2024 CCEA AS-Level Biology 1010 Practice Paper with Answers

(a) A peptide bond is formed, and water is released (in a condensation reaction).

(b) This describes the primary structure of a protein.

(c) Hydrogen bonds form regularly between the C=O group of one amino acid and the N-H group of another amino acid, occurring at regular intervals along the polypeptide backbone; these numerous hydrogen bonds pull the chain into a regularly coiled, helical shape (the alpha-helix), stabilising this secondary structure.

(d) Example: water has a high specific heat capacity, meaning a relatively large amount of energy is needed to raise its temperature; this helps to buffer organisms and aquatic habitats against rapid temperature fluctuations, providing a thermally stable environment for metabolic reactions. (Other valid answers: high latent heat of vaporisation, important for cooling by sweating/transpiration; cohesion, important for water transport in the xylem; water's role as a metabolite/reactant in reactions such as hydrolysis and photosynthesis.)

(a) [1] peptide bond; [1] water. (b) [1] primary structure. (c) [1] correctly identifies hydrogen bonding between C=O and N-H groups (or 'between adjacent turns of the chain'/'between amino acids at regular intervals'); [1] correctly links this to the regular coiling/stability of the helix. (d) [1] valid property named (high specific heat capacity, high latent heat of vaporisation, cohesion/surface tension, or role as a reactant); [1] correct, clearly explained biological significance of that property.

(a) The general formula for carbohydrates is \( (CH_2O)_n \). Glucose is a monosaccharide, a single sugar unit; maltose is a disaccharide, formed from two monosaccharide units (two glucose molecules) joined together.

(b) A glycosidic bond links the two monosaccharide units; it is formed by a condensation reaction (with the release of a water molecule).

(c) Cellulose molecules are long, straight, unbranched chains of beta-glucose; many of these parallel chains are held together side-by-side by numerous hydrogen bonds, bundling them into strong microfibrils, which gives cellulose great tensile strength suitable for a structural cell-wall component.

(a) [1] correct general formula \( (CH_2O)_n \); [1] correct distinction (glucose = single sugar unit/monomer, maltose = two joined units). (b) [1] glycosidic bond; [1] condensation (reaction). (c) [1] correctly identifies long, straight/unbranched chains (of beta-glucose); [1] correctly identifies hydrogen bonding between parallel chains forming strong microfibrils/fibres.

(a) Percentage change \( = \dfrac{\text{final mass}-\text{initial mass}}{\text{initial mass}}\times100 \). 0.0 M: \( \dfrac{5.85-5.20}{5.20}\times100=+12.5\% \). 0.2 M: \( +6.7\% \). 0.4 M: \( 0.0\% \). 0.6 M: \( -6.7\% \). 0.8 M: \( -12.5\% \). 1.0 M: \( -17.3\% \).

(b) Points plotted accurately with suitable linear scales on both axes, occupying at least half the grid; a smooth curve (or straight line, since this data set is very close to linear) of best fit drawn through the points.

(c) The isotonic point is where the line/curve crosses the x-axis, i.e. where percentage change in mass is zero — reading from the graph (or from the table directly, since one data point falls exactly at zero), this is at a sucrose concentration of 0.4 mol dm^-3. At this concentration, the water potential of the sucrose solution exactly equals the water potential of the potato cells, so there is no net movement of water into or out of the cells, and hence no change in mass.

(d) Distilled water has a water potential of 0 kPa, the highest possible value (least negative), which is higher than (less negative than) the water potential of the potato cells (which is lowered by dissolved solutes in the cell sap). Water therefore moves, by osmosis, down its water potential gradient from the distilled water (high/less negative water potential) into the potato cells (lower/more negative water potential), across the partially permeable cell surface membranes, causing the cells to gain water and increase in mass.

Final answer: (a) as shown above; (c) isotonic sucrose concentration \( \approx0.4\ \text{mol dm}^{-3} \).

(a) [1] correct method shown for at least one value; [1] correct values for 0.0, 0.2 M rows (+12.5%, +6.7%); [1] correct values for 0.6, 0.8, 1.0 M rows (-6.7%, -12.5%, -17.3%; 0.4 M given as 0.0% needs no calculation). (b) [1] suitable linear scales, at least half the grid used; [1] all 6 points plotted accurately (±1 small square, ECF from (a)); [1] single smooth curve or best-fit line drawn (not dot-to-dot); [1] line/curve consistent with the general trend of the data. (c) [1] correctly reads x-intercept (percentage change = 0) as approximately 0.4 mol dm^-3; [1] correct reasoning that this is where water potential of solution equals that of the cells (isotonic), so no net water movement. (d) [1] correctly identifies distilled water has a higher (less negative) water potential than the potato cells; [1] correctly explains osmosis occurs down the water potential gradient into the cells, across the partially permeable membrane, causing a mass increase.

(a) The amount of DNA in the cell doubles, as each chromosome is replicated to form two identical sister chromatids.

(b) Metaphase (chromosomes line up at the equator); anaphase (chromatids separate and move to opposite poles).

(c) The G1/S checkpoint checks that the cell's DNA is undamaged (and that conditions/cell size/nutrients are adequate) before the cell is allowed to proceed into the S phase and replicate its DNA; if damage is detected, the cell cycle is halted (allowing time for repair, or triggering programmed cell death if damage is irreparable).

(d) If damaged DNA is not detected and repaired at this checkpoint, the cell will proceed to replicate this damaged (mutated) DNA and pass the mutation(s) on to daughter cells through further mitotic divisions. If the mutation(s) occur in genes controlling cell division (e.g. genes normally limiting growth), the resulting cells may divide uncontrollably, forming an increasing mass of abnormal cells — a tumour.

(e) Mitosis in eukaryotes produces two genetically identical diploid daughter cells with a defined nuclear envelope reforming around each set of chromosomes, whereas binary fission in bacteria (which lack a true nucleus) produces two genetically identical daughter cells from a single circular chromosome without the formation of a spindle apparatus or distinct mitotic stages.

(a) [1] correctly states DNA amount doubles/DNA is replicated. (b) [1] metaphase; [1] anaphase. (c) [1] correctly identifies checking for DNA damage (or cell readiness/size/nutrients) before allowing replication; [1] correctly identifies the consequence (cycle halted for repair, or apoptosis triggered, if damage found). (d) [1] correctly identifies damaged/mutated DNA is replicated and passed to daughter cells; [1] correctly links this to mutations accumulating in genes controlling cell division; [1] correctly concludes this can lead to uncontrolled cell division/tumour formation. (e) [1] valid, correctly explained difference (e.g. presence of a nuclear envelope/spindle in eukaryotic mitosis, absent in bacterial binary fission; or reference to a true nucleus vs no nucleus).

(a) The graph rises steeply at first (roughly proportionally) as substrate concentration increases from low values, but the rate of increase progressively slows, and the curve eventually levels off to a plateau (maximum rate) at higher substrate concentrations.

(b) At low substrate concentrations, there are more free (unoccupied) enzyme active sites available than substrate molecules, so increasing substrate concentration increases the frequency of successful enzyme-substrate collisions, increasing rate roughly proportionally. As substrate concentration increases further, a progressively greater proportion of the fixed number of active sites present (since enzyme concentration is constant) become occupied at any instant, so each further increase in substrate concentration has a smaller effect on rate. Eventually, at high substrate concentration, essentially all active sites are occupied/saturated at all times, and the rate becomes limited only by how quickly each enzyme molecule can process (catalyse the breakdown of) a substrate molecule and become free again — the rate reaches a maximum and further increases in substrate concentration have no further effect.

(c) The initial rate is used because, as the reaction proceeds, the substrate concentration falls (as it is used up) and products accumulate, so the rate itself changes over time; measuring only the initial rate (over the first few seconds, when substrate concentration is still very close to its starting value) ensures a fair, reliable comparison between the different starting substrate concentrations tested, each measured under equivalent (near-constant, starting) conditions.

(d) Any two of: immobilised enzymes can easily be recovered/removed from the reaction mixture (e.g. by filtering out the beads) and reused multiple times, reducing cost; the product is not contaminated by the enzyme (no need for a separate, costly purification step to remove the enzyme from the product); immobilised enzymes are often more stable (more resistant to changes in temperature/pH) than free enzymes in solution; they allow continuous-flow industrial processes (substrate solution passed continuously over/through immobilised enzyme) rather than batch processing.

(e) A disadvantage is that immobilisation can reduce the rate of reaction compared with a free enzyme in solution, because substrate molecules must first diffuse to the active site (which may be less accessible when trapped within a bead/matrix), and immobilisation may also slightly alter the enzyme's tertiary structure and reduce its activity.

(f) Any one valid variable: temperature; pH; enzyme (catalase) concentration; volume of enzyme/substrate solution used; the source/preparation of the catalase (e.g. same batch).

(a) [1] correctly describes initial steep/proportional rise; [1] correctly describes rate levelling off to a plateau/maximum at higher concentrations. (b) [1] correctly identifies excess free active sites at low substrate concentration; [1] correctly links this to rate increasing roughly proportionally at low concentration; [1] correctly identifies active sites becoming progressively occupied/saturated as concentration rises; [1] correctly explains the plateau as all active sites being saturated/rate limited by enzyme turnover, not substrate availability. (c) [1] correctly identifies substrate concentration/rate changes over the course of the reaction; [1] correctly explains initial rate gives a fair, standardised comparison across different starting concentrations. (d) [1] each for any two valid, distinct advantages (reusability/recovery; product purity/no contamination; increased stability; enables continuous-flow processing); max [2]. (e) [1] valid disadvantage identified (reduced accessibility/diffusion limitation, or altered enzyme structure/reduced activity); [1] correct supporting explanation. (f) [1] valid controlled variable named; [1] correct reasoning as to why it must be controlled (to ensure any difference in rate is due only to substrate concentration).

(a) The membrane consists mainly of a bilayer of phospholipid molecules, each with a hydrophilic (polar) phosphate 'head' and a hydrophobic (non-polar) fatty acid 'tail'; the two layers arrange themselves with their hydrophilic heads facing outward, towards the aqueous environment on each side, and their hydrophobic tails facing inward, away from water. Proteins are embedded within this bilayer at varying depths (some spanning the full width as integral/transmembrane proteins, others only partially embedded as peripheral proteins), scattered throughout the membrane like a 'mosaic'; both the phospholipids and the proteins are able to move sideways within the layer (the membrane is described as 'fluid').

(b) Similarity: both simple diffusion and facilitated diffusion involve the net movement of molecules down their concentration gradient (from high to low concentration) and neither requires metabolic energy (ATP). Difference: simple diffusion involves molecules passing directly through the phospholipid bilayer itself (suitable for small, non-polar/lipid-soluble molecules), whereas facilitated diffusion requires specific membrane protein channels or carrier proteins to allow the movement of larger or polar/charged molecules (e.g. glucose, ions) that cannot readily cross the hydrophobic bilayer directly.

(c) Active transport moves substances against their concentration gradient (from an area of low concentration to an area of higher concentration), which cannot happen spontaneously; energy (from ATP hydrolysis) must therefore be supplied to change the shape of the carrier protein and drive this 'uphill' movement, whereas diffusion is a passive process, moving substances down their concentration gradient, which occurs spontaneously without any energy input.

(d) Ciliated epithelial cells possess cilia (hair-like projections) on their free surface, which beat rhythmically to waft a layer of mucus (produced by neighbouring goblet cells) — together with any trapped particles and pathogens — up and away from the lungs towards the throat, protecting the delicate gas-exchange surfaces from debris and infection. Squamous epithelial cells, found in the alveoli, are extremely thin (flattened, one cell thick), which minimises the diffusion distance for respiratory gases (oxygen and carbon dioxide) between the air in the alveolus and the blood in the surrounding capillaries, maximising the rate of gaseous exchange.

(a) [1] correctly describes the phospholipid bilayer; [1] correctly identifies hydrophilic heads facing outward, hydrophobic tails facing inward; [1] correctly describes proteins embedded at varying depths (integral/peripheral); [1] correctly identifies components can move within the layer (fluidity). (b) [1] correct similarity (both passive, down concentration gradient, no ATP required); [1] correct identification that simple diffusion crosses the bilayer directly; [1] correct identification that facilitated diffusion requires a channel or carrier protein. (c) [1] correctly identifies active transport moves substances against the concentration gradient; [1] correctly explains this requires energy from ATP (as it does not occur spontaneously), unlike diffusion which is passive/spontaneous. (d) [1] correctly identifies cilia on ciliated epithelium; [1] correctly explains their function (wafting mucus/trapped particles away from the lungs); [1] correctly identifies squamous epithelium as very thin/flattened; [1] correctly links this to a short diffusion distance/efficient gas exchange.

An indicative full-mark response would include:

Nucleus: bounded by a double membrane (nuclear envelope) containing nuclear pores, which allow the passage of large molecules such as messenger RNA (mRNA) and ribosomal subunits between the nucleus and the cytoplasm. The nucleus contains chromatin (DNA associated with protein), which carries the genetic code for the amino acid sequence of proteins, and a nucleolus, which manufactures ribosomal RNA (rRNA) and assembles ribosomal subunits. In protein synthesis, the DNA code for the required protein is transcribed into mRNA within the nucleus, which then passes out through a nuclear pore into the cytoplasm.

Rough endoplasmic reticulum (RER): a system of flattened, membrane-bound sacs (cisternae), studded on their outer (cytoplasmic) surface with ribosomes, giving it a 'rough' appearance. The mRNA leaving the nucleus attaches to ribosomes on the RER, where the protein is synthesised (translated) and threaded into the lumen (interior) of the RER as it is made; the RER folds and processes the protein and forms small membrane-bound vesicles that pinch off, transporting the newly-made protein towards the Golgi apparatus.

Golgi apparatus: a stack of flattened, curved, membrane-bound sacs. Transport vesicles from the RER fuse with the Golgi apparatus, delivering the protein; the Golgi apparatus further processes/modifies the protein (e.g. by adding carbohydrate groups to form glycoproteins) and then packages the finished protein into new secretory vesicles, which pinch off from the far (trans) face of the Golgi apparatus.

Mitochondrion: bounded by a double membrane, the inner one folded into cristae (increasing surface area) and enclosing the matrix; this is the site of aerobic respiration, producing ATP. Protein synthesis, vesicle formation/movement and secretion are all active, energy-demanding processes, so mitochondria in a secretory cell supply the large quantities of ATP required to power these processes (e.g. active transport of materials, movement of vesicles along the cytoskeleton, and exocytosis).

How they work together: the sequence in a secretory cell is: DNA in the nucleus is transcribed into mRNA, which moves to ribosomes on the RER; the protein is synthesised on the RER and enters its lumen; vesicles bud off from the RER and carry the protein to the Golgi apparatus for further processing/modification and packaging into secretory vesicles; these secretory vesicles move to and fuse with the cell surface membrane, releasing the finished protein from the cell by exocytosis. Mitochondria supply the ATP needed to power the active transport, vesicle movement (along cytoskeletal tracks) and exocytosis involved at every stage of this pathway. Secretory cells specialised for this role typically contain especially large numbers of ribosomes, RER, Golgi apparatus and mitochondria to support high rates of protein production and secretion.

Levels-of-response (QWC) mark scheme, out of 15 marks.

Level 3 (11–15 marks): Detailed, accurate, well-organised account of the structure AND function of all four named organelles (nucleus, RER, Golgi apparatus, mitochondrion), correctly linking each structural feature to its function, AND a clear, correctly-sequenced explanation of how they work together in the secretory pathway (transcription in nucleus → translation on RER → processing/packaging in Golgi → secretion by exocytosis, with mitochondria supplying ATP throughout). Correct, precise specialist terminology throughout (e.g. cisternae, cristae, exocytosis, transcription); coherent structure; accurate spelling, punctuation and grammar.

Level 2 (6–10 marks): Good coverage of most organelles' structure and function, with the general sequence of the secretory pathway broadly correct, but with some omissions (e.g. one organelle covered in less depth, or the role of the mitochondrion not fully integrated) or minor inaccuracies; generally clear communication with occasional lapses.

Level 1 (1–5 marks): Limited or fragmented account (e.g. only 1–2 organelles described, or organelles listed without functional detail or without any attempt to link them into a sequence); weak terminology or organisation.

0 marks: No creditable response.

Full marks at Level 3 require substantive, accurate coverage of all four organelles AND a correctly-sequenced account of how they cooperate in protein secretion, not simply four separate descriptions.

(i) surface area to volume ratio — C. (ii) diffusion gradient — A. (iii) Fick's law — B. (iv) counter-current system — E. (v) ventilation — D.

[1] each for each correctly matched pair; max [5].

(a) Any two of: xylem vessel elements are dead cells with no cell contents (cytoplasm/organelles), providing an unobstructed, hollow tube for water flow; their end walls have broken down completely, so individual elements join end-to-end to form a long, continuous, open tube; their walls are thickened and impregnated with lignin, which is waterproof and provides mechanical strength to resist the inward pressure (collapse) caused by the tension of moving water/withstand the pull of transpiration.

(b) Water evaporates from the surface of mesophyll cells in the leaf and diffuses out through the stomata (transpiration), which lowers the water potential of the mesophyll cells; this pulls water out of the adjacent xylem vessels in the leaf, into the mesophyll cells, by osmosis. Because water molecules are strongly attracted to each other (cohesion, due to hydrogen bonding between polar water molecules), the loss of water from the top of the xylem column pulls up the whole unbroken, continuous column of water beneath it (creating tension, a pulling force, that is transmitted right down the xylem to the roots). This tension draws water into the xylem from the root cells, which in turn draws water into the root from the soil, maintaining a continuous flow of water up the plant against gravity, driven ultimately by transpiration at the top.

(a) [1] each for any two valid structural features (dead cells with no contents/hollow tube; end walls broken down forming continuous tube; lignified walls for waterproofing/strength); max [2]. (b) [1] correctly identifies transpiration/evaporation from mesophyll cells lowers their water potential, drawing water from the xylem; [1] correctly identifies cohesion between water molecules (hydrogen bonding); [1] correctly explains this cohesion causes the whole continuous water column to be pulled up under tension; [1] correctly links this tension to water being drawn into the roots from the soil, maintaining continuous flow.

(a) Minute ventilation = tidal volume × breathing rate \( = 0.4\times16 = 6.4\ \text{dm}^3\text{min}^{-1} \)

(b) Exercise minute ventilation \( = 0.9\times24 = 21.6\ \text{dm}^3\text{min}^{-1} \). Percentage increase \( = \dfrac{21.6-6.4}{6.4}\times100 = 237.5\% \).

Final answer: resting minute ventilation \( =6.4\ \text{dm}^3\text{min}^{-1} \); exercise minute ventilation \( =21.6\ \text{dm}^3\text{min}^{-1} \), an increase of 237.5%.

(a) [1] correct method (TV × rate); [1] \( 6.4\ \text{dm}^3\text{min}^{-1} \) with correct unit. (b) [1] correct exercise minute ventilation, \( 21.6\ \text{dm}^3\text{min}^{-1} \); [1] correct percentage-increase method; [1] \( 237.5\% \) (accept 237–238%, ECF).

(a) Any two of: high blood pressure (hypertension); smoking; a diet high in saturated fat/high blood cholesterol (particularly LDL cholesterol); lack of physical exercise/obesity; genetic factors/family history; increasing age; diabetes.

(b) The build-up of atheroma damages and roughens the smooth inner lining (endothelium) of the artery wall; this rough, damaged surface can trigger the blood clotting (coagulation) cascade, causing platelets to adhere to the site and release substances that activate a series of clotting factors, ultimately converting the soluble plasma protein fibrinogen into insoluble fibrin threads; these fibrin threads form a mesh that traps blood cells and platelets, forming a solid blood clot (thrombus) at the site of the atheroma.

(c) The coronary arteries supply oxygenated blood (and glucose) to the cardiac (heart) muscle itself. If a thrombus forms in (or an atheroma-narrowed section of) a coronary artery and blocks it, blood flow to the area of heart muscle supplied by that artery is cut off or severely reduced; the affected cardiac muscle cells are then deprived of oxygen (become ischaemic) and can no longer respire aerobically to produce sufficient ATP, so they stop contracting properly and, if the blockage persists, the muscle cells die (myocardial infarction), which can seriously impair the heart's ability to pump blood and may be fatal.

(a) [1] each for any two valid risk factors; max [2]. (b) [1] correctly identifies atheroma damages/roughens the endothelium; [1] correctly identifies this triggers the clotting cascade/platelet adhesion; [1] correctly identifies conversion of fibrinogen to fibrin forming the clot. (c) [1] correctly identifies coronary arteries supply the heart muscle itself with oxygen; [1] correctly explains the blockage cuts off oxygen supply to part of the heart muscle; [1] correctly explains the resulting death of oxygen-starved cardiac muscle cells and the impact on heart function.

(a) The number of new cases remains relatively stable (fluctuating slightly, around 42–48) for the first three months, then falls steadily and substantially from month 4 onwards, dropping from 38 cases in month 4 to just 6 cases by month 8.

(b) The water treatment plant likely removes or kills the bacterial pathogen from the water supply (e.g. through filtration, chlorination or other disinfection processes), reducing the number of people exposed to (and infected by) the pathogen through contaminated drinking water, which would explain the observed fall in new cases beginning in the same month the plant was installed.

(c) Other contributing factors could include: a public health campaign educating residents about hygiene measures (e.g. handwashing, boiling water) introduced around the same time; seasonal changes affecting pathogen survival/transmission (e.g. warmer months increasing bacterial growth, so a change in season could independently reduce cases); or increasing population immunity as more people who were exposed earlier develop immunity. Because the water treatment plant was introduced at the same time as (or possibly alongside) other changes, and the data only show a correlation (cases fell after the plant was installed) rather than being from a controlled experiment, it is difficult to be certain the treatment plant itself (rather than one of these other, coincidental factors) directly caused the fall — correlation does not necessarily prove causation.

(d) As human populations grow and urbanise, larger numbers of people live at higher densities, often sharing the same water sources; this increases the opportunity for a pathogen to spread from an infected individual to many others via a shared, contaminated water supply, and increases the chance that inadequate sanitation infrastructure (e.g. sewage contaminating drinking water sources) will affect a large number of people simultaneously, increasing the risk and scale of outbreaks.

(e) Suggestions could include: comparing case numbers with a similar town that did NOT install a treatment plant over the same period (a control for comparison); testing water samples directly, before and after treatment, for the presence/concentration of the pathogen, to directly confirm the plant removes it; or continuing to monitor cases for a longer period after month 8 to confirm the low rate is sustained (not a temporary fluctuation).

(a) [1] correctly describes the initial roughly stable/fluctuating phase (months 1–3); [1] correctly describes the subsequent steady decline from month 4 onwards. (b) [1] correctly links the water treatment plant to removal/destruction of the pathogen from the water supply; [1] correctly links this to reduced exposure/infection and hence falling case numbers. (c) [1] valid alternative factor identified (e.g. hygiene campaign, seasonal change, rising immunity); [1] correct explanation of why this factor could independently explain some of the fall; [1] correctly identifies the general principle that correlation alone does not establish causation given other possible explanations exist. (d) [1] correctly identifies increased population density/shared water sources with urbanisation; [1] correctly links this to greater opportunity for pathogen spread to more people; [1] correctly notes risk of sanitation infrastructure being overwhelmed/inadequate for larger populations. (e) [1] each for any valid, clearly explained suggestion (untreated control town/area for comparison; direct water pathogen testing before/after treatment; extended monitoring period); max [3], at least one mark requires clear justification of how the suggestion would help establish causation.

(a) Habitat A, N=40: \( \Sigma(n/N)^2 = (15/40)^2+(12/40)^2+(8/40)^2+(5/40)^2 = 0.1406+0.0900+0.0400+0.0156 = 0.2863 \). \( D = 1-0.2863 = 0.714 \) (3 s.f.).

(b) Habitat B, N=40: \( \Sigma(n/N)^2 = (32/40)^2+(4/40)^2+(3/40)^2+(1/40)^2 = 0.6400+0.0100+0.0056+0.0006 = 0.6563 \). \( D = 1-0.6563 = 0.344 \) (3 s.f.).

(c) Habitat A has a much higher value of D (0.714) than Habitat B (0.344), indicating Habitat A has greater biodiversity (a more even distribution of individuals across its species, i.e. no single species dominates). Habitat B, by contrast, is dominated by Species 1 (32 of the 40 individuals, 80%), giving it low diversity despite having the same number of species and the same total number of individuals as Habitat A. A possible ecological reason: Habitat B may have undergone some disturbance or contain conditions that particularly favour Species 1 (e.g. Species 1 may be a fast-growing coloniser species dominating a recently-disturbed or less stable habitat), out-competing the other species, whereas Habitat A may be a more stable, established habitat allowing a more even balance between species.

(d) Species richness is simply the number of different species present in a habitat (which is the same in both habitats here — 4 species each), without any reference to how the individuals are distributed between those species. Species diversity, however, also takes into account the relative abundance (evenness) of each species — as shown by the very different D values, a habitat with the same species richness can have very different diversity depending on whether individuals are spread evenly (as in Habitat A) or dominated by one species (as in Habitat B).

(e) Species.

(f) DNA/molecular classification directly reflects evolutionary relationships (genetic similarity/shared ancestry) between organisms, and is not affected by convergent evolution (where unrelated organisms independently evolve similar physical features due to similar environmental pressures), which can mislead classification based purely on physical appearance.

Final answer: \( D_A=0.714 \), \( D_B=0.344 \); Habitat A is more biodiverse.

(a) [1] correct method (Σ(n/N)² set up); [1] correct individual squared-fraction terms; [1] correct sum \( \approx0.286 \); [1] \( D=0.714 \) (accept 0.71–0.72). (b) [1] correct method; [1] correct individual squared-fraction terms; [1] correct sum \( \approx0.656 \); [1] \( D=0.344 \) (accept 0.34–0.35). (c) [1] correctly identifies Habitat A has higher diversity (higher D); [1] correctly links this to a more even distribution of individuals/no dominant species in A vs. dominance by Species 1 in B; [1] any valid, plausible ecological reason suggested for the difference. (d) [1] correct definition of species richness (number of species present, same in both habitats here); [1] correct definition of species diversity (accounts for relative abundance/evenness, differs between the habitats despite equal richness). (e) [1] species. (f) [1] valid reason (reflects genetic/evolutionary relationships directly; unaffected by convergent evolution/misleading physical similarity).

(a) The waxy cuticle is impermeable (or nearly so) to water; a thicker cuticle further reduces the amount of water that can evaporate directly through the leaf's outer surface (cuticular transpiration), so most water loss is restricted to passing through the stomata, which can be more tightly controlled.

(b) Sunken stomata, located within pits, create a small pocket of still air below the leaf surface; water vapour diffusing out of the stomata builds up within this trapped, still air pocket, raising the local humidity and reducing the water vapour concentration gradient between the leaf's internal air spaces and the air immediately outside the stomata, so the rate of diffusion (and hence transpiration) is reduced.

(c) Rolling the leaf so the lower surface (bearing the stomata) faces inward encloses a small pocket of still, humid air within the rolled leaf, around the stomata; as in (b), this raises local humidity near the stomata, reducing the diffusion gradient for water vapour and therefore reducing the rate of transpiration; it may also reduce the leaf's exposure to wind, which would otherwise sweep away humid air and increase the rate of water loss.

(d) A smaller total leaf surface area directly reduces the total area available for water to evaporate from (both cuticular evaporation and stomatal transpiration are reduced in proportion to the exposed surface area), so overall water loss from the plant is reduced, even though this comes at some cost to the plant's photosynthetic capacity (less surface area for light absorption/gas exchange for photosynthesis).

(a) [1] correctly identifies the cuticle is (largely) impermeable/waterproof; [1] correctly explains a thicker cuticle further reduces water loss directly through the leaf surface (cuticular transpiration). (b) [1] correctly identifies sunken stomata trap a pocket of still, humid air; [1] correctly explains this reduces the water vapour concentration gradient, reducing diffusion/transpiration rate. (c) [1] correctly identifies rolling traps still, humid air around the stomata (reducing the diffusion gradient); [1] correctly identifies reduced exposure to wind/air movement, which would otherwise remove humid air and increase water loss. (d) [1] correctly identifies reduced surface area directly reduces the area for evaporation/transpiration; [1] correctly notes the trade-off with reduced photosynthetic capacity, or otherwise gives a fully-reasoned explanation of the reduced water loss.

An indicative full-mark response would include:

Cardiac cycle: during atrial systole, both atria contract simultaneously, pushing a final volume of blood into the ventricles (which are relaxed, in diastole) through the open atrioventricular (AV) valves (tricuspid on the right, bicuspid/mitral on the left). During ventricular systole, both ventricles then contract; as ventricular pressure rises above atrial pressure, the AV valves are forced shut (preventing backflow of blood into the atria), and once ventricular pressure exceeds the pressure in the aorta/pulmonary artery, the semilunar valves are forced open, ejecting blood into the aorta (from the left ventricle) and pulmonary artery (from the right ventricle). During diastole, both atria and ventricles relax; as ventricular pressure falls below that in the aorta/pulmonary artery, the semilunar valves close (preventing backflow into the ventricles); as ventricular pressure falls further, below atrial pressure, the AV valves open again and blood begins to passively flow from the atria (which are meanwhile filling with returning blood) into the ventricles, before the cycle repeats.

Electrical coordination: the cycle is initiated and coordinated by a wave of electrical excitation (action potentials) that originates at the sinoatrial node (SAN), the heart's natural pacemaker, located in the wall of the right atrium; this excitation spreads across both atria, causing them to contract (atrial systole). The excitation reaches the atrioventricular node (AVN), where it is briefly delayed (allowing the atria to finish contracting and empty into the ventricles before the ventricles contract); the excitation then passes rapidly down specialised conducting fibres, the Bundle of His, through the septum between the ventricles, and out along the Purkyne fibres into the ventricle walls, causing the ventricles to contract from the base (apex) upwards, efficiently squeezing blood up and out through the arteries.

Double, closed circulation: the mammalian circulatory system is described as 'double' because blood passes through the heart twice on each complete circuit of the body — once through the right side of the heart to the lungs (pulmonary circulation) to be oxygenated, then back to the left side of the heart to be pumped out to the rest of the body (systemic circulation). This double circulation allows blood returning from the lungs to be re-pressurised by the left side of the heart before being sent around the (much longer, higher-resistance) systemic circuit, maintaining a higher blood pressure and faster flow rate to body tissues than a single circulation could achieve — important for efficient, rapid delivery of oxygen and nutrients, particularly to a large, active, warm-blooded (endothermic) mammal with a high metabolic/oxygen demand. The system is 'closed' because blood is always contained within blood vessels (arteries, capillaries and veins) rather than bathing tissues directly, allowing blood pressure to be maintained and blood to be directed efficiently to where it is needed; the extensive capillary networks throughout the tissues provide a very large surface area, with capillary walls only one cell thick, for efficient diffusion/exchange of oxygen, nutrients and waste products between the blood and body cells.

Levels-of-response (QWC) mark scheme, out of 15 marks.

Level 3 (11–15 marks): Detailed, accurate, correctly-sequenced account of the full cardiac cycle (atrial systole, ventricular systole, diastole) with correct explanation of AV and semilunar valve action at each stage; AND a correct, complete account of the electrical coordination pathway (SAN → atria → AVN with delay → Bundle of His → Purkyne fibres → ventricles); AND a clear, well-reasoned explanation of why a double, closed circulation efficiently supplies tissues (re-pressurisation between circuits; closed system maintains pressure; capillary structure for exchange). Correct, precise specialist terminology throughout; coherent structure; accurate spelling, punctuation and grammar.

Level 2 (6–10 marks): Good coverage of most of the cardiac cycle and electrical coordination, with some omissions (e.g. AVN delay not mentioned, or valve action only partially explained) or minor inaccuracies; double circulation explained but with less developed reasoning; generally clear communication with occasional lapses.

Level 1 (1–5 marks): Limited or fragmented account (e.g. names structures/valves without describing the sequence of events, or omits the electrical coordination or double-circulation explanation entirely); weak terminology or organisation.

0 marks: No creditable response.

Full marks at Level 3 require substantive, accurate coverage of ALL THREE elements: the cardiac cycle with valve action, the electrical coordination pathway, and the double/closed circulation explanation.

(a) \( \text{actual size} = \dfrac{\text{image size}}{\text{magnification}} = \dfrac{45\ \text{mm}}{300} = 0.15\ \text{mm} = 150\ \mu\text{m} \)

(b) Converting to consistent units: \( 60\ \text{mm} = 60\,000\ \mu\text{m} \). \( \text{magnification} = \dfrac{\text{image size}}{\text{actual size}} = \dfrac{60\,000\ \mu\text{m}}{20\ \mu\text{m}} = 3000 \), i.e. ×3000.

Final answer: (a) \( 150\ \mu\text{m} \); (b) ×3000.

(a) [1] correct method (image size ÷ magnification), with correct unit conversion; [1] \( 150\ \mu\text{m} \) (accept 0.15 mm). (b) [1] correct unit conversion (60 mm = 60 000 μm) and correct method (image size ÷ actual size); [1] ×3000.

(a) Any two of: only the outlines/boundaries of tissue areas should be drawn (as clear, continuous, unbroken, single lines), not individual cells; the diagram should be drawn to scale, with correct proportions between the different tissue regions; the diagram should be given a suitable title, magnification stated, and clearly labelled with straight, non-crossing label lines that touch (but do not have arrowheads into) the structure identified; the drawing should be large enough to show detail clearly, drawn in pencil with clean, unbroken lines and no shading/colouring.

(b) From outside inward: epidermis, cortex, endodermis, vascular tissue (xylem and phloem, at the centre).

(c) A low-power plan diagram is intended to show only the overall arrangement, relative size, shape and position of the different tissue regions within the whole specimen; individual cells are far too small to be resolved accurately or usefully at low power (their detail would only be seen properly at high power), so attempting to draw them on a low-power plan would be inaccurate, would clutter the diagram, and would not appropriately represent what is actually visible at that magnification.

(a) [1] each for any two valid conventions (tissue boundaries only/no individual cells; drawn to scale/correct proportions; correct labelling technique; clean unbroken lines, no shading); max [2]. (b) [1] for at least three tissue layers in the correct relative order; [1] for the fully correct order including vascular tissue at the centre (both marks require the full stated sequence to be correct, epidermis→cortex→endodermis→vascular tissue). (c) [1] correctly identifies individual cells are not resolvable/appropriate at low-power magnification; [1] correctly explains the purpose of a plan diagram is to show overall tissue arrangement/proportion, not cellular detail.

(a) Any valid safety precaution, e.g.: wear appropriate personal protective equipment (gloves, eye protection); use scalpels/sharp instruments carefully, cutting away from the body/hands; wash hands thoroughly after handling the specimen; follow the school/institution's specific biological specimen handling and disposal procedures.

(b) The wall of the left ventricle is noticeably much thicker (more muscular) than the wall of the right ventricle, when comparing a cross-section cut through both ventricles.

(c) This vessel is the pulmonary vein (returning oxygenated blood from the lungs to the left atrium). Like other veins, it has a relatively thin wall (compared with an artery of similar diameter) with a large lumen, because it carries blood at low pressure (having already passed through the capillary bed of the lungs); a thin wall requires less material/energy to produce and offers less resistance to the low-pressure flow.

(d) The aorta must withstand the very high pressure of blood ejected from the left ventricle, and smooth out the pulsatile flow into a steadier flow; its wall therefore contains a thick layer of smooth muscle and (particularly) a large amount of elastic tissue, which allows the vessel wall to stretch as a surge of blood is forced in during ventricular systole, and then to recoil elastically during diastole, helping to maintain blood flow and pressure between heartbeats and protecting the vessel (and downstream, smaller vessels) from the full force of the pressure surge.

(a) [1] any valid, relevant safety precaution. (b) [1] correctly identifies comparing wall thickness; [1] correctly states left ventricle wall is thicker. (c) [1] correctly identifies pulmonary vein; [1] correctly links thin wall/low pressure to it being a vein; [1] correctly explains this is appropriate given blood is at low pressure having passed through lung capillaries. (d) [1] correctly identifies thick muscular/elastic wall; [1] correctly explains stretching to accommodate the pressure surge from ventricular systole; [1] correctly explains elastic recoil helps maintain flow/pressure between heartbeats.

(a) A concentrated spot of the pigment extract should be applied to a pencil-drawn origin (baseline) line, positioned above the level of the solvent in the chromatography tank/beaker (so the spot itself does not dissolve directly into the solvent); the spot should be applied, allowed to dry, and then reapplied several times to the same small spot, to concentrate the pigment and produce a small, concentrated, well-defined origin spot (rather than one large, dilute, spread-out spot), which gives sharper, better-separated bands after the chromatogram has run.

(b) \( R_f = \dfrac{\text{distance travelled by pigment}}{\text{distance travelled by solvent front}} = \dfrac{7.2}{9.0} = 0.80 \)

(c) Different pigments have different chemical structures/solubilities and therefore differ in how strongly they are attracted to (adsorbed onto) the stationary phase (the chromatography paper) versus how soluble they are in the moving solvent (the mobile phase). A pigment that is more soluble in the solvent and/or less strongly attracted to the paper will be carried further up the paper by the moving solvent (travelling a greater distance, giving a higher Rf value), while a pigment that is less soluble in the solvent and/or binds more strongly to the paper will travel a shorter distance (lower Rf value); this differential movement is what allows the different pigments in the mixture to separate into distinct spots/bands.

Final answer: \( R_f = 0.80 \).

(a) [1] correctly identifies applying the spot on a pencil (baseline) line above the solvent level; [1] correctly identifies applying/drying/reapplying multiple times to the same spot; [1] correctly explains this concentrates the spot, giving a small, well-defined origin for sharper separation. (b) [1] correct method (distance pigment / distance solvent); [1] \( R_f=0.80 \) (no unit). (c) [1] correctly identifies pigments differ in solubility in the solvent (mobile phase) and/or attraction/adsorption to the paper (stationary phase); [1] correctly links greater solubility/lower attraction to greater distance travelled (higher Rf); [1] correctly concludes this differential movement produces separation into distinct spots.

(a) A quadrat (e.g. divided into a grid of 100 smaller squares) should be placed at each marked sampling point along the transect; the student estimates the percentage of the quadrat's area covered by the grass species (e.g. by counting the number of the 100 grid squares in which the species is present, or in which more than half the square is covered, giving a direct percentage), recording this value for each position along the transect.

(b) Mean \( = \dfrac{5+15+35+55+70+80}{6} = \dfrac{260}{6} = 43.3\% \) (3 s.f.)

(c) The percentage cover of the grass species increases steadily and substantially with increasing distance from the sea, from only 5% at the seaward edge (0 m) up to 80% at 10 m inland. A possible explanation: exposure to salt spray and wind (and the instability/mobility of loose sand) is likely to be greatest closest to the sea, both of which can inhibit plant establishment and growth, while further inland the sand becomes more stable and less saline, allowing this grass species to establish and cover a greater proportion of the ground.

(d) A kite diagram shows how the abundance (e.g. percentage cover) of one or more species varies along an environmental gradient (such as a transect), with the width of each 'kite' shape at a given point along the transect representing the abundance of that species there. An advantage over a simple bar chart is that a kite diagram allows the abundance patterns of several different species to be plotted together on the same transect axis and visually compared directly (e.g. to see where the ranges of different species overlap or where one species replaces another), which would be far harder to interpret from a set of separate bar charts.

(a) [1] correctly identifies placing a (gridded) quadrat at each transect point; [1] correctly describes estimating percentage cover (e.g. via counting occupied grid squares). (b) [1] correct method (sum ÷ 6); [1] \( 43.3\% \) (accept 43.3–43.4%). (c) [1] correctly describes the increasing trend with distance from sea; [1] valid environmental factor identified (salt spray, wind exposure, sand stability); [1] correct reasoning linking the factor to the observed trend. (d) [1] correctly describes a kite diagram as showing abundance/cover along a transect/gradient; [1] correctly identifies the advantage of directly comparing multiple species' distributions on one shared axis.

(a) Zeroing the colorimeter with a blank ensures that any absorbance due to the colour/turbidity of the reagents or solvent alone (not due to the substance being measured) is subtracted out, so that the readings taken for the standards and samples reflect only the absorbance caused by the reducing sugar concentration itself. The blank should contain all the same reagents (e.g. colour-development reagent, buffer, water) as the test samples, but with no glucose/reducing sugar present (i.e. distilled water in place of the glucose solution).

(b) Points plotted accurately (±1 small square) with suitable linear scales occupying at least half the grid; single straight best-fit line drawn through (or very close to) the origin.

(c) The data are exactly linear: gradient \( = \dfrac{0.45-0.00}{10-0} = 0.045 \) (absorbance units per mmol dm^-3).

(d) Using the gradient (since the line passes through the origin): \( \text{concentration} = \dfrac{\text{absorbance}}{\text{gradient}} = \dfrac{0.30}{0.045} = 6.67\ \text{mmol dm}^{-3} \) (3 s.f.) (equivalently, reading directly from the graph at absorbance = 0.30 gives approximately the same value).

(e) Since the sample was diluted 4-fold before measurement, the original (undiluted) concentration was 4 times greater: \( 6.67\times4 = 26.7\ \text{mmol dm}^{-3} \) (3 s.f.).

(f) Small variations in colorimeter calibration/settings (e.g. the specific wavelength/filter used, lamp brightness, or slight differences between individual colorimeters), or differences in how carefully solutions are prepared/reagents pipetted between students, could shift the relationship between absorbance and concentration; re-preparing the calibration curve under the same conditions as the actual measurements ensures the calibration remains valid and accurate for that specific set of results.

Final answer: (c) gradient = 0.045; (d) diluted sample = 6.67 mmol dm^-3; (e) original sample = 26.7 mmol dm^-3.

(a) [1] correctly explains the blank corrects for background absorbance not due to the substance being measured; [1] correctly states the blank should contain the same reagents/solvent but no glucose. (b) [1] suitable linear scales, at least half the grid used; [1] all 6 points plotted accurately (±1 small square); [1] single straight best-fit line, through/near the origin. (c) [1] triangle/points span at least half the line; [1] correct gradient, 0.045 (accept 0.044–0.046). (d) [1] correct method (absorbance ÷ gradient, or correct graphical reading); [1] correct substitution; [1] \( 6.67\ \text{mmol dm}^{-3} \) (accept 6.5–6.9, ECF from (c)). (e) [1] correctly identifies the need to multiply by the dilution factor (×4); [1] \( 26.7\ \text{mmol dm}^{-3} \) (ECF from (d)). (f) [1] valid reason identified (equipment/calibration variation between colorimeters/days, or variation in technique between students); [1] correct reasoning linking this to the need to re-validate the calibration relationship for reliable results.