2023 CCEA A-Level Biology 1010 模擬試題連答案詳解

Glucose and amino acids are actively reabsorbed by co-transport with sodium ions across the epithelium of the region of the nephron with numerous microvilli and mitochondria: the proximal convoluted tubule.

1 mark: proximal convoluted tubule (accept PCT).

P660 and P730 are the two interconvertible forms of the light-sensitive plant pigment phytochrome, which controls photoperiodic responses such as flowering.

1 mark: phytochrome.

A molecule capable of stimulating production of a specific, complementary antibody is termed an antigen.

1 mark: antigen.

The maximum population size an environment can sustain indefinitely is its carrying capacity.

1 mark: carrying capacity.

Primary succession begins in a previously unoccupied habitat, such as bare rock, where no soil or seed bank is present. Secondary succession occurs where an existing community has been destroyed (for example by fire or flooding) but soil and/or a seed bank remain, so colonisation and succession proceed considerably faster than in primary succession.

1 mark: primary succession begins in a previously unoccupied habitat (e.g. bare rock) with no soil/seed bank; 1 mark: secondary succession occurs where an existing community has been destroyed (e.g. by fire/flood) but soil and/or a seed bank remain, so it proceeds faster than primary succession; max 2.

The resting potential is the stable, polarised state of the axon membrane maintained when no impulse is being conducted, with the inside of the axon negative relative to the outside. The action potential is the rapid depolarisation (and reversal of membrane polarity) that occurs once a threshold stimulus has been reached, followed by repolarisation back to the resting state.

1 mark: resting potential = polarised state of the membrane maintained when no impulse is being conducted (inside negative relative to outside); 1 mark: action potential = rapid depolarisation/reversal of membrane potential when a threshold stimulus is reached, followed by repolarisation; max 2.

A pioneer community consists of the first colonising species able to tolerate the harsh abiotic conditions of an unoccupied habitat. A climax community is the final, stable stage of succession, in dynamic equilibrium with the prevailing environment, showing no further net change in species composition.

1 mark: pioneer community = first species to colonise a bare/unoccupied habitat, tolerant of extreme abiotic conditions; 1 mark: climax community = stable end-point of succession in equilibrium with the environment, showing no further net change; max 2.

r-selected species produce many small offspring with little or no parental care and reach reproductive maturity quickly, exploiting unstable or unpredictable environments. K-selected species produce few, larger offspring, invest heavily in parental care and mature slowly, typically existing near the carrying capacity of a stable environment.

1 mark: r-selected species = large numbers of small offspring, little/no parental care, rapid maturation, adapted to unstable/unpredictable environments; 1 mark: K-selected species = few, larger offspring, extensive parental care, slow maturation, population maintained near carrying capacity in stable environments; max 2.

The arrival of an action potential at the synaptic bulb causes synaptic vesicles containing acetylcholine to move towards, and fuse with, the pre-synaptic membrane. Acetylcholine is released into the synaptic cleft by exocytosis and diffuses across the cleft down its concentration gradient. It then binds to specific receptor sites on the post-synaptic membrane, causing a localised depolarisation and generating an excitatory post-synaptic potential (EPSP). If the EPSP reaches the threshold value, an action potential is generated in the post-synaptic membrane. Acetylcholinesterase then hydrolyses the acetylcholine, terminating the response.

1 mark: action potential arrival causes synaptic vesicles to move to/fuse with the pre-synaptic membrane; 1 mark: acetylcholine released into the synaptic cleft by exocytosis; 1 mark: acetylcholine diffuses across the cleft and binds to specific receptors on the post-synaptic membrane; 1 mark: binding causes depolarisation/generation of an excitatory post-synaptic potential (EPSP); 1 mark: if threshold reached an action potential is generated in the post-synaptic membrane / acetylcholinesterase hydrolyses acetylcholine to terminate the response; max 5.

A larger-diameter axon offers less internal resistance to the local (ionic) currents that flow ahead of an action potential, so depolarisation of the adjacent membrane occurs more rapidly, increasing conduction speed. In a myelinated axon, the myelin sheath acts as an electrical insulator, so ion exchange (and generation of an action potential) can occur only at the nodes of Ranvier; the local current therefore jumps from node to node (saltatory conduction), which is much faster than the continuous depolarisation of successive small sections of membrane required in an unmyelinated axon.

1 mark: larger diameter axon = less internal resistance to local current flow; 1 mark: this speeds depolarisation of adjacent membrane; 1 mark: myelin sheath insulates the axon so depolarisation/ion exchange can only occur at the nodes of Ranvier; 1 mark: current therefore jumps from node to node (saltatory conduction); 1 mark: this is faster than the continuous depolarisation of successive sections of membrane in an unmyelinated axon; max 5.

The epithelial cells of the proximal convoluted tubule are cuboidal with numerous microvilli on their luminal surface, greatly increasing the surface area available for reabsorption. The cells contain large numbers of mitochondria, providing the ATP required for active transport of glucose, amino acids and sodium ions from the filtrate into the surrounding capillaries. Basal infoldings of the cell membrane further increase the surface area for active transport of ions into the blood, and tight junctions between adjacent cells prevent leakage of reabsorbed solutes back into the filtrate.

1 mark: microvilli on the luminal surface increase surface area for reabsorption; 1 mark: numerous mitochondria provide ATP for active transport of glucose/amino acids/sodium ions; 1 mark: basal infoldings increase surface area for transport of ions into the surrounding capillaries; 1 mark: tight junctions between cells prevent leakage of reabsorbed solutes back into the filtrate; max 4.

As filtrate flows down the descending limb, which is permeable to water, water moves out by osmosis into the surrounding medulla tissue, concentrating the filtrate towards the apex of the loop. As filtrate flows up the ascending limb, which is impermeable to water, sodium and chloride ions move out into the surrounding medulla tissue. This creates and maintains a salt gradient in the medulla that increases in concentration towards the apex. This gradient allows water to be reabsorbed osmotically from the descending limb, the distal convoluted tubule and the collecting duct as urine passes back down through the medulla, producing urine that is more concentrated than the blood plasma.

1 mark: descending limb permeable to water, so water leaves filtrate by osmosis into the medulla, concentrating filtrate towards the apex; 1 mark: ascending limb impermeable to water, sodium/chloride ions leave into the surrounding medulla tissue; 1 mark: this creates/maintains a salt (solute) gradient in the medulla, increasing towards the apex; 1 mark: gradient allows osmotic reabsorption of water from the descending limb/distal convoluted tubule/collecting duct, concentrating the urine; max 4.

In darkness, P730 slowly converts back to P660, so the P730 level gradually falls throughout the night. A short-day plant flowers only when P730 has fallen below a threshold level, which requires an uninterrupted dark period longer than the critical night length. A flash of red light delivered during the night is absorbed by P660, rapidly converting it back to P730 and resetting the P730 level to a high concentration. Because P730 is not then removed within a single continuous dark period, the plant does not experience an effective night longer than the critical length, so flowering is not induced.

1 mark: P730 is converted (slowly) back to P660 during darkness/P730 level falls during the night; 1 mark: short-day plant flowers only when P730 falls below a threshold, requiring an uninterrupted dark period longer than the critical night length; 1 mark: flash of red light rapidly converts P660 back to P730 (red light absorbed by P660); 1 mark: this resets/restores P730 level, so the effective dark period is interrupted/not long enough to remove P730, preventing flowering; max 4.

Auxins promote growth by stimulating elongation of existing cells, acting mainly in the region just behind the growing tip. Gibberellins promote growth mainly by stimulating elongation of the internodal regions of the stem, increasing the distance between nodes and so increasing overall stem length.

1 mark: auxins promote elongation of (individual) cells, typically near the growing tip; 1 mark: gibberellins promote elongation of the internodal regions of the stem; 1 mark: both increase overall stem length but via different regions/mechanisms; max 3.

Active immunity results from an individual's own immune system being stimulated, naturally by infection or artificially by vaccination, to produce specific antibodies, T-cells and memory cells against a particular antigen. Passive immunity involves the direct transfer of ready-made antibodies from another source, for example placental or colostral transfer, into an individual, without stimulating that individual's own immune system. Because no memory cells are produced in passive immunity, and the transferred antibodies are gradually broken down, the protection is short-lived; active immunity is long-lasting because memory cells persist and can rapidly divide to produce more antibody if the same antigen is encountered again.

1 mark: active immunity = individual's own immune system produces antibodies/T-cells/memory cells (following infection or vaccination); 1 mark: passive immunity = ready-made antibodies transferred from another source (e.g. placenta/colostrum/antiserum) without stimulating the individual's own immune system; 1 mark: no memory cells are formed in passive immunity; 1 mark: transferred antibodies are broken down/not replenished, so protection is short-term, whereas memory cells in active immunity give long-term protection; max 4.

The transplanted kidney carries foreign antigens on its cell surfaces because the donor is not genetically identical to the recipient. These foreign antigens are recognised by the recipient's immune system, stimulating the production of specific T-cells and B-cells that attack and destroy the transplanted tissue. Immunosuppressant drugs inhibit DNA replication in dividing lymphocytes, inactivating B-cell and T-cell responses, so the foreign antigens on the transplanted kidney are not attacked and rejection is prevented.

1 mark: transplanted tissue carries foreign antigens (donor not genetically identical to recipient); 1 mark: these are recognised by the recipient's immune system, stimulating specific T-cells/B-cells that attack and destroy the tissue; 1 mark: immunosuppressant drugs inhibit DNA replication in lymphocytes / inactivate B- and T-cell responses so the foreign antigens are not attacked; max 3.

If donor red blood cells carry an antigen that is not present on the recipient's own red blood cells, the recipient's plasma will contain the complementary antibody. This antibody binds to the antigen on the donor red blood cells, causing agglutination (clumping) of the donor red blood cells. The resulting clumps of agglutinated red blood cells are large enough to block small blood vessels or capillaries, restricting the flow of blood and oxygen to tissues supplied by that vessel.

1 mark: donor red blood cells carry an antigen for which the recipient's plasma contains the complementary antibody; 1 mark: antibody binds to antigen causing agglutination (clumping) of donor red blood cells; 1 mark: clumps of agglutinated cells block small blood vessels/capillaries, restricting blood/oxygen flow to tissues; max 3.

During the lag phase, growth rate is low as individuals are becoming established, so the population increases only slowly. During the exponential (log) phase, resources are abundant and not limiting, so the population grows at its maximum rate (biotic potential). During the stationary phase, growth rate falls to approximately zero because increasing competition for the now-limited resources, and/or accumulation of toxic waste products, causes the birth rate to approximately equal the death rate, so the population size levels off at the carrying capacity.

1 mark: lag phase = growth rate low, population becoming established; 1 mark: exponential phase = resources not limiting, population grows at maximum rate/biotic potential; 1 mark: stationary phase = growth rate falls to ~zero as competition for limited resources and/or waste accumulation increases; 1 mark: stationary phase = birth rate approximately equals death rate, population levels off at carrying capacity; max 4.

r-selected species produce very large numbers of offspring, so even if the survival rate of each individual offspring is low, a large total number of individuals will survive to colonise the new habitat. r-selected species also reach reproductive maturity very quickly, allowing the population to increase in size rapidly once individuals arrive in the new habitat, before conditions change or space becomes limiting.

1 mark: produce large numbers of offspring, so despite low individual survival, many individuals can colonise the habitat; 1 mark: rapid attainment of reproductive maturity or short generation time, allowing rapid population increase; 1 mark for a valid second distinct feature (e.g. efficient/wide dispersal mechanism) with explanation; max 3.

Where two species have very similar ecological niches, they require the same limited resources, such as food, space or light. One species is likely to be marginally more efficient at exploiting these resources. This more efficient species will out-compete the less efficient species for the shared, limited resources, so the less efficient species will have a lower survival and/or reproductive rate. Over time, this leads to a decline in the population of the less efficient species until it is eliminated from that habitat, a process known as competitive exclusion.

1 mark: similar niches mean the two species compete for the same limited resources; 1 mark: one species is likely to be more efficient at exploiting the shared resource(s); 1 mark: the less efficient species has reduced survival/reproduction and is progressively eliminated from the habitat (competitive exclusion); max 3.

Immediately after the fire, bare ground with residual soil and often a surviving seed bank is exposed. Pioneer species, such as fast-growing herbaceous plants and grasses that tolerate the open, exposed conditions, colonise the area first. As these pioneers grow and die, they add organic matter to the soil, which is broken down by decomposers, improving soil fertility. These changed conditions allow less tolerant species, such as shrubs, to become established, out-competing many of the original pioneers. Over successive seral stages, larger, longer-lived species, eventually including trees, become dominant, each stage further modifying the environment and allowing the next community to establish, with an increase in species diversity at each stage. This continues until a climax community, in this case woodland, is reached, which is in stable equilibrium with the prevailing climate. Because soil and a seed bank were already present, the process is faster and involves fewer seral stages than a primary succession on bare rock.

1 mark: pioneer species colonise the burnt area rapidly, exploiting the presence of soil/a surviving seed bank; 1 mark: pioneers modify the abiotic environment (e.g. adding organic matter, improving soil fertility); 1 mark: modified conditions allow less tolerant species to become established, out-competing/replacing earlier colonisers; 1 mark: succession proceeds through further seral stages with increasing species diversity, each stage altering conditions for the next; 1 mark: process culminates in a climax community in equilibrium with the environment, and is faster than primary succession because soil/seed bank were already present; max 5.

In the early stages of succession, species diversity is low because only pioneer species tolerant of the open, nutrient-poor conditions of the abandoned field are able to establish. As succession proceeds, pioneers modify the abiotic environment, increasing organic matter and nutrient availability, allowing new species with different requirements to colonise, so diversity increases through the early-to-middle seral stages. In the later stages, as taller, more competitive species such as shrubs and trees become dominant, they may out-compete and eliminate some earlier, less competitive species, for example through shading, so diversity may fall slightly as the community approaches the climax stage. The climax community represents the final, stable seral stage; it is in dynamic equilibrium with the prevailing climate and local conditions, and shows no further net change in species composition over time unless the habitat is disturbed.

1 mark: diversity low at the pioneer stage as only tolerant species can colonise; 1 mark: pioneers modify the abiotic environment, allowing further species to establish; 1 mark: diversity increases through early/middle seral stages as more species colonise; 1 mark: diversity may fall in later stages as dominant, competitive species (e.g. trees) out-compete/eliminate earlier species; 1 mark: climax community = stable, final seral stage in equilibrium with the environment/climate, no further net change unless disturbed; max 5.

Volume reabsorbed \( = 180 - 1.5 = 178.5 \text{ dm}^3 \). Percentage reabsorbed \( = \frac{178.5}{180} \times 100 = 99.2\% \) (to 1 d.p.).

1 mark: correct volume reabsorbed \( (178.5 \text{ dm}^3) \) or correct set-up of the percentage calculation; 1 mark: correct substitution \( \left( \frac{178.5}{180} \times 100 \right) \); 1 mark: correct final answer, 99.2% (accept 99.1-99.2%; own figure rule applies); max 3.

\( N = \frac{n_1 \times n_2}{m_2} = \frac{84 \times 96}{24} = \frac{8064}{24} = 336 \) woodlice.

1 mark: correct identification of \( n_1 = 84 \), \( n_2 = 96 \), \( m_2 = 24 \) substituted into the Lincoln index formula; 1 mark: correct substitution, \( \frac{84 \times 96}{24} \); 1 mark: correct final answer, 336 (own figure rule applies); max 3.

Total \( N = 18+42+12+28 = 100 \). \( \Sigma \left( \frac{n}{N} \right)^2 = \left( \frac{18}{100} \right)^2 + \left( \frac{42}{100} \right)^2 + \left( \frac{12}{100} \right)^2 + \left( \frac{28}{100} \right)^2 = 0.0324 + 0.1764 + 0.0144 + 0.0784 = 0.3016 \). \( D = 1 - 0.3016 = 0.70 \) (to 2 d.p.).

1 mark: correct total, \( N = 100 \), and correct method, \( \Sigma \left( \frac{n}{N} \right)^2 \) calculated for all four species; 1 mark: correct value of \( \Sigma \left( \frac{n}{N} \right)^2 = 0.3016 \) (or equivalent correct working shown); 1 mark: correct final answer, \( D = 0.70 \) (accept 0.69-0.70; own figure rule applies); max 3.

Convert units: distance \( = 60 \text{ mm} = 0.06 \text{ m} \); time \( = 2.0 \text{ ms} = 0.002 \text{ s} \). Conduction velocity \( = \frac{\text{distance}}{\text{time}} = \frac{0.06}{0.002} = 30 \text{ m s}^{-1} \).

1 mark: correct conversion of units \( (0.06 \text{ m and } 0.002 \text{ s}) \); 1 mark: correct substitution into \( v = \frac{d}{t} \); 1 mark: correct final answer, \( 30 \text{ m s}^{-1} \) with correct unit (own figure rule applies); max 3.

When plotted, the data show a smooth decline in % P730 remaining with time in darkness. Reading directly from the table/graph, the percentage of P730 remaining falls to 50% at 6 hours of continuous darkness.

1 mark: correctly plotted axes/scale with % P730 remaining on the y-axis and time in darkness on the x-axis, points plotted accurately and joined with a smooth curve; 1 mark: correct reading taken from the curve at the point where % P730 = 50; 1 mark: correct final answer, 6 hours (accept 5.5-6.5 hours to allow for graph-reading tolerance); max 3.

A strong response should cover the following, using named structures throughout: (1) Sensory receptor - for example rods and cones in the retina of the mammalian eye, which convert light energy into a generator potential; if this reaches threshold, a nerve impulse is initiated in the associated sensory neurone. Convergence of rods allows summation of sub-threshold stimuli (increasing sensitivity but decreasing acuity), while cones provide colour vision and high visual acuity. (2) Transmission of the nerve impulse - the resting axon membrane is polarised (inside negative relative to outside); a threshold stimulus causes depolarisation and generation of an action potential, which is propagated along the axon by local circuits of current flow; in a myelinated axon, depolarisation is restricted to the nodes of Ranvier, producing rapid saltatory conduction; a refractory period follows each action potential, ensuring impulses travel in one direction and imposing a maximum frequency of firing. (3) Synaptic transmission - on arrival at a synapse, the impulse causes exocytosis of neurotransmitter (e.g. acetylcholine) from the pre-synaptic membrane; the neurotransmitter diffuses across the synaptic cleft and binds to receptors on the post-synaptic membrane, generating an excitatory post-synaptic potential (EPSP) that may trigger a further action potential if threshold is reached; acetylcholinesterase then breaks down the neurotransmitter, preventing continuous stimulation. (4) Effector - for example voluntary (skeletal) muscle, in which arrival of an impulse at the neuromuscular junction leads, via the sliding filament mechanism, to myosin heads attaching to actin filaments (in the presence of calcium ions) and pulling them over the myosin, shortening the sarcomere and producing contraction, using ATP. A full response links these stages into a coherent account of how the nervous system detects a stimulus and brings about a co-ordinated, appropriate response, using accurate specialist vocabulary and correct spelling, punctuation and grammar throughout.

Banded mark scheme (18 marks total), assessed with Quality of Written Communication. Band 3 (13-18 marks): accurate, detailed and well-organised account covering receptor structure/function, generation and propagation of the nerve impulse (including the role of myelination/saltatory conduction), synaptic transmission, and effector response (e.g. muscle contraction via the sliding filament mechanism), using named examples and accurate specialist terminology throughout, with high-quality written communication (correct spelling, punctuation, grammar and logical structure). Band 2 (7-12 marks): a mostly accurate account covering several of the required stages (receptor/transmission/synapse/effector) but with less detail, fewer named examples, or some gaps in the sequence; generally appropriate terminology; adequate written communication with occasional errors that do not significantly hinder meaning. Band 1 (1-6 marks): limited, fragmentary or largely descriptive points about the nervous system with little reference to named structures or the correct sequence of events; inconsistent or absent use of specialist terminology; noticeable errors in spelling, punctuation or grammar that may hinder meaning. 0 marks: no relevant content, or response does not address co-ordination and control by the nervous system.

A non-coding section of DNA removed from the primary transcript before translation is called an intron.

1 mark: intron.

Transcription is the synthesis of a complementary strand of mRNA from a DNA template strand in the nucleus, catalysed by RNA polymerase. Translation is the synthesis of a polypeptide from the mRNA code at a ribosome, in which tRNA molecules bring specific amino acids into position according to the sequence of codons.

1 mark: transcription = synthesis of an mRNA strand complementary to a DNA template strand (in the nucleus), catalysed by RNA polymerase; 1 mark: translation = synthesis of a polypeptide at a ribosome by 'reading' the mRNA codons, with tRNA delivering specific amino acids; max 2.

A short, known-sequence length of DNA used to locate a complementary sequence is called a DNA probe.

1 mark: DNA probe.

PCR uses a heat-stable DNA polymerase (for example Taq polymerase) to extend the primers and synthesise new complementary DNA strands.

1 mark: (heat-stable) DNA polymerase (accept Taq polymerase).

A marker gene, such as an antibiotic-resistance gene, is used to identify cells that have successfully taken up the donor gene. A vector, such as a bacterial plasmid or bacteriophage, is the agent used to carry or transfer the donor gene into a recipient cell.

1 mark: marker gene = used to identify cells that have been successfully transformed, e.g. antibiotic resistance or fluorescent marker gene; 1 mark: vector = agent used to carry/insert the donor gene into the recipient cell, e.g. plasmid or bacteriophage; max 2.

Somatic-cell gene therapy introduces a functional gene into body (non-reproductive) cells, so the change is not passed on to offspring and treatment may need repeating. Germ-line gene therapy alters the genes in reproductive cells, gametes or embryos, so the change is heritable and passed on to future generations, raising greater ethical concerns.

1 mark: somatic-cell therapy = functional gene introduced into body/non-reproductive cells only, not passed to offspring; 1 mark: germ-line therapy = alteration to reproductive cells/gametes/embryo, heritable/passed to offspring, raising additional ethical issues; max 2.

mRNA binds to a ribosome, with the first codon positioned at the peptidyl (P) site. A tRNA molecule carrying an anticodon complementary to the first mRNA codon, and bonded to its specific amino acid, binds at this site. A second tRNA, carrying the next specific amino acid, binds by complementary base pairing at the adjacent aminoacyl (A) site. A peptide bond forms between the two amino acids, catalysed by the ribosome. The ribosome then translocates along the mRNA by one codon, transferring the growing polypeptide to the tRNA now in the P site and freeing the A site for the next tRNA. This cycle repeats, extending the polypeptide chain, until a stop codon is reached, at which point the completed polypeptide is released from the ribosome.

1 mark: mRNA binds to the ribosome with codons exposed at the P and A sites; 1 mark: tRNA molecules bind by complementary base pairing between anticodon and mRNA codon, each tRNA carrying its specific amino acid; 1 mark: peptide bond forms (condensation) between adjacent amino acids at the ribosome; 1 mark: ribosome translocates along the mRNA by one codon, transferring the polypeptide and freeing the A site for the next tRNA; 1 mark: process repeats until a stop codon is reached, releasing the completed polypeptide; max 5.

The DNA sample, together with primers, free nucleotides and heat-stable DNA polymerase, is heated to around 95 degrees C, breaking the hydrogen bonds between the two DNA strands and separating them. The mixture is then cooled to around 50-60 degrees C, allowing short DNA primers, complementary to sequences at each end of the target region on each strand, to anneal. The temperature is then raised to around 70-75 degrees C, at which heat-stable DNA polymerase extends each primer using the free nucleotides, producing two new double-stranded DNA molecules. This three-step cycle of denaturation, annealing and extension is repeated many times, with the amount of target DNA doubling with each cycle, giving an exponential increase in the amount of the target DNA sequence.

1 mark: mixture heated (~95 degrees C) to separate/denature the two DNA strands by breaking hydrogen bonds; 1 mark: mixture cooled to allow primers to anneal to complementary sequences at each end of the target region on each strand; 1 mark: temperature raised, heat-stable DNA polymerase extends the primers using free nucleotides to synthesise new complementary strands; 1 mark: three-step cycle (denaturation/annealing/extension) is repeated many times; 1 mark: amount of target DNA doubles with each cycle, giving an exponential increase in DNA amount; max 5.

A sample of an individual's DNA is cut into fragments using restriction endonucleases, which cut the DNA at specific base sequences. The resulting fragments vary in length between individuals because of variation in the number of repeat sequences. These fragments are separated according to size using gel electrophoresis: the negatively charged fragments are loaded into wells in a gel and an electric current is applied, causing them to move through the gel towards the positive electrode, with smaller fragments moving further than larger fragments. Fluorescent or radioactively labelled DNA probes, complementary to specific repeat sequences, are then used to locate particular fragments, producing a pattern of bands unique to that individual, which constitutes their genetic fingerprint.

1 mark: DNA cut into fragments using restriction endonucleases, which cut at specific base sequences; 1 mark: fragment length varies between individuals due to variation in repeat sequences (e.g. microsatellites/VNTRs); 1 mark: fragments separated according to size by gel electrophoresis, moving through the gel under an electric current (smaller fragments travel further); 1 mark: labelled DNA probes used to locate/visualise specific fragments; 1 mark: resulting band pattern is unique to the individual, forming their DNA/genetic fingerprint; max 5.

The human gene of interest is obtained, for example using a restriction endonuclease to cut it from chromosomal DNA, or using reverse transcriptase to synthesise complementary DNA from the corresponding mRNA. A compatible restriction endonuclease is used to cut open a vector, typically a bacterial plasmid, producing matching sticky ends. The donor gene is inserted into the cut plasmid and joined using the enzyme DNA ligase, forming a recombinant plasmid. This recombinant plasmid is introduced into a recipient bacterial cell, such as Escherichia coli, a process called transformation. A marker gene, for example for antibiotic resistance, present on the plasmid is used to identify successfully transformed cells, by growing the bacteria on a medium containing the relevant antibiotic so only transformed cells survive, producing large numbers of transformed cells capable of expressing the human gene.

1 mark: donor gene obtained using a restriction endonuclease (from chromosomal DNA) or reverse transcriptase (to make cDNA from mRNA); 1 mark: vector (e.g. plasmid) cut open using the same/a compatible restriction endonuclease to produce complementary sticky ends; 1 mark: donor gene inserted into the vector and joined using DNA ligase to form a recombinant plasmid; 1 mark: recombinant plasmid introduced into recipient bacterial cells (transformation); 1 mark: marker gene (e.g. antibiotic resistance) used to identify/select successfully transformed cells; max 5.

Photosystem II absorbs light energy, exciting electrons in chlorophyll, and these excited electrons are passed along a chain of electron carriers to photosystem I; as they pass along the chain, energy is released and used to synthesise ATP from ADP and inorganic phosphate. The electrons lost from photosystem II are replaced by electrons derived from the photolysis of water, which also releases protons and oxygen. Photosystem I also absorbs light energy, exciting further electrons, which, together with electrons arriving from photosystem II, combine with hydrogen ions from the dissociation of water to reduce NADP to NADPH.

1 mark: PSII absorbs light, exciting electrons that pass along an electron transport/carrier chain to PSI, releasing energy used to synthesise ATP (photophosphorylation); 1 mark: electrons lost from PSII are replaced by electrons from the photolysis of water, which also releases oxygen and protons; 1 mark: PSI absorbs light, exciting further electrons; 1 mark: electrons (from PSI, via the chain) combine with H+ to reduce NADP to NADPH; max 4.

Carbon dioxide diffuses into the stroma and combines with ribulose bisphosphate, catalysed by the enzyme rubisco, in a reaction called carbon fixation. This produces an unstable six-carbon intermediate that immediately breaks down into two molecules of glycerate phosphate. Glycerate phosphate is then reduced to triose phosphate, using reduced NADP and ATP supplied by the light-dependent stage. \( \frac{5}{6} \) of the triose phosphate produced is recycled, using further ATP, to regenerate ribulose bisphosphate, allowing the cycle to continue; the remaining \( \frac{1}{6} \) is used in the synthesis of hexose sugars and other organic compounds needed by the plant.

1 mark: CO2 combines with RuBP (5C), catalysed by rubisco, forming an unstable 6C intermediate that breaks down to two molecules of GP (3C); 1 mark: GP reduced to triose phosphate (TP) using NADPH and ATP from the light-dependent stage; 1 mark: \( \frac{5}{6} \) of TP produced is recycled (using ATP) to regenerate RuBP; 1 mark: \( \frac{1}{6} \) of TP is used to synthesise hexose sugars/other organic compounds; max 4.

A decrease in light intensity directly reduces the rate of the light-dependent stage, because fewer photons are available to excite electrons in photosystems I and II. This results in a reduced rate of photophosphorylation and reduced production of reduced NADP. Because the light-independent stage depends on a continuous supply of ATP and NADPH from the light-dependent stage, a reduced supply slows down the reduction of glycerate phosphate and the regeneration of ribulose bisphosphate. As a result, the overall rate of photosynthesis decreases as light intensity decreases, until at very low light intensities light becomes the limiting factor.

1 mark: lower light intensity reduces the rate of the light-dependent stage as fewer electrons in PSI/PSII are excited; 1 mark: this reduces the rate of ATP synthesis (photophosphorylation) and NADPH production; 1 mark: light-independent stage depends on ATP/NADPH from the light-dependent stage (e.g. to reduce GP to TP / regenerate RuBP), so its rate also falls; 1 mark: overall rate of photosynthesis decreases as light intensity decreases, i.e. light becomes limiting; max 4.

A concentrated leaf pigment extract is applied as a small spot near the base of a strip of chromatography paper, which is placed upright in a shallow layer of solvent, with the pigment spot above the solvent level. As the solvent rises up the paper by capillary action, the different pigments are carried up the paper at different rates depending on their solubility in the solvent and their affinity for the paper; more soluble pigments travel further, so the pigments separate into distinct spots. The distance travelled by each pigment and by the solvent front are measured, and the Rf value is calculated as \( R_f = \frac{\text{distance travelled by the pigment}}{\text{distance travelled by the solvent front}} \). Because the Rf value of a given pigment in a given solvent is constant, it can be compared with published Rf values for known pigments to identify each spot.

1 mark: pigment extract spotted near the base of the paper, paper placed in solvent below the level of the spot, solvent rises by capillary action; 1 mark: pigments separate because they differ in solubility in the solvent/affinity for the paper, more soluble pigments travelling further; 1 mark: \( R_f = \frac{\text{distance travelled by pigment}}{\text{distance travelled by solvent front}} \); 1 mark: Rf value is constant for a given pigment/solvent and can be compared to known/reference values to identify the pigment; max 4.

Gross photosynthesis is the total rate at which a plant fixes carbon dioxide and produces organic compounds by photosynthesis. Net photosynthesis is the rate of gross photosynthesis minus the rate at which the plant simultaneously uses up organic compounds by respiration, so it represents the actual net gain of organic material by the plant. The compensation point is the light intensity at which the rate of gross photosynthesis exactly equals the rate of respiration, so there is no net exchange of CO2 or O2 by the plant. Net photosynthesis can be determined by measuring the net rate of CO2 uptake by an illuminated plant; because respiration is occurring simultaneously and releasing CO2, this measured value represents net, not gross, photosynthesis.

1 mark: gross photosynthesis = total rate of carbon fixation/organic compound production by photosynthesis; 1 mark: net photosynthesis = gross photosynthesis minus (simultaneous) respiration, i.e. actual net CO2 uptake/organic gain; 1 mark: compensation point = light intensity at which rate of photosynthesis equals rate of respiration (no net CO2/O2 exchange); 1 mark: net photosynthesis measured directly as net CO2 uptake in the light, because respiration is occurring simultaneously and offsetting some of the CO2 fixed; max 4.

Glycolysis occurs in the cytoplasm and is common to both aerobic and anaerobic respiration. Glucose is first phosphorylated, using 2 ATP, and converted to fructose bisphosphate, a six-carbon compound, which is then split into two molecules of triose phosphate. Each triose phosphate molecule is oxidised to pyruvate, and this conversion produces reduced NAD and generates ATP. Because two triose phosphate molecules are produced and processed per glucose molecule, and 2 ATP were used at the start of the pathway, the overall net yield of glycolysis per glucose molecule is 2 ATP and 2 reduced NAD.

1 mark: glycolysis occurs in the cytoplasm, common to aerobic and anaerobic respiration; 1 mark: glucose phosphorylated (using 2 ATP) and converted to fructose bisphosphate, which splits into two triose phosphate molecules; 1 mark: each triose phosphate is converted to pyruvate, producing reduced NAD and generating ATP; 1 mark: net yield per glucose molecule = 2 ATP and 2 reduced NAD; max 4.

Reduced NAD and reduced FAD, produced during glycolysis and the Krebs cycle, are oxidised at the electron transport chain on the inner mitochondrial membrane, releasing electrons and protons. These electrons pass along a series of carriers at progressively lower energy levels; as electrons pass from one carrier to the next, energy is released and used to synthesise ATP at specific points along the chain. Oxygen acts as the final electron acceptor at the end of the chain, combining with electrons and protons to form water; without oxygen to accept these electrons, the chain would become blocked and further ATP production via the chain would stop.

1 mark: reduced NAD/FAD from glycolysis and the Krebs cycle are oxidised at the electron transport chain (on the inner mitochondrial membrane/cristae), releasing electrons; 1 mark: electrons pass along a chain of carriers at progressively lower energy levels, releasing energy; 1 mark: this energy is used to synthesise ATP (oxidative phosphorylation) at specific points on the chain; 1 mark: oxygen is the final electron/hydrogen acceptor, combining with electrons and H+ to form water, allowing the chain (and Krebs cycle) to continue; max 4.

During intense exercise, when oxygen cannot be supplied to muscle tissue fast enough to sustain aerobic respiration, muscle cells respire anaerobically: pyruvate produced by glycolysis is converted to lactate, regenerating NAD so glycolysis and ATP production can continue in the absence of oxygen. This represents extra ATP generated without immediate oxygen use, and lactate accumulates, creating an oxygen debt: the additional oxygen that will be required afterwards. After exercise, elevated breathing and heart rate persist, and the extra oxygen taken in is used to oxidise the accumulated lactate and to resynthesise the ATP and phosphocreatine stores that were depleted during the exercise.

1 mark: during intense exercise oxygen supply cannot meet demand, so muscle cells respire anaerobically, converting pyruvate to lactate to regenerate NAD and allow glycolysis/ATP production to continue; 1 mark: this represents extra ATP generated without immediate oxygen use, and lactate accumulates, creating an oxygen debt; 1 mark: after exercise, extra oxygen is taken in (elevated breathing/heart rate persists); 1 mark: this oxygen is used to oxidise/metabolise the accumulated lactate and to resynthesise depleted ATP/phosphocreatine stores; max 4.

Parental genotypes: Tt (tall) x tt (dwarf). Parental gametes: T, t (from the Tt parent) and t, t (from the tt parent).

Punnett square:
t t
T Tt Tt
t tt tt

Offspring genotypes: Tt, Tt, tt, tt (1 Tt : 1 tt). Offspring phenotypes: 2 tall (Tt) : 2 dwarf (tt), an expected phenotypic ratio of 1 tall : 1 dwarf.

1 mark: correct parental gametes identified (T and t from the heterozygous parent; t from the homozygous recessive parent); 1 mark: correct genetic diagram/Punnett square showing offspring genotypes Tt, Tt, tt, tt; 1 mark: correct phenotypic ratio, 1 tall : 1 dwarf; max 3.

Parental genotypes: Bb x Bb. Gametes: B, b from each parent.

Punnett square:
B b
B BB Bb
b Bb bb

Offspring genotypes: 1 BB : 2 Bb : 1 bb. Offspring phenotypes: 3 black (BB or Bb) : 1 white (bb). Probability that a randomly selected offspring is white \( = \frac{1}{4} \) (0.25).

1 mark: correct genetic diagram/Punnett square showing genotypes BB, Bb, Bb, bb; 1 mark: correct phenotypic ratio, 3 black : 1 white; 1 mark: correct probability of a white offspring, \( \frac{1}{4} \) (accept 0.25 or 25%); max 3.

Let X-H = normal allele, X-h = haemophilia allele. Mother's genotype: X-H X-h (carrier). Father's genotype: X-H Y (unaffected). Gametes: mother produces X-H or X-h; father produces X-H or Y.

Punnett square:
X-H Y
X-H X-H X-H X-H Y
X-h X-H X-h X-h Y

Offspring: X-H X-H (unaffected female), X-H Y (unaffected male), X-H X-h (carrier female), X-h Y (haemophiliac male). Of the sons (X-H Y and X-h Y, in a 1:1 ratio), half will have haemophilia. Expected proportion of sons with haemophilia \( = \frac{1}{2} \) (0.5).

1 mark: correct genotypes assigned using sex-linked notation (e.g. X-H X-h mother, X-H Y father) and correct gametes identified; 1 mark: correct genetic diagram/Punnett square showing all four offspring genotypes; 1 mark: correct identification that \( \frac{1}{2} \) of sons are affected (X-h Y); max 3.

Parental genotypes: RrYy x RrYy. By the law of independent assortment, each parent produces four types of gamete in equal proportions: RY, Ry, rY, ry. Combining these gametes in a 4x4 Punnett square gives 16 equally likely offspring combinations, comprising 9 round yellow (R_Y_), 3 round green (R_yy), 3 wrinkled yellow (rrY_) and 1 wrinkled green (rryy). Expected phenotypic ratio = 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green.

1 mark: correct identification of the four gamete types (RY, Ry, rY, ry) produced by each parent; 1 mark: correctly constructed 4x4 Punnett square (or equivalent) combining the gametes; 1 mark: correct identification of the four phenotype classes and their offspring numbers out of 16; 1 mark: correct final ratio, 9:3:3:1; max 4.

Expected numbers (ratio 9:3:3:1, total \( = 160 \)): purple smooth \( = \frac{9}{16} \times 160 = 90 \); purple wrinkled \( = \frac{3}{16} \times 160 = 30 \); yellow smooth \( = \frac{3}{16} \times 160 = 30 \); yellow wrinkled \( = \frac{1}{16} \times 160 = 10 \).

\( \chi^2 = \frac{(82-90)^2}{90} + \frac{(30-30)^2}{30} + \frac{(32-30)^2}{30} + \frac{(16-10)^2}{10} = \frac{64}{90} + \frac{0}{30} + \frac{4}{30} + \frac{36}{10} = 0.71 + 0 + 0.13 + 3.60 = 4.44 \) (to 2 d.p.).

1 mark: correct expected values calculated for all four classes (90, 30, 30, 10); 1 mark: correct substitution of all four \( \frac{(O-E)^2}{E} \) terms into the formula; 1 mark: correct individual terms and running total shown; 1 mark: correct final answer, \( \chi^2 = 4.44 \) (accept 4.4; own figure rule applies); max 4.

Null hypothesis: there is no significant difference between the observed and expected offspring numbers, i.e. any difference is due to chance, and the results are consistent with a 9:3:3:1 ratio. Because the calculated value of \( \chi^2 \) (4.44) is less than the critical value (7.81) at p = 0.05 for 3 degrees of freedom, the null hypothesis is accepted: there is no significant difference between the observed and expected results, so the data are consistent with the expected 9:3:3:1 dihybrid ratio and independent assortment of the two genes.

1 mark: correct null hypothesis stated (no significant difference between observed and expected numbers / data consistent with expected 9:3:3:1 ratio); 1 mark: correct degrees of freedom identified (3), consistent with the critical value used; 1 mark: correct comparison, calculated \( \chi^2 \) (4.44) is less than the critical value (7.81) at p = 0.05; 1 mark: correct conclusion, null hypothesis accepted / observed and expected results not significantly different; max 4.

A strong response should compare the two phyla directly across each feature: (1) Symmetry - both phyla are bilaterally symmetrical, but Platyhelminthes (e.g. planarian, liver fluke) are additionally flattened dorso-ventrally, whereas Annelida (e.g. earthworm, lugworm) are round in transverse section. (2) Segmentation - Platyhelminthes show no segmentation, whereas Annelida are metamerically segmented, with the body divided into a series of repeating segments. (3) Gut structure - Platyhelminthes have a single opening to the gut (mouth only, no anus), so waste must leave via the mouth and the gut cannot show full regional specialisation; Annelida have a gut with both a mouth and an anus, allowing one-way flow of food and regional specialisation along its length, which is more efficient for digestion. (4) Skeletal support - Platyhelminthes have no specialised skeletal system and are supported only by body tissue (parenchyma); Annelida possess a hydrostatic skeleton formed by the fluid-filled segmental body cavities, against which muscles act. A full response should also note the functional consequence of these differences: the combination of segmentation and a hydrostatic skeleton allows Annelida more precise, localised and efficient movement than the simple body-tissue support of Platyhelminthes, and the through-gut of Annelida allows more efficient digestion via regional specialisation than the single-opening gut of Platyhelminthes.

Banded mark scheme (18 marks total), assessed with Quality of Written Communication. Band 3 (13-18 marks): accurate, detailed and well-organised direct comparison covering symmetry, segmentation, gut structure and skeletal support for both phyla, using appropriate specialist terminology (e.g. metameric segmentation, hydrostatic skeleton, dorso-ventrally flattened) accurately and consistently, with a clear, logically structured comparison (not simply two separate descriptions) and accurate spelling, punctuation and grammar. Band 2 (7-12 marks): accurate but less complete or less well-organised comparison, covering most of the required features for both phyla with generally appropriate terminology; some structure to the comparison, though it may be more descriptive (parallel accounts) than explicitly comparative; occasional lapses in spelling, punctuation or grammar that do not significantly hinder meaning. Band 1 (1-6 marks): some accurate but limited or fragmentary points about one or both phyla, with limited direct comparison and inconsistent or absent use of specialist terminology; poorly organised with noticeable errors in spelling, punctuation or grammar that may hinder meaning. 0 marks: no relevant content, or response does not address the comparison requested.

Aseptic technique aims to prevent contamination of the culture by unwanted microorganisms and to prevent the culture escaping into the environment. Suitable techniques include flaming the inoculating loop before and after use to sterilise it, flaming the neck of the culture vessel as it is opened and closed, working close to a lit Bunsen burner flame to create a sterile updraught, and minimising the time the Petri dish lid is left open.

1 mark each for any two of: flaming the inoculating loop before and after use; flaming the neck of the culture bottle/test tube; working close to a lit Bunsen burner flame (to create an updraught reducing contamination); minimising the time the Petri dish lid is open; sterilising the agar/media before use; max 2.

A solvent-only control allows the effect of the solvent to be distinguished from the effect of the dissolved plant extract. If the solvent disc produces no zone of inhibition, any inhibition seen around a plant extract disc can be confidently attributed to the antimicrobial properties of the plant extract rather than the solvent used to dissolve it.

1 mark: to check whether the solvent itself has any antimicrobial/inhibitory effect on the bacteria; 1 mark: so that any inhibition zone around the true extract discs can be attributed to the plant extract itself (a valid comparison/control); max 2.

A haemocytometer is a specialised microscope slide with an etched grid of known area and known depth, and therefore known volume. A sample of the yeast suspension is introduced beneath the coverslip, and the number of cells within a defined number of grid squares is counted under a microscope; because the volume of the grid is known, this count can be converted into a cell density.

1 mark: haemocytometer (accept counting chamber); 1 mark: allows cells within a grid/chamber of known volume to be counted so cell density can be calculated; max 2.

Potassium hydroxide solution absorbs the carbon dioxide released by the respiring seeds; without it, the CO2 released would replace much of the O2 taken up, and little change in gas volume would be detected. Because the measured change in volume is then due only to O2 uptake, this allows the rate of oxygen consumption to be determined. The control tube contains an equal volume of inert material instead of respiring seeds, so any change in the level of the manometer fluid in the control must be due to changes in temperature or atmospheric pressure rather than respiration, and this value is used to correct the reading from the experimental tube.

1 mark: potassium hydroxide absorbs the CO2 produced, so volume/pressure change reflects only O2 uptake; 1 mark: control (with beads, no respiring material) accounts for/corrects for changes in temperature/atmospheric pressure that are not due to respiration; max 2.

Redox indicators such as methylene blue and DCPIP act as artificial hydrogen acceptors and are blue in their oxidised form. As the reaction being investigated proceeds, for example dehydrogenase activity in respiring tissue, the indicator becomes reduced and loses its blue colour, turning colourless. This colour loss over time can be used as a measure of the rate of the reaction being studied.

1 mark: colour change from blue to colourless as the indicator is reduced; 1 mark: this demonstrates that hydrogen/electrons are being released from the substrate and accepted by the indicator (acting as an artificial hydrogen acceptor), showing that respiration/photosynthesis is occurring; max 2.

Gel electrophoresis of DNA involves both an electrical hazard, since a current is passed through the buffer solution to move the DNA fragments through the gel, and a chemical hazard, since staining agents used to visualise DNA bands are frequently mutagenic. Appropriate precautions include disconnecting the power supply before opening the tank or handling the gel, and wearing gloves when handling stains and the gel itself.

1 mark each for any two of: wearing gloves when handling stains/samples (e.g. mutagenic/hazardous stains); ensuring the power supply is switched off and disconnected before opening the tank/removing the gel (risk of electric shock); handling the buffer/gel carefully to avoid spillage; wearing eye protection; max 2.

Insect mouthparts are small and delicate, so careful technique is required. This includes using a binocular dissecting microscope or hand lens throughout to view the structures clearly as they are separated, using fine instruments such as fine forceps and a mounted needle to gently tease apart the mouthpart components without tearing or losing small structures, and keeping the specimen appropriately moist on a dissecting mat to prevent it drying out and becoming brittle during the procedure.

1 mark each for any two of: use of a binocular microscope/hand lens to magnify the small structures while working; use of fine dissecting instruments (fine forceps, mounted needle, fine scissors) to carefully separate structures without damage; working on a dissecting board/mat/dish to stabilise the specimen; keeping the specimen moist to prevent it drying out and becoming brittle during dissection; max 2.

A manometer, consisting of a capillary tube containing a coloured liquid connected to the sealed respirometer chamber, is used. As oxygen is taken up by the respiring organism, and the CO2 produced is absorbed by potassium hydroxide, the total gas volume in the sealed system falls, drawing the liquid along the capillary tube towards the respiration chamber. The distance moved by the liquid along a calibrated scale, combined with the known bore of the tube, allows the volume of oxygen taken up in a given time to be calculated.

1 mark: manometer (a liquid-filled capillary/graduated tube connected to the respirometer) is used; 1 mark: movement of the liquid along the graduated capillary tube (as gas volume in the sealed chamber changes) is measured/used to calculate the volume of gas absorbed; max 2.

Using aseptic technique throughout, spread a sterile agar plate evenly with a standardised inoculum of a bacterial suspension of known density, using a sterile spreader. Using sterile forceps, place discs impregnated with each antibiotic to be tested at equally spaced positions around the plate, to prevent overlap of inhibition zones, and label the position of each disc on the base of the plate. Include a control disc soaked in sterile water or solvent only, instead of an antibiotic, to check that the disc material itself has no antimicrobial effect. Seal the lid of the plate with tape, leaving air holes, and incubate at a suitable, safe temperature, such as 25 degrees C, for a standardised time, such as 24-48 hours, keeping all other variables, such as agar type/volume, incubation temperature and time, and inoculum density, constant between plates. After incubation, without opening the lid, measure the diameter of the clear zone of inhibition surrounding each antibiotic disc using a ruler. Repeat the whole investigation, for example three times, to obtain replicate measurements for each antibiotic, and calculate a mean zone diameter for each antibiotic to allow valid comparison of antimicrobial effectiveness.

1 mark: aseptic technique used throughout (e.g. flaming loop/spreader, working near a Bunsen flame); 1 mark: agar plate seeded evenly with a standardised volume/density of bacterial suspension using a sterile spreader; 1 mark: antibiotic discs placed at equally spaced positions (to avoid overlapping inhibition zones) using sterile forceps, and labelled; 1 mark: control disc included (e.g. soaked in sterile water/solvent only, no antibiotic); 1 mark: other variables kept constant/standardised (e.g. incubation temperature, time, agar type, inoculum density) and plate incubated at a safe temperature (e.g. 25 degrees C) for a fixed time; 1 mark: diameter of the zone of inhibition around each disc measured (without opening the lid) using a ruler; 1 mark: investigation repeated to obtain replicates and a mean zone diameter calculated for each antibiotic for valid comparison; max 7.

Prepare a standard yeast suspension, using a fixed mass of dried yeast dissolved in a fixed volume of glucose solution of known, standard concentration, and place a fixed volume of this suspension in a test tube fitted with a delivery tube leading to an inverted, graduated measuring cylinder filled with water and standing in a trough of water, so that gas produced can be collected and its volume measured. Place the yeast/glucose mixture in a water bath set to the first test temperature and allow time for the mixture to reach this temperature before starting timing. Measure the volume of gas collected over a fixed time interval, then repeat using fresh yeast/glucose mixtures at a range of different temperatures, keeping the mass/concentration of yeast, volume and concentration of glucose solution, and time allowed for gas collection constant at every temperature, so that temperature is the only variable being changed. Repeat each temperature at least twice and calculate a mean volume of gas produced per unit time at each temperature, to allow a reliable rate of respiration to be calculated and plotted against temperature.

1 mark: standard/fixed mass and concentration of yeast suspension in a fixed, known volume and concentration of glucose solution prepared; 1 mark: suitable apparatus described for collecting and measuring the volume of gas produced (e.g. delivery tube to an inverted measuring cylinder/gas syringe over water); 1 mark: mixture equilibrated to each test temperature (using a water bath) before timing/gas collection begins; 1 mark: volume of gas collected over a fixed, standardised time interval; 1 mark: range of different temperatures tested while keeping all other variables (yeast mass/concentration, glucose volume/concentration, collection time) constant; 1 mark: investigation repeated at each temperature to obtain replicate readings; 1 mark: mean rate of gas production calculated (e.g. volume per unit time) at each temperature, to be plotted against temperature; max 7.

Using pitfall traps, or by direct searching under logs and leaf litter, placed at random locations across the habitat, capture as large a sample of woodlice as practicable. Count and record the number of individuals captured in this first sample. Mark each captured individual with a small, harmless mark, for example a small dot of non-toxic, quick-drying paint on the dorsal surface, that will not affect its survival or behaviour and will not wash off before the next sampling. Release the marked individuals back into the exact locations from which they were captured, and allow sufficient time, such as 24-48 hours, for the marked individuals to disperse fully and randomly mix back into the wider population before it is re-sampled. Using the same method, take a second sample of woodlice from the same area, and count how many of this second sample are marked. Use the Lincoln index to calculate the estimated population size. For this estimate to be valid, several assumptions must be made: the marked individuals must have fully and randomly redistributed themselves throughout the population before the second sample is taken; the population must be closed, with no significant immigration, emigration, births or deaths, between the two sampling occasions; the mark must not affect the individuals' survival, behaviour or their chance of being recaptured; and marked individuals must be as likely to be captured in the second sample as unmarked individuals.

1 mark: suitable capture method described (e.g. pitfall traps/direct searching at random locations) for the first sample; 1 mark: individuals in the first sample counted and marked with a method that does not harm/is not readily lost (e.g. a small non-toxic paint mark); 1 mark: marked individuals released back into the original habitat and allowed sufficient time to redistribute/mix randomly before re-sampling; 1 mark: second sample taken by the same method, and the number of marked individuals recaptured recorded; 1 mark: Lincoln index formula correctly stated/used to calculate the population estimate; 1 mark for a valid assumption (e.g. no significant births/deaths/migration between samples, or that the population is closed); 1 mark for a further valid assumption (e.g. marking does not affect survival/behaviour/capture probability, or marked individuals redistribute fully and randomly); max 7.

Convert units so image and actual size are in the same units: image width \( = 45 \text{ mm} = 45\,000 \ \mu\text{m} \). Magnification \( = \frac{\text{image size}}{\text{actual size}} = \frac{45\,000}{30} = 1500 \). In standard form, magnification \( = 1.5 \times 10^{3} \) (x1500).

1 mark: correct unit conversion \( (45 \text{ mm} = 45\,000\ \mu\text{m}) \); 1 mark: correct substitution into \( \text{magnification} = \frac{\text{image size}}{\text{actual size}} \); 1 mark: correct answer, x1500; 1 mark: correctly expressed in standard form, \( 1.5 \times 10^{3} \); max 4.

Actual size \( = \frac{\text{image size}}{\text{magnification}} = \frac{25}{50} = 0.5 \text{ mm} \). Converting to \( \mu \text{m} \): \( 0.5 \times 1000 = 500\ \mu\text{m} \).

1 mark: correct rearrangement, \( \text{actual size} = \frac{\text{image size}}{\text{magnification}} \); 1 mark: correct substitution, \( \frac{25}{50} \); 1 mark: correct answer in mm, 0.5 mm; 1 mark: correctly converted to \( \mu \text{m} \), 500 \( \mu \text{m} \); max 4.

Mean \( = \frac{2.1+2.4+2.0+2.6+2.3+2.2+2.5+2.1+2.3+2.4}{10} = \frac{22.9}{10} = 2.29 \text{ mm} \).

Median: arrange values in order: 2.0, 2.1, 2.1, 2.2, 2.3, 2.3, 2.4, 2.4, 2.5, 2.6. With 10 values, the median is the mean of the 5th and 6th values \( = \frac{2.3 + 2.3}{2} = 2.3 \text{ mm} \).

1 mark: correct sum of values (22.9) leading to correct mean, 2.29 mm; 1 mark: values correctly arranged in ascending order; 1 mark: correct identification of the 5th and 6th values (2.3 and 2.3) and correct median, 2.3 mm; 1 mark: both final answers given to an appropriate/consistent number of decimal places with correct units (mm); max 4.

Mean \( = \frac{12+15+11+22+14+13+16+13}{8} = \frac{116}{8} = 14.5 \text{ cm} \). Median, data in order 11, 12, 13, 13, 14, 15, 16, 22, \( = \frac{13+14}{2} = 13.5 \text{ cm} \). The value 22 cm is an outlier, considerably higher than the rest of the data set, which lies mainly between 11 and 16 cm. Because the mean uses the numerical value of every data point, this single outlier disproportionately increases the mean away from the bulk of the data. The median, being based on rank order rather than numerical values, is much less affected by this outlier, so the median gives a value more representative of most of the seedlings in this sample.

1 mark: correct mean calculated, 14.5 cm; 1 mark: correct median calculated, 13.5 cm; 1 mark: 22 cm identified as an outlier/anomalous result relative to the rest of the data; 1 mark: explanation that the mean is disproportionately affected/skewed by this outlier (as it uses every numerical value) whereas the median (based on rank order) is much less affected, so is more representative here; max 4.

A negative correlation means that as one variable, resting heart rate, increases, the other variable, 400 m running time, tends to decrease, or vice versa; in this context, individuals with a lower resting heart rate tended to have a faster 400 m running time. However, a correlation, even a clear one, does not by itself demonstrate causation: there may be a third factor, for example overall fitness level or amount of aerobic training, that independently influences both resting heart rate and running speed, so it cannot be concluded from the scatter diagram alone that a lower resting heart rate directly causes a faster running time.

1 mark: negative correlation = as one variable increases the other tends to decrease (correctly applied to this context, i.e. lower heart rate associated with faster/shorter running time); 1 mark: correlation does not necessarily indicate causation; 1 mark: valid explanation, e.g. a third/confounding variable (such as overall fitness or training level) could independently affect both variables, so cause cannot be confirmed from a scatter diagram alone; max 4.

Actual diameter \( = \text{number of eyepiece units} \times \text{calibration value} = 18 \times 2.5 = 45\ \mu\text{m} \).

1 mark: correct method (number of eyepiece units x calibration value); 1 mark: correct substitution, \( 18 \times 2.5 \); 1 mark: correct final answer, 45 \( \mu \text{m} \); max 3.