metabolic activity, blood flow or organ function. (d) Natural background radiation arises from a number of sources, including radon gas (released from the radioactive decay of uranium naturally present in rocks and soil, and which can accumulate indoors), cosmic rays from space, radioactivity within rocks, soil and building materials, and naturally occurring radioactive isotopes present in food and drink. (e) The medical use of radiation, including diagnostic X-rays, CT scans, and radioactive tracers administered for diagnosis or treatment, itself emits or involves ionising radiation. Across the whole population, the cumulative use of these medical procedures adds a measurable, and now significant, contribution to the total average radiation dose received by the population, alongside natural background sources; this medical contribution must therefore be taken into account (alongside natural background) when considering a population's total average radiation exposure, and can also contribute to measured background count rates within a hospital environment specifically. Final answer: (a) a strong magnetic field aligns and causes precession of hydrogen nuclei; a matched RF pulse tips them out of alignment (resonance); as they relax back into alignment they emit a detectable signal that differs by tissue type, used to build the image. (b) no ionising radiation used. (c) gamma-emitting tracer administered; collimator selects only perpendicular photons (giving spatial resolution); scintillation crystal converts photons to light, detected by photomultiplier tubes and processed into an image of radioactivity distribution. (d) e.g. radon gas/cosmic rays/rocks and soil/food. (e) diagnostic and therapeutic uses of radiation add a significant, measurable contribution to the population's total average radiation dose/background count.

(a) [1] a strong external magnetic field is applied to the patient; [1] correctly identifies hydrogen nuclei/protons as the nuclei involved, which align with/precess in the field; [1] correctly describes the RF pulse being applied at a matching/resonant frequency, tipping protons out of alignment; [1] correctly describes protons relaxing back into alignment and emitting a detectable RF signal; [1] correctly explains that differences in this signal between tissue types allow an image to be constructed. (b) [1] correctly states MRI does not use ionising radiation (or another valid, correctly explained advantage e.g. better soft tissue contrast). (c) [1] correctly identifies a gamma-emitting radiopharmaceutical is administered to the patient; [1] correctly describes the collimator as allowing only photons travelling in a specific (perpendicular) direction through; [1] correctly links the collimator to giving spatial resolution/mapping the origin of photons; [1] correctly describes the scintillation crystal converting gamma photons to light; [1] correctly describes photomultiplier tubes detecting/amplifying this light and a computer building an image of radioactivity distribution. (d) [1] a valid source of natural background radiation (e.g. radon gas, cosmic rays, rocks/soil, food). (e) [1] correctly explains that medical use of radiation (diagnostic or therapeutic) adds a significant, measurable contribution to the population's total/average background radiation dose. Maximum 13 marks.

(a) In PET imaging, a patient is administered a radiopharmaceutical labelled with a radioisotope that decays by positron emission (beta-plus decay). When a positron is emitted from a nucleus within the body, it travels only a very short distance through the surrounding tissue before colliding with a nearby electron. Because a positron and an electron are matter and antimatter equivalents of each other, when they collide they undergo annihilation, in which both particles are completely converted into energy; this energy is released as two gamma photons, each with an energy of 511 keV, emitted travelling in almost exactly opposite directions (approximately 180 degrees apart) from the point of annihilation, in order to conserve momentum. The patient is surrounded by a ring of gamma-ray detectors; because the two photons from a single annihilation event travel in opposite directions and arrive at two detectors on opposite sides of the ring at effectively the same instant, the scanner can identify these coincident detection events and, from the line connecting the two detectors involved, determine that the annihilation occurred somewhere along that line. By recording very large numbers of these coincidence events from many different angles around the patient, a computer can reconstruct the precise three-dimensional locations of the annihilation events, building up a detailed image showing where the radiopharmaceutical (and hence a process such as metabolic activity or blood flow) is concentrated within the body. (b) Rubidium-82 is a radiopharmaceutical used specifically in PET imaging; it is rapidly taken up by heart muscle, and so is used in PET perfusion imaging of the heart, showing the distribution of blood flow to different regions of the heart muscle. (c) Brachytherapy is a method of treating cancer in which a sealed radioactive source is placed directly inside the body, either within the tumour itself or immediately adjacent to it (rather than directing radiation at the tumour from an external source outside the body, as in external beam radiotherapy). Because radiation intensity decreases rapidly with distance from a source (following the inverse square law), placing the source very close to, or within, the tumour allows the tumour itself to receive a very high radiation dose, while healthy tissue only a short distance further away receives a much lower dose; this is an advantage over external beam radiotherapy, in which the radiation beam must pass through a greater thickness of healthy tissue on its way to (and, depending on the technique, potentially beyond) the tumour, exposing more healthy tissue to a significant dose. (d) External beam radiotherapy commonly uses high-energy X-rays generated by a linear accelerator (linac), directed at the tumour from outside the body (gamma rays from a source such as cobalt-60 are also an accepted example of radiation used in external beam radiotherapy). Final answer: (a) a positron emitted in the body annihilates with a nearby electron, producing two 511 keV gamma photons travelling in opposite directions, detected in coincidence by a ring of detectors and used to reconstruct a 3D image of tracer distribution. (b) rubidium-82, used for PET perfusion imaging of the heart. (c) brachytherapy places a sealed source inside/next to the tumour, giving a high localised dose to the tumour while sparing more distant healthy tissue (inverse square law), unlike external beam radiotherapy which must pass through healthy tissue to reach the tumour. (d) high-energy X-rays (from a linear accelerator) or gamma rays (e.g. cobalt-60).

(a) [1] correctly identifies that a positron is emitted from the radiopharmaceutical/isotope within the body; [1] correctly describes the positron colliding with an electron and undergoing annihilation; [1] correctly states that annihilation produces two gamma photons of equal energy (511 keV each); [1] correctly states the two photons travel in (almost) exactly opposite directions; [1] correctly describes coincidence detection by a ring of detectors being used to locate annihilation events and build a 3D image. (b) [1] correctly names rubidium-82 (or another valid PET radiopharmaceutical); [1] correctly states its use (e.g. cardiac/heart perfusion imaging for rubidium-82). (c) [1] correctly describes brachytherapy as placing a sealed source inside or next to the tumour; [1] correctly links this to a high dose delivered to the tumour with reduced dose to healthy tissue further away (inverse square law); [1] correctly contrasts this with external beam radiotherapy needing to pass through healthy tissue to reach the tumour. (d) [1] a valid type of radiation used in external beam radiotherapy (e.g. high-energy X-rays/linac, or gamma rays/cobalt-60). Maximum 11 marks.

(a) Specific acoustic impedance, Z, of a tissue is defined as the product of the density of the tissue and the speed of sound within that tissue: Z = (rho)c, where rho is density (kg m^-3) and c is the speed of sound in that tissue (m s^-1). (b) Using Z = (rho)c: Z = 1050 x 1540 = 1 617 000 = 1.617x10^6 kg m^-2 s^-1 (to 4 s.f.). (c) Substituting the given values into R = [(Z2-Z1)/(Z2+Z1)]^2: R = [(1.70x10^6 - 1.50x10^6)/(1.70x10^6 + 1.50x10^6)]^2 = [(0.20x10^6)/(3.20x10^6)]^2 = [0.0625]^2 = 0.00390625, which rounds to R = 0.00391 (or 0.391%, to 3 s.f.). (d) The intensity reflection coefficient, R, represents the fraction of incident intensity reflected at the boundary; the remaining fraction is transmitted across the boundary. Fraction transmitted = 1 - R = 1 - 0.00391 = 0.99609, so the percentage transmitted = 0.99609 x 100 = 99.6% (to 3 s.f.). This shows that, because fat and muscle have fairly similar acoustic impedances, only a very small fraction of the ultrasound intensity (about 0.4%) is reflected at this boundary, with the vast majority (99.6%) transmitted onward to be reflected at deeper boundaries and contribute to the image. (e) Without a coupling gel, there would be a thin layer of air between the ultrasound probe and the patient's skin. Air has a very low acoustic impedance compared with skin/soft tissue, creating a very large impedance mismatch at an air-skin boundary; using the reflection coefficient equation, such a large mismatch means R would be very close to 1, meaning almost all of the ultrasound intensity would be reflected straight back at the skin surface and would not enter the body at all. The coupling gel, which has an acoustic impedance much closer to that of skin/soft tissue than air does, excludes air from between the probe and skin and provides a much smaller impedance mismatch, allowing ultrasound to be transmitted efficiently into (and out of) the body with minimal unwanted reflection/energy loss at the skin surface. (f) While ultrasound itself is defined as sound with a frequency above 20 kHz, medical diagnostic ultrasound imaging typically uses frequencies between about 1 MHz and 18 MHz. Deep structures such as the liver and kidney are typically imaged using lower frequencies within this range, around 1-6 MHz, because lower-frequency ultrasound is attenuated (absorbed/scattered) less as it travels through tissue, allowing it to penetrate more deeply into the body and still produce a usable echo from a deep structure; however, lower frequencies give poorer resolution of fine detail. Superficial structures such as the thyroid gland or breast, which do not require the beam to penetrate deeply, are instead imaged using higher frequencies, around 7-18 MHz, which cannot penetrate as deeply (and so are less suitable for deep structures) but can resolve much finer structural detail, giving a sharper image where deep penetration is not required. (g) An ultrasonic A-scan (amplitude scan) produces a simple one-dimensional output: a graph, along a single line through the body, of the amplitude of each returning echo plotted against depth (or equivalently the time delay of the echo), showing the position and relative reflecting strength of each boundary encountered along that single line, but with no two-dimensional spatial/image information. A B-scan (brightness scan) builds a genuine two-dimensional cross-sectional image by combining information from many individual A-scan lines swept across an area (for example, by using an array of transducer elements, or mechanically/electronically sweeping the beam); rather than displaying each echo as a point on an amplitude-against-depth graph, each echo is instead converted into a single bright dot on the image, with the position of the dot corresponding to the depth of the reflecting boundary and the brightness of the dot corresponding to the amplitude (strength) of the echo, and the dots from many adjacent scan lines are combined to build up a full two-dimensional image of a cross-section through the tissue. Final answer: (a) Z = (rho)c. (b) Z = 1.617x10^6 kg m^-2 s^-1. (c) R = 0.00391 (0.391%). (d) 99.6% of the intensity is transmitted. (e) gel excludes air (which would otherwise cause almost total reflection at the skin due to a large impedance mismatch), allowing efficient transmission of ultrasound into the body. (f) 1-18 MHz typically, with deep structures (1-6 MHz) using lower frequencies for greater penetration (less attenuation) at the cost of resolution, and superficial structures (7-18 MHz) using higher frequencies for better resolution where deep penetration is not needed. (g) an A-scan is a 1D plot of echo amplitude against depth along a single line; a B-scan combines many such lines, converting each echo into a bright dot (brightness = amplitude, position = depth) to build a 2D cross-sectional image.

(a) [1] correctly defines Z as the product of density and speed of sound in the tissue; [1] correctly states the equation Z = (rho)c. (b) [1] correct substitution into Z = (rho)c; [1] correct final value, 1.617x10^6 kg m^-2 s^-1 (accept equivalent rounding, e.g. 1.62x10^6). (c) [1] correct substitution of Z1 and Z2 into the given equation; [1] correct numerator (0.20x10^6) and denominator (3.20x10^6) evaluated; [1] correct ratio squared (0.0625^2); [1] correct final value, R = 0.00391 (or 0.391%), to an acceptable degree of precision. (d) [1] correctly identifies that transmitted fraction = 1 - R; [1] correct final answer, 99.6% (ECF from the candidate's R value in (c)). (e) [1] correctly identifies that air would otherwise be present between probe and skin, causing (near-)total reflection due to a large impedance mismatch; [1] correctly explains that gel has an impedance closer to skin/soft tissue, reducing reflection and allowing efficient transmission into the body. (f) [1] correct overall frequency range for medical diagnostic ultrasound (approximately 1-18 MHz); [1] correctly identifies deep structures use lower frequencies (approx. 1-6 MHz) for greater penetration/less attenuation; [1] correctly identifies superficial structures use higher frequencies (approx. 7-18 MHz) for better resolution; [1] correctly explains the trade-off (lower frequency = greater penetration but poorer resolution, and vice versa). (g) [1] correctly describes an A-scan as a 1D plot of echo amplitude against depth/time along a single line; [1] correctly describes a B-scan as combining multiple scan lines into a 2D image; [1] correctly explains that each echo is shown as a bright dot, with brightness representing amplitude and position representing depth. Maximum 19 marks.

(a) In order of increasing frequency, the regions of the electromagnetic spectrum are: radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays. (b) All electromagnetic waves travel at the same speed in a vacuum, 3x10^8 m/s (the speed of light, c). (c) Rearranging v = f(lambda) to make wavelength the subject: lambda = v/f = (3x10^8)/(100x10^6) = (3x10^8)/(1x10^8) = 3.0 m. (d) All electromagnetic waves are transverse waves, and all travel at the same speed (3x10^8 m/s) when travelling through a vacuum; they also all transfer energy without transferring matter, and can all, in principle, undergo reflection, refraction, diffraction and interference. Any two of these properties should be credited. (e) Using v = f(lambda), rearranged to make frequency the subject: f = v/lambda = 1540/(0.77x10^-3) = 1540/0.00077 = 2 000 000 = 2.0x10^6 Hz (2.0 MHz). Final answer: (a) radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, gamma rays; (b) 3x10^8 m/s; (c) 3.0 m; (d) e.g. all transverse, all travel at the same speed in a vacuum; (e) 2.0x10^6 Hz.

(a) [1] correct order of at least 5 regions; [1] fully correct order of all 7 regions (radio, microwave, infrared, visible, ultraviolet, X-ray, gamma). (b) [1] for 3x10^8 m/s. (c) [1] correct rearrangement of v = f(lambda); [1] correct substitution; [1] correct final answer, 3.0 m. (d) [1] mark for each valid shared property, to a maximum of 2 (e.g. transverse waves; same speed in a vacuum; transfer energy not matter; can reflect/refract/diffract/interfere). (e) [1] correct substitution/rearrangement; [1] correct final answer, 2.0x10^6 Hz (2.0 MHz). Maximum 10 marks.

(a) Level of response guidance (indicative content, to be marked using the 3-band rubric below): Generation - a radio signal is generated using a transmitting dipole antenna, which is connected to a source of alternating current (or alternating voltage) at the desired radio frequency; this alternating current causes electrons within the antenna to oscillate back and forth at that same frequency, and this oscillating movement of charge generates an oscillating electric field (and an associated oscillating magnetic field) that radiates outward from the antenna as an electromagnetic (radio) wave, travelling at the frequency of the original alternating current. Detection - a separate receiving antenna, placed some distance away, intercepts this travelling oscillating electric field as the radio wave passes it; the changing electric field exerts a changing force on the free electrons within the receiving antenna, inducing a small oscillating current (an alternating EMF) within it, at the same frequency as the original transmitted signal; this tiny induced signal is then amplified and processed (decoded) by the receiving device's circuitry to recover the information that was originally encoded onto the transmitted radio wave. Application to wireless/Bluetooth - technologies such as wireless internet (Wi-Fi) and Bluetooth apply this same basic principle of generating and detecting radio waves using antennas, but operate at specific, standardised radio frequencies (in the microwave part of the radio spectrum) and encode digital data (such as audio, video or other information) onto the radio wave, allowing two devices, each containing a small transmitting/receiving antenna, to exchange data wirelessly over short to medium distances without a physical cable connection. Final answer: an oscillating current in a transmitting dipole antenna generates a radiating oscillating electric field (radio wave); this induces a small oscillating current in a receiving antenna, which is amplified and decoded; wireless and Bluetooth apply this principle at specific frequencies to exchange data wirelessly between devices. (b) The attenuation (weakening) of a radio signal as it travels can be categorised as path loss, which describes the general reduction in signal strength as a radio wave travels and interacts with its environment (for example due to obstacles, reflection or absorption along its path), and free space loss, which describes the reduction in signal strength that occurs simply because the wave's energy spreads out over an increasingly large area as it travels outward from the transmitter, even with no obstacles present, meaning the signal intensity naturally decreases with distance from the source.

(a) Level of response mark scheme (6 marks, 3 bands). Excellent (5-6 marks): accurately describes both generation (oscillating current in a dipole antenna produces a radiating oscillating field) and detection (the travelling field induces a corresponding oscillating current in a receiving antenna), and discusses how this principle underlies wireless/Bluetooth technology; accurate specialist vocabulary; clear, coherent, well-structured writing. Good (3-4 marks): describes generation and/or detection with reasonable accuracy and makes some reference to wireless/Bluetooth technology, though may lack full detail on both sides of the process; mostly accurate vocabulary; generally clear writing with minor errors. Basic (1-2 marks): gives a limited or partially correct description of antenna transmission/reception, with little or no link to wireless/Bluetooth technology; writing may be list-like or contain errors that hinder meaning. 0 marks: no creditable response. (b) [1] path loss; [1] free space loss. Maximum 8 marks.

(a) Total internal reflection can only occur when light is travelling within a more optically dense material (with a higher refractive index) and reaches a boundary with a less optically dense material (with a lower refractive index); in addition, the angle of incidence at that boundary, measured from the normal, must be greater than the critical angle for that particular pair of materials. If either of these two conditions is not met (for example, if light travels from a less dense to a more dense material, or if the angle of incidence is smaller than the critical angle), refraction (with some partial reflection) occurs instead of total internal reflection. (b) To determine the critical angle experimentally using a semi-circular glass block, a ray of light is directed into the flat face of the block, aimed so that it travels through the glass and reaches the centre of the curved face; because the ray hits the curved surface travelling along a radius, it meets this second surface at normal incidence there, meaning it does not refract as it exits (avoiding any additional, unwanted refraction at the curved surface that would complicate the measurement). The angle of incidence at the flat face (inside the glass) is then gradually increased, and the behaviour of the ray at the curved surface is observed. As the angle of incidence increases, the corresponding angle of refraction (as the ray would exit into the air) also increases; the critical angle is identified as the specific angle of incidence at which the angle of refraction reaches exactly 90 degrees, meaning the refracted ray grazes along the curved surface itself; for any angle of incidence greater than this, no refracted ray emerges at all, and total internal reflection occurs instead. (c) A fibre-optic cable consists of a central core made of glass or plastic, surrounded by an outer layer called the cladding, which is made of a similar but different material with a lower refractive index than the core. Light is introduced into the end of the core at a suitable angle; because the core has a higher refractive index than the surrounding cladding, and the light strikes the core-cladding boundary at an angle of incidence greater than the critical angle for that boundary, the light undergoes total internal reflection repeatedly at the core-cladding boundary as it travels along the fibre, effectively bouncing along the length of the core and being guided along the fibre with very little loss of light, even around gentle bends in the cable. (d) A single-mode fibre has a much narrower core diameter than a multi-mode fibre, so narrow that it only allows light to travel along essentially a single path (a single 'mode') straight down the centre of the fibre, whereas a multi-mode fibre has a wider core that allows light to travel along many different paths/angles (multiple modes) simultaneously. Single-mode fibre is preferred for long-distance communication (for example, undersea or long-distance telecommunications links), because in a multi-mode fibre, light travelling via different path lengths (different modes) arrives at the far end at slightly different times, causing the signal to spread out and distort (a problem called modal dispersion) over long distances; because single-mode fibre only allows one path, it avoids this problem and preserves signal quality over much greater distances. Final answer: (a) light must travel from a more dense to a less dense material, with angle of incidence greater than the critical angle. (b) direct light into the flat face aimed at the centre of the curved face, increase the angle of incidence, and identify the critical angle as the angle of incidence at which the angle of refraction reaches 90 degrees. (c) a higher-refractive-index core surrounded by lower-refractive-index cladding; light undergoes repeated TIR at the core-cladding boundary, guiding it along the fibre. (d) single-mode fibre has a much narrower core (single light path) than multi-mode fibre (multiple paths); single-mode is preferred for long-distance communication, since it avoids modal dispersion/signal distortion over long distances.

(a) [1] light must travel from a more optically dense/higher refractive index material to a less dense one; [1] the angle of incidence must be greater than the critical angle. (b) [1] correctly describes directing the ray into the flat face aimed at the centre of the curved face (avoiding refraction at the curved surface); [1] correctly describes varying/increasing the angle of incidence; [1] correctly identifies the critical angle as the angle of incidence at which the angle of refraction becomes 90 degrees (refracted ray grazes the surface). (c) [1] correctly describes the core (higher refractive index) surrounded by cladding (lower refractive index); [1] correctly explains that repeated total internal reflection at the core-cladding boundary guides light along the fibre. (d) [1] correctly identifies that single-mode fibre has a narrower core allowing only a single light path, compared with a wider, multi-path core in multi-mode fibre; [1] a valid application (e.g. long-distance communication) correctly linked to avoiding modal dispersion/signal distortion. Maximum 9 marks.

(a)(i) The amplitude of a wave is its maximum displacement from the equilibrium (rest) position, so the amplitude is 3.0 mm. (ii) The time period is the time taken for one complete oscillation; since 5 complete oscillations take 0.020 s, the time period T = 0.020/5 = 4.0x10^-3 s (0.0040 s). (iii) Frequency is the reciprocal of the time period: f = 1/T = 1/(4.0x10^-3) = 250 Hz. (b)(i) Since 3 complete wave cycles occur over a distance of 4.2 m, the wavelength (the length of one complete cycle) is lambda = 4.2/3 = 1.4 m. (ii) Using v = f(lambda): v = 250 x 1.4 = 350 m/s. (c) Phase difference can be calculated from path difference using the proportion: phase difference (in degrees) = (path difference / wavelength) x 360 degrees. Substituting the given values: phase difference = (0.35/1.4) x 360 = 0.25 x 360 = 90 degrees. (This corresponds to a path difference of exactly one quarter of a wavelength, giving a phase difference of one quarter of a full cycle, 360/4 = 90 degrees, or pi/2 radians.) (d) A standing (stationary) wave is formed when two waves that are identical (having the same frequency, wavelength, amplitude and wave speed) travel in opposite directions along the same line and meet, so that they continuously superpose (overlap) with each other; unlike a normal progressive wave, the resulting wave pattern does not travel onward but instead appears to stay in a fixed position, with the amplitude at each point building up due to this superposition (an effect called resonance). At certain fixed points along a standing wave, called nodes, the two component waves are always in antiphase (exactly out of step) with each other, so they undergo destructive interference at that point at all times, meaning the amplitude at a node is permanently zero (no displacement ever occurs there). At other fixed points, called antinodes, the two component waves are always exactly in phase with each other, undergoing constructive interference, so the amplitude at an antinode reaches the maximum possible value (twice the amplitude of each individual component wave), with the medium oscillating with maximum amplitude at that point. Final answer: (a) amplitude 3.0 mm; T = 4.0x10^-3 s; f = 250 Hz. (b) wavelength 1.4 m; speed 350 m/s. (c) phase difference = 90 degrees. (d) a standing wave forms when two identical waves travelling in opposite directions superpose; nodes are points of permanently zero amplitude (destructive interference), antinodes are points of maximum amplitude (constructive interference).

(a)(i) [1] for 3.0 mm. (ii) [1] for 4.0x10^-3 s (0.0040 s), with correct working (0.020/5). (iii) [1] correct method (1/T); [1] correct final answer, 250 Hz (ECF from (a)(ii)). (b)(i) [1] correct method (distance/number of cycles); [1] correct final answer, 1.4 m. (ii) [1] correct substitution into v = f(lambda) using the candidate's own values; [1] correct final answer, 350 m/s (ECF from (a)(iii) and (b)(i)). (c) [1] correct method/formula (path difference / wavelength x 360); [1] correct substitution; [1] correct final answer, 90 degrees. (d) [1] correctly describes two identical waves (same frequency/wavelength/amplitude) travelling in opposite directions; [1] correctly describes superposition/interference producing a stationary (non-travelling) pattern; [1] correctly defines a node as a point of permanently zero/minimum amplitude (destructive interference); [1] correctly defines an antinode as a point of maximum amplitude (constructive interference). Maximum 15 marks.

(a) The threshold intensity for human hearing, I0, is 1x10^-12 W m^-2. Sound intensity is measured in units of watts per square metre, W m^-2. (b) Substituting into the given equation: dB level = 10 log10(I/I0) = 10 log10((1x10^-6)/(1x10^-12)) = 10 log10(1x10^6) = 10 x 6 = 60 dB. (c) Rearranging is not needed, as the equation is already given in the form I = I0 x 10^(dB level/10). Substituting: I = (1x10^-12) x 10^(85/10) = (1x10^-12) x 10^8.5 = (1x10^-12) x (3.162x10^8) = 3.16x10^-4 W m^-2 (to 3 s.f.). (d) Human hearing is sensitive across an extremely wide range of intensities, from the threshold of hearing (1x10^-12 W m^-2) up to intensities many orders of magnitude greater (such as the threshold of pain, around 1 W m^-2, a factor of 10^12 times greater); stating such a huge range of values directly in W m^-2 would require using very large or very small numbers, or standard form, over and over, making comparisons cumbersome. Because the decibel scale is logarithmic, it compresses this enormous range of intensities into a much smaller, more manageable range of numbers (0 dB to around 120 dB across the full range of human hearing), which is both more practical to work with and better reflects the way loudness is subjectively perceived by the human ear (which itself responds roughly logarithmically to changes in intensity). Final answer: (a) I0 = 1x10^-12 W m^-2; units W m^-2. (b) 60 dB. (c) 3.16x10^-4 W m^-2. (d) human hearing spans an enormous range of intensities, and a logarithmic (decibel) scale compresses this into a small, manageable range of numbers that also better reflects how loudness is perceived.

(a) [1] I0 = 1x10^-12 W m^-2; [1] units of intensity, W m^-2. (b) [1] correct substitution into the given equation; [1] correct evaluation of log10(1x10^6) = 6; [1] correct final answer, 60 dB. (c) [1] correct substitution into the given equation; [1] correct evaluation of 10^8.5; [1] correct final answer, 3.16x10^-4 W m^-2 (accept equivalent rounding). (d) [1] correctly identifies that human hearing covers an extremely wide range of intensities; [1] correctly explains that a logarithmic scale compresses this into a smaller, more manageable range (and/or better reflects perceived loudness). Maximum 10 marks.

(a) (i) The cornea is the transparent, curved front surface of the eye, and is responsible for refracting (bending) most of the light entering the eye, providing the majority of the eye's overall focusing power. (ii) The lens is a flexible, transparent structure that provides fine adjustment of the eye's focusing power, allowing the light to be focused precisely onto the retina for objects at different distances (a process called accommodation). (iii) The ciliary muscles are a ring of muscle surrounding the lens; they contract or relax to change the shape (curvature) and therefore the focusing power of the lens, enabling accommodation. (iv) The retina is the light-sensitive layer at the back of the eye, containing rod and cone cells, which detect the focused image falling on it and convert this light information into nerve impulses sent to the brain via the optic nerve. (b) To focus on a near object, the ciliary muscles contract; this reduces the tension in the suspensory ligaments holding the lens in place, allowing the lens's own natural elasticity to pull it into a more rounded, thicker shape, which increases its curvature and therefore its focusing power (needed to bend light from a nearby, strongly diverging source enough to focus it on the retina). To focus on a distant object, the ciliary muscles relax; this increases the tension in the suspensory ligaments, pulling the lens into a thinner, flatter shape, which decreases its curvature and focusing power (since light from a distant object is only weakly diverging/close to parallel, and needs less bending to be focused on the retina). (c) The lens must produce a virtual image of an object held at the normal near point (0.25 m) at the person's own near point (1.00 m), since this is the closest distance at which they can naturally focus. Using the lens equation with a consistent sign convention (both the real object distance and the resulting virtual image distance taken as negative, since both lie on the same side of the lens as the incoming light): 1/f = 1/v - 1/u = 1/(-1.00) - 1/(-0.25) = -1.00 + 4.00 = +3.00 m^-1. Since optical power P = 1/f (with f in metres), the required power is P = +3.0 D. The positive value confirms this must be a converging (convex) lens, which is consistent with correcting long-sightedness (hyperopia). Final answer: (a) cornea - main refraction of incoming light; lens - fine focusing adjustment (accommodation); ciliary muscles - change lens shape/power; retina - detects the focused image (rods/cones) and converts it to nerve impulses. (b) ciliary muscles contract to make the lens more rounded/powerful for near objects, and relax to make it thinner/less powerful for distant objects. (c) +3.0 D converging lens.

(a) [1] mark for each correct function, to a maximum of 4: cornea (main refraction of light entering the eye); lens (fine focusing/accommodation); ciliary muscles (change lens shape/power); retina (detects image via rods/cones, converts to nerve impulses). (b) [1] correctly describes ciliary muscles contracting for a near object; [1] correctly links this to the lens becoming more rounded/thicker/more powerful; [1] correctly describes the opposite (ciliary muscles relax, lens thinner/less powerful) for a distant object. (c) [1] correct substitution into the lens equation with a consistent sign convention; [1] correct working (-1.00 + 4.00); [1] correct final answer, +3.0 D (converging lens), with the sign/type of lens correctly identified. Maximum 10 marks.

(a) The lens equation, using the real-is-positive sign convention, is 1/f = 1/u + 1/v, where u is the object distance, v is the image distance, and f is the focal length of the lens (all measured in the same units, typically metres for calculating power). (b) An illuminated object (such as an object with a cross-wire or an arrow-shaped hole illuminated from behind) is set up on an optical bench, together with the converging lens under test and a white screen, arranged in a straight line. The screen is placed a suitable distance from the lens, and either the lens or the screen is moved along the bench until a sharp, clearly focused, real image of the object is seen on the screen. Once the image is sharply focused, the distance from the object to the lens (u) and the distance from the lens to the screen (v) are measured. This process is repeated for several different values of u (by moving the object to new positions and each time refocusing the image on the screen), and each pair of u and v values is substituted into the lens equation, 1/f = 1/u + 1/v, to calculate a value of f; taking a mean of the calculated f values improves the reliability of the final result and reduces the effect of random measurement error. Once a reliable value of f has been found, the lens's optical power can be calculated as P = 1/f, with f expressed in metres, giving the power in dioptres. (c)(i) Substituting the given values (with both u and v positive, since this is a real object and a real image, using the real-is-positive convention) into the lens equation, first converting to metres: u = 0.40 m, v = 0.24 m. 1/f = 1/u + 1/v = 1/0.40 + 1/0.24 = 2.500 + 4.167 = 6.667 m^-1. Therefore f = 1/6.667 = 0.15 m (15 cm). (ii) Optical power P = 1/f = 1/0.15 = 6.7 D (to 2 s.f.). (d) A person's far point is the greatest distance at which they can focus clearly; a corrective lens for myopia needs to take light from a genuinely distant object (which can be treated as coming from infinity, since the rays are essentially parallel by the time they reach the eye) and instead produce a virtual image at the person's own far point (2.0 m), which is a distance the myopic eye's own lens system CAN focus on. Using 1/f = 1/v - 1/u, with the object at infinity (so 1/u = 1/infinity = 0) and the image being virtual, formed on the same side of the lens as the object, so v is taken as negative: v = -2.0 m. 1/f = 1/(-2.0) - 0 = -0.5 m^-1. Power P = 1/f = -0.5 D. Because the power is negative, this must be a diverging (concave) lens, which is the type of lens always required to correct short-sightedness (myopia), since a diverging lens spreads out (diverges) incoming light rays before they reach the eye, effectively moving the point they appear to originate from closer to the eye, matching the myopic eye's shorter far point. Final answer: (a) 1/f = 1/u + 1/v. (b) form a real image of an illuminated object on a screen for several object distances, measure u and v each time, use 1/f=1/u+1/v to find f (take a mean), then P=1/f. (c) f = 0.15 m (15 cm); P = 6.7 D. (d) P = -0.5 D, a diverging (concave) lens.

(a) [1] for 1/f = 1/u + 1/v (or an equivalent correctly stated form of the lens equation). (b) [1] correctly describes setting up an illuminated object, lens and screen on an optical bench; [1] correctly describes adjusting the screen/lens position until a sharp focused image is obtained, and measuring u and v; [1] correctly describes repeating for multiple object distances and finding a mean value of f using the lens equation; [1] correctly states that power is then found using P = 1/f. (c)(i) [1] correct substitution of u and v (in metres) into the lens equation; [1] correct final value, f = 0.15 m (15 cm). (ii) [1] correct method (P = 1/f); [1] correct final value, 6.7 D (accept 6.67 D). (d) [1] correctly identifies that the object is treated as being at infinity (1/u = 0); [1] correctly identifies the required image is virtual, formed at the person's far point, with v correctly signed as negative; [1] correct substitution into the lens equation; [1] correct final value, P = -0.5 D; [1] correctly identifies a diverging/concave lens is required, with valid reasoning. Maximum 14 marks.

(a) (i) The pinna (outer, visible part of the ear) collects and funnels sound waves from the surrounding air into the auditory canal. (ii) The tympanic membrane (eardrum), at the end of the auditory canal, vibrates in response to the pressure variations of an incoming sound wave, transmitting these vibrations to the ossicles of the middle ear. (iii) The ossicles (the malleus, incus and stapes, three small connected bones in the middle ear) transmit the vibrations from the tympanic membrane across the middle ear to the oval window, amplifying the pressure of the vibrations as they do so. (iv) The oval window is a membrane-covered opening between the middle ear and the inner ear; vibration of the stapes against the oval window transmits the vibrations into the fluid contained within the cochlea. (v) The cochlea is a fluid-filled, coiled (spiral-shaped) organ in the inner ear; it contains thousands of tiny hair cells along its length which are stimulated by the resulting fluid vibrations, converting these mechanical vibrations into electrical nerve impulses that are sent to the brain via the auditory nerve. (vi) The Eustachian tube connects the middle ear cavity to the back of the throat (pharynx); its function is to equalise air pressure on either side of the tympanic membrane, allowing the eardrum to vibrate freely and correctly in response to sound, and preventing damage from pressure differences (for example, during changes in altitude). (b) Sound waves initially travel through air, which has a relatively low acoustic impedance, but must ultimately be transmitted into the fluid contained within the cochlea, which has a much higher acoustic impedance (fluid is much denser than air). If sound simply passed directly from air to the cochlear fluid without amplification, the large impedance mismatch between these two media would mean that most of the sound energy would simply be reflected at this boundary rather than being transmitted into the fluid, resulting in a substantial loss of hearing sensitivity. To overcome this, the ossicles amplify the pressure of the vibration in two ways: firstly, the three connected bones act as a lever system, providing a mechanical advantage that increases the force of the vibration; secondly, and more significantly, the vibrations collected over the relatively large surface area of the tympanic membrane are transmitted via the ossicles and concentrated onto the much smaller surface area of the oval window, and because pressure is force divided by area, concentrating the same force onto a much smaller area substantially increases the pressure delivered to the cochlear fluid, helping to overcome the impedance mismatch and ensure efficient transmission of sound energy into the inner ear. (c) The semicircular canals are three fluid-filled canals within the inner ear, each oriented in a different plane (roughly perpendicular to one another), positioned close to, but functioning separately from, the cochlea. They contain hair cells that detect the movement of the fluid within the canals; when the head rotates, the fluid within the canals lags behind the surrounding bony structure due to inertia, causing relative movement of the fluid past the hair cells, which are stimulated and send nerve impulses to the brain. Because the three canals are oriented in different planes, this allows rotational movement (angular acceleration) of the head to be detected in any direction. This information is used by the brain to help maintain balance and a sense of equilibrium/spatial orientation, and is not involved in the detection of sound itself. (d) The auditory threshold at a given frequency is the minimum intensity of sound at that specific frequency that can just be detected by the average human ear. This threshold is not the same at all frequencies: it is at its lowest (meaning the ear is most sensitive, requiring the least intensity to be detected) in the mid-frequency range, around 3-4 kHz (corresponding to the natural resonance frequency of the auditory canal), and the threshold increases (meaning the ear becomes progressively less sensitive, requiring a greater intensity to be detected) at both lower frequencies and higher frequencies moving away from this most sensitive region. Final answer: (a) pinna - collects sound; tympanic membrane - vibrates, transmits to ossicles; ossicles - transmit/amplify vibrations to oval window; oval window - transmits vibration into cochlear fluid; cochlea - converts vibrations to nerve impulses via hair cells; Eustachian tube - equalises pressure across the eardrum. (b) needed to overcome the air-to-fluid impedance mismatch (which would otherwise cause most sound energy to be reflected); achieved via ossicle lever action and concentrating force from the large tympanic membrane onto the small oval window, increasing pressure. (c) detect rotational head movement via fluid movement past hair cells in three differently oriented canals, providing balance information (not hearing). (d) minimum detectable intensity at a given frequency; lowest (most sensitive) around 3-4 kHz, increasing (less sensitive) at lower and higher frequencies.

(a) [1] mark for each correct function, to a maximum of 6: pinna (collects/funnels sound); tympanic membrane (vibrates, transmits to ossicles); ossicles (transmit/amplify vibration to oval window); oval window (transmits vibration into cochlear fluid); cochlea (converts vibration to nerve impulses via hair cells); Eustachian tube (equalises pressure across the eardrum). (b) [1] correctly identifies the air-to-fluid impedance mismatch as the reason amplification is needed (and/or that without it, most sound energy would be reflected); [1] correctly describes the lever action of the ossicles; [1] correctly describes the concentration of force from the larger tympanic membrane onto the smaller oval window, increasing pressure. (c) [1] correctly identifies the semicircular canals detect rotational movement/acceleration of the head (via hair cells responding to fluid movement); [1] correctly links this to balance/equilibrium, distinct from hearing. (d) [1] correct definition of auditory threshold (minimum detectable intensity at a given frequency); [1] correctly describes the threshold being lowest around 3-4 kHz and increasing at lower/higher frequencies. Maximum 13 marks.

(a) The loudness of a sound, measured in phons, is defined by comparing it with a standard reference tone of frequency 1 kHz; the loudness of the test sound in phons is numerically equal to the sound intensity level, in decibels, of a 1 kHz tone that is judged to sound equally loud. This is established experimentally by playing the test tone (at its particular frequency and intensity) to a panel of listeners, and separately playing them a 1 kHz reference tone whose intensity level (in dB) can be adjusted; the reference tone's dB level is adjusted until the panel of listeners judges it to sound equally as loud as the test tone. The dB level of the 1 kHz reference tone at this point of equal perceived loudness is then taken as the loudness of the original test tone, expressed in phons. (b) The human ear's sensitivity to sound is not the same at all frequencies; as established by the variation of auditory threshold with frequency, the ear is most sensitive around 3-4 kHz and less sensitive at both lower and higher frequencies. This means that two sounds with exactly the same physical intensity (in W m^-2), but at different frequencies, will not generally be perceived as equally loud by a listener; a sound at a frequency where the ear is more sensitive will be perceived as louder than a sound of identical intensity at a frequency where the ear is less sensitive. Because loudness depends on this frequency-dependent perception by the listener's ear and brain, rather than on the physical intensity of the sound alone, and can also vary somewhat between individual listeners, loudness is described as a subjective measure, in contrast to intensity (in W m^-2) or intensity level (in dB), which are objective, directly measurable physical quantities. (c) An equal loudness curve is a graph, plotted against frequency, of the sound intensity level (in dB) that is required at each frequency to produce a constant perceived loudness (that is, a single, fixed value in phons) across the whole range of audible frequencies. Because the ear's sensitivity varies with frequency, the shape of an equal loudness curve is not a flat, horizontal line, but instead a U-shaped or bowl-shaped curve: the curve reaches its lowest point (requiring the least intensity level to achieve the given loudness) at around 3-4 kHz, where the ear is most sensitive, and rises (requiring progressively greater intensity levels to achieve the same perceived loudness) towards both lower frequencies and higher frequencies, moving away from this most sensitive region. (d) A basic hearing aid consists of three main components: a microphone, which detects incoming sound and converts it into an electrical signal; an amplifier, which increases the size (amplitude) of this electrical signal; and a loudspeaker, which converts the amplified electrical signal back into a louder sound wave delivered to the ear. Final answer: (a) loudness in phons is defined relative to an equally-loud 1 kHz reference tone's dB level, found by listener comparison. (b) hearing sensitivity varies with frequency, so equal intensity does not mean equal perceived loudness, making loudness subjective. (c) equal loudness curves plot the dB level needed at each frequency for constant perceived loudness; U-shaped/bowl-shaped, minimum around 3-4 kHz, rising at lower and higher frequencies. (d) microphone, amplifier and loudspeaker.

(a) [1] correctly defines phon loudness as numerically equal to the dB level of an equally loud 1 kHz reference tone; [1] correctly describes playing the test tone to a panel of listeners; [1] correctly describes adjusting the 1 kHz reference tone's dB level until judged equally loud; [1] correctly states that this matched dB level gives the loudness in phons. (b) [1] correctly identifies that the ear's sensitivity varies with frequency; [1] correctly explains that two sounds of equal intensity but different frequency are not perceived as equally loud as a result; [1] correctly concludes that loudness therefore depends on perception (subjective) rather than intensity alone (objective). (c) [1] correctly describes an equal loudness curve as showing the dB level needed at each frequency for a constant perceived loudness (phon value); [1] correctly describes the curve's minimum around 3-4 kHz; [1] correctly describes the curve rising towards both lower and higher frequencies (U-shaped/bowl-shaped). (d) [1] all three of microphone, amplifier and loudspeaker correctly named. Maximum 11 marks.

(a) A single DNA nucleotide is made up of three components: a deoxyribose sugar (a five-carbon, or pentose, sugar), a phosphate group, and one of four nitrogenous bases (adenine, cytosine, guanine or thymine). (b) DNA base pairing follows a specific, complementary rule: adenine always pairs with thymine (A-T), held together by two hydrogen bonds, and cytosine always pairs with guanine (C-G), held together by three hydrogen bonds. This specific pairing (a purine base always pairing with a pyrimidine base of the correct complementary type) allows the two strands of the DNA double helix to fit together with a constant width along its length. Final answer: (a) deoxyribose, phosphate group, nitrogenous base; (b) A pairs with T, C pairs with G.

(a) [1] mark for each correctly named component, to a maximum of 3: deoxyribose (sugar); phosphate (group); nitrogenous base. (b) [1] correctly states A pairs with T; [1] correctly states C pairs with G. Maximum 5 marks.

(a) A triplet is a sequence of three bases, and each triplet codes for one specific amino acid (the building block from which proteins/polypeptides are assembled). (b) Because the base sequence of a gene is read by the cell in consecutive, non-overlapping triplets starting from a fixed point, deleting a single base from the middle of the sequence shifts every triplet that follows it along the sequence by one base; this is called a frameshift mutation, and it means essentially every triplet (and therefore every amino acid coded for) from the point of the deletion onwards is changed, usually producing a completely different, non-functional protein, or introducing a premature stop codon that truncates the protein. In contrast, a substitution mutation only changes the identity of a single base within a single triplet, so it can only ever change (at most) the single amino acid coded for by that one triplet, or, due to the degeneracy of the genetic code (where more than one triplet can code for the same amino acid), may have no effect on the amino acid sequence at all (a silent mutation); this means a substitution mutation is generally far more limited in its potential effect on the protein than a deletion mutation, which disrupts the entire reading frame downstream of the mutation. (c) Since I^A is dominant to I^O, a person with the genotype I^A I^O has the blood group A phenotype (the I^O allele is recessive and not expressed when paired with the dominant I^A allele). The term 'multiple alleles' describes a situation where more than two different alleles of a single gene exist within a population, in contrast to simple inheritance patterns involving only two alleles (e.g. one dominant, one recessive). The ABO blood group gene illustrates this, since three different alleles (I^A, I^B and I^O) exist for this gene within the human population, and different combinations of these three alleles (taking two at a time, since each individual only inherits one allele from each parent) produce the four possible ABO blood group phenotypes (A, B, AB and O); however, any single individual can still only ever carry two of these three alleles at once (one on each of their two homologous chromosomes). Final answer: (a) three bases; codes for one amino acid. (b) a deletion causes a frameshift affecting the whole downstream reading frame/nearly all subsequent amino acids, while a substitution affects at most one amino acid (or none, due to code degeneracy). (c) blood group A; multiple alleles means more than two alleles of a gene exist in a population (here, three: I^A, I^B, I^O), though each individual carries only two.

(a) [1] correctly states three bases per triplet; [1] correctly states a triplet codes for one amino acid. (b) [1] correctly identifies a deletion causes a frameshift, altering the reading frame for all subsequent triplets; [1] correctly links this to changing most/all of the amino acid sequence downstream of the mutation (or introducing a premature stop codon); [1] correctly explains a substitution only affects a single triplet/amino acid (and may have no effect at all, due to the degenerate genetic code), making its effect more limited by comparison. (c) [1] correctly states blood group A; [1] correctly links this to I^A being dominant over I^O; [1] correct definition of multiple alleles (more than two alleles of a gene existing in a population); [1] correctly applies this to the ABO gene, correctly noting an individual carries only two of the three alleles at a time. Maximum 9 marks.

(a) A DNA probe is a short length of single-stranded DNA, typically only a few tens of bases long, whose base sequence is complementary to part of the base sequence of the specific gene or allele being screened for. The probe is chemically labelled, for example with a radioactive isotope or a fluorescent marker, so that its presence can later be detected. To screen a patient, a sample of the patient's DNA is first separated into single strands (denatured), and the labelled probe is mixed with this single-stranded DNA under suitable conditions; if the patient's DNA sample contains the specific target sequence, the probe will bind (hybridise) to it via complementary base pairing between the probe and the matching region of the patient's DNA. After allowing time for hybridisation, any probe that has not bound is washed away, leaving only probe that is bound to a genuinely matching sequence, if present. The sample is then examined for the label used (e.g. radioactivity detected on X-ray film, or fluorescence observed under UV light); a positive signal indicates that the patient's DNA does contain the specific target sequence (for example, a disease-associated allele), while no signal indicates it is absent. (b) DNA sequencing is used to determine the precise order of bases within a candidate gene or a known disease-associated mutation, providing the sequence information needed to design a DNA probe whose own sequence is exactly complementary to this specific target. The polymerase chain reaction (PCR) is used to amplify (make many copies of) a target region of a patient's DNA from what may initially be only a very small original sample; this ensures there is enough DNA present in the sample being tested for a probe (or other screening technique) to produce a clear, reliably detectable result. (c)(i) If genetic screening shows that both parents-to-be are carriers of the same recessive disease-causing allele (heterozygous, showing no symptoms themselves but each carrying one copy of the defective allele), genetic counselling uses this information, together with a genetic (Punnett square) diagram, to inform the parents of the probability that a future child could inherit the defective allele from both parents and be affected by the condition; for two carrier parents of a standard recessive condition, this probability is 1 in 4 (25%) for each pregnancy. This information helps the parents make informed reproductive decisions, such as whether to proceed with a pregnancy, to use prenatal genetic testing, or to consider options such as IVF with pre-implantation genetic screening. (ii) If a cancer patient's tumour cells are screened and found to carry a specific, identified oncogene (a gene, when mutated or expressed abnormally, that can drive cancer development), this genetic information can be used by clinicians in genetic counselling and treatment planning to select the treatment most likely to be effective against that specific genetic profile of cancer, since different oncogenes can respond very differently to different drugs or targeted therapies; this allows a more personalised, targeted choice of treatment than would be possible without this genetic information. Final answer: (a) a DNA probe is a short, labelled, single-stranded DNA sequence that hybridises (binds) specifically to a complementary target sequence in a patient's DNA, with a detectable signal confirming the target is present. (b) sequencing identifies the target sequence to design the probe; PCR amplifies the patient's DNA sample so enough is present for reliable screening. (c)(i) informs carrier parents of the probability (e.g. 1 in 4) of an affected child, supporting informed reproductive decisions; (ii) identifying a specific oncogene helps select the most appropriate/targeted cancer treatment.

(a) [1] correctly describes a DNA probe as a short, single-stranded DNA sequence; [1] correctly states the probe's sequence is complementary to the target gene/allele; [1] correctly states the probe is labelled (radioactively or fluorescently) for detection; [1] correctly describes hybridisation/binding of the probe to a matching sequence in the patient's (denatured, single-stranded) DNA; [1] correctly describes washing away unbound probe; [1] correctly describes detecting the label to confirm presence/absence of the target sequence. (b) [1] correctly explains the role of DNA sequencing in identifying the target sequence to design the probe; [1] correctly explains the role of PCR in amplifying the patient's DNA sample to give enough material for reliable screening. (c)(i) [1] correctly explains that screening identifies carrier status in both parents; [1] correctly links this to informing parents of the probability of an affected child (e.g. 1 in 4) to support reproductive decision-making. (ii) [1] correctly explains that identifying a specific oncogene provides genetic information about the cancer; [1] correctly links this to selecting the most appropriate/targeted treatment for that patient. Maximum 12 marks.

(a) A single PCR cycle consists of three main steps. Denaturation occurs at a high temperature, typically around 95C; this heat energy breaks the hydrogen bonds holding the two complementary strands of the DNA double helix together, separating the double-stranded DNA into two single strands. Annealing then occurs at a lower temperature, typically around 50-65C; at this temperature, short, single-stranded DNA primers (designed to be complementary to specific sequences flanking the target region) are able to bind, by complementary base pairing, to the matching sequences on each of the now single-stranded template DNA strands. Extension then occurs at an intermediate-to-high temperature, typically around 72C (the optimum temperature for the enzyme used); the enzyme Taq polymerase binds at each primer and synthesises a new, complementary DNA strand by adding free DNA nucleotides, one at a time, that are complementary to the corresponding base on the single-stranded template, extending the new strand in one direction along the template. This full three-step cycle (denaturation, annealing, extension) is then repeated many times in succession, with each cycle approximately doubling the number of copies of the target DNA sequence present. (b) Taq polymerase is the enzyme responsible for catalysing the synthesis of the new complementary DNA strand during the extension step, by joining together free nucleotides in the correct sequence, complementary to the template strand. Taq polymerase (originally isolated from the bacterium Thermus aquaticus, which naturally lives in very hot environments such as hot springs) is used specifically because it is heat-stable (thermostable), meaning it is not denatured (does not lose its structure/function) at the very high temperature (around 95C) used repeatedly during the denaturation step of each cycle. A standard human DNA polymerase enzyme would be denatured and permanently lose its activity at this high temperature, meaning fresh enzyme would need to be added after every single cycle, making the reaction far too slow, impractical and expensive to automate; because Taq polymerase survives repeated heating, the whole PCR reaction can instead be run automatically through many repeated cycles using a single addition of enzyme at the start. (c) In PCR, assuming 100% efficiency, the number of DNA molecules approximately doubles with each complete cycle, since each of the two original strands acts as a template for a new complementary strand. Starting with 1 double-stranded DNA molecule, after n cycles the number of molecules present is 1 x 2^n. After 5 cycles: number of molecules = 1 x 2^5 = 1 x 32 = 32 molecules. Final answer: (a) denaturation (~95C, strands separate); annealing (~50-65C, primers bind by base pairing); extension (~72C, Taq polymerase synthesises new complementary strands). (b) Taq polymerase catalyses new strand synthesis; used because it is heat-stable and survives repeated heating to ~95C, unlike human DNA polymerase, which would need replacing every cycle. (c) 32 molecules.

(a) [1] correctly names denaturation with a correct approximate temperature (around 90-96C); [1] correctly describes strands separating/hydrogen bonds breaking during denaturation; [1] correctly names annealing with a correct approximate temperature (around 50-65C); [1] correctly describes primers binding by complementary base pairing during annealing; [1] correctly names extension with a correct approximate temperature (around 70-75C); [1] correctly describes Taq polymerase synthesising new complementary strands by adding nucleotides during extension. (b) [1] correctly states Taq polymerase's role in catalysing new strand synthesis; [1] correctly explains it is heat-stable/thermostable and survives the high denaturation temperature, unlike human DNA polymerase which would need replacing every cycle. (c) [1] correct method shown (doubling per cycle, or 2^5); [1] correct final answer, 32 molecules. Maximum 10 marks.

(a) In gel electrophoresis, DNA samples are loaded into small wells cut into one end of a gel (typically made of agarose), which is submerged in a conducting buffer solution, and an electric current is passed through the gel from one end to the other. Because the phosphate groups in DNA's sugar-phosphate backbone are negatively charged, DNA fragments are attracted towards the positive electrode (anode) and migrate through the gel towards it when the current is applied. The gel itself acts as a molecular sieve, made up of a mesh-like matrix of pores; smaller DNA fragments are able to move through the pores of this matrix relatively easily and quickly, while larger DNA fragments are more impeded/slowed down by the matrix as they move through it. As a result, over a fixed period of time, smaller DNA fragments travel further through the gel (from the well towards the positive electrode) than larger DNA fragments, meaning that a mixture of DNA fragments of different sizes becomes separated out into distinct bands positioned according to fragment size, with the smallest fragments furthest from the well and the largest fragments closest to the well. (b) Using the given formula: Rf = (distance travelled by fragment) / (distance travelled by solvent front) = 3.6/4.8 = 0.75. (c) A smaller DNA fragment has a larger Rf value (having travelled further, closer to the solvent front), while a larger DNA fragment has a smaller Rf value (having travelled a shorter distance). This relationship exists because, as described in part (a), smaller fragments experience less resistance moving through the pores of the gel matrix and are therefore able to migrate further through the gel in a given time than larger fragments, which are more strongly impeded by the matrix. (d) A sample of a patient's DNA, containing the region of a specific gene of interest (such as a known oncogene), can be amplified using PCR and then run through gel electrophoresis, producing a pattern of bands corresponding to the sizes of DNA fragments present in that region. This band pattern can be compared directly against the band pattern produced from a known, normal (healthy) reference sample of the same gene region. If the patient's DNA contains a mutation in the oncogene (for example, an insertion or deletion of bases altering fragment length, or a change to a base sequence recognised by a restriction enzyme used to cut the DNA before running the gel), this can result in a band appearing at an unexpected size/position (or an extra or missing band) compared with the normal reference pattern; identifying such a difference can indicate the presence of the mutated oncogene, helping to screen the patient for an increased cancer risk, or to confirm a diagnosis, without needing to fully sequence the entire gene. Final answer: (a) DNA fragments (negatively charged) migrate through a gel matrix towards the positive electrode under an electric current; smaller fragments move further/faster than larger fragments, separating the sample by size. (b) Rf = 0.75. (c) smaller fragments have larger Rf/travel further, because they move more easily through the gel matrix's pores than larger fragments. (d) comparing the patient's amplified DNA band pattern with a normal reference pattern; a differently sized/positioned band can indicate a mutated oncogene, helping to screen for cancer risk.

(a) [1] correctly describes DNA samples loaded into wells in a gel, with an electric current applied; [1] correctly states DNA is negatively charged and migrates towards the positive electrode; [1] correctly describes the gel matrix acting as a molecular sieve/mesh of pores; [1] correctly states smaller fragments move through the gel more easily/quickly than larger fragments; [1] correctly concludes this separates fragments by size over time (smaller fragments travel further). (b) [1] correct substitution into the given Rf formula; [1] correct final answer, 0.75. (c) [1] correctly states smaller fragments have a larger Rf/travel further (and larger fragments a smaller Rf/travel less far); [1] correctly explains this in terms of smaller fragments experiencing less resistance moving through the gel matrix. (d) [1] correctly describes comparing a patient's (PCR-amplified) DNA band pattern against a normal reference pattern; [1] correctly explains that a difference in band size/position can indicate a mutated oncogene, aiding cancer screening. Maximum 11 marks.

(a) Level of response guidance (indicative content, to be marked using the 4-band rubric below): The semi-conservative model - during DNA replication, the enzyme DNA helicase moves along the DNA molecule, unwinding the double helix and breaking the hydrogen bonds holding the two complementary polynucleotide strands together, separating them into two single strands. Each of these original (parental) strands then acts as a template: free DNA nucleotides in the surrounding nucleoplasm are attracted to, and pair with, the exposed bases on each template strand by specific complementary base pairing (A with T, C with G). The enzyme DNA polymerase then catalyses a condensation reaction, joining adjacent nucleotides together to form a new, continuous sugar-phosphate backbone along each template strand. The result of this process is two new DNA double helix molecules, each one made up of one original (parental) strand and one newly synthesised strand; because half of each new molecule is 'conserved' from the original molecule, this model of replication is called semi-conservative. Meselson and Stahl's experiment - in 1958, Meselson and Stahl designed an experiment to distinguish between three competing theories of DNA replication that existed at the time: the conservative model (which proposed the original double helix stayed completely intact, with an entirely separate, entirely new molecule made alongside it), the dispersive model (which proposed that both strands of both resulting molecules would contain a scattered mixture of original and new DNA segments), and the semi-conservative model. They grew E. coli bacteria for many generations in a growth medium containing only the heavy nitrogen isotope, 15N, so that all of the nitrogen atoms within the bases of the bacterial DNA (all newly incorporated during this time) became heavy, making this DNA denser than normal DNA. They then transferred these bacteria into a growth medium containing only the normal, lighter nitrogen isotope, 14N, and allowed the bacteria to replicate their DNA for one, and then a second, generation, extracting a DNA sample after each generation. Each DNA sample was separated by density using density-gradient centrifugation, a technique that separates molecules into distinct bands according to their mass. After one generation of replication in the 14N medium, all of the DNA formed a single band at a density exactly intermediate between fully heavy (15N/15N) DNA and fully light (14N/14N) DNA; this result immediately ruled out the conservative model, since that model predicted two separate bands after one generation (one fully heavy, unreplicated original DNA, and one fully light, entirely new DNA), rather than the single intermediate band that was actually observed. After a second generation of replication in the 14N medium, two distinct bands appeared: one still at the intermediate density (as seen after generation one), and a second, new band at the fully light density. This result is exactly what the semi-conservative model predicts, since each intermediate (hybrid) molecule from the first generation contains one heavy strand and one light strand; when this hybrid molecule itself replicates, each of its two strands (one heavy, one light) acts as a separate template for a new light strand, producing one hybrid (intermediate) molecule and one fully light molecule from each original hybrid molecule, giving the observed mixture of intermediate and fully light bands. This result also ruled out the dispersive model, since that model predicts that, however many generations of replication occur, all of the resulting DNA molecules would still contain some scattered heavy DNA mixed throughout every molecule, and so would all still appear as a single band, of a density that gradually becomes lighter with each generation, rather than splitting into two clearly separate bands (one intermediate, one fully light) as was actually observed. Final answer: DNA replication is semi-conservative, with each new DNA molecule containing one original (template) strand and one newly synthesised strand, produced via helicase unwinding the double helix and DNA polymerase joining new complementary nucleotides to each template strand; Meselson and Stahl's density-gradient centrifugation results (a single intermediate band after one generation, then both an intermediate and a fully light band after a second generation) matched the predictions of the semi-conservative model and ruled out both the conservative and dispersive models. (b) Meiosis produces genetically different gametes through two main features. Independent segregation (assortment) of homologous chromosomes occurs during meiosis I: each pair of homologous chromosomes lines up at the equator of the cell and separates independently of every other pair, meaning that the particular combination of maternal and paternal chromosomes ending up in each resulting gamete is random and differs between different gametes produced by the same individual, generating many different possible chromosome combinations. Crossing over (genetic recombination) occurs during prophase I, when homologous chromosomes pair up closely together (forming a bivalent) and exchange corresponding sections of DNA between non-sister chromatids at points of contact called chiasmata; this creates new combinations of alleles along a chromosome that were not present on either original parental chromosome, further increasing genetic variation among the gametes produced. (c) Meiosis produces haploid cells (gametes), each containing only one copy of each chromosome (half the normal diploid chromosome number of the parent cell). This is important for sexual reproduction because, at fertilisation, two haploid gametes (one from each parent) fuse together to form a diploid zygote, restoring the full, normal diploid chromosome number (one complete set inherited from each parent); if gametes were not haploid (i.e. if meiosis did not halve the chromosome number before fertilisation), the chromosome number would double with every generation, which would not be sustainable.

(a) Level of response mark scheme (9 marks, 4 bands). Excellent (7-9 marks): accurately describes the semi-conservative mechanism (helicase unwinding/breaking hydrogen bonds, template strands, complementary base pairing of new nucleotides, DNA polymerase joining nucleotides, resulting hybrid molecules) AND accurately describes the Meselson-Stahl experimental method (15N/14N labelling, density-gradient centrifugation) and correctly explains how BOTH the generation-1 result (single intermediate band, ruling out conservative) and the generation-2 result (intermediate + fully light bands, ruling out dispersive) support the semi-conservative model; accurate specialist vocabulary throughout; clear, coherent, well-structured writing. Good (4-6 marks): describes the semi-conservative mechanism with reasonable accuracy and gives a generally correct account of the Meselson-Stahl experiment and at least one of the two key results, though may be less complete or less precise in linking results to ruling out the alternative models; mostly accurate vocabulary; generally clear writing with minor errors. Basic (1-3 marks): gives a limited or partially correct description of DNA replication and/or the Meselson-Stahl experiment, with little development or explanation of how the results distinguish between the models; writing may be list-like or contain errors that hinder meaning. 0 marks: no creditable response. (b) [1] mark for each correctly described feature, to a maximum of 3: independent segregation/assortment of homologous chromosomes [1-2 depending on detail]; crossing over/recombination between non-sister chromatids at chiasmata [1-2 depending on detail] (award up to 3 marks total across both features, with at least one mark requiring correct linkage to increased genetic variation). (c) [1] correctly states haploid; [1] correctly explains that fusion of two haploid gametes at fertilisation restores the diploid number, preventing it doubling each generation. Maximum 14 marks.

(a) To produce human insulin using genetically modified bacteria, the gene coding for human insulin is first isolated and cut out using a restriction enzyme; the same restriction enzyme is used to cut open a bacterial plasmid (a small circular piece of DNA), producing complementary 'sticky ends' on both the insulin gene and the opened plasmid. The insulin gene is then inserted into the plasmid, and the enzyme DNA ligase is used to join the gene into the plasmid, sealing the sugar-phosphate backbone and forming a single recombinant plasmid. This recombinant plasmid is introduced into bacterial cells (commonly E. coli), a process which produces genetically modified bacteria now carrying the human insulin gene. These modified bacteria are then cultured (grown) on a large industrial scale in fermenters/bioreactors; as the bacteria grow and divide, they express the inserted human gene using their own protein synthesis machinery, producing human insulin, which is then extracted from the culture and purified for medical use; Humulin is the licensed drug produced this way. Two advantages of producing insulin this way, rather than extracting it from natural animal sources such as cattle or pigs, are: firstly, bacterially produced human insulin is chemically identical to naturally occurring human insulin, whereas insulin extracted from cattle, dogs or pigs is not identical to human insulin, and can cause adverse immune reactions in some patients; secondly, genetically modified bacteria can be grown very rapidly and in very large quantities in fermenters, allowing much larger amounts of insulin to be produced, more reliably and at lower production cost, than the difficult and laborious process of extracting relatively small amounts of insulin from animal sources (extraction of the hormone from animals is very difficult); using bacterially produced insulin also raises fewer ethical or religious objections than using an animal-derived product. (b) In this approach, the human gene coding for the desired therapeutic protein (such as human serum albumin) is inserted into the genome of an animal (creating a transgenic/genetically modified animal), often using genetic engineering techniques designed so that the inserted gene is expressed specifically within the cells of the animal's mammary glands. As a result, the genetically modified animal naturally produces and secretes the human protein directly into its milk, alongside its normal milk proteins; because milk can be collected repeatedly, in large volumes, without harming the animal, the human therapeutic protein can then be extracted and purified from the milk on a large, ongoing commercial scale, providing an efficient and plentiful supply of an otherwise scarce therapeutic protein, such as human serum albumin used to treat burns patients. (c) A major risk associated with obtaining factor VIII from pooled human blood serum is the potential transmission of blood-borne viruses present in donor blood, such as HIV or hepatitis viruses; historically, this risk was realised on a large scale, resulting in many haemophiliacs contracting HIV/AIDS after being treated with contaminated factor VIII derived from pooled donor blood. Genetically engineered factor VIII, produced using genetically modified cells or organisms carrying the human factor VIII gene, avoids this specific risk because its production does not involve collecting, pooling or processing any human donor blood at all, removing the route by which a blood-borne viral infection could be transmitted to the patient receiving the treatment. Final answer: (a) the human insulin gene is inserted into a bacterial plasmid (using restriction enzymes and DNA ligase) and introduced into bacteria, which are cultured to express and produce insulin, identical to human insulin and produced in far greater quantity/at lower cost than from animal extraction. (b) inserting the human gene into an animal's genome so it is expressed in mammary gland cells, allowing the protein to be extracted from the animal's milk. (c) risk of viral transmission (e.g. HIV/hepatitis) from pooled donor blood; genetically engineered factor VIII avoids this because it is not derived from human blood at all.

(a) [1] correctly describes the insulin gene being inserted into a bacterial plasmid (using restriction enzymes/DNA ligase); [1] correctly describes the plasmid being introduced into bacteria, which are cultured to express/produce insulin; [1] mark for each of two valid, distinct advantages correctly explained, to a maximum of 3 marks total for this part (e.g. identical to human insulin/fewer adverse reactions; larger quantities/lower cost/more reliable production; fewer ethical/religious objections). (b) [1] correctly describes inserting the human gene into the animal's genome, targeted to mammary gland expression; [1] correctly describes the protein being secreted in milk and extracted/purified from it. (c) [1] correctly identifies the risk of viral transmission (e.g. HIV/hepatitis) from pooled donor blood; [1] correctly explains that genetically engineered factor VIII avoids this because it is not derived from human blood. Maximum 9 marks.

(a)(i) Since neither parent shows symptoms of cystic fibrosis, but they have a child (their daughter) who is affected, both parents must carry one copy of the recessive allele without being affected themselves; therefore both parents must have the heterozygous genotype, Ff. (ii) Genetic diagram: Parents: Ff (father) x Ff (mother). Gametes from father: F or f. Gametes from mother: F or f. Combining these gametes in a Punnett square gives four possible offspring genotype combinations: FF, Ff, Ff and ff, each equally likely. This gives a genotypic ratio among the offspring of 1 FF : 2 Ff : 1 ff (i.e. 1/4 FF, 1/2 Ff, 1/4 ff). (iii) The affected daughter has the genotype ff (homozygous recessive), since cystic fibrosis is a recessive condition and can only be expressed in an individual carrying two copies of the recessive allele. It is possible for two unaffected parents to have an affected child because both parents, despite being phenotypically unaffected themselves, are heterozygous carriers (Ff) of the recessive allele; an unaffected carrier shows the normal/unaffected phenotype because their one dominant F allele is sufficient to mask the effect of their one recessive f allele, but each carrier parent still has a chance of passing on their recessive f allele to any given child; if a child happens to inherit the recessive f allele from both parents, becoming ff, they will be affected by the condition, even though neither parent shows any symptoms themselves. (iv) From the genotypic ratio calculated in (a)(ii), the probability of a child being Ff (heterozygous, and therefore a carrier who is not affected by the condition, since F is dominant) is 2 out of the 4 equally likely possible outcomes, i.e. 2/4 = 1/2 = 50%. (b)(i) Cystic fibrosis is caused by a defective gene that fails to produce a properly functioning protein needed for normal cell function (in the epithelial cells lining the lungs and other organs). In principle, stem cell technology could offer a treatment approach distinct from gene therapy: rather than directly repairing or replacing the specific faulty gene within a patient's own existing cells (as gene therapy attempts to do), stem cells (which are capable of dividing and differentiating into specialised cell types) could potentially be used to generate new, healthy epithelial or lung cells that carry a normal, functioning copy of the gene; these new, healthy cells could then, in principle, be introduced into the patient's affected tissue (such as the lining of the lungs) to replace or supplement the patient's own damaged or malfunctioning cells, potentially restoring more normal tissue function. (ii) Embryonic stem cells are pluripotent, meaning they have the potential to differentiate into almost any type of specialised cell found in the body (though not, on their own, into an entire new organism); adult stem cells are generally described as multipotent, meaning their potential to differentiate is more limited, typically only being able to differentiate into a narrower range of related cell types associated with the particular tissue in which they are found (for example, bone marrow stem cells mainly differentiating into different types of blood cell). This means embryonic stem cells offer a wider range of potential therapeutic applications than adult stem cells, though adult stem cells avoid some of the ethical concerns associated with embryonic stem cells. (iii) A key ethical issue associated with the use of embryonic stem cells is that they are typically obtained from early-stage embryos (for example, embryos left over from IVF treatment), and extracting the stem cells results in the destruction of the embryo; this raises significant ethical concerns for some people about the moral status of the embryo and whether it is ethically acceptable to use and destroy a potential human life, even at a very early stage of development, for research or therapeutic purposes. Final answer: (a)(i) both parents Ff; (ii) offspring genotype ratio 1 FF : 2 Ff : 1 ff; (iii) affected daughter is ff, possible because both unaffected parents are heterozygous carriers; (iv) probability of an unaffected carrier child = 1/2 (50%). (b)(i) stem cells could generate new, healthy cells carrying a functional gene copy to replace/supplement damaged cells, rather than repairing the existing faulty gene as gene therapy does; (ii) embryonic stem cells are pluripotent (can become almost any cell type), adult stem cells are multipotent (more limited range); (iii) destruction of the embryo to obtain embryonic stem cells raises concerns about the moral status of the embryo.

(a)(i) [1] mark for each parent correctly identified as Ff, to a maximum of 2. (ii) [1] correct gametes shown for each parent (F, f); [1] correct Punnett square/diagram showing all four offspring combinations; [1] correct genotypes identified (FF, Ff, Ff, ff); [1] correct genotypic ratio stated, 1 FF : 2 Ff : 1 ff. (iii) [1] correctly states the affected daughter's genotype as ff; [1] correct explanation that both unaffected parents are heterozygous carriers, each capable of passing on the recessive allele. (iv) [1] correct identification that carrier-but-unaffected corresponds to genotype Ff; [1] correct final probability, 1/2 (50%) (accept 2/4). (b)(i) [1] correctly identifies the general stem cell approach as generating new, healthy cells (rather than repairing the existing gene); [1] correctly links this to replacing/supplementing damaged cells in the affected tissue (e.g. lung/epithelial cells); [1] for a valid, correctly explained distinction from gene therapy's direct gene-repair approach. (ii) [1] correctly identifies embryonic stem cells as pluripotent (wide differentiation potential); [1] correctly identifies adult stem cells as multipotent (more limited differentiation potential). (iii) [1] correctly identifies a valid ethical issue (e.g. destruction of the embryo/moral status of the embryo); [1] for a further correctly explained point/valid distinct ethical issue. Maximum 17 marks.

(a) Each parent is heterozygous for both genes (RrYy), and assuming the two genes assort independently (are not epistatic and are not linked), each parent produces four types of gamete in equal proportions: RY, Ry, rY and ry. Combining these gametes from both parents in a fully labelled Punnett square (a 4x4 grid) gives 16 equally likely offspring genotype combinations. Grouping these genotypes by phenotype: any offspring with at least one dominant R allele and at least one dominant Y allele is round and yellow (9 out of 16 combinations); any offspring with at least one dominant R allele but homozygous recessive yy is round and green (3 out of 16); any offspring homozygous recessive rr but with at least one dominant Y allele is wrinkled and yellow (3 out of 16); and the offspring that is homozygous recessive for both genes, rryy, is wrinkled and green (1 out of 16). This gives the classic dihybrid cross phenotypic ratio of 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green. (b) Epistasis occurs when the allele(s) present at one gene locus mask, suppress or otherwise modify the phenotypic expression of a different gene at a separate locus, meaning the final observed phenotype does not simply reflect the independent expression of both genes as in a standard dihybrid cross. A commonly used example of epistasis is coat colour in some mammals, which can be controlled by (at least) two separate genes: one gene (for example, C/c) controls whether any pigment is deposited in the fur at all, while a separate gene (for example, B/b) controls which colour (e.g. black versus brown) is produced if pigment is deposited. If an individual is homozygous recessive for the first gene (cc), no pigment is produced at all, and the individual is albino (white), regardless of which alleles it has at the second gene (B/b); in this case, the cc genotype at the first locus is epistatic to (masks the effect of) the second gene, since the second gene's alleles cannot produce their normal effect (black or brown colouration) when no pigment is being produced in the first place. (c)(i) For a total of 320 offspring, the expected number in each category is found by multiplying the total by each fraction of the expected 9:3:3:1 ratio (which totals 9+3+3+1 = 16 parts): expected round yellow = (9/16) x 320 = 180; expected round green = (3/16) x 320 = 60; expected wrinkled yellow = (3/16) x 320 = 60; expected wrinkled green = (1/16) x 320 = 20. (ii) Calculating (O-E)^2/E for each category: round yellow: (185-180)^2/180 = 25/180 = 0.139; round green: (62-60)^2/60 = 4/60 = 0.067; wrinkled yellow: (58-60)^2/60 = 4/60 = 0.067; wrinkled green: (15-20)^2/20 = 25/20 = 1.250. Summing these four values: chi-squared = 0.139 + 0.067 + 0.067 + 1.250 = 1.52 (to 3 s.f.). (iii) Since the calculated chi-squared value (1.52) is less than the given critical value (7.82) for p = 0.05 with 3 degrees of freedom, the difference between the observed and expected results is not statistically significant; this means there is no significant evidence against the expected 9:3:3:1 ratio, so the observed data supports (is consistent with) the expected 9:3:3:1 ratio, with any difference between observed and expected counts likely being due to chance. Final answer: (a) 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green. (b) epistasis is when one gene's alleles mask/modify the expression of a different gene, e.g. an albino (cc) genotype at a pigment-deposition gene masking the effect of a separate pigment-colour gene. (c)(i) expected 180, 60, 60, 20; (ii) chi-squared = 1.52; (iii) supports the 9:3:3:1 ratio, since 1.52 is less than the critical value of 7.82 (difference not statistically significant).

(a) [1] correctly identifies the gametes produced by each parent (RY, Ry, rY, ry); [1] correct/complete Punnett square (or equivalent genetic diagram) shown; [1] correct grouping of genotypes into the four phenotype categories; [1] correct final ratio, 9:3:3:1. (b) [1] correct definition of epistasis (one gene's allele(s) masking/modifying the expression of a different gene); [1] a valid, distinct example given (e.g. albinism gene masking a separate colour gene); [1] correct explanation of how the example demonstrates epistasis (e.g. cc genotype produces no pigment regardless of the colour gene's alleles). (c)(i) [1] correct method (fraction x total); [1] all four expected values correct (180, 60, 60, 20). (ii) [1] correct calculation of at least two of the four (O-E)^2/E terms; [1] all four terms correctly calculated; [1] correct final chi-squared value, 1.52 (accept values rounding to this from correct working). (iii) [1] correctly concludes the data supports the 9:3:3:1 ratio, with correct reasoning (calculated value less than the critical value, so not statistically significant). Maximum 13 marks.