State the location within or outside a eukaryotic cell where each of the following stages of aerobic respiration occurs: Glycolysis, Link reaction, Krebs cycle, Oxidative phosphorylation.

• 1. Oxidative phosphorylation: Inner mitochondrial membrane (cristae).
• 2. Glycolysis: Mitochondrial matrix.
• 3. Krebs cycle: Mitochondrial matrix.
• 4. Glycolysis: Cytoplasm (cytosol).
• 5. Link reaction: Mitochondrial matrix.

Explain why ATP synthesis by oxidative phosphorylation stops in anaerobic conditions (absence of oxygen).

• 1. Electron carriers (like NADH, FADH₂) cannot pass electrons to the ETC and therefore remain reduced; NAD⁺ and FAD are not regenerated.
• 2. Without oxygen, the Krebs cycle accelerates significantly to compensate for the lack of ATP, quickly depleting acetyl-CoA.
• 3. Oxygen serves as the crucial final electron acceptor at the end of the electron transport chain (ETC).
• 4. Consequently, the pumping of protons (H⁺ ions) from the matrix to the intermembrane space by the ETC complexes ceases.
• 5. Without oxygen to accept electrons, the entire electron transport chain backs up and stops functioning.
• 6. The proton gradient (proton motive force) across the inner mitochondrial membrane dissipates or cannot be established/maintained.
• 7. Without the flow of protons back through ATP synthase driven by the gradient, ATP synthesis via chemiosmosis stops.

Some cancer cells rely heavily on lactate fermentation even when oxygen is present. Explain how inhibiting the enzyme that catalyses the final step… (pyruvate to lactate) would reduce ATP production in these cells.

• 1. Glycolysis, the main ATP source in these cells, requires a constant supply of NAD⁺ to continue operating.
• 2. Inhibiting lactate dehydrogenase prevents the conversion of pyruvate to lactate.
• 3. Since glycolysis produces ATP via substrate-level phosphorylation, slowing or stopping glycolysis directly reduces the cell’s ATP production.
• 4. Inhibiting lactate dehydrogenase directly blocks the activity of ATP synthase in the mitochondria.
• 5. The conversion of pyruvate to lactate (fermentation) is essential for regenerating NAD⁺ from NADH produced during glycolysis. Inhibiting this step prevents NAD⁺ regeneration.

Identify one similarity and one difference between paracrine cell signalling (local diffusion) and endocrine cell signalling (hormones in blood).

• 1. Difference: Endocrine signals (hormones) typically travel long distances via the bloodstream to reach target cells throughout the body; paracrine signals diffuse only over short distances to affect nearby cells.
• 2. Similarity: Both signalling pathways primarily utilize lipid-soluble signalling molecules that can easily diffuse across cell membranes.
• 3. Similarity: Both involve a signalling molecule (ligand) being released (secreted) by a signalling cell and binding to specific receptors on target cells to elicit a response.

Explain why a neurotransmitter such as acetylcholine (ACh) could be described as a paracrine signalling molecule.

• 1. ACh travels long distances within the bloodstream to coordinate responses in multiple organs simultaneously.
• 2. ACh acts specifically on receptors located on the immediately adjacent postsynaptic cell membrane.
• 3. ACh is released by a presynaptic neurone and diffuses across the very short distance (nanometers) of the synaptic cleft to reach its target.

Experiments investigated a signalling molecule released by muscle cells after palmitate exposure… Results: Control ≈ 1.0; Expt A (medium) ≈ 1.5; Expt B (boiled medium) ≈ 2.6; Expt C (lipid part) ≈ 1.5. Outline conclusions.

• 1. Conclusion: The signalling molecule is likely lipid-soluble or associated with lipids, as activity similar to the whole medium was recovered in the lipid fraction (Expt C ≈ 1.5, similar to Expt A ≈ 1.5).
• 2. Conclusion: The signalling molecule is likely a protein, as boiling significantly increased its activity (Expt B ≈ 2.6 compared to Expt A ≈ 1.5).
• 3. Conclusion: Palmitate-exposed muscle cells release a signalling molecule into the medium that induces stress gene expression in recipient cells (Expt A shows higher expression than Control).
• 4. Conclusion: The signalling molecule appears to be heat-stable (not easily denatured by boiling), as boiling the medium did not reduce its signalling effect (Expt B shows activity ≥ Expt A).

If stress gene expression inhibits protein translation in developing muscle cells, suggest three ways the structure of these cells might differ from normal muscle cells.

• 1. Fewer and/or smaller sarcomeres (the basic contractile units containing actin and myosin).
• 2. Lower abundance of actin and myosin protein filaments within the cytoplasm.
• 3. Fewer mitochondria, as mitochondrial biogenesis requires synthesis of numerous proteins encoded by both nuclear and mitochondrial DNA.
• 4. An increased number of ribosomes attached to the endoplasmic reticulum to compensate for the inhibition.
• 5. Fewer and/or smaller myofibrils (the bundles of actin and myosin filaments that make up muscle fibers).
• 6. Reduced amounts of regulatory proteins like troponin and tropomyosin associated with the contractile filaments.

Describe the mechanism by which stomata open in response to sunlight.

• 1. Potassium ions (K⁺) move out of the guard cells, causing water potential inside to rise and water to follow by osmosis, leading to flaccidity.
• 2. Light (specifically blue light) activates proton pumps (H⁺-ATPases) in the guard cell plasma membrane, actively pumping H⁺ ions out of the cells using ATP.
• 3. The accumulation of K⁺ ions (and associated anions like malate) inside the guard cells significantly lowers (makes more negative) the water potential within these cells.
• 4. Water enters the guard cells from surrounding epidermal cells via osmosis, moving down the water potential gradient (from higher potential outside to lower potential inside).
• 5. Potassium ions (K⁺) move into the guard cells through specific ion channels, driven down the electrochemical gradient established by the proton pumps.
• 6. The influx of water increases the turgor pressure of the guard cells, causing them to bend or bow outwards (due to differential thickening of their cell walls), thereby opening the stomatal pore between them.

A graph shows that as ABA concentration increases from near 0 to ~250 µmol m⁻³, stomatal conductance decreases steeply (from >2.0 to <0.5 mol m⁻² s⁻¹). Above 250 µmol m⁻³ ABA, conductance remains very low. Describe this relationship.

• 1. At higher ABA concentrations (significantly above 250 µmol m⁻³), further substantial increases in ABA concentration cause a proportional, steep increase in stomatal conductance.
• 2. There is a clear negative correlation between ABA concentration and stomatal conductance: as ABA levels rise, conductance generally falls.
• 3. The relationship is non-linear; the effect of ABA is most pronounced at lower concentrations, causing a steep decrease in conductance initially.
• 4. At high ABA concentrations, stomatal conductance reaches a minimum low level, and further increases in ABA have minimal additional effect (saturation effect).

Suggest explanations for the relationship where increasing ABA concentration decreases stomatal conductance.

• 1. ABA binds to specific receptor proteins on the guard cell plasma membrane (or possibly inside the cell), triggering an intracellular signalling cascade, often involving second messengers like calcium ions (Ca²⁺) and reactive oxygen species.
• 2. The signalling cascade leads to the opening of anion channels and K⁺ efflux channels, causing the efflux (movement out) of potassium ions (K⁺) and anions (like Cl⁻, malate²⁻) from the guard cells.
• 3. The loss of solutes increases (makes less negative) the water potential inside the guard cells relative to the surrounding cells.
• 4. ABA binding directly activates the H⁺-ATPase proton pumps in the guard cell membrane, leading to K⁺ influx and increased turgor.
• 5. Water leaves the guard cells by osmosis, moving down the water potential gradient (from higher potential inside to lower potential outside), causing the cells to lose turgor and become flaccid.
• 6. The loss of turgor causes the guard cells to straighten, closing the stomatal pore between them, thus reducing conductance. ABA is often produced in response to drought stress, signalling the need to conserve water.

State where transcription factors bind on DNA to cause an increase in the rate of transcription.

• 1. Ribosome binding site (e.g., Shine-Dalgarno sequence in prokaryotes).
• 2. Terminator sequence.
• 3. Promoter region (often close to the transcription start site) or enhancer regions (which can be distant).
• 4. Start codon (AUG on mRNA).

Explain how the North American signal crayfish could have evolved resistance to crayfish plague.

• 1. Inheritance: Resistant crayfish that survived the plague passed the alleles conferring resistance to their offspring.
• 2. Selection Pressure: Exposure to the crayfish plague pathogen acted as a strong environmental pressure favouring resistant individuals.
• 3. Variation: Within the original crayfish population, random mutations likely generated genetic variation, with some individuals happening to possess alleles providing some level of resistance.
• 4. Directed Mutation: Individual crayfish that were exposed to the plague developed the necessary resistance mutations specifically in response to the need to survive.
• 5. Change in Allele Frequency: Over many generations of exposure and differential survival, the frequency of the resistance alleles increased significantly within the population.
• 6. Differential Survival/Reproduction: Crayfish with pre-existing resistance alleles were more likely to survive infection and reproduce compared to susceptible individuals.

Outline and explain the three main temperature steps that occur in one cycle of the polymerase chain reaction (PCR).

• 1. Annealing (e.g., 50–65°C): Heat-stable DNA polymerase actively binds to the template DNA and begins synthesizing new strands from available nucleotides.
• 2. Denaturation (e.g., 90–98°C): High temperature breaks the hydrogen bonds holding the two strands of the target DNA double helix together, separating them into single strands.
• 3. Extension/Elongation (e.g., 68–75°C, often 72°C): The temperature is raised to the optimal level for the heat-stable DNA polymerase (like Taq polymerase) to synthesize new complementary DNA strands, starting from the primers.
• 4. Annealing (e.g., 50–65°C): The temperature is lowered sufficiently to allow the short, single-stranded DNA primers to bind (anneal) specifically to their complementary sequences on the separated template strands.

Suggest how researchers identify suitable unique DNA sequences for designing PCR primers specific to a target species like A. pallipes.

• 1. By randomly synthesizing large numbers of short DNA primer sequences and testing each pair empirically until one successfully amplifies DNA only from A. pallipes samples.
• 2. By comparing available DNA sequence data (e.g., from genomic databases) of the target species (A. pallipes) with sequences from closely related species and non-target species using bioinformatics alignment tools.
• 3. By targeting highly conserved DNA regions, such as ribosomal RNA genes, that are known to be identical across all crayfish species and most crustaceans.
• 4. By sequencing parts of the genome (e.g., mitochondrial DNA or specific nuclear genes) from multiple individuals of A. pallipes and comparing these to sequences from other species expected to be in the environment, looking for regions unique to A. pallipes.

Standard qPCR curves show fluorescence increasing over cycles. Curves starting with higher DNA amounts cross a set threshold at earlier cycles. Four standards (A, B, C, D) cross the threshold progressively later. Which curve represents the highest starting concentration?

• 1. C
• 2. A
• 3. D
• 4. B

In a qPCR experiment, DNA from an eDNA sample reached threshold fluorescence at cycle 28. Standard curve C reached threshold at cycle 25. Given that DNA doubles each cycle, calculate how many times lower the starting concentration of the eDNA sample was compared to standard C.

• 1. 6 times lower (based on 2 multiplied by the difference in cycles: 2 * (28 – 25) = 6).
• 2. 9 times lower (based on squaring the difference in cycles: (28 – 25)² = 3² = 9).
• 3. 3 times lower (based directly on the difference in cycles: 28 – 25 = 3).
• 4. 8 times lower (since each cycle represents a doubling, a 3-cycle difference corresponds to 2³ = 8-fold difference).

To create genetically modified bacteria expressing GFP, what component (vector) is typically taken up by the bacteria, and what two essential genetic elements must this component contain besides the GFP gene itself?

• 1. Vector: Chromosome fragment. Element 1: GFP gene. Element 2: Terminator sequence.
• 2. Vector: Plasmid. Element 1: GFP gene. Element 2: Suitable promoter sequence.
• 3. Vector: Plasmid. Element 1: Promoter sequence. Element 2: Selectable marker (e.g., antibiotic resistance). (Note: assumes GFP gene is also present implicitly for expression context)
• 4. Vector: Ribosome. Element 1: GFP mRNA. Element 2: Promoter sequence.

Suggest why the gene for GFP is sometimes transferred along with another desired gene in genetic engineering.

• 1. GFP provides strong resistance to common antibiotics, allowing easy selection of transformed cells on antibiotic plates.
• 2. GFP acts as a visual reporter or marker gene; its fluorescence allows researchers to easily identify cells or organisms that have successfully taken up the plasmid (and thus likely the desired gene) by simply observing them under appropriate light.
• 3. GFP encodes an essential chaperone protein required for the correct folding and expression of the co-transferred desired gene’s protein product.
• 4. The bright fluorescence of the GFP protein directly interacts with and enhances the activity or stability of the protein product from the desired gene.

Name areas A (outside the thylakoid) and B (inside the thylakoid) within the chloroplast.

• 1. A: Intermembrane space, B: Stroma
• 2. A: Cytoplasm, B: Matrix
• 3. A: Stroma, B: Thylakoid lumen / Thylakoid space
• 4. A: Thylakoid lumen, B: Stroma

Name the specific biochemical process involving PSI, electron carriers C, and ATP synthase E that produces ATP using light energy but does not involve PSII. Name the protein labelled E.

• 1. Process: Calvin cycle. E: Rubisco.
• 2. Process: Cyclic photophosphorylation. E: ATP synthase / ATP synthetase.
• 3. Process: Non-cyclic photophosphorylation. E: NADP reductase.
• 4. Process: Chemiosmosis. E: Proton pump.

During non-cyclic photophosphorylation, a reduced hydrogen carrier (labelled D) is produced. Identify product D and describe its specific role in the Calvin cycle.

• 1. Identity of D: FADH₂. Role: Donates electrons to regenerate RuBP in the final stage of the cycle.
• 2. Identity of D: Reduced NADP (NADPH). Role: Provides reducing power (high-energy electrons and H⁺) used to reduce glycerate-3-phosphate (GP) to triose phosphate (TP) during the reduction phase of the cycle.
• 3. Identity of D: NADH. Role: Used directly by ATP synthase to produce ATP required for carbon fixation in the stroma.
• 4. Identity of D: ATP. Role: Directly provides the energy required for the initial carbon fixation step catalysed by RuBisCO.

Data (MJ m⁻² yr⁻¹): Tropical forest ≈ 38, Temperate forest ≈ 24, Snow forest ≈ 15. Describe these results and suggest how limiting factors explain the differences.

• 1. Description: Net primary productivity (indicative of energy transfer) is highest in snow forests and lowest in tropical forests, likely due to adaptation to cold.
• 2. Description: Energy transfer through photosynthesis decreases significantly from tropical regions towards higher latitudes (temperate and snow forests). Explanation: Factors like lower average annual temperature and/or reduced light intensity and duration at higher latitudes limit the overall rate of photosynthesis compared to the consistently warm and high-light conditions in the tropics.
• 3. Description: All three forest types exhibit very similar rates of energy transfer per unit area per year. Explanation: This is because atmospheric CO₂ concentration, the primary substrate, is relatively uniform globally.

Data (MJ m⁻² yr⁻¹): Grasslands (Tropical ≈ 12, Temperate ≈ 9) and Desert (<5) show lower energy transfer rates than forests. Suggest reasons for this.

• 1. Reason: Forests typically have a much higher density of plant life and greater total leaf area index (biomass) per unit ground area compared to most grasslands and especially deserts, allowing them to intercept more sunlight.
• 2. Reason: Water availability is often a major limiting factor in deserts and many grasslands, restricting photosynthesis because plants close stomata to conserve water, limiting CO₂ uptake.
• 3. Reason: Deserts generally experience optimal temperatures and consistent sunlight throughout the year, allowing maximum photosynthetic rates whenever water is present.
• 4. Reason: Nutrient availability (e.g., nitrogen, phosphorus) in the soil might be lower or more limiting in certain grassland or desert ecosystems compared to productive forests.

Homologous chromosomes share the same genes at the same loci. State one other structural feature shared by a pair of homologous chromosomes visible during meiosis.

• 1. Same number of mitochondria associated with their centromeres.
• 2. Same overall size / length.
• 3. Identical alleles for every gene along their length.
• 4. Same characteristic banding pattern when subjected to specific staining techniques (e.g., G-banding).
• 5. Same shape, determined by the position of the centromere (e.g., both metacentric, both acrocentric).

Albinism (lack of pigment) can be caused by homozygous recessive mutations of the TYR gene, which codes for the enzyme tyrosinase involved in melanin production. Explain why a person homozygous recessive for TYR shows albinism.

• 1. The mutated recessive alleles code for an overactive form of the tyrosinase enzyme, which rapidly consumes all melanin precursors, preventing pigment formation.
• 2. Being homozygous recessive means the individual inherited two copies of a mutated allele for the TYR gene (one from each parent).
• 3. The enzyme tyrosinase plays a critical, early role in the biochemical pathway that synthesizes melanin pigments.
• 4. These mutated alleles typically result in the production of a non-functional or inactive tyrosinase enzyme, or potentially no enzyme at all.
• 5. Without functional tyrosinase enzyme activity, the melanin synthesis pathway is blocked, and melanin pigment cannot be produced, resulting in the characteristic lack of pigmentation (albinism).

In rabbits, black fur (B) is dominant to brown (b), but allele F is required for pigment production (ff results in white fur). Complete the genetic diagram for a cross between two black rabbits heterozygous for both genes (BbFf × BbFf) and state the ratio of offspring phenotypes.

• 1. Gametes that can be produced by each BbFf parent: BF, Bf, bF, bf.
• 2. Parental Genotypes for the cross: BbFf × BbFf.
• 3. Offspring Phenotype Ratio expected: 9 Black : 3 Brown : 4 White.
• 4. Offspring Phenotype Ratio expected: 9 Black : 7 White (where brown and white phenotypes are indistinguishable).
• 5. Punnett square analysis yields expected genotype proportions of 9 B_F_ : 3 bbF_ : 3 B_ff : 1 bbff. Phenotypes: B_F_ = Black; bbF_ = Brown; B_ff = White; bbff = White.

Data: Cattle (0.08 mutations/Mb/yr, 20yr lifespan); Ferret (0.20, 8yr); Horse (0.05, 30yr); Human (0.02, 75yr); Mouse (0.28, 3.6yr). Describe the relationship shown.

• 1. There is no clear or discernible relationship between the somatic mutation rate per year and the maximum lifespan across these species.
• 2. There is a strong positive correlation: species with higher mutation rates per year tend to exhibit significantly longer lifespans.
• 3. There appears to be a negative correlation or inverse relationship: species with higher mutation rates per megabase per year generally have shorter lifespans (e.g., compare Mouse/Ferret to Human/Horse).

Using the relationship described (inverse correlation between mutation rate/year and lifespan), explain how these species might differ in their potential rate of adaptation if environmental selection pressures change.

• 1. Species with lower mutation rates per year (like humans and horses) are likely to adapt much faster because their genomes are more stable, allowing beneficial adaptations to persist without being disrupted by new mutations.
• 2. All species likely adapt at roughly the same rate over geological time, as the intensity of the selection pressure is the primary determinant, outweighing differences in mutation rate or lifespan.
• 3. Species with higher mutation rates per year and shorter generation times (often associated with shorter lifespans, like mice) generate more genetic variation per unit of time and undergo more generations of selection in that time, giving them a greater potential capacity to adapt more quickly to changing environments.
• 4. Species with longer lifespans (like horses and humans) have a greater potential for adaptation because individuals have more time within their own lifetime to develop physiological or learned adaptations to environmental changes.

Compare the typical timescales needed for selective breeding versus genetic engineering to significantly increase the frequency of a rare desirable allele in a cattle population. Give a reason.

• 1. Comparison: Both methods generally require a similar number of generations and therefore take roughly the same amount of calendar time to achieve significant allele frequency changes in long-generation animals like cattle.
• 2. Comparison: Genetic engineering techniques generally allow for much faster increases in allele frequency compared to the much slower process of traditional selective breeding.
• 3. Reason for Selective Breeding Speed: Progress is gradual, requiring repeated cycles of identifying superior individuals, mating them, and evaluating offspring over many generations, limited by the animal’s natural reproductive cycle length.
• 4. Reason for Genetic Engineering Speed: Allows for direct introduction or selection of the allele, potentially combined with advanced reproductive technologies (like cloning, marker-assisted selection with IVF, sperm/egg sorting) to rapidly multiply desired genotypes, potentially achieving large changes in 1-2 generations.
• 5. Comparison: Selective breeding is generally considered faster because it utilizes natural mating processes, whereas genetic engineering involves complex and time-consuming laboratory procedures before application.

State two features of dairy cattle that could potentially be improved using genetic engineering.

• 1. Milk yield / overall quantity of milk produced per lactation.
• 2. Ability to perform photosynthesis using sunlight and CO₂.
• 3. Resistance to specific diseases (e.g., mastitis, respiratory infections).
• 4. Milk quality attributes (e.g., higher protein content, altered fat composition, reduced allergens).
• 5. Development of functional wings for flight capability.

Outline three roles of the International Union for Conservation of Nature (IUCN) in the conservation of species.

• 1. Assessing the global conservation status of species and assigning them to risk categories (e.g., Critically Endangered, Vulnerable) published in the IUCN Red List of Threatened Species.
• 2. Directly managing and enforcing international treaties and regulations governing the trade in endangered species and their products.
• 3. Providing scientific information, expertise, and policy advice to governments, international conventions (like CITES, CBD), and conservation organizations.
• 4. Raising public awareness about biodiversity loss and conservation issues through publications, campaigns, and educational materials.
• 5. Supporting and facilitating on-the-ground conservation action through its network of members, commissions, and partnerships.

Grizzly bears and polar bears… can interbreed… to produce fertile ‘pizzly bear’ offspring. Discuss whether they should be classified as separate species, referring to three different species concepts.

• 1. Morphological Species Concept: Based on their significant, distinct physical differences (morphology) adapted to different environments, they would likely be classified as separate species.
• 2. Biological Species Concept (BSC): Because they can interbreed in nature and produce fertile offspring, indicating a lack of complete reproductive isolation, they might be considered the same species (or subspecies) under a strict BSC definition.
• 3. Ecological Species Concept: Given that they occupy distinct ecological niches defined by their different habitats (arctic ice vs temperate forests/tundra) and primary diets (seals vs varied omnivorous), they would likely be classified as separate species.
• 4. Phylogenetic Species Concept: Because they undoubtedly share a relatively recent common ancestor, they must definitively be classified as the same species under this concept.

Calculate Simpson’s Index of Diversity (D) for a spectacled bear population using D = 1 – Σ(n/N)², given the sum of the (n/N)² values [Simpson’s Dominance Index] is 0.272.

• 1. D = 0.272 (This is the Dominance Index, not the Diversity Index)
• 2. D = 1 / 0.272 ≈ 3.68 (This calculates the reciprocal of the Dominance Index, sometimes used as another diversity measure, but not the one defined by the formula D = 1 – Σ(n/N)²)
• 3. D = 1 – 0.272 = 0.728
• 4. D = 1 + 0.272 = 1.272 (Incorrect formula application)

The calculated Simpson’s Index of Diversity (D) value for the disturbed habitat population was 0.728, while the D value for an undisturbed population was 0.833. Explain what the difference in D values indicates about the bears’ diet in the two habitats.

• 1. The higher D value (0.833) found in the undisturbed habitat indicates a lower dietary diversity, meaning the bears consumed fewer types of food or relied heavily on one type.
• 2. The lower D value (0.728) found in the disturbed habitat indicates a higher dietary diversity compared to the undisturbed habitat.
• 3. The higher D value (0.833) for the undisturbed population indicates greater dietary diversity (the bears consumed a wider variety of food items, and/or consumed them in more even proportions) compared to the bears in the disturbed habitat (D=0.728).
• 4. The D values directly represent the total number of individual bears sampled in each habitat.

State two differences between discontinuous variation and continuous variation, other than their genetic basis.

• 1. Phenotypic Categories: Discontinuous variation falls into distinct, separate categories (e.g., blood groups A, B, AB, O); Continuous variation shows a range or spectrum of phenotypes with intermediates (e.g., height).
• 2. Environmental Influence: Phenotypes in discontinuous variation are usually minimally affected by environmental factors; phenotypes in continuous variation are often significantly influenced by the environment.
• 3. Genetic Basis: Discontinuous variation typically involves multiple genes (polygenic); Continuous variation typically involves a single gene with major effects.
• 4. Graphical Representation: Discontinuous variation data is often plotted using bar charts; Continuous variation data is typically plotted using histograms, often approximating a normal distribution curve.
• 5. Environmental Influence: Continuous variation phenotypes are largely determined by genetics with little environmental input; Discontinuous variation phenotypes are heavily modified by environmental conditions.

Harlequin ladybirds show several distinct colour patterns (e.g., conspicua, spectabilis, succinea) in populations. Suggest the genetic basis for this type of variation.

• 1. It represents continuous variation, likely controlled by the cumulative effects of many genes (polygenic inheritance), each with a small additive effect, interacting with environmental factors.
• 2. It represents discontinuous variation, where individuals fall into distinct phenotypic categories. This pattern is likely controlled by one or a few major genes, possibly involving multiple alleles at a single locus or epistatic interactions between loci.
• 3. It results purely from environmental factors experienced during development (e.g., diet, temperature) affecting pigment synthesis, with no genetic component involved.

Adult ladybird body size shows continuous variation (5mm to 8mm). State two environmental factors during the larval stage that might affect the final adult body size.

• 1. Availability, quantity, and nutritional quality of the food supply accessible to the larva during its growth period.
• 2. The specific allele(s) inherited by the larva for the gene controlling its adult colour pattern (e.g., conspicua vs succinea allele).
• 3. Ambient temperature experienced during larval development, affecting metabolic rate and growth duration.
• 4. The genetically determined number of spots that will appear on the adult ladybird’s elytra.

Describe how the binding of sucrose to a chemoreceptor cell leads to the generation of an action potential in the associated sensory neurone.

• 1. Sucrose molecules bind to specific transmembrane receptor proteins located on the membrane of the chemoreceptor cell.
• 2. This binding triggers a conformational change in the receptor, leading to the opening of specific ion channels (often permeable to Na⁺ or Ca²⁺) in the chemoreceptor cell membrane.
• 3. The influx of positive ions causes a depolarisation (making the inside less negative) of the chemoreceptor cell membrane, known as a receptor potential.
• 4. If this receptor potential is strong enough to reach a threshold level, it triggers the release of neurotransmitter molecules from the chemoreceptor cell into the synaptic cleft separating it from the sensory neurone.
• 5. Neurotransmitter molecules diffuse across the cleft and bind to specific ligand-gated ion channels (often Na⁺ channels) on the postsynaptic membrane of the sensory neurone, causing them to open.
• 6. The resulting influx of Na⁺ into the sensory neurone causes a depolarisation (postsynaptic potential). If this depolarisation reaches the threshold potential for the sensory neurone, it triggers the opening of voltage-gated sodium channels, initiating an action potential that propagates along the sensory axon.
• 7. Sucrose binding directly triggers a full, propagating action potential within the chemoreceptor cell itself, which then transmits directly to the sensory neurone.