2022 CCEA AS-Level Biology 1010 模擬試題連答案詳解

(a) Two monosaccharides join together by a condensation reaction (releasing a molecule of water), forming a glycosidic bond between them.
(b) Starch is a polymer of α-glucose monomers. Cellulose is a polymer of β-glucose monomers.

(a) 2 marks: 1 mark for 'condensation reaction', 1 mark for 'glycosidic bond'. (b) 2 marks: 1 mark for α-glucose (starch), 1 mark for β-glucose (cellulose).

(a) Amino acids are linked together by peptide bonds, forming the primary structure (the sequence of amino acids in a polypeptide).
(b) The α-helix secondary structure is maintained by hydrogen bonds between different parts of the polypeptide chain.
(c) The tertiary structure (the overall 3D folding of a polypeptide) is maintained by a combination of: hydrogen bonds, ionic bonds, disulfide bonds, and hydrophobic interactions between R-groups (any two of these are acceptable).

(a) 1 mark: peptide bond. (b) 1 mark: hydrogen bond(s). (c) 2 marks: 1 mark each for any two of hydrogen bonds, ionic bonds, disulfide bonds, hydrophobic interactions.

(a) The active site is the specific region on the surface of an enzyme molecule, formed by the folding of the polypeptide chain(s), which has a shape complementary to its substrate and is where the substrate binds to form an enzyme-substrate complex.
(b) The lock-and-key hypothesis proposes that the enzyme's active site has a rigid, fixed shape that is already exactly complementary to the shape of its substrate, like a key fitting a lock. The induced-fit hypothesis proposes instead that the active site is flexible, and changes shape slightly as the substrate binds, moulding itself more closely around the substrate to form a better fit — the substrate induces this change in shape.

(a) 1 mark: correct definition referencing the specific region/shape that binds the substrate. (b) 3 marks: 1 mark for correct description of lock-and-key (fixed, complementary shape), 1 mark for correct description of induced-fit (active site changes shape as substrate binds), 1 mark for a clear point of contrast drawn between the two.

(a) Prokaryotic cells (e.g. bacteria) have naked, circular DNA that is not enclosed within a nuclear envelope (unlike the membrane-bound nucleus of eukaryotic cells); they have smaller ribosomes than eukaryotic cells; and they lack membrane-bound organelles such as mitochondria and endoplasmic reticulum (any two valid distinguishing features accepted).
(b) Examples of membrane-bound organelles found only in eukaryotic cells include: mitochondria, (rough or smooth) endoplasmic reticulum, Golgi apparatus, lysosomes, and (in plant cells) chloroplasts.

(a) 2 marks: 1 mark each for any two valid distinguishing features (e.g. naked/circular DNA without nuclear envelope; smaller ribosomes; absence of membrane-bound organelles; presence of a cell wall of different composition). (b) 2 marks: 1 mark each for any two correctly named membrane-bound organelles (e.g. mitochondria, endoplasmic reticulum, Golgi apparatus, lysosomes, chloroplasts).

(a) Osmosis is the net movement of water molecules from a region of higher (less negative) water potential to a region of lower (more negative) water potential, through a partially permeable membrane.
(b) A plant cell placed in a solution of very low (very negative) water potential loses water and undergoes plasmolysis (the protoplast shrinks away from the cell wall). An animal cell placed in the same conditions loses water and undergoes crenation (the cell shrinks and its surface becomes shrivelled/crinkled, since it has no cell wall to maintain its shape).

(a) 2 marks: correct reference to net movement of water [1], correct reference to higher to lower water potential through a partially permeable membrane [1]. (b) 2 marks: 1 mark for 'plasmolysis' (plant cell), 1 mark for 'crenation' (animal cell).

(a) DNA replication takes place during the S phase (synthesis phase) of interphase.
(b) The four stages of mitosis, in order, are: prophase, metaphase, anaphase, telophase.

(a) 1 mark: 'S phase' (accept 'synthesis phase'). (b) 3 marks: all four stages named in the correct order (prophase, metaphase, anaphase, telophase); deduct 1 mark for each stage named incorrectly or out of order, to a minimum of 0.

(a) \( \text{actual size} = \frac{\text{image size}}{\text{magnification}} = \frac{40 \text{ mm}}{20\,000} = 0.002 \text{ mm} = 2.0 \text{ μm} \)

(b) The magnification of an image is only valid for the specific image size it was originally calculated for; magnification is defined as \(\frac{\text{image size}}{\text{actual size}}\), so once the image is enlarged (making the image size larger) while the actual size of the mitochondrion obviously stays the same, the true magnification of the new, larger image must also increase — the old stated value of ×20 000 no longer applies.
New magnification: \( \text{magnification} = \frac{\text{new image size}}{\text{actual size}} = \frac{60 \text{ mm}}{0.002 \text{ mm}} = 30\,000 \), i.e. ×30 000.

(a) 3 marks: correct formula (image size/magnification) [1], correct unit conversion [1], answer 2.0 μm [1]. (b) 3 marks: correct explanation that actual size is unchanged while image size has increased [1], correct link to magnification therefore also increasing [1], correct new magnification ×30 000 (ecf) [1].

(a) \( 1 \text{ unit} = \frac{0.50 \text{ mm}}{20} = 0.025 \text{ mm} = 25 \text{ μm} \)

(b) \( \text{cell length} = 8 \times 25 = 200 \text{ μm} \)

(c) The eyepiece graticule scale is a fixed physical scale that does not itself change, but the total magnification of the microscope changes when a different objective lens is used. This means the amount of actual (real-world) distance represented by each graticule division is different at ×40 magnification than it was at ×10 magnification (each graticule unit now corresponds to a smaller real distance, since the image is magnified more). The student must recalibrate the graticule against the stage micrometer at the new (×40) magnification to find the correct conversion factor before taking further measurements.

(a) 2 marks: correct method (division) [1], answer 25 μm [1]. (b) 2 marks: correct method [1], answer 200 μm (ecf) [1]. (c) 2 marks: correctly identifies the graticule scale is fixed but total magnification has changed [1], correctly explains this changes the real distance per graticule unit, requiring recalibration [1].

(a) Water potential is the algebraic sum of the solute and pressure potentials:
\( \psi_{cell} = \psi_s + \psi_p = (-900) + (+350) = -550 \text{ kPa} \)

(b) The solution has a water potential of −400 kPa, which is higher (less negative) than the cell's water potential of −550 kPa. Water moves by osmosis from a region of higher (less negative) water potential to a region of lower (more negative) water potential, across the partially permeable cell membrane. Therefore, net water movement is from the solution into the cell.

(c) As water enters the cell, the cell's pressure potential increases (becomes more positive), as the increasing volume of cell contents presses outward against the relatively rigid, inextensible cellulose cell wall, which in turn exerts an inward (wall) pressure back on the cell contents. This continues until the cell's water potential rises to equal that of the external solution (−400 kPa), at which point net water movement stops (equilibrium is reached).

(a) 2 marks: correct formula ψ=ψs+ψp [1], answer −550 kPa [1]. (b) 2 marks: correct direction (solution into cell) [1], correct explanation referencing movement from higher to lower water potential [1]. (c) 2 marks: correctly states pressure potential increases [1], correct explanation referencing the cell wall resisting expansion/exerting back-pressure [1].

(a) As substrate concentration increases, the rate of reaction also increases, but the size of the increase becomes progressively smaller at higher substrate concentrations; the rate begins to level off (plateau) at the higher substrate concentrations shown (e.g. only a small increase from [S]=8 to [S]=16).

(b) At high substrate concentrations, the (fixed) number of enzyme molecules present, and therefore the fixed number of active sites, becomes the limiting factor. Essentially all of the available active sites are occupied (saturated) with substrate molecules at any given moment, so adding more substrate cannot increase the rate any further, since there are no free active sites available for the additional substrate molecules to bind to — the rate becomes limited by how quickly each occupied active site can complete a catalytic cycle and become free again, not by substrate availability.

(c) The rate of reaction at [S]=16 mmol dm⁻³ would be expected to increase (roughly double), because doubling the enzyme concentration doubles the number of available active sites; since substrate remains in excess at this concentration, more of these newly available active sites can now bind substrate and catalyse the reaction simultaneously, increasing the overall rate.

(a) 2 marks: correctly describes rate increasing [1], correctly describes rate of increase slowing/levelling off at high [S] [1]. (b) 2 marks: correctly identifies active sites becoming saturated/all occupied [1], correctly links this to enzyme (not substrate) becoming the limiting factor [1]. (c) 2 marks: correctly predicts rate increases (roughly doubles) [1], correct reasoning referencing more available active sites with excess substrate [1].

(a) \( \text{mitotic index} = \frac{\text{number of cells in mitosis}}{\text{total number of cells counted}} \times 100 = \frac{20}{250}\times100 = 8.0\% \)

(b) Prophase is typically the longest of the four stages of mitosis in terms of the time taken to complete it. Since the cells in the sample were all 'frozen' at a single moment in time (when the tissue was fixed and stained), the proportion of cells observed in each stage roughly reflects the proportion of the total time spent in that stage. Because prophase takes the longest, more cells are likely to be 'caught' in this stage than in the shorter stages (metaphase, anaphase, telophase) at any given snapshot in time.

(c) Root tips contain a region of meristematic tissue in which cells are actively and continuously dividing (undergoing repeated mitosis) to drive growth of the root. Mature, differentiated tissue (such as a fully grown leaf) contains mostly cells that have stopped dividing and are carrying out their specialised functions, so very few (if any) cells would be observed undergoing mitosis in such tissue, making root tips a much better choice for reliably observing the stages of mitosis.

(a) 2 marks: correct formula [1], answer 8.0% [1]. (b) 2 marks: correctly identifies prophase as the longest stage [1], correctly links this to a greater proportion of cells being observed in that stage at a given moment [1]. (c) 2 marks: correctly identifies root tips as meristematic/actively dividing tissue [1], correct contrast with mature/differentiated tissue where cells are not dividing [1].

(a) \( \text{total surface area} = 0.65 \times 8 \times 20 = 104 \text{ m}^2 \)

(b) The very large surface area (104 m² in this case) means that a large amount of nutrient molecules can be in contact with, and diffuse across, the epithelium at any given moment, increasing the overall rate of absorption (by Fick's law, rate of diffusion is proportional to surface area). At the same time, the rich blood supply provided by the capillary network in the submucosa continuously carries absorbed nutrients away from the site of absorption into general circulation; this prevents nutrient concentration from building up in the blood immediately beneath the epithelium, keeping the concentration of nutrients in the blood much lower than in the gut lumen. Together, the large surface area and the constant removal of absorbed nutrients by the blood supply maintain a steep concentration (diffusion) gradient across the epithelium, sustaining a high rate of absorption.

(a) 2 marks: correct method (multiplying all three factors) [1], answer 104 m² [1]. (b) 4 marks: correctly explains large surface area increases rate of diffusion/absorption (reference to Fick's law or equivalent) [2], correctly explains rich blood supply removes absorbed nutrients, maintaining a low blood concentration [1], correctly links both factors to maintaining a steep diffusion gradient [1].

Structure of the plasma membrane:
The plasma membrane is described by the fluid mosaic model. Its basic structure is a phospholipid bilayer, in which the hydrophilic phosphate 'heads' of the phospholipid molecules face outward (towards the aqueous cytoplasm and extracellular fluid), while the hydrophobic fatty acid 'tails' face inward, away from water, forming the core of the membrane. The membrane is described as 'fluid' because the phospholipid molecules, and other components, are able to move laterally within the layer. Embedded within this phospholipid bilayer are proteins: intrinsic (integral) proteins span part or all of the way through the membrane, while extrinsic (peripheral) proteins are attached to only one surface. Many of these proteins, along with some lipids, have carbohydrate chains attached to their outer surface, forming glycoproteins and glycolipids; these often act as cell-surface receptors or recognition markers (antigens). Cholesterol molecules are also interspersed within the phospholipid bilayer (particularly in animal cell membranes), where they help regulate membrane fluidity and contribute to membrane stability.

Movement of substances across the membrane:
Diffusion is the net passive movement of molecules from a region of higher concentration to a region of lower concentration, down a concentration gradient, requiring no input of metabolic energy. Small, non-polar or lipid-soluble molecules (such as oxygen and carbon dioxide) can diffuse directly through the phospholipid bilayer itself, moving between the phospholipid molecules.

Facilitated diffusion is also a passive process, moving substances down their concentration gradient without requiring energy, but it applies to larger or polar/charged molecules (such as glucose or ions) that cannot readily pass through the hydrophobic core of the phospholipid bilayer unaided. Instead, these substances move through specific channel proteins (forming hydrophilic pores through the membrane) or are carried across by specific carrier proteins, which bind the substance and undergo a conformational change to transport it across.

Osmosis is a special case of diffusion: the net movement of water molecules from a region of higher (less negative) water potential to a region of lower (more negative) water potential, across a partially permeable membrane. Water molecules are small enough to move directly through the phospholipid bilayer, and their movement can also be facilitated by specific channel proteins (aquaporins) that allow rapid passage of water molecules.

Active transport, in contrast to the three processes above, moves substances against their concentration gradient — from a region of lower concentration to a region of higher concentration. Because this movement is against the natural direction of diffusion, it requires an input of metabolic energy, in the form of ATP. Active transport uses specific carrier proteins in the membrane; the carrier protein binds the substance to be transported on one side of the membrane, and ATP is hydrolysed to provide the energy needed to change the shape of the carrier protein, moving the bound substance across the membrane and releasing it on the other side.

Indicative content — membrane structure (up to 8 points): phospholipid bilayer; hydrophilic heads face outward/hydrophobic tails face inward; fluid nature (lateral movement of components); intrinsic (integral) proteins; extrinsic (peripheral) proteins; glycoproteins/glycolipids as receptors/antigens; cholesterol present (animal cells) and its role in stability/fluidity; overall reference to the fluid mosaic model. Indicative content — transport mechanisms (up to 8 points): diffusion defined (net movement, high to low concentration, no energy); diffusion occurs directly through the bilayer for small/lipid-soluble molecules; facilitated diffusion defined and correctly distinguished from simple diffusion (uses channel or carrier proteins, still passive/down gradient); osmosis correctly defined in terms of water potential across a partially permeable membrane; role of aquaporins in osmosis; active transport defined (against concentration gradient); active transport requires ATP/energy; active transport uses carrier proteins that change shape. Level 1 (1-5 marks): basic, list-like points made on structure and/or transport, limited accuracy or detail; QWC basic. Level 2 (6-11 marks): a reasonable range of correct points made on both membrane structure and transport mechanisms, with some explanation, though not fully comprehensive or with some imprecision; QWC generally clear, competent use of terminology. Level 3 (12-16 marks): a wide-ranging, accurate and well-explained account covering membrane structure in detail and all four transport mechanisms with correct, precise distinctions between them (e.g. passive vs active, simple vs facilitated diffusion), written in clear, well-organised continuous prose with accurate, extensive use of specialist vocabulary throughout. Responses that are list-like, or that omit either the structure section or the transport section entirely, cannot access Level 3.

(a) The volume of water taken up corresponds to the volume of the cylinder of air/tube through which the bubble has moved:
\( \text{radius} = \frac{1.0}{2} = 0.5 \text{ mm} \)
\( V = \pi r^2 l = \pi \times (0.5)^2 \times 45 = 35.3 \text{ mm}^3 \)

(b) \( \text{rate} = \frac{35.3}{5} = 7.07 \text{ mm}^3\text{ min}^{-1} \)

(c) Light intensity: increasing light intensity generally increases the rate of water uptake, because stomata open wider (or more stomata open) in response to greater light intensity, allowing more water vapour to diffuse out and be lost through transpiration, drawing more water up through the xylem. Air movement (wind speed): increasing air movement around the leaf increases the rate of water uptake, because moving air sweeps away the layer of water vapour that would otherwise build up just outside the stomata, maintaining a steeper diffusion gradient for water vapour to diffuse out, increasing transpiration. (Other valid factors: temperature — increasing temperature increases the rate, as evaporation and diffusion of water vapour occur faster; humidity — increasing humidity decreases the rate, as it reduces the diffusion gradient for water vapour loss.)

(d) It is assumed that essentially all of the water taken up by the cut shoot is lost by transpiration from the leaves, with only a negligible amount being retained within the plant for photosynthesis, growth, or maintaining cell turgor (i.e. that water uptake is a valid proxy/estimate for transpiration rate).

(a) 3 marks: correct formula for cylinder volume [1], correct substitution [1], answer 35.3 mm³ [1]. (b) 2 marks: correct method (division by time) [1], answer 7.07 mm³ min⁻¹ [1]. (c) 3 marks: two different, valid factors named [1], correct predicted effect for factor 1 with reasoning [1], correct predicted effect for factor 2 with reasoning [1]. (d) 2 marks: correctly identifies the assumption that water uptake ≈ water lost by transpiration (negligible use elsewhere in the plant).

(a) \( \text{total cycle} = 0.1+0.3+0.4 = 0.8 \text{ s} \)

(b) \( \text{heart rate} = \frac{60}{0.8} = 75 \text{ beats per minute} \)

(c) The wave of electrical excitation that triggers each heartbeat originates in the sino-atrial (SA) node in the wall of the right atrium and spreads across both atria first, causing them to contract (atrial systole) and push the remaining blood into the ventricles. The wave then reaches the atrioventricular (AV) node, where it is briefly delayed, before being conducted rapidly down the bundle of His and Purkinje fibres to stimulate contraction of the ventricles (ventricular systole). This delay and sequence ensures the ventricles are as full as possible (having received the extra blood pushed in by atrial contraction) before they contract to pump blood out of the heart.

(d) During ventricular systole, pressure in the ventricles rises sharply and becomes higher than the pressure in the atria; this pressure difference forces the AV valves (tricuspid on the right, bicuspid/mitral on the left) shut, preventing blood from flowing backward from the ventricles into the atria, so that blood is instead forced forward, out through the semilunar valves into the arteries. The chordae tendinae are tough, inelastic tendinous cords that connect the cusps (flaps) of the AV valves to the papillary muscles in the ventricle wall. As the papillary muscles contract (along with the rest of the ventricle wall) during ventricular systole, they keep the chordae tendinae taut; this prevents the high ventricular pressure from pushing the valve cusps too far and turning them inside-out (prolapsing) back into the atria, ensuring the valve remains closed and holds firm against the pressure.

(e) The SA node is located in the wall of the right atrium. It acts as the heart's natural pacemaker, spontaneously generating a wave of electrical excitation at a regular rate, which initiates each heartbeat by triggering the atria to contract.

(a) 1 mark: 0.8 s. (b) 2 marks: correct formula [1], answer 75 bpm (ecf) [1]. (c) 2 marks: correct reference to sequential spread of excitation (SA node → atria → AV node/delay → ventricles) [1], correct link to maximising ventricular filling before ventricular contraction [1]. (d) 3 marks: correctly explains AV valves prevent backflow into atria [1], correctly explains the role of chordae tendinae/papillary muscles in preventing valve inversion [2]. (e) 2 marks: correct location (wall of right atrium) [1], correct role as initiator of the heartbeat/pacemaker [1].

(a) \( \text{volume} = 24 \text{ mm} \times 0.02 \text{ cm}^3\text{mm}^{-1} = 0.48 \text{ cm}^3 \)

(b) \( \text{rate} = \frac{0.48}{10} = 0.048 \text{ cm}^3\text{min}^{-1} \)

(c) The potassium hydroxide solution absorbs the carbon dioxide produced by the woodlice's respiration as it is released. Without this, the CO2 released would replace much of the volume of O2 removed from the gas space (since respiration consumes O2 and produces CO2 in roughly similar volumes), meaning very little overall volume/pressure change would be recorded. By absorbing the CO2, the only volume change remaining is due to the O2 being consumed and not replaced, allowing O2 consumption specifically to be measured.

(d) The experiment should be repeated using the same woodlice and apparatus, but with the potassium hydroxide replaced by an equal volume of water (or another substance that does not absorb CO2). This second experiment measures the net volume change in the respirometer, which reflects the difference between the volume of O2 consumed and the volume of CO2 produced. Since the volume of O2 consumed has already been found (from the first experiment, with KOH present), the volume of CO2 produced can be calculated by comparing it with this net volume change (CO2 produced = O2 consumed − net volume change observed without KOH, taking the direction of fluid movement into account).

(a) 3 marks: correct method (multiply movement by conversion factor) [1], correct substitution [1], answer 0.48 cm³ [1]. (b) 2 marks: correct method (division by time) [1], answer 0.048 cm³ min⁻¹ [1]. (c) 2 marks: correctly explains KOH absorbs CO2 produced [1], correctly explains this isolates O2 consumption as the measured volume change [1]. (d) 3 marks: correctly describes repeating without KOH (e.g. with water) [1], correctly identifies this measures the net volume change (O2 consumed minus CO2 produced) [1], correctly describes comparing the two results to find CO2 production [1].

(a) \( \Sigma n(n-1) = 45(44)+30(29)+15(14)+10(9) = 1980+870+210+90 = 3150 \)
\( N(N-1) = 100(99) = 9900 \)
\( D = \frac{9900}{3150} = 3.14 \)

(b) A high value of D indicates greater species diversity (using this form of Simpson's Index); a low value of D indicates lower diversity.

(c) Simpson's Index takes into account both species richness (the number of different species present) and species evenness (how evenly the individuals are distributed among those species), giving a more complete and meaningful measure of diversity. Simply counting the number of species ignores relative abundance — for example, a habitat with 4 species where one species makes up 97% of all individuals would be considered less diverse in practice than a habitat with 4 roughly equally abundant species, but a simple species count would treat both habitats as equally 'diverse' (both having 4 species).

(d) A limited number of quadrats may not adequately represent the whole habitat, especially if the species being sampled are patchily/unevenly distributed, and if quadrat placement is not truly random, this could introduce sampling bias (for example, if the student unconsciously avoids placing quadrats in awkward or inaccessible areas). This source of error could be reduced by using a larger number of quadrats, and by using a formal random method to select quadrat positions (for example, generating random coordinates using random numbers along two measuring tapes set up at right angles across the habitat, rather than placing quadrats by eye).

(a) 5 marks: correct calculation of Σn(n-1) with correct working shown [2], correct calculation of N(N-1) [1], correct final answer D=3.14 [2]. (b) 1 mark: correctly states 'high' value indicates greater diversity. (c) 2 marks: correctly identifies that Simpson's Index accounts for relative abundance/evenness as well as richness [1], with a valid illustrative point or clear reasoning [1]. (d) 2 marks: valid source of error identified (e.g. non-random placement, insufficient number of quadrats, patchy distribution) [1] with a correct, relevant method of reduction [1].

(a) Rainfall washes (leaches) some of the applied inorganic fertiliser, containing nitrate and phosphate ions, off the field and into the nearby stream as run-off. These mineral ions are normally present in only limited amounts in natural waterways and act as limiting nutrients for the growth of algae and other aquatic plants; the sudden abundant supply of these nutrients removes this limitation, allowing algae (and other aquatic plants) to grow and reproduce extremely rapidly, producing a dense algal bloom, often covering the surface of the water.

(b) The dense algal bloom at the water surface blocks light from penetrating to submerged plants growing lower down in the water, preventing them from photosynthesising; these light-deprived plants, and some of the algae themselves as the bloom ages, die. This creates a large amount of dead organic matter in the water. Saprophytic (decomposer) bacteria and fungi feed on this dead organic matter, respiring aerobically as they break it down, and in doing so consume large quantities of dissolved oxygen from the water — this increases the biological oxygen demand (BOD), which is a measure of the amount of oxygen used by microorganisms decomposing organic matter in water. As dissolved oxygen levels in the water fall (due to this high BOD), there is insufficient oxygen available to support the aerobic respiration needs of fish and other aquatic animals; as a result, fish and other oxygen-demanding organisms suffocate and die, or move away from the affected area if possible, causing the observed decline in the fish population.

(c) Applying fertiliser only at the recommended rate needed by the crop (avoiding excess application), and/or avoiding fertiliser application immediately before periods of heavy rainfall (when run-off risk is highest), and/or maintaining buffer strips of vegetation (e.g. grass margins) alongside waterways to intercept and absorb nutrients in run-off before they reach the water, would all help reduce this risk.

(a) 3 marks: correctly identifies fertiliser (nitrate/phosphate) entering the stream via run-off [1], correctly identifies these as previously limiting nutrients [1], correctly links to rapid algal growth/bloom [1]. (b) 4 marks: correctly explains algal bloom blocks light, causing plant/algal death [1], correctly explains decomposer/saprophytic microorganisms respiring aerobically on dead matter, consuming oxygen (raising BOD) [2], correctly explains resulting low dissolved oxygen causes fish death/decline [1]. (c) 3 marks: a valid, specific agricultural practice named and correctly explained as reducing fertiliser run-off (e.g. applying at recommended rate; avoiding application before rain; buffer strips along waterways).

(a) Plants structurally adapted to survive in habitats with a limited water supply are called xerophytes.

(b) Any three of the following, each with a correct explanation:
• Rolled or curled leaves (or leaves with hairs, spines, or sunken stomata in pits/grooves) trap a layer of humid air close to the stomata, reducing the concentration (diffusion) gradient for water vapour between the leaf's air spaces and the outside atmosphere, which slows the rate of diffusion of water vapour out of the leaf.
• Sunken stomata (positioned in pits below the leaf surface) similarly trap a pocket of still, humid air immediately above the stomatal pore, reducing the diffusion gradient and hence the rate of transpirational water loss.
• A reduced leaf surface area (for example, needle-shaped leaves, as in many conifers) reduces the total area available for water vapour to diffuse out of the leaf, directly reducing water loss.
• A thick, waxy cuticle on the leaf surface provides an additional waterproof barrier, reducing water loss through the cuticle itself (the minor transpiration route).
• Succulent tissue capable of storing water within the plant allows the plant to survive periods without external water supply by drawing on these internal reserves.

(c) Water evaporates from the surface of mesophyll cells within the leaf's air spaces and diffuses out through open stomata (transpiration); as water is lost from these mesophyll cells, they draw replacement water, by osmosis, from the xylem vessels within the leaf veins, creating a region of lower pressure (a 'pulling' or tension force) at the top of the water column within the xylem. Because water molecules are polar, they form hydrogen bonds with one another; this creates strong cohesive forces between water molecules, holding them together to form a continuous, unbroken column of water extending from the roots, up through the xylem vessels, all the way to the leaves. As a result of this cohesion, the tension created at the top of the column (in the leaf) is transmitted all the way down the water column, pulling more water up through the xylem from the roots, to replace the water lost by transpiration.

(a) 1 mark: xerophytes. (b) 6 marks: 2 marks for each of three distinct, correctly explained xerophytic adaptations (1 mark for correctly naming the adaptation, 1 mark for a correct explanation of how it reduces water loss). (c) 3 marks: correctly links transpiration (water loss from mesophyll cells/stomata) to tension/negative pressure in the leaf xylem [1], correctly explains cohesion between water molecules (via hydrogen bonding) maintaining a continuous water column [1], correctly links this to water being pulled up the xylem from the roots to replace water lost [1].

(a) Arteries and veins are both blood vessels with a wall structure consisting of squamous endothelium lining the lumen, surrounded by layers of elastic fibres, smooth muscle and fibrous (collagen) tissue, though the relative proportions of these tissues differ considerably between the two vessel types, reflecting their different functions.

Arteries carry blood away from the heart at high pressure, and this pressure fluctuates with each heartbeat (pulsatile flow). Artery walls are therefore relatively thick and contain a high proportion of elastic fibres; these allow the artery wall to stretch (distend) as a surge of blood is pumped in during ventricular systole, and then to recoil elastically during diastole, which helps smooth out the pulsatile flow into a steadier flow and helps maintain blood pressure between heartbeats. Arteries also contain a substantial layer of smooth muscle, which can contract (vasoconstriction) or relax (vasodilation) to alter the diameter of the vessel, helping to regulate the volume of blood supplied to different organs according to demand. The narrow lumen of an artery helps maintain the high pressure of the blood within it.

Veins carry blood back towards the heart at much lower and steadier pressure. Vein walls are thinner overall than artery walls, with relatively less elastic tissue and smooth muscle (since high pressure resistance and fine control of vessel diameter are less critical), but they do contain a substantial amount of fibrous (collagen) tissue, which provides support and helps prevent the vein from being damaged or collapsing, especially as many veins are compressed by surrounding skeletal muscles. Veins have a much larger lumen than arteries of a comparable size, which reduces resistance to flow and helps compensate for the low pressure driving the blood. Veins also contain valves at intervals along their length; since the pressure of blood in the veins is low and could otherwise flow backward (particularly against gravity, for example in the legs), these valves ensure blood flow continues in one direction only, back towards the heart, particularly assisted by the squeezing action of surrounding skeletal muscles during movement.

(b) Blood clotting is triggered when a blood vessel is damaged and platelets come into contact with the exposed tissue (or collagen) at the site of injury. This contact, together with damaged tissue cells, causes platelets to break down and release a substance called thromboplastin (thrombokinase) at the wound site. Thromboplastin, in the presence of calcium ions and various other clotting factors (including vitamin K-dependent factors), catalyses the conversion of the inactive plasma protein prothrombin into its active form, thrombin. Thrombin then acts as an enzyme that catalyses the conversion of fibrinogen — a soluble plasma protein that is normally present in the blood plasma — into fibrin, which is insoluble. The fibrin molecules produced form a mesh of fine threads across the wound site; this fibrin mesh traps red blood cells and platelets flowing past, forming a blood clot that seals the wound and prevents further blood loss, and eventually helps form a scab as it dries.

(a) Level-based (9 marks): Level 1 (1-3 marks): basic, largely undeveloped description of structure with limited/no functional link. Level 2 (4-6 marks): reasonable description of structural differences between arteries and veins (e.g. wall thickness, elastic tissue, muscle, lumen size, valves), with some correct functional explanation. Level 3 (7-9 marks): detailed, accurate description of the histological structure of both arteries and veins, with a clear, correctly reasoned explanation of how each structural feature relates to function (elastic recoil/pulsatile flow, vasoconstriction/dilation, valve function, lumen size), in clear, well-organised prose. (b) 6 marks (indicative content, marking points): platelets/damaged tissue trigger the release of thromboplastin at the wound site [1]; thromboplastin (with calcium ions/other factors) converts prothrombin to thrombin [1]; correct identification that prothrombin/thrombin are involved in this specific order [1]; thrombin converts fibrinogen to fibrin [1]; correct reference to fibrinogen being soluble and fibrin insoluble [1]; fibrin forms a mesh trapping blood cells to form the clot [1].

(a) \( \%\text{ change} = \frac{1.95-2.00}{2.00}\times100 = -2.5\% \)

(b) Percentage change in mass is plotted against the solute potential of the surrounding solution for all five results, and a best-fit line (or smooth curve) is drawn through the points. The point at which this best-fit line crosses the x-axis (i.e. where percentage change in mass = 0%) is then read off; the value of solute potential at this point gives an estimate of the average water potential of the potato tissue.

(c) At the point where percentage change in mass is zero, there has been no net movement of water into or out of the potato cells, meaning the water potential of the external solution at that point is exactly equal to the average water potential of the potato cells (assuming the pressure potential of the external solution is zero, so its water potential equals its solute potential). This is significant because it allows the (otherwise difficult to measure directly) water potential of the plant tissue to be estimated experimentally, without needing to know the individual solute and pressure potentials of the cells.

(a) 3 marks: correct formula [1], correct substitution [1], answer −2.5% [1]. (b) 2 marks: correctly describes plotting the graph and drawing a best-fit line [1], correctly describes reading the x-intercept (where y=0) [1]. (c) 2 marks: correctly identifies no net water movement / solution ψ = cell ψ at this point [1], correctly explains the significance (this value estimates the water potential of the tissue) [1].

(a) Incipient plasmolysis is the point at which the protoplast (cell surface membrane and cytoplasm) of a plant cell just begins to separate from the cell wall — the earliest, just-visible stage of plasmolysis.

(b) 50% of cells are recorded as plasmolysed at a solute potential of −600 kPa; this is taken as the point at which, on average across the population of cells, incipient plasmolysis is occurring (with some cells slightly plasmolysed and others not quite, averaging out to the point of incipient plasmolysis).

(c) At the point where 50% of cells are plasmolysed, the population of cells is, on average, exactly at incipient plasmolysis — the very beginning of plasmolysis, where the protoplast has just started to pull away from the cell wall but the cell wall is exerting essentially no inward pressure on the protoplast. Since pressure potential arises from the cell wall pushing back against the cell contents, at this point the average pressure potential of the cells is taken to be zero. Using the water potential equation, \( \psi_{cell} = \psi_s + \psi_p \), when \(\psi_p = 0\), \( \psi_{cell} = \psi_s \) (the cell's water potential equals its solute potential). Furthermore, since the cells at this point are in equilibrium with the external solution (no further net water movement, as the cells are just at the boundary of losing water), the water potential of the cells must equal the water potential of the external solution at that point; therefore, the solute potential of the cells can be taken as equal to the solute potential of the external solution at 50% plasmolysis (−600 kPa).

(a) 2 marks: correct definition referencing the protoplast just beginning to separate from the cell wall. (b) 2 marks: correctly reads/identifies −600 kPa from the data (at 50% plasmolysis) [2]. (c) 3 marks: correctly explains why pressure potential is zero at incipient plasmolysis [1], correctly applies ψ=ψs+ψp to show ψcell=ψs at this point [1], correctly links equilibrium with the external solution to allow the cell's solute potential to be estimated as equal to the solution's solute potential [1].

(a) Each dilution step halves the concentration (equal volumes of solution and water are mixed):
\( \text{concentration after 1 step} = \frac{1.0}{2} = 0.50 \text{ mol dm}^{-3} \)

(b) \( \text{concentration after 4 steps} = \frac{1.0}{2^4} = \frac{1.0}{16} = 0.0625 \text{ mol dm}^{-3} \)

(c) Serial dilution is quicker and requires far less glucose overall than preparing each concentration separately, since only a single, accurately prepared stock solution is needed, from which all other concentrations are derived; this also reduces the cumulative error that could otherwise arise from separately weighing out several different (often very small) masses of glucose, and easily produces an evenly-spaced (logarithmic) range of concentrations.

(a) 2 marks: correct method (halving) [1], answer 0.50 mol dm⁻³ [1]. (b) 3 marks: correct method (repeated halving/division by 2⁴) [2], answer 0.0625 mol dm⁻³ [1]. (c) 2 marks: valid advantage stated (e.g. quicker, less glucose needed, reduced cumulative weighing error, only one solution needs accurate preparation) with correct justification.

(a) Converting to consistent units: \( 90 \text{ mm} = 90\,000 \text{ μm} \)
\( \text{magnification} = \frac{\text{drawing size}}{\text{actual size}} = \frac{90\,000}{45} = 2000 \), i.e. ×2000

(b) Without a stated magnification (or an accompanying scale bar), it is impossible for anyone examining the drawing to determine the true, actual size of the structure being shown, since a drawing carries no inherent, obvious sense of scale. This information is essential for correctly interpreting biological structures, comparing them with other specimens, and communicating scientific findings accurately and reproducibly.

(c) A carefully made line drawing allows the person drawing it to selectively include only the structures they can clearly identify and interpret, and to omit background debris, staining artefacts, or out-of-focus regions that would be visible (and potentially confusing or misleading) in a photograph. Clear, unambiguous outlines and labels can be added directly to a drawing, making key structures and boundaries easier to identify than they might be in an unlabelled photographic image.

(a) 3 marks: correct unit conversion [1], correct formula (drawing size/actual size) [1], answer ×2000 [1]. (b) 2 marks: correctly explains that true size cannot be determined without magnification/scale bar [1], correctly links this to the importance for scientific interpretation/comparison [1]. (c) 2 marks: valid, well-explained reason (e.g. selective inclusion of clearly interpreted structures; omission of artefacts/debris; clarity of labelling).

(a) A transect (a line transect, or a belt transect for more detailed percentage cover data) is the most appropriate technique. This is because the investigation is specifically concerned with how percentage cover changes in a systematic way with distance from the tree — that is, along a defined gradient — rather than simply assessing the overall abundance of moss across the whole area. A transect, laid out running away from the tree, allows the student to sample systematically at set distances along this gradient, directly relating percentage cover to distance. Random sampling would scatter sampling points across the area without regard to distance from the tree, making it much harder to detect and describe any pattern related specifically to that gradient.

(b) At each sampling point along the transect, a quadrat (typically divided into a grid of, for example, 100 smaller squares) is placed on the ground. The student estimates the percentage cover of the moss species within the quadrat by counting how many of the small grid squares contain the moss (or are more than half covered by it), and expressing this as a percentage of the total number of squares in the quadrat.

(c) Light intensity would be a useful abiotic factor to measure at each point, since light intensity is likely to vary systematically with distance from the tree (for example, being more shaded closer to the trunk, beneath a denser canopy), and moss growth is often influenced by light availability. Measuring light intensity alongside percentage cover at each point would help the student determine whether any pattern observed in moss cover is actually related to changing light levels (rather than, or in addition to, distance from the tree as such), giving a more meaningful interpretation of the results. (Other valid answers, such as soil moisture or humidity, with appropriate justification, are also acceptable.)

(a) 3 marks: correctly identifies transect as the appropriate technique [1], correct justification referencing the systematic gradient being investigated [2]. (b) 2 marks: correctly describes placing a (gridded) quadrat at intervals along the transect [1], correctly describes estimating percentage cover using the grid squares [1]. (c) 2 marks: valid abiotic factor named (e.g. light intensity, soil moisture, humidity) [1] with a correct, relevant explanation of its usefulness [1].

(a) \( R_f = \frac{\text{distance moved by spot}}{\text{distance moved by solvent front}} = \frac{8.4}{14.0} = 0.60 \)

(b) \( R_f = \frac{10.2}{14.0} = 0.73 \)

(c) Spot B is most likely to be leucine, because its calculated Rf value (0.73) exactly matches the known Rf value of leucine (0.73) under the same conditions, whereas spot A's Rf value (0.60) does not match.

(d) The chromatography should be carried out in a sealed or covered container (tank), to keep the atmosphere inside saturated with solvent vapour; this prevents the solvent from evaporating unevenly as it moves up the paper, ensuring the solvent front moves at a consistent rate and giving accurate, reproducible Rf values. (Other valid precautions include: applying only a small, concentrated spot of sample at the origin; ensuring the solvent level in the tank is below the level of the spotted origin; using pencil, not pen, to mark the origin.)

(a) 2 marks: correct formula [1], answer 0.60 [1]. (b) 2 marks: correct formula [1], answer 0.73 [1]. (c) 2 marks: correctly identifies spot B [1], correct reasoning (Rf values match exactly) [1]. (d) 1 mark: valid precaution stated (e.g. sealed/covered tank; origin above solvent level; small/concentrated spot; pencil for origin).

(a) The absorbance decreases steadily and continuously from 0.80 at \(t=0\) to 0.18 at \(t=8\) minutes, but then remains unchanged (constant at 0.18) between \(t=8\) and \(t=10\) minutes. This levelling off (plateau) suggests that the reaction has finished by \(t=8\) minutes — that is, essentially all of the starch present has been broken down (hydrolysed) by the amylase, so there is no further starch available to produce a colour change with iodine, and absorbance stops decreasing.

(b) Starch forms a characteristic blue-black colour with iodine solution; a solution with a high concentration of starch therefore has a high absorbance (of the appropriate wavelength of light). As the amylase enzyme progressively breaks down (hydrolyses) the starch present into smaller sugar molecules (such as maltose), the concentration of starch remaining decreases, so the intensity of the blue-black colour formed with iodine decreases, and therefore the absorbance recorded by the colorimeter also decreases over the course of the reaction.

(c) Before taking any readings, the colorimeter should be calibrated (blanked/zeroed) using a reference ('blank') sample — for example, a cuvette containing only the reagent(s) used, such as iodine solution mixed with water instead of the starch/amylase mixture, or distilled water — and this reading should be set as zero absorbance (or 100% transmission). This ensures that all subsequent absorbance readings taken during the experiment reflect only the absorbance due to the substance being investigated (in this case, the remaining starch-iodine colour), rather than including any background absorbance from the reagents, cuvette, or solvent themselves. An appropriate colour filter (matched to the colour of the solution being measured) should also be selected.

(a) 3 marks: correctly describes the steady decrease in absorbance [1], correctly describes the plateau after t=8 min [1], correctly concludes the reaction is complete (no starch remaining) [1]. (b) 2 marks: correctly explains starch-iodine forms a coloured complex with high absorbance [1], correctly links decreasing starch concentration (via hydrolysis) to decreasing absorbance [1]. (c) 2 marks: correctly describes calibrating/blanking the colorimeter with a reagent blank before use [1], correct additional detail (e.g. setting this as zero absorbance, or selecting an appropriate filter) [1].