2023 CCEA GCSE Science Double Award 1370 模擬試題連答案詳解

The mercury sits midway between the two 1°C graduations, so the reading is estimated to the nearest 0.5°C, giving 34.5°C.

1 mark: correct value 34.5; 1 mark: unit °C given and reading shown to one decimal place (0.5°C precision). Reject 34°C or 35°C (fails to estimate between graduations). [2]

Titre = final reading − initial reading = 27.65 − 2.40 = 25.25 cm³.

1 mark: correct subtraction shown (27.65 − 2.40); 1 mark: correct answer 25.25 cm³ to 2 decimal places with unit. [2]

The plunger is exactly on a graduation, so the reading is 46 cm³. Accuracy is improved by viewing the scale at eye level (avoiding parallax) and checking the syringe is not leaking.

1 mark: correct reading 46 cm³ with unit; 1 mark: valid precaution, e.g. read at eye level / avoid parallax error / check for leaks / take reading at eye level with syringe held horizontally; accept any one sensible precaution. [2]

Two divisions above 3.0 N, with each division worth 0.2 N, gives 3.0 + (2 × 0.2) = 3.4 N.

1 mark: correct working (3.0 + 0.4); 1 mark: correct final value 3.4 with unit N. [2]

Although the stopwatch display resolves to 0.01 s, the student's own reaction time when starting and stopping the watch introduces an uncertainty of a few tenths of a second, so the reading should be rounded to the nearest whole second (42 s).

1 mark: sensible rounded reading, 42 s (accept 42.4 s if precision to 1 d.p. is justified); 1 mark: correct explanation referring to human reaction time exceeding the stopwatch's display resolution. [2]

The colour lies closer to the pH 4 reference swatch, so pH 4 is recorded. Universal indicator relies on the observer's judgement of colour, which is subjective and cannot distinguish fine differences (e.g. pH 4.2 vs pH 4.4), unlike a calibrated pH probe.

1 mark: pH 4 recorded with justification (closer match); 1 mark: valid limitation, e.g. subjective/imprecise colour judgement vs numerical probe reading; accept reference to colour-blindness affecting reliability. [2]

Volume of mineral = 58.0 − 40.0 = 18.0 cm³. Density = mass ÷ volume = 47.6 ÷ 18.0 = 2.644... = 2.64 g/cm³ (3 s.f.).

1 mark: volume by displacement correctly found (18.0 cm³); 1 mark: correct substitution into density = mass ÷ volume; 1 mark: correct numerical answer 2.64; 1 mark: correct unit g/cm³ and answer given to 3 s.f. e.c.f. applies if the volume in mark 1 is used consistently. [4]

Moles of NaOH = concentration × volume (in dm³) = 0.100 × (25.0/1000) = 0.00250 mol. Since the mole ratio NaOH:HCl is 1:1, moles of HCl = 0.00250 mol. Concentration of HCl = moles ÷ volume = 0.00250 ÷ (22.5/1000) = 0.111 mol/dm³ (3 s.f.).

1 mark: moles of NaOH correctly calculated (0.00250 mol); 1 mark: correct use of 1:1 mole ratio to find moles HCl; 1 mark: correct substitution concentration = moles ÷ volume in dm³; 1 mark: correct final answer 0.111 mol/dm³ (accept 0.110-0.112 range for rounding). e.c.f. applies throughout. [4]

Energy supplied = power × time = 40.0 × 300 = 12000 J. Temperature change = 44.8 − 20.0 = 24.8°C. Specific heat capacity = energy ÷ (mass × temperature change) = 12000 ÷ (0.500 × 24.8) = 12000 ÷ 12.4 = 967.7 = 968 J/(kg°C) (3 s.f.).

1 mark: energy correctly calculated (12000 J, from power × time); 1 mark: temperature change correctly found (24.8°C); 1 mark: correct substitution into the given formula; 1 mark: correct final answer in range 960-975 J/(kg°C) with correct unit. e.c.f. applies if an earlier value is carried through consistently.; 1 mark: final answer correctly rounded to 3 significant figures with the unit stated separately from the numerical value. [5]

Increasing acid concentration raises the number of acid particles per unit volume, increasing collision frequency and therefore rate. Crushing the marble increases its surface area, exposing more particles to collision with acid particles, also increasing rate. Decreasing temperature reduces the average kinetic energy of particles, so fewer collisions have energy exceeding the activation energy, decreasing rate.

Row 1: 1 mark for 'increases'; 1 mark for reason referencing increased particle concentration / frequency of collisions. Row 2: 1 mark for 'increases'; 1 mark for reason referencing increased surface area / more collision sites. Row 3: 1 mark for 'decreases'; 1 mark for reason referencing lower kinetic energy / fewer particles reaching activation energy. Accept collision theory language throughout; reject vague answers such as 'more reaction happens' without a mechanism. [6]

Air movement continually replaces humid air next to the leaf with drier air, steepening the diffusion gradient and increasing transpiration. Higher humidity reduces this gradient, slowing water loss by diffusion. Petroleum jelly physically blocks the stomata (concentrated on the lower epidermis in most plants), directly reducing the pathway for water vapour to escape.

Row 1: 1 mark 'increases'; 1 mark reason referencing steeper diffusion gradient from moving air removing water vapour. Row 2: 1 mark 'decreases'; 1 mark reason referencing reduced concentration gradient for diffusion. Row 3: 1 mark 'decreases'; 1 mark reason referencing blocked stomata reducing water vapour loss. Accept 'evaporation' interchangeably with diffusion of water vapour where used correctly. [6]

Plotting extension against load gives a straight-line pattern through the origin, (1.0, 2.1), (2.0, 4.0), (3.0, 6.2) and (4.0, 8.1), consistent with Hooke's law (extension proportional to load). The point (5.0, 11.5) lies well above this trend and is anomalous, likely because the spring has been stretched beyond its elastic limit; it should be plotted but excluded when drawing the line of best fit.

1 mark: sensible linear scales chosen using more than half the grid on both axes with no awkward multiples; 1 mark: axes correctly labelled with quantity and unit; 1 mark: all six points plotted accurately to within half a small square; 1 mark: single ruled straight line of best fit drawn through the origin and the five consistent points; 1 mark: anomalous point (5.0 N) correctly identified and not used in drawing the line; 1 mark: line has a roughly even scatter of points above and below it (best-fit judgement). [6]

Using the origin (0.0, 0.0) and the point (4.0, 8.1) on the line: gradient = Δy ÷ Δx = (8.1 − 0.0) ÷ (4.0 − 0.0) = 2.03 ≈ 2.0 cm/N. The gradient represents how much the spring extends for each newton of load applied — i.e. the reciprocal of the spring's stiffness (extension per unit force).

1 mark: two points read correctly from the line, at least half the line's length apart; 1 mark: correct Δy and Δx values shown; 1 mark: correct gradient calculation (value in range 1.9-2.1); 1 mark: correct unit cm/N (or m/N if converted, provided consistent); 1 mark: correct physical interpretation (extension per unit load / measure of spring's flexibility). e.c.f. applies if points are misread but gradient method is correct. [5]

The data show a negative, non-linear relationship: bubble count falls sharply between 10 cm and 30 cm (from 48 to 13) then more gradually between 30 cm and 50 cm (13 to 4). Because light intensity decreases with the square of distance from the lamp, moving the pondweed further away sharply reduces the light energy available to drive the light-dependent reactions of photosynthesis, which in turn limits the rate of oxygen production.

1 mark: correct trend described (bubbles/rate decreases as distance increases); 1 mark: trend qualified as non-linear / decreasing rate of decrease (steeper fall at short distances); 1 mark: explanation linking distance to reduced light intensity and hence reduced rate of the light-dependent reaction / photosynthesis. Reject explanations that omit the light-intensity link. [3]

Reaction rate is inversely related to time taken, so as time falls from 96 s to 23 s over the concentration range, rate is increasing with concentration. Comparing 0.5 mol/dm³ (96 s) with 1.0 mol/dm³ (46 s) shows an approximate halving of time when concentration doubles, and comparing 1.0 mol/dm³ (46 s) with 2.0 mol/dm³ (23 s) shows the same pattern, suggesting rate (1/time) is approximately proportional to concentration over this range.

1 mark: correct trend (time decreases / rate increases as concentration increases); 1 mark: quantitative comparison cited from table (e.g. specific time values or ratio); 1 mark: valid conclusion about proportionality (rate roughly proportional to concentration, evidenced by the halving pattern). e.c.f. for a numerically consistent but differently-worded ratio argument. [3]

Dividing resistance by length at each row gives a constant value of 0.055 Ω/cm (1.1/20, 2.2/40, 3.3/60, 4.4/80, 5.5/100 all equal 0.055), confirming that resistance increases in direct proportion to length — consistent with R = ρL/A where resistivity and cross-sectional area are constant.

1 mark: correct trend stated (resistance increases as length increases); 1 mark: relationship correctly identified as directly proportional (not just 'increases'); 1 mark: supporting evidence given, e.g. resistance/length ratio constant at 0.055 Ω/cm, or doubling length doubles resistance. [3]

The shortest time (40 s, fastest rate) occurs at pH 7, with times increasing symmetrically at both more acidic (pH 3, 5) and more alkaline (pH 9, 11) values. This U-shaped pattern indicates an optimum pH around 7 for this amylase; deviation from the optimum in either direction alters the shape of the enzyme's active site (denaturation/reduced substrate binding), reducing the rate of reaction.

1 mark: correct overall trend described (rate fastest/time shortest at pH 7, slower either side); 1 mark: optimum pH correctly identified as approximately 7 with supporting time value; 1 mark: explanation referencing active site shape change / reduced enzyme-substrate binding away from optimum pH. [3]

The uncovered beaker loses the most heat (lowest final temperature, 52°C) since heat escapes freely by conduction, convection and radiation from the exposed surface. Each additional covering reduces one or more of these loss mechanisms; cotton wool traps insulating air pockets reducing conduction/convection, and the shiny foil layer reduces radiative loss, so the combined covering retains the most heat, giving the highest final temperature (68°C).

1 mark: correct trend described (higher final temperature with better/more insulation); 1 mark: comparison quantified using the 52°C vs 68°C values; 1 mark: valid explanation referencing reduced conduction/convection (trapped air in cotton wool) and/or reduced radiation (foil), for the difference between the two extremes. [3]

As external sucrose concentration rises, percentage mass change falls linearly, from +12% at 0.0 mol/dm³ to −18% at 0.8 mol/dm³, passing through zero between the 0.2 mol/dm³ (+4%) and 0.4 mol/dm³ (−3%) readings — closer to 0.4 mol/dm³. At this crossing point, the external solution's water potential equals the potato cells' water potential, so this concentration approximates the concentration of the cell sap (roughly 0.35-0.4 mol/dm³).

1 mark: correct trend (mass change decreases from positive to negative as concentration increases); 1 mark: zero-change point correctly located between 0.2 and 0.4 mol/dm³, nearer 0.4; 1 mark: correct link to cell sap concentration = external concentration at zero mass change, with a numerical estimate in the range 0.3-0.4 mol/dm³. e.c.f. for a consistent estimate from an accurately read crossing point. [3]

Cooling time rises from 3 minutes at 1 cm to 95 minutes at 8 cm, increasing much faster than the side length itself (an 8-fold increase in side length gives roughly a 30-fold increase in cooling time). This is explained by surface area to volume ratio: surface area scales with side² while volume scales with side³, so as cubes get larger the surface area to volume ratio falls, meaning proportionally less surface is available to radiate/conduct heat away relative to the volume of material storing that heat, so larger cubes cool more slowly.

1 mark: correct trend (cooling time increases sharply/disproportionately with side length); 1 mark: quantitative reference to the data (e.g. 3 min at 1 cm vs 95 min at 8 cm); 1 mark: correct explanation referencing decreasing surface area to volume ratio as size increases. [3]

Dividing each acceleration value by its corresponding force gives values close to 0.48-0.51 (0.24/0.5=0.48, 0.51/1.0=0.51, 0.73/1.5=0.49, 1.02/2.0=0.51, 1.24/2.5=0.50), which are approximately constant. This confirms acceleration is directly proportional to force when mass is constant, in agreement with Newton's second law, F = ma.

1 mark: correct trend (acceleration increases as force increases); 1 mark: relationship identified as approximately directly proportional, supported by a calculated ratio (≈0.5 m/s² per N); 1 mark: correct link to Newton's second law / F = ma at constant mass. [3]

Breathing rate more than doubles immediately after exercise, peaking at 34 bpm at the 1-minute mark, reflecting the increased demand for oxygen (and removal of carbon dioxide) generated by muscle activity during exercise. It then decreases steadily over the next 8 minutes as the body repays its oxygen debt, reaching the pre-exercise resting rate of 16 bpm by 9 minutes, indicating recovery.

1 mark: correct initial trend (sharp rise from rest to peak at 1 minute, with values quoted); 1 mark: correct recovery trend (steady/gradual fall back towards resting rate); 1 mark: recognition that the rate returns to the original resting value (16 bpm) by 9 minutes, indicating full recovery. [3]

Time falls from 180 s at 20°C to 17 s at 60°C. Comparing successive 10°C intervals, the time roughly halves each time (180→95→52→29→17, ratios of about 0.53, 0.55, 0.56, 0.59), meaning the rate (1/time) roughly doubles for each 10°C rise. This is consistent with the general rule that reaction rate approximately doubles for every 10°C increase in temperature, reflecting the exponential increase in the proportion of particles with energy ≥ activation energy predicted by collision theory.

1 mark: correct overall trend (time decreases / rate increases with temperature); 1 mark: pattern quantified — time approximately halves for each 10°C rise (or rate approximately doubles), with reference to specific values; 1 mark: link made to collision theory / proportion of particles exceeding activation energy increasing with temperature. [3]

Dividing paperclips lifted by number of turns gives approximately constant values (3/10=0.30, 7/20=0.35, 10/30=0.33, 14/40=0.35, 17/50=0.34), indicating an approximately proportional relationship. This is consistent with the magnetic field strength of a solenoid being proportional to the number of turns (for constant current), so a stronger field produces a greater lifting force.

1 mark: correct trend (paperclips lifted increases as number of turns increases); 1 mark: relationship identified as approximately directly proportional, with supporting ratio calculation; 1 mark: correct link to magnetic field strength of a solenoid depending on number of turns (at constant current). [3]

As temperature rises from 15°C to 35°C, the time for the ring to appear falls from 210 s to 108 s, roughly halving. Higher temperature gives gas particles greater average kinetic energy, so they move faster and diffuse more quickly through the air in the tube; the ammonia and hydrogen chloride particles therefore meet and react (forming solid ammonium chloride) sooner at higher temperatures.

1 mark: correct trend (time decreases as temperature increases, values cited); 1 mark: explanation referencing increased kinetic energy of particles at higher temperature; 1 mark: explanation correctly links increased kinetic energy to faster diffusion / faster particle movement, causing earlier meeting/reaction of the two gases. [3]

Cut equal-sized (e.g. 1 cm³) pieces of fresh liver using a cork borer to standardise surface area, and use the same mass of liver in each trial. Prepare a range of hydrogen peroxide concentrations (e.g. 5%, 10%, 15%, 20%, 25%) by diluting a stock solution with distilled water, keeping total volume constant. Add the liver to a measured volume of hydrogen peroxide in a conical flask connected by a delivery tube to an inverted measuring cylinder or gas syringe, and record the volume of oxygen collected after a fixed time (e.g. 60 seconds), starting the stopwatch the instant the liver is added. Keep temperature constant by conducting all trials in the same room or using a water bath at a set temperature, since temperature also affects enzyme activity. Repeat each concentration at least three times and calculate a mean volume of oxygen to improve reliability and identify anomalies. Wear eye protection throughout, since hydrogen peroxide is an irritant and the oxygen released is under pressure from the reaction.

Level of Response mark scheme (4-band): Band A (5-6 marks): a full, logically ordered method covering all four bullet points with correct scientific terminology (e.g. independent/dependent/controlled variable, fair test, enzyme, substrate) and few or no errors of grammar/spelling that would impede understanding. Band B (3-4 marks): most elements present but method may be incomplete, poorly sequenced, or control of one variable is vague; generally sound use of scientific terms with occasional communication errors. Band C (1-2 marks): only partial method given, e.g. independent variable identified but little detail on measurement or control; weak use of terminology or communication significantly hinders clarity. Band D (0 marks): no relevant, creditable content. Indicative content: dilution series of H₂O₂ to vary independent variable; standardised liver mass/surface area (cork borer); gas syringe/inverted-cylinder collection of O₂ as dependent variable, over a fixed, timed interval; control of temperature (constant room/water bath) and liver mass/type as key confounding variables; repeats and mean calculation for reliability; eye protection as safety precaution (irritant chemical / pressurised gas). [6]

Set up a circuit with a cell, ammeter (in series), voltmeter (in parallel across the test wire) and the resistance wire clamped along a metre rule with a fixed crocodile clip at one end. Vary the length of wire in the circuit (the independent variable) by moving the second crocodile clip to set lengths, e.g. 10 cm, 20 cm, 30 cm, 40 cm, 50 cm, measured using the metre rule. At each length, close the switch briefly, record the current (A) from the ammeter and potential difference (V) from the voltmeter, then calculate resistance using R = V ÷ I (the dependent variable). Keep the wire's material and diameter (cross-sectional area) constant by using a single continuous piece of the same gauge wire throughout, and keep the wire's temperature approximately constant by only closing the switch for the short time needed to take each reading, since resistance increases with temperature. Repeat each length reading at least twice and calculate a mean resistance to improve reliability. As a safety precaution, switch off the circuit between readings (or use a low-voltage supply and keep contact times short) to prevent the wire overheating and causing a burn.

Level of Response mark scheme (4-band): Band A (5-6 marks): complete, well-sequenced method addressing all four bullet points, using correct terminology (independent/dependent/controlled variable, ammeter, voltmeter, series/parallel, resistance = V/I) with clear, largely error-free communication. Band B (3-4 marks): most elements present, but method may lack detail on how a variable is controlled, or sequencing is imperfect; generally clear use of scientific terms. Band C (1-2 marks): only a basic outline given, e.g. states length is varied and resistance is measured with little further detail; weak terminology or clarity. Band D (0 marks): no relevant, creditable content. Indicative content: fixed and moveable crocodile clip to vary wire length against a metre rule (independent variable); ammeter in series and voltmeter in parallel to obtain I and V; resistance calculated using R = V/I as the dependent variable; control of wire material/gauge (same wire) and temperature (brief switch-on times) as key variables; repeats and mean for reliability; safety precaution of switching off between readings / short contact time to avoid overheating/burns. [6]

Prepare a range of salt (sodium chloride) solution concentrations by diluting a stock solution with distilled water, e.g. 0.0, 0.2, 0.4, 0.6, 0.8, 1.0 mol/dm³, keeping total solution volume constant in each boiling tube. Use a cork borer to cut potato cylinders of equal diameter, trim them to equal length with a scalpel and ruler, blot each cylinder dry and record its initial mass on a balance. Immerse one cylinder in each concentration for a fixed time (e.g. 24 hours), then remove, blot dry again and re-weigh to obtain the final mass. Calculate percentage change in mass (the dependent variable) using ((final mass − initial mass) ÷ initial mass) × 100 for each concentration. Control variables by using potato from the same potato/source, cutting all cylinders to the same dimensions, keeping the immersion time identical for every tube, and keeping all tubes at the same room temperature (or in a water bath), since temperature also affects the rate of osmosis. Repeat each concentration with at least three cylinders and calculate a mean percentage change to improve reliability. As a safety precaution, cut carefully away from the hand when using the cork borer and scalpel to avoid injury.

Level of Response mark scheme (4-band): Band A (5-6 marks): full, logically sequenced method covering all four bullet points, with correct terminology (independent/dependent/controlled variable, osmosis, water potential, percentage change in mass) and clear, largely error-free communication. Band B (3-4 marks): most elements present but with gaps in detail, e.g. control of cylinder size or timing not fully specified; generally appropriate terminology. Band C (1-2 marks): only a basic outline, e.g. states solutions of different concentration are used and potato is weighed, with little further detail; weak communication or terminology. Band D (0 marks): no relevant, creditable content. Indicative content: dilution series of salt solution to vary concentration (independent variable); equal-sized potato cylinders cut with cork borer; mass measured before and after fixed immersion time; percentage change in mass calculated as (final − initial)/initial × 100 (dependent variable); control of cylinder size, immersion time and temperature; repeats and mean for reliability; safety precaution using cork borer/scalpel with care. [6]

The data show a curve typical of Newton's law of cooling: the rate of temperature fall is greatest when the temperature difference between the water and its surroundings is greatest (steep gradient near t = 0), and the curve becomes progressively less steep as the water cools and approaches room temperature, giving a smooth, concave curve rather than a straight line.

1 mark: sensible linear scales chosen, using more than half the available grid on both axes; 1 mark: axes correctly labelled with quantity and unit (time/min, temperature/°C); 1 mark: all seven points plotted accurately to within half a small square; 1 mark: single smooth curve drawn (not straight-line segments, not a ruled line) passing through or close to all points; 1 mark: curve correctly shows a decreasing gradient over time (steep initially, levelling off later), consistent with cooling behaviour.; 1 mark: curve correctly shown levelling off asymptotically towards room temperature rather than becoming perfectly flat. [6]

The curve is steepest near t = 0 s, when reactant concentration (and therefore rate) is highest, and its gradient decreases steadily as the reaction proceeds and reactants are used up, becoming flat (zero gradient) once the volume reaches a plateau at 52 cm³ from around 100 s onwards, indicating the reaction has finished (a reactant, here the calcium carbonate, is fully used up).

1 mark: sensible linear scales chosen, using more than half the available grid on both axes; 1 mark: axes correctly labelled with quantity and unit (time/s, volume/cm³); 1 mark: all seven points plotted accurately to within half a small square, including the origin; 1 mark: single smooth curve drawn (not straight-line segments) through or close to all points; 1 mark: curve correctly shows decreasing gradient over time and a clear horizontal plateau once the reaction finishes (from 100-120 s). [5]

The curve should show the classic sigmoid growth pattern: a shallow 'lag phase' as the small starting population grows slowly (days 0-1), a steep exponential 'log phase' once there are enough cells reproducing rapidly with abundant nutrients (days 1-4), and a levelling 'stationary phase' from around day 5 onwards as nutrients become limited and/or waste products accumulate, capping further growth at approximately 7.6 × 10⁶ cells per cm³.

1 mark: sensible linear scales chosen, using more than half the available grid on both axes; 1 mark: axes correctly labelled with quantity and unit (day, cell density / ×10⁶ per cm³); 1 mark: all eight points plotted accurately to within half a small square; 1 mark: single smooth curve drawn (not straight-line segments) through or close to all points; 1 mark: curve correctly shows the sigmoid (S-shaped) pattern — slow start, steep middle rise, and a clear plateau from day 5-7. [5]

A burette delivers precise, variable volumes of liquid; because most liquids used form a concave meniscus, readings are taken from the bottom of the curve, viewed at eye level to avoid parallax error.

1 mark: burette named correctly; 1 mark: correct reading convention (bottom of meniscus, at eye level). [2]

Superheating can cause a liquid to boil suddenly and violently ('bumping'), ejecting hot liquid from the tube; pointing the open end away from people prevents anyone being scalded if this occurs.

1 mark: correct precaution stated (point tube away from people, use a test-tube holder, do not look directly into the tube, etc. — any one valid precaution); 1 mark: correct reason linking the precaution to the risk of sudden boiling/spitting causing injury. [2]

A volumetric pipette is manufactured and calibrated to deliver a single specified volume (e.g. 25.0 cm³) with high precision (typically ±0.06 cm³ or better), whereas a measuring cylinder has wider graduations and a larger tolerance, making it less accurate for delivering an exact fixed volume.

1 mark: correct apparatus named (volumetric/graduated pipette, accept 'pipette'); 1 mark: correct reason — smaller tolerance/higher precision than a measuring cylinder for a single fixed volume. [2]

Chlorine gas is toxic and irritating if inhaled, so the experiment should be carried out with good ventilation or in a fume cupboard; hydrogen gas is highly flammable, so naked flames must be kept away from the collected gas to avoid ignition.

1 mark: valid precaution relating to chlorine's toxicity (fume cupboard/ventilation); 1 mark: valid precaution relating to hydrogen's flammability (no naked flames nearby). [2]

An electronic balance reading to 0.01 g gives the required precision; taring (zeroing) the empty weighing boat or container first, or weighing the container before and after adding the solid ('weighing by difference'), removes the mass of the container from the reading and reduces systematic error.

1 mark: correct apparatus named (electronic/top-pan balance); 1 mark: valid technique to improve accuracy (tare/zero the container, or weigh by difference). [2]

Concentrated acids and alkalis are corrosive substances; a splash reaching the eyes can cause severe, potentially permanent, damage, so safety goggles must be worn at all times when handling them.

1 mark: correct precaution stated (wear eye protection/goggles, or other valid precaution e.g. gloves, working over a sink); 1 mark: correct reason referencing the corrosive nature of the chemicals and risk of injury (e.g. to eyes/skin). [2]

A test-tube holder clamps firmly around the tube and can be held safely at a distance from the flame throughout heating; tongs are designed for solid objects and do not provide a secure enough grip on a cylindrical tube for sustained heating, risking the tube slipping and causing a spillage or breakage hazard.

1 mark: correct apparatus named (test-tube holder); 1 mark: valid reason why tongs are unsuitable (insecure grip / risk of dropping the tube). [2]

A gas syringe collects gas directly and its plunger position gives a direct, accurate volume reading; unlike water displacement, it does not require the gas to be insoluble in water, so it can be used for soluble gases such as carbon dioxide or ammonia without loss of accuracy.

1 mark: correct apparatus named (gas syringe); 1 mark: valid advantage (works for water-soluble gases / no gas dissolves and is lost, or gives a direct reading without needing to invert a measuring cylinder). [2]

Microscope lamps can be bright enough to cause eye strain or discomfort with prolonged direct viewing, so the light intensity should be kept as low as needed to see the specimen clearly, and breaks should be taken during extended observation.

1 mark: valid precaution (avoid prolonged direct viewing of a bright light source / use lowest suitable light intensity); 1 mark: valid reason referencing eye strain/discomfort from prolonged bright light exposure. [2]

The tripod provides a stable metal frame that holds the beaker well clear of the flame, while the wire gauze placed on top spreads the heat evenly across the base of the beaker and prevents localised overheating or cracking of glassware, and stops the beaker from toppling.

1 mark: correct apparatus named (tripod, accept 'tripod and gauze'); 1 mark: correct safety/functional role (stable support above flame / even heat distribution preventing glassware cracking or toppling). [2]

Disposable gloves prevent direct skin contact with biological material and any preservative chemicals used, while a lab coat or apron protects clothing and skin from contamination or splashes during the dissection.

1 mark each for two valid items of PPE, e.g. gloves, lab coat/apron (any two correct, excluding eye protection which is excluded by the question). [2]

A dropping pipette delivers liquid one drop at a time, giving fine control over small volumes; this is especially useful near a titration's endpoint, where adding solution too quickly could easily overshoot the exact point of colour change.

1 mark: correct apparatus named (dropping/teat pipette); 1 mark: valid application (fine control near titration endpoint to avoid overshooting). [2]

The blue, roaring flame burns at a much higher temperature and, because it produces very little visible light, can be difficult to see against a bright background, increasing the risk of accidental burns or of flammable materials being ignited nearby if the burner is left unattended in this state.

1 mark: correct precaution (only use roaring flame while actively heating; revert to safety flame/turn off when not needed); 1 mark: correct reason (flame is hotter and hard to see, increasing burn/fire risk). [2]

A micrometer screw gauge measures to a resolution of about 0.01 mm, using a calibrated screw thread, whereas a ruler's smallest division is typically 1 mm and cannot be read precisely enough, nor easily aligned, to measure the diameter of a thin wire accurately.

1 mark: correct apparatus named (micrometer / micrometer screw gauge; accept digital callipers as an alternative with justification); 1 mark: correct reason referencing much higher precision/resolution than a ruler. [2]

Flammable liquids such as ethanol give off vapour that can ignite even without the liquid itself being directly exposed to a flame; keeping stock bottles closed and at a safe distance from any Bunsen burner or other naked flame prevents vapour from igniting and a flame from 'travelling back' to the main container, which could cause a serious fire.

1 mark: correct precaution (keep flammable liquid/its container away from and/or closed near naked flames); 1 mark: correct reason referencing flammable vapour igniting and risk of fire spreading back to the source. [2]

Plant cells possess a cellulose cell wall, a permanent (sap) vacuole and, in photosynthetic cells, chloroplasts, none of which are present in animal cells. The cell wall's function is to provide mechanical strength and prevent the cell bursting when turgid.

1 mark: two correct structures named (any two of cell wall, chloroplast, permanent vacuole); 1 mark: correct function given for one named structure. [2]

Photosynthesis combines carbon dioxide and water, using light energy absorbed by chlorophyll, to produce glucose, with oxygen released as a by-product.

1 mark: correct word equation (reactants carbon dioxide and water; products glucose and oxygen); 1 mark: oxygen correctly identified as the by-product gas released. [2]

Adding iodine solution to a food sample tests for starch; if starch is present, the iodine solution changes colour from its usual orange-brown to blue-black.

1 mark: correct test named (iodine test); 1 mark: correct positive colour change described (orange/brown to blue-black). [2]

The optimum is the specific temperature or pH value at which an enzyme's rate of reaction is at its maximum.

1 mark: 'optimum' (temperature/pH) correctly stated. [1]

Excess heat energy breaks the intermolecular bonds that maintain the enzyme's specific 3D tertiary structure. This permanently changes the shape of the active site, so the substrate molecule no longer fits (loss of the 'lock and key' fit), and the enzyme is denatured and cannot catalyse the reaction.

1 mark: reference to bonds breaking / change in tertiary structure or active site shape (denaturation); 1 mark: link to substrate no longer fitting/binding the active site. [2]

The diaphragm, a sheet of muscle beneath the lungs, contracts and flattens during inhalation, increasing thoracic volume.

1 mark: 'diaphragm' correctly named.; 1 mark: correct reference to the diaphragm's position beneath the lungs, separating thorax from abdomen. [2]

Alveoli are adapted for efficient diffusion of oxygen and carbon dioxide by having a very large total surface area (millions of alveoli), extremely thin (one-cell-thick) walls that minimise diffusion distance, and a dense capillary network that maintains a steep concentration gradient by constantly removing oxygen and delivering carbon dioxide.

1 mark each for two valid adaptations, e.g. large surface area, thin walls/short diffusion distance, good blood supply/steep concentration gradient, moist lining (any two). [2]

Hormones are chemical messengers produced by endocrine glands (e.g. the pituitary, thyroid, adrenal glands and pancreas) and are secreted directly into the bloodstream, through which they are transported to specific target organs bearing the appropriate receptors.

1 mark: endocrine glands correctly identified as the source; 1 mark: correct transport method stated (via the blood/bloodstream). [2]

A food chain is a diagram showing the linear feeding relationships between organisms in a habitat, with arrows indicating the direction of energy transfer from producer through successive consumers. A valid woodland example is: oak leaves (producer) → caterpillar (primary consumer) → blue tit (secondary consumer).

1 mark: correct definition referencing transfer of energy/biomass between feeding organisms; 1 mark: valid food chain example with at least three organisms and correctly directed arrows. [2]

A proton carries a relative charge of +1 and a relative mass of 1. An electron carries a relative charge of −1 but has a relative mass that is negligible compared with a proton (about 1/1836).

1 mark: correct proton charge (+1) and mass (1); 1 mark: correct electron charge (−1) and mass (very small/negligible, accept 1/1836 or 0). [2]

Number of neutrons = mass number − atomic number = 35 − 17 = 18.

1 mark: correct answer 18.; 1 mark: method correctly shown (mass number minus atomic number) rather than the answer alone. [2]

Group 1 elements are known as the alkali metals. Reactivity increases down the group because the outer electron is further from the nucleus and increasingly shielded by inner shells, so it is more easily lost.

1 mark: 'alkali metals' correctly named; 1 mark: correct trend stated (reactivity increases down the group). [2]

Sodium chloride consists of a metal (sodium) and a non-metal (chlorine) which transfer electrons to form oppositely charged ions held together by ionic bonds.

1 mark: 'ionic (bonding)' correctly stated.; 1 mark: correct reference to the transfer of electrons from sodium to chlorine forming the ions. [2]

In solid sodium chloride, Na⁺ and Cl⁻ ions are arranged in a giant regular lattice, held together by strong electrostatic forces of attraction between oppositely charged ions extending throughout the structure. Melting requires breaking a very large number of these strong ionic bonds simultaneously, which requires a large input of energy, giving a high melting point.

1 mark: reference to giant ionic lattice structure with strong electrostatic forces between oppositely charged ions; 1 mark: link to large amount of energy needed to overcome/break these forces throughout the structure. [2]

Diamond has each carbon atom covalently bonded to four neighbours in a rigid tetrahedral lattice, giving it exceptional hardness since many strong covalent bonds must be broken to deform it. Graphite has each carbon bonded to only three others, forming flat hexagonal layers with weak intermolecular forces between the layers; these layers can slide over one another, making graphite soft and a good lubricant, and its delocalised electrons (one per carbon, not used in bonding) allow it to conduct electricity, unlike diamond.

1 mark: correct structural difference (diamond = 4 bonds per carbon/3D lattice vs graphite = 3 bonds per carbon/layered structure); 1 mark: correctly linked property difference (hardness, conductivity, or lubrication) explained by the structural difference. [2]

Nanoparticles are defined as particles with at least one dimension in the range of approximately 1 to 100 nanometres.

1 mark: correct range given (1-100 nm, accept values within this range). [1]

Adding dilute hydrochloric acid to a carbonate causes visible effervescence as carbon dioxide gas is released, alongside the formation of a soluble chloride salt and water.

1 mark: correct observation (effervescence/fizzing/bubbles); 1 mark: gas correctly identified as carbon dioxide. [2]

Average speed is defined as the distance travelled divided by the time taken for the journey.

1 mark: correct equation (speed = distance ÷ time), accept in words or symbols.; 1 mark: equation correctly given in symbol form as well as words (speed = distance / time). [2]

Constant speed produces a straight-line graph with a constant, non-zero gradient (steeper for higher speed). A horizontal section (zero gradient) means distance is not changing over that time interval, i.e. the object is stationary.

1 mark: correct shape described (straight line / constant gradient) for constant speed; 1 mark: correct interpretation of horizontal section (object stationary/at rest). [2]

Newton's First Law states that an object's motion will not change unless a resultant force acts on it — it remains at rest or continues at constant velocity otherwise.

1 mark: correct statement referencing no change in motion / rest or constant velocity unless acted on by a resultant force.; 1 mark: correct reference to 'resultant' or 'unbalanced' force specifically (not just 'a force'). [2]

The resultant force is found by combining all individual forces acting on a body into one equivalent overall force. For forces acting in opposite directions, they are subtracted: 10 N − 6 N = 4 N, acting in the direction of the larger (10 N) force.

1 mark: correct definition of resultant force (single combined/overall force with the same effect); 1 mark: correct value 4 N with direction correctly identified (direction of the larger force). [2]

Density is defined as an object's mass divided by its volume.

1 mark: correct equation (density = mass ÷ volume).; 1 mark: equation correctly given in symbol form as well as words (density = mass / volume). [2]

According to kinetic theory, gas particles are widely spaced with large volumes of empty space between them and move freely and randomly, with only very weak forces of attraction between them, so reducing the volume simply pushes the particles closer together through the available empty space. Liquid particles, by contrast, are already close together (touching), so there is very little empty space left to remove, making liquids far harder to compress.

1 mark: correct description of gas particle arrangement (widely spaced, large empty space between particles); 1 mark: correct contrast with liquid particles (already close together, little space to compress), explaining the difference in compressibility. [2]

Radioactive decay can release alpha particles, beta particles, or gamma rays (electromagnetic radiation), depending on the isotope and type of decay.

1 mark: all three types correctly named (alpha, beta, gamma). [1]

Gamma radiation is uncharged, high-energy electromagnetic radiation with (effectively) no mass, so it interacts weakly with the atoms of any material it passes through and requires several centimetres of lead (or metres of concrete) to be significantly absorbed. Alpha particles, by contrast, are relatively large, heavy and carry a +2 charge, so they interact strongly and frequently with atoms (causing intense ionisation) and are absorbed within a few centimetres of air or by a sheet of paper.

1 mark: gamma correctly described as having no mass and no charge (or being uncharged electromagnetic radiation); 1 mark: correct comparison explaining why this reduces interaction with matter compared to (charged, more massive) alpha particles, giving greater penetrating power. [2]

Magnesium and oxygen combine in a 2:1:2 ratio to form magnesium oxide. Balancing the equation ensures equal numbers of Mg and O atoms on each side.

1 mark: correct formulae (Mg, O₂, MgO); 1 mark: correctly balanced (coefficients 2, 1, 2); 1 mark: correct state symbols (s), (g), (s).; 1 mark: conservation of mass confirmed - equal numbers of Mg and O atoms shown on both sides of the equation. [4]

This is a neutralisation reaction in which the acid and alkali react in a 1:1 ratio to form a salt (sodium chloride) and water.

1 mark: correct formulae for all reactants and products; 1 mark: equation correctly balanced (already balanced 1:1:1:1, so full balancing credit for correct formulae here); 1 mark: correct state symbols (aq), (aq), (aq), (l).; 1 mark: reaction correctly classified as a neutralisation (acid + alkali -> salt + water). [4]

Calcium carbonate reacts with sulfuric acid to form calcium sulfate, water and carbon dioxide. Formulae for each species must be correctly determined from ionic charges before balancing.

1 mark: all formulae correct (CaCO₃, H₂SO₄, CaSO₄, H₂O, CO₂); 1 mark: equation correctly balanced (1:1:1:1:1); 1 mark: all four state symbols correct; 1 mark: correct identification that CaSO₄ is formed as a solid (low solubility), showing understanding beyond simple balancing. [4]

Balancing requires equal numbers of Fe and Cl atoms on both sides: 2 iron atoms react with 3 chlorine molecules (6 Cl atoms) to give 2 formula units of FeCl₃ (6 Cl atoms).

1 mark: correct formula FeCl₃ used with correct formulae for Fe and Cl₂; 1 mark: correctly balanced (2, 3, 2); 1 mark: correct state symbols (s), (g), (s).; 1 mark: formula of iron(III) chloride correctly justified from balancing Fe3+ and Cl- ionic charges. [4]

Zinc is more reactive than copper, so it displaces copper from copper(II) sulfate solution, forming zinc sulfate solution and copper metal. The blue colour of the Cu²⁺ ions fades as they are removed from solution.

1 mark: correct formulae for all species; 1 mark: equation correctly balanced (1:1:1:1); 1 mark: correct state symbols; 1 mark: correct observation (blue solution fades to colourless/pale, reddish-brown copper deposited). [4]

Complete combustion of a hydrocarbon-based fuel such as ethanol produces carbon dioxide and water; balancing carbon, hydrogen and then oxygen atoms in turn gives the coefficients shown.

1 mark: correct formulae for ethanol, oxygen, carbon dioxide and water; 1 mark: carbon and hydrogen atoms correctly balanced (2 CO₂, 3 H₂O); 1 mark: oxygen atoms correctly balanced overall (3 O₂ on the left balances 4+3=7 O atoms on the right — accept fractional intermediate working if final whole-number equation is correct).; 1 mark: oxygen atoms confirmed balanced on both sides (7 O atoms each side) as a final conservation-of-mass check. [4]

Alpha decay releases a helium nucleus (₂⁴He), reducing the mass number by 4 (226 − 4 = 222) and the atomic number by 2 (88 − 2 = 86), which corresponds to radon (Rn).

1 mark: alpha particle correctly represented as ₂⁴He; 1 mark: mass numbers correctly balanced on both sides (226 = 222 + 4); 1 mark: atomic numbers correctly balanced on both sides (88 = 86 + 2); 1 mark: daughter nucleus correctly identified as radon-222 (₈₆²²²Rn). [4]

In beta-minus decay, a neutron in the nucleus changes into a proton and an electron (beta particle) is emitted. The mass number is unchanged (14), while the atomic number increases by 1 (6 → 7), giving nitrogen-14.

1 mark: beta particle correctly represented as ₋₁⁰e (or ₋₁⁰β); 1 mark: mass numbers correctly balanced (14 = 14 + 0); 1 mark: atomic numbers correctly balanced (6 = 7 + (−1)); 1 mark: product correctly identified as nitrogen(-14). [4]

Convert units first: distance = 1.5 km = 1500 m; time = 5 minutes = 300 s. Speed = distance ÷ time = 1500 ÷ 300 = 5 m/s.

1 mark: distance correctly converted to metres (1500 m); 1 mark: time correctly converted to seconds (300 s); 1 mark: correct substitution into speed = distance ÷ time; 1 mark: correct final answer 5 m/s. e.c.f. applies for consistent unit errors.; 1 mark: final answer given to an appropriate number of significant figures (2 s.f.). [5]

(a) Acceleration = change in velocity ÷ time = (18 − 0) ÷ 6.0 = 3.0 m/s². (b) Using distance = average velocity × time, average velocity = (0 + 18)/2 = 9.0 m/s, so distance = 9.0 × 6.0 = 54 m.

1 mark: correct substitution for acceleration = Δv/t; 1 mark: correct answer 3.0 m/s²; 1 mark: correct method for distance (e.g. average velocity × time, or use of s = ut + ½at²); 1 mark: correct substitution shown; 1 mark: correct final answer 54 m. e.c.f. applies for part (b) using an incorrect acceleration from part (a) if the method is otherwise correct. [5]

Using F = ma, acceleration = force ÷ mass = 250 ÷ 50 = 5.0 m/s². Using v = u + at, with u = 0, velocity = 0 + (5.0 × 3.0) = 15 m/s.

1 mark: correct substitution into F = ma; 1 mark: correct acceleration 5.0 m/s²; 1 mark: correct use of v = u + at with the calculated acceleration; 1 mark: correct final velocity 15 m/s. e.c.f. applies throughout.; 1 mark: correct units given for both acceleration (m/s2) and final velocity (m/s). [5]

Volume = 4.0 × 5.0 × 2.0 = 40 cm³. Density = mass ÷ volume = 316 ÷ 40 = 7.9 g/cm³, which matches the density of iron rather than aluminium.

1 mark: correct volume calculated (40 cm³); 1 mark: correct substitution into density = mass ÷ volume; 1 mark: correct answer 7.9 g/cm³; 1 mark: correct identification as iron with reference to the matching value. e.c.f. applies if volume is calculated incorrectly but method is consistent.; 1 mark: comparison correctly made against both reference densities (aluminium and iron) before concluding. [5]

(a) GPE = mgh = 0.20 × 10 × 4.0 = 8.0 J. (b) Assuming GPE converts fully to KE: ½mv² = 8.0, so v² = (2 × 8.0) ÷ 0.20 = 80, v = √80 = 8.9 m/s (2 s.f.).

1 mark: correct substitution into GPE = mgh; 1 mark: correct answer 8.0 J; 1 mark: correct rearrangement of KE = ½mv² to find v² (using KE = GPE from part (a)); 1 mark: correct value of v² = 80; 1 mark: correct final answer 8.9 m/s. e.c.f. applies for part (b) using an incorrect GPE value from part (a). [5]

Power = energy transferred ÷ time = 138000 ÷ 60 = 2300 W. Some electrical energy is inevitably wasted as thermal energy heating the kettle's body and surrounding air (and a small amount as sound), rather than usefully heating the water.

1 mark: correct substitution into power = energy ÷ time; 1 mark: correct answer 2300 W (accept 2.3 kW); 1 mark: valid wasted energy form named (heat/thermal to surroundings, or sound); 1 mark: correct qualitative link that this energy does not usefully heat the water (not the desired output).; 1 mark: correct alternative unit stated (2.3 kW) alongside the answer in watts. [5]

Work done against gravity = gain in GPE = mgh. Rearranging: h = work done ÷ (m × g) = 45000 ÷ (150 × 10) = 45000 ÷ 1500 = 30 m.

1 mark: correct identification that work done = gain in GPE = mgh; 1 mark: correct rearrangement to make h the subject; 1 mark: correct substitution shown; 1 mark: correct final answer 30 m.; 1 mark: answer checked for sensible order of magnitude against the work done. [5]

Mr(CaCO₃) = 40 + 12 + (3 × 16) = 40 + 12 + 48 = 100. Moles = mass ÷ Mr = 20.0 ÷ 100 = 0.200 mol.

1 mark: correct Ar values used/added; 1 mark: correct Mr = 100; 1 mark: correct formula stated (moles = mass ÷ Mr); 1 mark: correct substitution 20.0 ÷ 100; 1 mark: correct final answer 0.200 mol. [5]

Mr(NH₄NO₃) = 14 + (4 × 1) + 14 + (3 × 16) = 14 + 4 + 14 + 48 = 80. Total mass of nitrogen = 2 × 14 = 28 (there are two N atoms). Percentage = (28 ÷ 80) × 100 = 35.0%.

1 mark: correct Mr of NH₄NO₃ calculated (80); 1 mark: total mass of nitrogen correctly identified as 28 (2 N atoms); 1 mark: correct substitution into percentage mass formula; 1 mark: correct final answer 35.0%.; 1 mark: percentage answer given to an appropriate number of significant figures (3 s.f.). [5]

Moles of Mg = mass ÷ Ar = 6.0 ÷ 24 = 0.25 mol. From the equation, the mole ratio Mg : H₂ is 1:1, so moles of H₂ produced = 0.25 mol. Mass of H₂ = moles × Mr = 0.25 × 2 = 0.50 g (Mr of H₂ = 2 × 1 = 2).

1 mark: correct moles of Mg calculated (0.25 mol); 1 mark: correct use of 1:1 mole ratio from the balanced equation; 1 mark: correct Mr of H₂ identified (2); 1 mark: correct substitution into mass = moles × Mr; 1 mark: correct final answer 0.50 g. e.c.f. applies throughout. [5]

Moles = concentration × volume (in dm³) = 0.200 × (50.0 ÷ 1000) = 0.200 × 0.0500 = 0.0100 mol.

1 mark: correct conversion of volume to dm³ (0.0500 dm³); 1 mark: correct formula stated (moles = concentration × volume); 1 mark: correct substitution shown; 1 mark: correct final answer 0.0100 mol. [4]

Number of half-lives elapsed = total time ÷ half-life = 32 ÷ 8 = 4. Activity halves each half-life: 640 → 320 (1) → 160 (2) → 80 (3) → 40 Bq (4).

1 mark: correct number of half-lives calculated (4); 1 mark: correct method of successive halving shown; 1 mark: activity after each half-life shown correctly (320, 160, 80); 1 mark: correct final answer 40 Bq; 1 mark: correct unit Bq retained throughout. [5]

At each trophic level, only a fraction of the energy available in an organism's biomass is passed on to the next consumer. A large proportion is lost as heat produced by respiration, particularly in maintaining body temperature in warm-blooded animals; further energy is used in movement and other life processes and is ultimately also lost as heat. Not all of an organism is eaten or digestible — parts such as bones, feathers, roots or fur are not consumed, and some ingested material passes through the gut undigested and is egested as faeces, representing further energy loss. As a result, typically only about 10% of the energy present at one trophic level is transferred to and incorporated into the biomass of the next. Because so much energy is lost at each stage, the total energy available becomes too small to support a further trophic level after four or five stages — there would not be enough biomass/energy to sustain a viable population of a fifth or sixth-level consumer, and predators at high trophic levels would need an impractically large hunting range to obtain sufficient food.

Level of Response mark scheme (4-band): Band A (5-6 marks): a full, well-sequenced explanation covering multiple energy loss mechanisms (respiration/heat, movement, undigested material/egestion, waste excretion) and a clear, correctly reasoned explanation of why food chains are short (insufficient energy remaining to support further levels), using correct terminology (trophic level, biomass, respiration, egestion) with clear communication. Band B (3-4 marks): most energy loss mechanisms identified with a reasonable explanation of shortened food chains, but explanation may be less developed or terminology used inconsistently. Band C (1-2 marks): basic statement that energy is lost (e.g. 'energy is lost as heat') with little further development; food chain length only vaguely addressed. Band D (0 marks): no relevant, creditable content. Indicative content: energy lost as heat via respiration/movement/maintaining body temperature; energy lost in undigested material (faeces/egestion) and inedible parts; energy lost in excretory waste (urine); approximately 10% transfer efficiency between trophic levels; insufficient energy/biomass remains after several transfers to support a further trophic level; practical limitations for top predators (foraging range/population size). [6]

The modern Periodic Table arranges all known elements in order of increasing atomic number. Elements are organised into horizontal rows called periods, across which the number of electron shells stays the same but the number of outer-shell (valence) electrons increases from left to right, and vertical columns called groups, within which every element has the same number of electrons in its outer shell. Because chemical properties are largely determined by the number and arrangement of outer-shell electrons, elements within the same group show similar chemical behaviour and typically form ions with the same charge — for example, Group 1 elements each have one outer electron, readily lose it to form 1+ ions, and become more reactive down the group as this electron becomes further from the nucleus and more shielded, so is more easily lost. Similarly, Group 7 elements each have seven outer electrons and readily gain one to form 1− ions, becoming less reactive down the group as the nucleus's attraction for an incoming electron weakens with increasing atomic radius and shielding. This systematic arrangement means that, given an element's group and period, a chemist can predict its number of outer electrons, likely ion charge, and broad trends in reactivity and metallic/non-metallic character without needing to test the element directly.

Level of Response mark scheme (4-band): Band A (5-6 marks): a full, well-sequenced description of arrangement by atomic number into periods and groups, correctly linking group number to outer-shell electron number and explaining how this predicts ion charge and reactivity trends (using at least one specific group example), with correct terminology and clear communication. Band B (3-4 marks): describes the periods/groups arrangement and links it to similar properties within a group, but with less developed or only partially correct explanation of why properties can be predicted (e.g. omits a worked group example or the reasoning behind a reactivity trend). Band C (1-2 marks): basic statement that the table is arranged by atomic number and/or into groups and periods, with little explanation of predictive value. Band D (0 marks): no relevant, creditable content. Indicative content: arrangement by increasing atomic (proton) number; periods = rows, groups = columns; elements in the same group share the same number of outer-shell electrons; similar outer-shell electron number gives similar chemical properties/reactions; specific example (e.g. Group 1 alkali metals, 1 outer electron, 1+ ions, reactivity increases down the group; or Group 7 halogens, 7 outer electrons, 1− ions, reactivity decreases down the group); use of position to predict reactivity/bonding/ion charge without direct testing. [6]

The half-life of a radioactive isotope is the time taken for the activity of a sample (or the number of undecayed radioactive nuclei present) to fall to half of its original value; it is a fixed, characteristic property of each isotope, unaffected by physical or chemical conditions. Activity (the rate of decay, measured in becquerels) is inversely related to half-life for a sample of a given size: a short half-life means the sample's nuclei are decaying rapidly, releasing a large number of ionising particles or rays per second, so its activity is high. A long half-life means decay happens much more slowly, spread over a very long period, giving a much lower activity for the same number of atoms present. Because biological damage from radiation depends on the amount of ionising radiation absorbed by tissue in a given time (the dose rate), a highly active, short-half-life source delivers far more ionising radiation to the body per second of exposure than a long-half-life source of similar size, causing greater cellular damage (e.g. DNA damage, increased cancer risk) for the same handling time. A long-half-life source, though it remains radioactive for far longer overall, poses a lower immediate handling risk because its activity — and therefore dose rate — is much lower at any given moment, even though careful long-term storage and disposal are still required.

Level of Response mark scheme (4-band): Band A (5-6 marks): a full, accurate definition of half-life and a clear, correctly reasoned explanation linking short half-life to high activity/dose rate and therefore greater immediate biological risk, with long half-life correctly linked to low activity/dose rate despite longer overall persistence, using correct terminology (activity, becquerel, ionising radiation, dose) and clear communication. Band B (3-4 marks): reasonably accurate definition of half-life and a broadly correct link between short half-life and higher danger, but reasoning about activity/dose rate less fully developed or not fully explained. Band C (1-2 marks): basic or partial definition of half-life (e.g. 'time for radioactivity to decrease') with minimal explanation of the danger comparison. Band D (0 marks): no relevant, creditable content. Indicative content: half-life = time for activity/number of undecayed nuclei to halve; fixed property of the isotope; short half-life = high activity = many decays per second = high dose rate = more ionising radiation absorbed per unit time = greater biological/DNA damage; long half-life = low activity = low dose rate = lower immediate risk despite very long persistence; distinction between total radiation emitted over the source's lifetime and the rate (danger) of exposure at any given moment. [6]

Osmosis is a special case of diffusion involving only water molecules, which pass through a partially permeable membrane (one that allows water but not larger solute molecules through) from a dilute (high water potential) solution to a more concentrated (lower water potential) solution.

1 mark: net movement of water molecules from high to low water potential; 1 mark: correct reference to a partially permeable membrane.; 1 mark: correct reference to net movement (not simply 'movement') of water molecules. [3]

The concentrated salt solution outside the cell has a lower water potential than the cell's cytoplasm and vacuole. Water therefore moves out of the cell, down the water potential gradient, by osmosis. As the cell loses water, the cytoplasm and cell membrane shrink and pull away from the rigid cell wall, a state called plasmolysis.

1 mark: water moves out of the cell by osmosis, correctly linked to a water potential gradient; 1 mark: correct description of the membrane/cytoplasm pulling away from the cell wall.; 1 mark: term 'plasmolysis' correctly used to name the state described. [3]

Arteries carry blood at high pressure directly from the heart, so they have thick walls containing muscle to withstand and help maintain this pressure, elastic fibres that allow the vessel to stretch and recoil with each heartbeat (smoothing pulsatile flow), and a relatively narrow lumen compared with veins.

1 mark each for three valid structural adaptations, e.g. thick/muscular walls, elastic tissue, narrow lumen (any three, max 3). [3]

The menstrual cycle is controlled by several hormones, including FSH and LH (from the pituitary gland) and oestrogen and progesterone (from the ovaries). Oestrogen, for example, causes the uterus lining to thicken and repair following menstruation, and triggers the LH surge that causes ovulation.

1 mark: two correctly named hormones (any two of FSH, LH, oestrogen, progesterone); 1 mark: correct role given for one of the named hormones.; 1 mark: correct role given for a second named hormone (e.g. progesterone maintaining the uterus lining, or LH triggering ovulation). [3]

A dominant allele masks the effect of a recessive allele and is expressed whenever it is present, whether in one or two copies (heterozygous or homozygous). A recessive allele is only expressed in the phenotype when no dominant allele is present, i.e. in the homozygous recessive genotype.

1 mark: correct description of dominant allele (expressed if present, masks recessive); 1 mark: correct description of recessive allele (only expressed when homozygous/no dominant allele present).; 1 mark: correct use of the terms 'genotype' and 'phenotype' within the explanation. [3]

Human body cells are diploid, containing 46 chromosomes (23 homologous pairs). Gametes (sperm and egg cells) are haploid, containing only 23 chromosomes, produced by meiosis, so that fertilisation restores the diploid number.

1 mark: correct number for body cell (46); 1 mark: correct number for gamete (23).; 1 mark: correct reference to meiosis producing haploid gametes from diploid body cells. [3]

Continuous variation covers a full, unbroken range of values within a population, typically influenced by both genes and environment (e.g. human height or body mass). Discontinuous variation occurs in distinct, separate categories with no intermediate forms, usually controlled by a single gene (e.g. human ABO blood group).

1 mark: correct distinction made (continuous = full range/no categories; discontinuous = distinct categories); 1 mark: two correct, appropriately matched human examples (one for each type).; 1 mark: correct reference to continuous variation being influenced by both genetic and environmental factors. [3]

Lymphocytes are white blood cells that produce antibodies, proteins that bind specifically to complementary antigens found on the surface of a particular pathogen. This binding can neutralise the pathogen, cause pathogens to clump together (agglutinate), or mark them so that phagocytes more readily engulf and destroy them.

1 mark: 'lymphocyte' correctly named; 1 mark: correct description of antibody action (binds specific antigen; neutralises/agglutinates/marks pathogen for destruction).; 1 mark: correct reference to the specific (complementary) shape match between antibody and antigen. [3]

A vaccine introduces a safe form of a pathogen's antigens (from a dead, inactivated, or weakened pathogen) into the body without causing disease. This triggers a primary immune response, in which lymphocytes produce antibodies specific to that antigen; some of these lymphocytes remain in the body afterwards as long-lived memory cells. If the real pathogen is encountered later, these memory cells enable a much faster and larger secondary immune response, producing antibodies quickly enough to destroy the pathogen before it can cause noticeable illness.

1 mark: correct description of the vaccine introducing antigens/a safe form of the pathogen; 1 mark: correct reference to production of antibodies and memory cells; 1 mark: correct explanation of the faster/stronger secondary response on re-exposure preventing illness. [3]

Deoxygenated blood returning to the heart enters the right atrium, passes into the right ventricle, and is then pumped out through the pulmonary artery towards the lungs for gas exchange.

1 mark: 'pulmonary artery' correctly named; 1 mark: correct flow sequence described (right atrium → right ventricle → pulmonary artery).; 1 mark: correct note that the pulmonary artery is unusual in carrying deoxygenated blood despite being an artery. [3]

The reactivity series lists metals in order of decreasing reactivity from top to bottom. Metals below hydrogen in the series, such as copper, silver and gold, do not react with dilute acids because they cannot displace hydrogen from the acid.

1 mark: correct trend (reactivity decreases down the series); 1 mark: valid unreactive metal named (copper, silver or gold).; 1 mark: correct reference to displacement reactions as evidence supporting the reactivity order. [3]

A catalyst provides an alternative reaction pathway with a lower activation energy, meaning a greater proportion of particle collisions have sufficient energy to react successfully, which increases the rate of reaction without the catalyst itself being used up.

1 mark: correct effect on activation energy (lowered/reduced); 1 mark: correct effect on rate (increased).; 1 mark: correct note that a catalyst is chemically unchanged/not used up at the end of the reaction. [3]

A reversible reaction can proceed in both the forward direction (reactants → products) and the backward direction (products → reactants). In a closed system, dynamic equilibrium is reached when the forward and backward reaction rates become equal, so although both reactions continue to occur, the overall (macroscopic) concentrations of reactants and products no longer change.

1 mark: correct definition of reversible reaction; 1 mark: correct definition of dynamic equilibrium (forward rate = backward rate, concentrations constant).; 1 mark: correct reference to a closed system being required for true dynamic equilibrium. [3]

Alkanes are saturated hydrocarbons with the general formula CₙH₂ₙ₊₂. The first member (n=1) is methane (CH₄) and the second (n=2) is ethane (C₂H₆).

1 mark: correct general formula CₙH₂ₙ₊₂; 1 mark: both first two members correctly named (methane, ethane).; 1 mark: correct molecular formula given for both named members (CH4 and C2H6). [3]

Bromine water is orange. Alkenes contain a C=C double bond and react with bromine water, decolourising it (turning colourless), whereas saturated alkanes have no double bond to react with bromine and so the bromine water remains orange.

1 mark: correct test named (bromine water); 1 mark: correct results for both alkene (decolourises) and alkane (stays orange/no change).; 1 mark: correct reference to the C=C double bond as the reason alkenes react with bromine water. [3]

In molten lead(II) bromide, Pb²⁺ ions are attracted to and gain electrons at the negative cathode, forming lead metal (reduction), while Br⁻ ions are attracted to and lose electrons at the positive anode, forming bromine gas (oxidation).

1 mark: correct product at the cathode (lead); 1 mark: correct product at the anode (bromine).; 1 mark: correct identification of the ion (Pb2+ or Br-) responsible for each product via oxidation/reduction. [3]

During rusting, iron reacts with oxygen and water. Iron atoms are oxidised, losing electrons to form iron ions, while oxygen atoms are reduced, gaining electrons to form oxide ions. Because oxidation (electron loss) and reduction (electron gain) both occur together, rusting is classified as a redox (reduction-oxidation) reaction.

1 mark: correct reference to iron losing electrons (oxidation); 1 mark: correct reference to oxygen gaining electrons (reduction), correctly identifying both processes occurring together as redox.; 1 mark: correct overall classification stated as a redox (reduction-oxidation) reaction. [3]

A glowing (not flaming) wooden splint is inserted into a test tube of the gas; if oxygen is present, it relights the splint because oxygen supports combustion.

1 mark: correct test described (glowing splint inserted); 1 mark: correct positive result (splint relights). [2]

In an exothermic reaction, more energy is released when new bonds form in the products than is required to break bonds in the reactants, so overall energy is transferred to the surroundings, increasing their temperature. In an endothermic reaction, more energy is needed to break bonds than is released forming new ones, so energy is absorbed from the surroundings, which cool down.

1 mark: correct description of exothermic reaction (energy released to surroundings, surroundings warm); 1 mark: correct description of endothermic reaction (energy absorbed from surroundings, surroundings cool). [2]

Reaction rate can be increased without a catalyst by raising the temperature (giving particles more kinetic energy and more frequent successful collisions), increasing the concentration of the solution (more particles per unit volume, increasing collision frequency), or increasing the surface area of the solid (e.g. by crushing it into smaller pieces, exposing more particles to collision).

1 mark each for three valid methods (temperature increase, concentration increase, surface area increase), max 3. [3]

The wave equation relates the three key quantities describing a wave: speed equals frequency multiplied by wavelength. Frequency, the number of complete waves passing a point per second, is measured in hertz.

1 mark: correct equation (v = f × λ, accept in words); 1 mark: correct unit of frequency (hertz/Hz). [2]

Transverse waves have particle or field oscillations at right angles to the direction the wave travels, as in light and other electromagnetic waves. Longitudinal waves have oscillations that are parallel to the direction of travel, causing regions of compression and rarefaction, as in sound waves.

1 mark: correct distinction (transverse = perpendicular oscillation; longitudinal = parallel oscillation) with a correct example for one type; 1 mark: correct example given for the other type. [2]

The law of reflection states that the angle of incidence (between the incoming ray and the normal) equals the angle of reflection (between the reflected ray and the normal). The normal is a construction line drawn perpendicular to the surface at the point of incidence, used as the reference from which both angles are measured.

1 mark: correct law of reflection stated (angle of incidence = angle of reflection); 1 mark: correct definition of the normal (line perpendicular/at 90° to the surface at the point of incidence). [2]

Ohm's law relates potential difference across a component to the current flowing through it and its resistance: V = IR. Resistance is measured in ohms.

1 mark: correct equation (V = IR, accept in words); 1 mark: correct unit of resistance (ohm/Ω). [2]

In both circuit types, conventional current flows around the circuit from the positive terminal of the supply to the negative terminal. In a series circuit, there is only one path, so the same current passes through every component. In a parallel circuit, the current divides at each junction between the available branches, so different branches may carry different currents, though the total current from the supply equals the sum of the individual branch currents.

1 mark: valid similarity stated (e.g. direction of conventional current, or that both are driven by the same supply); 1 mark: correct difference (series = same current throughout; parallel = current splits between branches). [2]

The magnetic field strength of a solenoid can be increased by increasing the current passing through it, by increasing the number of turns of wire in the coil (for the same current), or by inserting a soft iron core into the coil, which concentrates and strengthens the magnetic field.

1 mark each for two valid methods (increase current, increase number of turns, add an iron core), max 2. [2]

An orbit is the closed path an object such as a planet follows around a more massive body, maintained by the continuous pull of gravitational force acting towards the central body, which provides the centripetal force needed to keep the orbiting body moving in a curved path rather than a straight line.

1 mark: correct description of an orbit (curved path around a body due to gravity); 1 mark: force correctly named as gravity/gravitational force. [2]

Stars are massive bodies that generate their own energy, light and heat by nuclear fusion of hydrogen (and other elements) in their core. Planets do not undergo nuclear fusion and produce no light of their own; they are visible only because they reflect light from a nearby star, such as the Sun.

1 mark: correct description of a star (produces own light/heat via nuclear fusion); 1 mark: correct description of a planet (no own light produced; visible by reflecting starlight). [2]

When light travels from a less dense medium (air) into a denser medium (glass), it slows down; since frequency is unchanged, wavelength must also decrease. This change in speed causes the light ray to change direction, bending towards the normal — this bending is called refraction.

1 mark: correct statement that both speed and wavelength decrease on entering the denser medium; 1 mark: effect correctly named as refraction, with the ray bending towards the normal. [2]

The resistance of a wire increases with its length (a longer path for electrons to travel through), decreases as its cross-sectional area increases (a thicker wire provides more parallel pathways for charge flow), and depends on the material of the wire (different materials have different resistivities); resistance also generally increases with temperature.

1 mark each for three valid factors (length, cross-sectional area/thickness, material/resistivity, temperature), max 3. [3]

Energy needed to break bonds (reactants): H-H (436) + Cl-Cl (242) = 678 kJ/mol. Energy released forming bonds (products): 2 × H-Cl = 2 × 431 = 862 kJ/mol. Overall energy change = energy in − energy out = 678 − 862 = −184 kJ/mol. Since the value is negative, more energy is released forming new bonds than is used breaking old ones, so the reaction is exothermic.

1 mark: correct bonds broken identified (H-H, Cl-Cl) with correct sum (678 kJ/mol); 1 mark: correct bonds formed identified (2 × H-Cl) with correct sum (862 kJ/mol); 1 mark: correct method (energy change = bonds broken − bonds formed); 1 mark: correct substitution shown; 1 mark: correct final answer −184 kJ/mol with correct sign; 1 mark: correct conclusion (exothermic) consistent with the negative sign. e.c.f. applies throughout for a numerically consistent but incorrect intermediate value. [6]

Bonds broken: 4 × C-H (in CH₄) = 4 × 413 = 1652 kJ/mol; 2 × O=O (in 2O₂) = 2 × 498 = 996 kJ/mol. Total = 1652 + 996 = 2648 kJ/mol.

1 mark: correct number and type of C-H bonds identified (4 × C-H); 1 mark: correct C-H total (1652 kJ/mol); 1 mark: correct number and type of O=O bonds identified (2 × O=O); 1 mark: correct O=O total (996 kJ/mol); 1 mark: correct overall total 2648 kJ/mol. [5]

Energy change = energy to break bonds − energy released forming bonds = 2652 − 2346 = +306 kJ/mol. Since the result is positive, the reaction absorbs more energy than it releases overall, so it is endothermic.

1 mark: correct method (bonds broken − bonds formed); 1 mark: correct substitution shown; 1 mark: correct final answer +306 kJ/mol with correct sign; 1 mark: correct classification as endothermic; 1 mark: correct reason given (more energy absorbed breaking bonds than released forming bonds). [5]

Energy = mass of water × specific heat capacity × temperature change = 100 × 4.2 × 30 = 12 600 J.

1 mark: correct formula identified (energy = mass × specific heat capacity × temperature change); 1 mark: correct substitution of values; 1 mark: correct arithmetic; 1 mark: correct final answer 12600 J (or 12.6 kJ) with unit.; 1 mark: energy value correctly expressed to an appropriate number of significant figures with correct unit (J). [5]

(a) Power = potential difference × current = 230 × 8.0 = 1840 W. (b) Time = 15 minutes = 900 s. Energy = power × time = 1840 × 900 = 1 656 000 J.

1 mark: correct substitution into P = VI; 1 mark: correct power 1840 W; 1 mark: correct conversion of time to seconds (900 s); 1 mark: correct substitution into energy = power × time; 1 mark: correct final answer 1 656 000 J (accept 1.656 MJ). e.c.f. applies for part (b) using an incorrect power. [5]

Power = work done ÷ time taken = 3600 ÷ 12 = 300 W.

1 mark: correct formula identified (power = work ÷ time); 1 mark: correct substitution; 1 mark: correct arithmetic; 1 mark: correct final answer 300 W with unit.; 1 mark: correct unit (W) stated explicitly alongside the numerical answer. [5]

(a) Using Vₚ/Vₛ = Nₚ/Nₛ: Vₛ = Vₚ × (Nₛ/Nₚ) = 230 × (1000/200) = 230 × 5 = 1150 V. (b) For an ideal (100% efficient) transformer, power in = power out: Vₚ × Iₚ = Vₛ × Iₛ, so Iₛ = (Vₚ × Iₚ) ÷ Vₛ = (230 × 2.0) ÷ 1150 = 460 ÷ 1150 = 0.40 A.

1 mark: correct turns-ratio equation used (Vₚ/Vₛ = Nₚ/Nₛ); 1 mark: correct secondary voltage 1150 V; 1 mark: correct use of power conservation (Vₚ Iₚ = Vₛ Iₛ) for part (b); 1 mark: correct substitution shown; 1 mark: correct final secondary current 0.40 A. e.c.f. applies for part (b) using an incorrect secondary voltage from part (a). [5]

Convert power to kilowatts: 60 W = 0.060 kW. Energy (kWh) = power (kW) × time (hours) = 0.060 × 5.0 = 0.30 kWh.

1 mark: correct conversion of power to kW (0.060 kW); 1 mark: correct formula used (energy = power × time); 1 mark: correct substitution shown; 1 mark: correct final answer 0.30 kWh.; 1 mark: intermediate conversion of power to kilowatts shown as a distinct working step. [5]

Time in hours = 4 ÷ 60 = 0.0667 hours. Energy = power × time = 2.5 × 0.0667 = 0.167 kWh (3 s.f.). Cost = energy × price per kWh = 0.167 × 28 = 4.67p.

1 mark: correct conversion of time to hours (4/60 = 0.0667 h); 1 mark: correct energy in kWh (0.167 kWh, 3 s.f.); 1 mark: correct method (cost = energy × price per kWh); 1 mark: correct final answer 4.67p (accept 4.6-4.7p for rounding).; 1 mark: final cost answer rounded sensibly to the nearest 0.1p with correct unit (p). [5]

Parental genotypes: Tt (tall) × tt (short). Gametes from Tt parent: T or t; gametes from tt parent: t or t. Combining gametes gives offspring genotypes Tt, Tt, tt, tt — i.e. 2 Tt : 2 tt, simplifying to a 1:1 ratio. Since T is dominant, Tt offspring are tall and tt offspring are short, giving a phenotype ratio of 1 tall : 1 short.

1 mark: correct gametes identified for each parent (T, t from Tt; t, t from tt); 1 mark: correct Punnett square/cross diagram construction; 1 mark: correct offspring genotypes shown (Tt, Tt, tt, tt); 1 mark: correct genotype ratio stated (1 Tt : 1 tt); 1 mark: correct phenotype ratio stated (1 tall : 1 short). [5]

Parental genotypes: Ff × Ff. Gametes from each parent: F or f. Combining gametes: FF, Ff, Ff, ff — a 1:2:1 genotype ratio. Only the ff genotype results in cystic fibrosis, which occurs in 1 out of 4 possible offspring, giving a probability of 1/4 = 25%.

1 mark: correct gametes identified (F, f from each parent); 1 mark: correct Punnett square/cross diagram construction; 1 mark: correct offspring genotypes shown (FF, Ff, Ff, ff); 1 mark: ff correctly identified as the affected genotype (1 out of 4); 1 mark: correct final probability 25% (or 1 in 4). [5]

Gametes from the father: Iᴬ or Iᴼ; gametes from the mother: Iᴮ or Iᴼ. Combining gives four possible genotypes: IᴬIᴮ (group AB), IᴬIᴼ (group A), IᴮIᴼ (group B), and IᴼIᴼ (group O), so all four blood groups are possible in their children.

1 mark: correct gametes identified for each parent; 1 mark: all four correct genotype combinations shown; 1 mark: each genotype correctly matched to its resulting blood group; 1 mark: correct identification that all four blood groups (A, B, AB, O) are possible.; 1 mark: correct reasoning given for why IAIB shows co-dominance (both antigens expressed) rather than one masking the other. [5]

The colour-blind father has genotype XⁿY. The mother, homozygous normal, has genotype XᴺXᴺ. Daughters receive one X chromosome from each parent: Xᴺ (from mother) and Xⁿ (from father), giving genotype XᴺXⁿ. Since the normal allele (Xᴺ) is dominant, all daughters will have normal colour vision but will be carriers of the colour-blindness allele.

1 mark: correct parental genotypes identified (XⁿY father, XᴺXᴺ mother); 1 mark: correct daughter genotype derived (XᴺXⁿ); 1 mark: correct reasoning that Xᴺ is dominant over Xⁿ; 1 mark: correct conclusion that all daughters are unaffected carriers (normal vision).; 1 mark: correct note that only daughters receive an X chromosome from the father, explaining why the analysis applies specifically to daughters. [5]

The BbHh parent produces four equally likely gamete combinations: BH, Bh, bH, bh. The bbhh parent produces only one gamete combination: bh. Combining these gives four equally likely offspring genotypes: BbHh (black, short), Bbhh (black, long), bbHh (white, short), and bbhh (white, long), each occurring with equal (1/4) probability, giving a 1:1:1:1 phenotype ratio.

1 mark: correct four gamete types identified from the BbHh parent (BH, Bh, bH, bh); 1 mark: correct single gamete type identified from the bbhh parent (bh); 1 mark: correct four offspring genotypes derived (BbHh, Bbhh, bbHh, bbhh); 1 mark: correct final phenotype ratio stated (1 black short : 1 black long : 1 white short : 1 white long). [4]

Conventionally, heart diagrams are drawn as if facing the patient, so the patient's right side appears on the reader's left. The four chambers are: right atrium (upper, reader's left) and right ventricle (lower, reader's left) on the deoxygenated side, and left atrium (upper, reader's right) and left ventricle (lower, reader's right) on the oxygenated side. The left side of the heart (left atrium and left ventricle) handles oxygenated blood returning from the lungs and pumps it to the body.

1 mark: all four chambers correctly named and matched to position; 1 mark: correct left/right orientation understood (patient's right = reader's left convention); 1 mark: correct identification that the left side pumps oxygenated blood.; 1 mark: correct explanation of why the heart's left side has more muscular walls than the right (pumps blood at higher pressure around the whole body vs only to the lungs). [4]

Xylem vessels consist of dead, hollow cells with thickened, lignified walls, forming a continuous open tube well suited to the one-way, passive movement of water and dissolved minerals under tension, and providing structural support to the plant. Phloem sieve tubes are living cells (though lacking a nucleus), with perforated end walls called sieve plates that allow sugars in solution to move up or down as needed, and each sieve tube is supported by an adjacent companion cell that carries out cell processes (e.g. providing ATP for active transport) on its behalf.

1 mark: correct structural difference relating to living/dead cell status (xylem dead, phloem living); 1 mark: correct structural difference relating to wall structure (xylem lignified/no cross walls vs phloem sieve plates); 1 mark: correct link made to differing function (unidirectional water transport vs bidirectional sugar transport, or support role of companion cells).; 1 mark: correct reference to the role of companion cells in supporting the enucleate phloem sieve tube cells. [4]

On a reaction profile diagram for an exothermic reaction, the y-axis represents energy and the x-axis represents the progress of the reaction. The curve begins at the energy level of the reactants and rises to a peak before falling to the energy level of the products, which is drawn lower than the reactants' starting level, reflecting the overall release of energy. The activation energy is labelled as the vertical distance from the reactants' energy level up to the peak of the curve — the minimum energy needed for the reaction to occur. The overall energy change (ΔH, negative for an exothermic reaction) is labelled as the vertical distance between the reactants' level and the products' level.

1 mark: products correctly drawn at a lower energy level than reactants (consistent with exothermic); 1 mark: activation energy correctly identified as the energy difference between reactants and the peak of the curve; 1 mark: overall energy change correctly identified as the energy difference between reactants and products' levels.; 1 mark: correct note that a catalyst would lower the height of the activation energy peak without changing the reactants' or products' energy levels. [4]

An electrolysis cell for molten lead(II) bromide requires two electrodes, typically inert graphite (carbon) rods, immersed in the molten electrolyte, connected by wires to a direct current (d.c.) power source. Conventional current is taken to flow from the positive terminal of the power source, through the external circuit to the anode (positive electrode); inside the electrolyte, ions carry the current (Pb²⁺ migrating to the cathode, Br⁻ migrating to the anode); current then returns from the cathode (negative electrode) through the external circuit back to the negative terminal of the supply.

1 mark: correct identification of two (inert/graphite) electrodes immersed in the molten electrolyte; 1 mark: correct identification of a d.c. power source connected in the circuit; 1 mark: correct direction of conventional current in the external circuit (positive terminal → anode; cathode → negative terminal). [3]

On a transverse wave diagram, a horizontal line represents the undisturbed position of the medium. Amplitude is defined as the maximum displacement of a point on the wave from this undisturbed position, shown as the vertical distance from the rest line up to a crest (or down to a trough). Wavelength is defined as the distance between two identical, adjacent points on the wave that are in phase, most easily shown as the horizontal distance from one crest to the next crest (or one trough to the next trough).

1 mark: correct definition of amplitude (max displacement from rest/undisturbed position); 1 mark: correct description of where amplitude is measured on the diagram (rest line to crest/trough); 1 mark: correct definition of wavelength (distance between two adjacent identical/in-phase points); 1 mark: correct description of where wavelength is measured on the diagram (e.g. crest to crest). [4]

Around a bar magnet, magnetic field lines are drawn leaving the north pole, curving through the space around the magnet, and entering the south pole, before continuing inside the magnet back to the north pole to form complete closed loops (field lines never start or stop). The lines are closest together near the poles, where the field is strongest, and spread further apart with increasing distance from the magnet, showing the field weakening. A solenoid (current-carrying coil) produces an external field pattern that closely resembles that of a bar magnet, with one end acting as a north pole and the other as a south pole. The key difference is that inside the solenoid, the field lines run parallel to the coil's axis and are evenly spaced, indicating a strong, uniform magnetic field within the coil — a region a simple bar magnet has no equivalent of, since its own interior is solid material rather than an accessible uniform-field space.

1 mark: field lines correctly shown/described running from north to south pole outside the magnet, forming closed loops; 1 mark: correct description of field strength shown by line spacing (closer near poles = stronger); 1 mark: solenoid's external field correctly compared as similar in shape to a bar magnet's; 1 mark: correct identification of the key difference — the strong, uniform, parallel field inside the solenoid's coil, absent in a bar magnet. [4]

In meiosis, a diploid parent cell (here with 4 chromosomes, i.e. 2 homologous pairs) undergoes two successive divisions. The first division separates the homologous chromosome pairs, and the second division separates sister chromatids, ultimately producing four daughter cells (gametes), each containing only 2 chromosomes — half the original (parental) chromosome number, making them haploid. This halving is essential because gametes later fuse during fertilisation; if gametes retained the full diploid number, the resulting zygote would have double the normal chromosome number, and this would keep doubling every generation. Meiosis ensures that fertilisation restores the correct diploid chromosome number (4) in the offspring.

1 mark: correct description that the parent cell (diploid, 4 chromosomes) undergoes two divisions to produce four cells; 1 mark: correct final chromosome number in each gamete stated (2, i.e. half of 4, haploid); 1 mark: correct explanation of why halving is necessary (so fertilisation restores the normal diploid number rather than doubling each generation). [3]

But-2-ene has a four-carbon chain with a C=C double bond starting at the second carbon: CH₃-CH=CH-CH₃. Each end carbon (C1 and C4) is a CH₃ group; the two central carbons (C2 and C3) are each bonded to one hydrogen and are joined to each other by a double bond. The functional group present is the carbon-to-carbon double bond (C=C), which identifies but-2-ene as a member of the alkene homologous series.

1 mark: correct carbon skeleton with four carbons in a chain (CH₃-C-C-CH₃ backbone); 1 mark: double bond correctly placed between the second and third carbon atoms with correct number of hydrogens shown on each carbon; 1 mark: functional group correctly identified as the C=C double bond. [3]

The skin is the body's first, non-specific line of defence, forming a physical barrier that most pathogens cannot cross; when a cut occurs, this barrier is broken, allowing pathogens to enter. Blood clotting quickly seals the wound, both stemming blood loss and reducing further pathogen entry, and forms a scab that provides a temporary physical barrier while the skin heals. Any pathogens that do enter the body are targeted by non-specific white blood cells called phagocytes, which recognise general features common to many pathogens, engulf them by phagocytosis, and destroy them using digestive enzymes — this response is rapid but not targeted to a particular pathogen. If phagocytes alone are insufficient, the specific immune response is activated: lymphocytes recognise the unique antigens on the surface of the particular pathogen involved and respond by producing antibodies precisely complementary to those antigens, which bind to the pathogen, neutralising it and/or marking it for destruction by phagocytes. This specific response is slower to develop initially, but some lymphocytes remain afterwards as memory cells, allowing a much faster, stronger response if the same pathogen is encountered again in future. Together, the rapid but general non-specific defences and the slower but highly targeted, adaptable specific defences provide layered protection against infection following a skin injury.

Level of Response mark scheme (4-band): Band A (5-6 marks): a full, well-sequenced account covering the skin barrier, clotting, non-specific phagocyte action, and the specific lymphocyte/antibody/memory-cell response, correctly explaining how the two systems work together (speed vs specificity), using correct terminology (phagocyte, lymphocyte, antigen, antibody, memory cell) with clear communication. Band B (3-4 marks): most stages present (e.g. skin barrier, phagocytes, and antibodies) but the coordination between non-specific and specific defences less clearly explained, or memory cells omitted; generally appropriate terminology. Band C (1-2 marks): only one or two basic points made (e.g. 'the skin stops germs getting in' or 'white blood cells fight infection') with little development or accurate terminology. Band D (0 marks): no relevant, creditable content. Indicative content: skin as a physical, non-specific barrier; clotting sealing the wound and reducing pathogen entry; phagocytes engulfing pathogens generally (non-specific, rapid); lymphocytes recognising specific antigens; production of specific antibodies that bind/neutralise/mark the pathogen; memory cells providing long-term, faster future protection; overall coordination — fast general response backed up by a slower, precisely targeted and long-lasting response. [6]

According to collision theory, a reaction can only occur when particles collide with sufficient energy (at least the activation energy) and with the correct orientation. Increasing temperature increases the average kinetic energy of the reacting particles, so they move faster; this has two effects — collisions become more frequent because faster-moving particles meet more often, and, more importantly, a substantially greater proportion of collisions now have energy equal to or exceeding the activation energy, since the distribution of particle energies shifts and its 'high energy tail' grows disproportionately for even a modest temperature rise. As a rough rule, reaction rate approximately doubles for every 10°C rise in temperature, reflecting this exponential-like sensitivity. Increasing concentration, by contrast, increases the number of particles per unit volume, which increases the frequency of collisions in a roughly proportional way, but it does not change the energy of any individual collision or the proportion of particles with sufficient energy to react — the energy distribution of the particles is unaffected by concentration. Because temperature affects both collision frequency and, much more significantly, the proportion of successful (sufficiently energetic) collisions, while concentration affects only collision frequency, a given percentage increase in temperature typically produces a much larger increase in rate than the same percentage increase in concentration.

Level of Response mark scheme (4-band): Band A (5-6 marks): a full, well-reasoned explanation referencing activation energy, the distribution of particle energies, and correctly distinguishing the dual effect of temperature (frequency and proportion of successful collisions) from the single effect of concentration (frequency only), using correct terminology (activation energy, collision frequency, kinetic energy) with clear communication. Band B (3-4 marks): explains that higher temperature increases both collision frequency and particle energy, and that concentration only increases frequency, but without full development of why the temperature effect is proportionally larger. Band C (1-2 marks): basic statement that temperature increases rate because particles move faster/collide more, with little reference to activation energy or comparison with concentration. Band D (0 marks): no relevant, creditable content. Indicative content: collisions require energy ≥ activation energy to be successful; temperature increases average kinetic energy of particles; higher temperature increases both collision frequency and the proportion of particles with energy ≥ activation energy; concentration increases only collision frequency, not particle energy/proportion exceeding activation energy; approximate rule that rate doubles per 10°C rise; correct overall comparison explaining why temperature's effect is proportionally greater. [6]

A star like the Sun begins as a nebula — a large cloud of gas (mainly hydrogen) and dust in space. Gravity causes regions of this cloud to contract and collapse inwards; as the collapsing material gains gravitational potential energy loss converted to heat, the core heats up and forms a protostar. Once the core temperature and pressure become high enough, nuclear fusion of hydrogen nuclei into helium begins, releasing enormous amounts of energy. This fusion produces an outward radiation/thermal pressure that exactly balances the inward pull of gravity, and the star enters a long, stable period called the main sequence, during which the Sun currently exists. Eventually, the hydrogen fuel in the core is used up; without fusion to balance it, gravity causes the core to contract further while the outer layers expand and cool, and the star swells into a red giant. Once core temperatures rise high enough, helium fusion begins, producing heavier elements such as carbon and oxygen. For a Sun-like star, this stage is relatively brief; the star cannot generate the core temperatures needed to fuse elements beyond this point, so once fusion finally stops, gravity is no longer balanced and the outer layers of the star are gently ejected into space, forming an expanding shell of gas called a planetary nebula. What remains is the exposed, extremely dense core — a white dwarf — which no longer undergoes fusion and simply radiates away its remaining heat, gradually cooling over billions of years.

Level of Response mark scheme (4-band): Band A (5-6 marks): a full, correctly sequenced life cycle (nebula → protostar → main sequence → red giant → planetary nebula → white dwarf) with accurate explanation of gravity's role in collapse/contraction at each stage and fusion's role in providing outward pressure balancing gravity, using correct terminology throughout with clear communication. Band B (3-4 marks): most stages correctly sequenced and named, with a reasonable but less complete explanation of the gravity/fusion balance at one or more stages. Band C (1-2 marks): only one or two stages correctly identified (e.g. 'stars form from gas clouds' and/or 'the Sun will become a red giant') with little explanation of the underlying physics. Band D (0 marks): no relevant, creditable content. Indicative content: formation from a nebula (gas and dust) via gravitational collapse; hydrogen fusion beginning once core conditions are sufficient; main sequence stability from balance between fusion (outward) and gravity (inward); red giant stage once core hydrogen is exhausted, helium fusion beginning; ejection of outer layers as a planetary nebula; formation of a white dwarf that cools over time without further fusion. [6]