The binding of ADH to its receptor on a collecting duct cell activates structure A, which leads to the production of structure B by adenylyl cyclase. This triggers an enzyme cascade resulting in the insertion of structure C into the luminal membrane (P). Name structures A, B, and C.

• 1. B: cyclic AMP / cAMP
• 2. C: Aquaporin
• 3. A: Adenylyl cyclase
• 4. A: G protein

Suggest reasons why the water potential of the blood may decrease.

• 1. Increased solute concentration in blood (e.g., due to high blood glucose in diabetes).
• 2. Excessive sweating / Significant water loss through evaporation from lungs/skin without replacement.
• 3. Drinking large amounts of pure water.
• 4. Dehydration / Insufficient water intake relative to losses.
• 5. Ingesting salty food or drinks / High solute intake without sufficient water.

Diabetes insipidus can be caused by decreased secretion of ADH. Suggest two symptoms that would occur in a person with this condition.

• 1. Excessive thirst (polydipsia) / Persistent dry mouth due to water loss.
• 2. Production of large volumes of urine (polyuria) / Frequent need to urinate.
• 3. Production of small volumes of highly concentrated urine.
• 4. Fatigue, potentially related to dehydration or electrolyte imbalance.
• 5. Production of dilute (low concentration / low osmolarity) urine.

In nephrogenic diabetes insipidus (NDI), ADH molecules cannot bind to their receptor proteins on collecting duct cells. Explain the effect this has on the water permeability of the luminal cell surface membrane (P) of these cells.

• 1. Vesicles within the cytoplasm containing pre-formed aquaporin water channels (C) are not stimulated to move to and fuse with the luminal membrane (P).
• 2. The intracellular signalling cascade involving the G protein and cAMP production is activated continuously, even though ADH cannot bind.
• 3. Because ADH fails to bind to its receptor, the downstream intracellular signalling pathway (G protein, adenylyl cyclase, cAMP) is not initiated or activated.
• 4. Consequently, few or no additional aquaporin channels are inserted into the luminal membrane beyond the basal level.
• 5. The luminal membrane therefore has a significantly reduced or low permeability to water compared to when ADH is present and functional.

NDI caused by a faulty receptor gene is X-linked recessive. Explain why a man with NDI could not have inherited the condition from his father.

• 1. Males have one X chromosome and one Y chromosome (XY genotype).
• 2. Males inherit their Y chromosome from their father and their single X chromosome from their mother.
• 3. A son typically inherits X-linked conditions directly from his affected father.
• 4. Since the gene responsible for NDI (in this case) is located on the X chromosome, and a son does not inherit his father’s X chromosome, he cannot inherit an X-linked condition from his father.

State whether the main cause of variation for the following scenario is likely to be due to genetic factors (V<0xE1><0xB5><0x82>), environmental factors (V<0xE1><0xB5><0x83>), or a combination (V<0xE1><0xB5><0x82> + V<0xE1><0xB5><0x83>): Tomato plants grown in a glasshouse and grown outside vary in the yield of tomatoes they produce. Seventeen genes associated with tomato yield have been identified.

• 1. V<0xE1><0xB5><0x82> + V<0xE1><0xB5><0x83> (Combination of genetic and environmental factors)
• 2. V<0xE1><0xB5><0x83> (Environmental factors only)
• 3. V<0xE1><0xB5><0x82> (Genetic factors only)

State whether the main cause of variation for the following scenario is likely to be due to genetic factors (V<0xE1><0xB5><0x82>), environmental factors (V<0xE1><0xB5><0x83>), or a combination (V<0xE1><0xB5><0x82> + V<0xE1><0xB5><0x83>): New strawberry plants from the variety called Sweet Ann are made by asexual reproduction. The new plants grow to different sizes and produce different numbers of fruit.

• 1. V<0xE1><0xB5><0x82> + V<0xE1><0xB5><0x83> (Combination of genetic and environmental factors)
• 2. V<0xE1><0xB5><0x83> (Environmental factors only)
• 3. V<0xE1><0xB5><0x82> (Genetic factors only)

State whether the main cause of variation for the following scenario is likely to be due to genetic factors (V<0xE1><0xB5><0x82>), environmental factors (V<0xE1><0xB5><0x83>), or a combination (V<0xE1><0xB5><0x82> + V<0xE1><0xB5><0x83>): The domestic cat has a blood group system with three possible blood types: A, B and AB, determined by antigens on red blood cells.

• 1. V<0xE1><0xB5><0x82> + V<0xE1><0xB5><0x83> (Combination of genetic and environmental factors)
• 2. V<0xE1><0xB5><0x83> (Environmental factors only)
• 3. V<0xE1><0xB5><0x82> (Genetic factors only)

State whether the main cause of variation for the following scenario is likely to be due to genetic factors (V<0xE1><0xB5><0x82>), environmental factors (V<0xE1><0xB5><0x83>), or a combination (V<0xE1><0xB5><0x82> + V<0xE1><0xB5><0x83>): Over 50 genes have variants associated with excessive weight gain in humans. Other risk factors include diet and exercise.

• 1. V<0xE1><0xB5><0x82> + V<0xE1><0xB5><0x83> (Combination of genetic and environmental factors)
• 2. V<0xE1><0xB5><0x83> (Environmental factors only)
• 3. V<0xE1><0xB5><0x82> (Genetic factors only)

State whether the main cause of variation for the following scenario is likely to be due to genetic factors (V<0xE1><0xB5><0x82>), environmental factors (V<0xE1><0xB5><0x83>), or a combination (V<0xE1><0xB5><0x82> + V<0xE1><0xB5><0x83>): Resting heart rate in humans varies. Factors influencing it include biological sex, family history, smoking, and medication.

• 1. V<0xE1><0xB5><0x82> + V<0xE1><0xB5><0x83> (Combination of genetic and environmental factors)
• 2. V<0xE1><0xB5><0x83> (Environmental factors only)
• 3. V<0xE1><0xB5><0x82> (Genetic factors only)

Name a spontaneous, random event occurring in cells that is the ultimate source of new genetic variation.

• 1. Crossing Over
• 2. Mutation
• 3. Independent Assortment

Other than mutation, describe three features of sexual reproduction that contribute to the production of genetically different offspring.

• 1. Asexual reproduction (cloning).
• 2. Independent assortment: The random alignment and separation of homologous chromosome pairs during meiosis I (and sister chromatids in meiosis II, if crossing over occurred) leads to different combinations of chromosomes in gametes.
• 3. Random fusion of gametes / Random fertilisation: The combination of genetic material from two parents occurs randomly when a sperm fertilises an egg, creating unique zygote genotypes.
• 4. Crossing over: The exchange of segments of genetic material between homologous chromosomes during prophase I of meiosis creates new combinations of alleles along a single chromosome.
• 5. Random mating (in a population context): The non-selective choice of mates further shuffles alleles within the population’s gene pool over generations.

In Primula plants, petal colour involves gene T/t (T allows pigment, t prevents) and gene D/d (D inhibits pigment production, d allows). A cross between two plants with genotype TtDd was performed. State the genotypes resulting from this cross that would produce blue petals (requiring allele T and lacking allele D).

• 1. TTdd
• 2. TtDd
• 3. Ttdd
• 4. TTDD

Using the results of the TtDd x TtDd cross (where T allows pigment, t prevents; D inhibits pigment, d allows; blue requires T and dd), state the expected phenotypic ratio of non-blue to blue petals.

• 1. 9 : 3 : 3 : 1
• 2. 13 : 3
• 3. 3 : 1
• 4. 9 : 7

Name the type of gene interaction illustrated by genes T/t and D/d affecting Primula petal colour (where T allows pigment, t prevents; D inhibits pigment production, d allows).

• 1. Codominance
• 2. Dominant Inhibition Epistasis
• 3. Recessive Epistasis
• 4. Incomplete Dominance

Discuss the possible roles of the proteins coded for by gene T/t and gene D/d in the control of the production of the blue pigment malvidin (T allows pigment, t prevents; D inhibits pigment, d allows).

• 1. Functional Protein: It’s likely that dominant alleles (T, D) code for functional proteins, while recessive alleles (t, d) code for non-functional proteins or no protein.
• 2. Enzyme Role (Gene T): Allele T might code for a functional enzyme required at some step in the biochemical pathway synthesising malvidin. Allele t would code for a non-functional version of this enzyme.
• 3. Inhibitor Role (Gene D): Allele D could code for a protein that actively inhibits the malvidin synthesis pathway, perhaps by binding to and inhibiting the enzyme produced by T, or another enzyme in the pathway. Allele d would code for a non-functional inhibitor, allowing the pathway to proceed if T is functional.
• 4. Transcription Factor Role (Gene T): Alternatively, Protein T could be a transcription factor required to activate the expression of genes encoding the enzymes of the malvidin pathway.
• 5. Final Product Role (Gene D): The protein coded by allele D is the blue pigment malvidin itself, and allele d fails to produce it.
• 6. Transcription Factor Inhibition (Gene D): Alternatively, Protein D could be a repressor protein that inhibits the function of transcription factor T, or binds to promoter regions to block transcription of pathway enzyme genes.

Explain the relationship between the F8 gene, the protein factor VIII, and the inherited condition haemophilia.

• 1. The F8 gene, located on the X chromosome (making it sex-linked), contains the instructions for producing the protein known as clotting factor VIII.
• 2. Deficiency or non-functionality of factor VIII protein leads to haemophilia A, a condition characterized by abnormally enhanced and rapid blood clotting.
• 3. Mutations in the F8 gene, typically inherited as recessive alleles (X<0xE1><0xB5><0xAB>), result in the production of insufficient or non-functional factor VIII protein.
• 4. Factor VIII plays a critical role as a cofactor in the intrinsic pathway of the blood coagulation cascade, necessary for proper clot formation.
• 5. Due to its X-linked inheritance, haemophilia A predominantly affects males, who have only one X chromosome.

Outline the role of restriction endonuclease in genetic engineering techniques involving plasmids.

• 1. Seals gaps or nicks in the sugar-phosphate backbone of DNA by forming phosphodiester bonds.
• 2. Synthesizes a DNA strand using an mRNA molecule as a template.
• 3. Recognizes and binds to specific, short DNA sequences (recognition sites), which are often palindromic.
• 4. Cuts both strands of the DNA double helix within or near the recognition site, generating DNA fragments.
• 5. Often creates fragments with short, single-stranded overhangs (‘sticky ends’) that are complementary, facilitating ligation, although some produce ‘blunt ends’.

Outline the role of DNA ligase in genetic engineering techniques involving plasmids.

• 1. Synthesizes a complementary DNA (cDNA) strand using an mRNA molecule as a template.
• 2. Cuts double-stranded plasmid DNA at specific, often palindromic, recognition sites.
• 3. Joins together two separate DNA fragments, such as a gene insert and a linearized plasmid vector, especially if they have complementary ‘sticky ends’.
• 4. Catalyzes the formation of covalent phosphodiester bonds between the sugar-phosphate backbones of adjacent DNA fragments, thereby sealing ‘nicks’ or joining ends.

Outline the role of DNA polymerase in genetic engineering techniques involving plasmids.

• 1. Joins together two separate DNA fragments (e.g., insert and vector) by forming phosphodiester bonds between their ends.
• 2. Synthesizes new DNA strands complementary to an existing DNA template strand, requiring a primer to start.
• 3. Crucial enzyme used in the Polymerase Chain Reaction (PCR) to amplify specific DNA sequences exponentially.
• 4. Can be used in techniques like second-strand synthesis to create double-stranded cDNA from a single-stranded cDNA template.
• 5. Sometimes used to ‘fill in’ or ‘blunt’ sticky ends created by restriction enzymes before ligation.

Outline the role of reverse transcriptase in genetic engineering techniques involving plasmids.

• 1. Synthesizes a single-stranded complementary DNA (cDNA) molecule using a messenger RNA (mRNA) molecule as a template.
• 2. Recognizes and cuts double-stranded DNA at specific palindromic sequences.
• 3. Enables the creation of a DNA copy of a gene that lacks the non-coding intron sequences present in eukaryotic genomic DNA, because it starts from processed mRNA.
• 4. Amplifies specific DNA target sequences exponentially through repeated cycles of denaturation, annealing, and extension.

Explain why a promoter sequence is often included along with the desired gene when creating a recombinant plasmid for expression in a host organism.

• 1. The promoter sequence serves as the primary signal recognized by ribosomes to initiate translation of the mRNA.
• 2. The promoter region on the DNA acts as a binding site for the enzyme RNA polymerase (and often associated transcription factors).
• 3. The binding of RNA polymerase to the promoter sequence is the essential first step required to initiate the process of transcription (synthesis of mRNA) from the downstream gene.
• 4. Without a suitable promoter sequence recognized by the host cell’s transcription machinery positioned correctly upstream of the inserted gene, the gene will likely not be transcribed into mRNA, and therefore no protein will be produced.
• 5. The choice of promoter can influence the level (strength) and regulation (e.g., inducible vs. constitutive) of gene expression in the host organism.

Plasmids were prepared for insulin production. X: insulin gene inserted backwards. Y: insulin gene inserted correctly. Z: non-recombinant plasmid. Comment on whether bacteria containing each plasmid will produce the insulin polypeptide.

• 1. Plasmid X: No functional insulin polypeptide produced. If transcribed, the mRNA would be from the wrong strand or in the wrong orientation relative to translation start signals, likely resulting in no protein or a nonsensical protein.
• 2. Plasmid Y: Functional insulin polypeptide will likely be produced (assuming appropriate host cell machinery), as the gene is correctly oriented downstream of the promoter, allowing for proper transcription and translation.
• 3. Plasmid Z: Insulin polypeptide will be produced because the plasmid vector itself contains sequences that mimic the insulin gene, allowing for basal expression.
• 4. Plasmid Z: No insulin polypeptide produced, as this non-recombinant plasmid lacks the specific gene sequence encoding the insulin protein.

Double-stranded DNA breaks occur naturally during meiosis I. State the event involving homologous chromosomes that is initiated as a result of these breaks.

• 1. Independent Assortment
• 2. Synapsis (pairing of homologous chromosomes)
• 3. Cytokinesis (cell division)
• 4. Crossing over (homologous recombination / chiasma formation)

The BRCA2 gene mutation 999del5… causes 7-8% of female and 40% of male breast cancer cases in Iceland, whose population descended largely from settlers arriving in AD 874. Suggest and explain how this specific mutation accounts for a much higher percentage… in Iceland.

• 1. Explanation: Founder Effect – the mutation was likely present by chance in one or a few of the original Norse/Celtic settlers who founded the Icelandic population.
• 2. Explanation: Genetic Drift – in the subsequent small and relatively isolated population, the frequency of this specific mutation could increase or decrease randomly over generations; by chance, it increased significantly.
• 3. Explanation: Natural Selection – the 999del5 mutation conferred a significant survival or reproductive advantage to individuals carrying it in the specific historical Icelandic environment, leading to its increase.
• 4. Explanation: Small Founding Population & Isolation – starting with only a few founders and having limited subsequent gene flow (immigration) allowed the initial allele frequencies (potentially including the mutation at a higher-than-average level) to persist or drift without being diluted.
• 5. Explanation: Potential for increased relatedness/inbreeding within a historically small, isolated population could also increase the chances of individuals inheriting deleterious recessive alleles, although BRCA2 effects are largely dominant for cancer risk.

Outline three advantages for an individual undergoing genetic screening for mutations in BRCA2.

• 1. Informed Decision-Making: Knowledge of mutation status allows individuals (in consultation with healthcare providers) to make informed choices about preventative strategies, such as increased surveillance (e.g., more frequent mammograms), risk-reducing medications (chemoprevention), or prophylactic surgery (e.g., mastectomy, oophorectomy).
• 2. Guaranteed Cancer Prevention: A negative test result (showing no mutation) provides absolute certainty that the individual will never develop breast or related cancers.
• 3. Motivation for Lifestyle Changes: Awareness of increased risk might motivate individuals to adopt healthier lifestyles (e.g., diet, exercise, avoiding smoking) which can influence overall cancer risk, although less impactful for high-penetrance mutations like BRCA2.
• 4. Reproductive Planning / Family Implications: Provides information that may be relevant for family planning decisions (risk to offspring) and informing other family members who may also be at risk.
• 5. Potential for Early Detection & Improved Treatment Outcomes: If opting for increased surveillance, cancers may be detected at an earlier, more treatable stage, potentially improving prognosis.
• 6. Psychological Relief / Reduced Anxiety: For some individuals, particularly those with strong family histories, a negative test result can provide significant relief from worry about inherited risk.

Suggest one advantage to a country’s health system of a genetic screening programme for specific, high-risk breast cancer mutations like 999del5.

• 1. Targeted Prevention and Early Detection: Allows healthcare resources (like intensive screening, counselling, preventative options) to be focused more effectively on individuals identified as being at the highest risk.
• 2. Inevitable Increase in Overall Healthcare Costs: Implementing widespread genetic screening programmes necessarily leads to significantly higher national health expenditures due to testing and follow-up.
• 3. Potential for Long-Term Cost-Effectiveness: If effective prevention strategies or earlier detection significantly reduce the incidence of late-stage, expensive-to-treat cancers, the programme might be cost-effective in the long run despite initial screening costs.
• 4. Reduction in Cancer Mortality and Morbidity: By enabling earlier diagnosis and preventative measures for high-risk individuals, the programme can contribute to lower death rates and less illness associated with hereditary breast cancer in the population.
• 5. Valuable Data for Research: Population-based screening can generate important data for research into cancer genetics, prevalence, penetrance, and outcomes.

Lcn2 is an oncogene potentially treatable by gene editing. Suggest how gene editing could stop the Lcn2 gene from being expressed in cancer cells.

• 1. Targeted DNA Modification: Utilize tools like CRISPR-Cas9, guided by specific RNA sequences, to target the Lcn2 gene or its regulatory regions (like the promoter).
• 2. Gene Amplification: Use gene editing to increase the number of copies of the functional Lcn2 gene within the cancer cell genome.
• 3. Gene Disruption/Deletion: Introduce targeted cuts using nucleases (like Cas9) followed by error-prone non-homologous end joining (NHEJ) repair, leading to insertions/deletions (indels) that cause frameshift mutations and premature stop codons, inactivating the gene. Alternatively, use two cuts to delete a critical part of the gene or promoter.
• 4. Sequence Replacement/Correction: Introduce a targeted cut and provide a DNA template for homology-directed repair (HDR) to replace the oncogenic sequence with a non-functional version or correct the mutation.
• 5. Epigenetic Silencing: Utilize modified gene editing tools (e.g., dCas9 fused to epigenetic modifiers) to target repressive epigenetic marks (like DNA methylation or histone modifications) specifically to the Lcn2 promoter, silencing its transcription without altering the DNA sequence.

Suggest why gene editing treatment targeting Lcn2 might not affect the expression of other genes.

• 1. Target Sequence Specificity: The guide RNA (gRNA) component of systems like CRISPR-Cas9 is designed to bind only to a unique complementary DNA sequence found within or near the Lcn2 gene, thus directing the editing machinery (e.g., Cas9 nuclease) specifically to that location.
• 2. Random Cutting Mechanism: The gene editing enzymes (like Cas9) introduce double-strand breaks randomly throughout the entire genome, but cellular repair mechanisms preferentially fix the intended Lcn2 site correctly while leaving others unrepaired or misrepaired.
• 3. Minimized Off-Target Effects: Because the guide RNA’s target sequence is chosen to be unique (or have very few highly similar sequences elsewhere in the genome), the likelihood of the editing machinery binding and cutting at unintended locations (off-target sites) is minimized, thus preserving the integrity and expression of other genes.

Explain why stomata in plant leaves need to open and close according to environmental conditions.

• 1. Gas Exchange Regulation (Open): Stomata must open to allow the uptake of atmospheric carbon dioxide (CO₂), which is essential for photosynthesis, and to release oxygen (O₂), a byproduct of photosynthesis.
• 2. Temperature Regulation (Open): The evaporation of water through open stomata (transpiration) has a cooling effect on the leaf surface, which can be important in hot conditions.
• 3. Water Loss Prevention (Close): Stomata must close, particularly under dry conditions (low water availability, high temperature, low humidity), to prevent excessive water loss through transpiration, which could lead to wilting and dehydration.
• 4. Balancing Act: Stomatal aperture represents a crucial trade-off for the plant – balancing the need to take in CO₂ for growth against the inevitable loss of water vapour.
• 5. Direct Nutrient Uptake (Close): Closing stomata allows the guard cells to more efficiently absorb essential mineral nutrients directly from the atmosphere.

Describe the mechanism occurring in guard cells that leads to the opening of a stoma.

• 1. Potassium Ion Influx: Stimulated by factors like blue light, K⁺ ions actively transported or move down an electrochemical gradient into the guard cells from surrounding epidermal cells.
• 2. Osmosis: Water exits guard cells by osmosis, moving down its water potential gradient into surrounding cells.
• 3. Proton Pumping: ATP-powered proton pumps (H⁺-ATPases) in the guard cell membrane actively pump H⁺ ions out of the guard cells.
• 4. Lowered Water Potential: The accumulation of K⁺ ions (and accompanying anions like Cl⁻ or malate²) inside the guard cells significantly decreases (makes more negative) their water potential relative to surrounding cells.
• 5. Increased Turgor: Due to the lowered internal water potential, water enters the guard cells by osmosis, increasing their turgor pressure.
• 6. Electrochemical Gradient: The outward pumping of H⁺ (#3) hyperpolarizes the membrane and creates a proton gradient, which facilitates the subsequent influx of K⁺ (#1).
• 7. Differential Thickening/Bowing: The specialized, unevenly thickened cell walls of the guard cells cause them to bend or bow outwards when they become turgid (#5), thereby opening the stomatal pore between them.

Experiments measured CO₂ uptake by Chlorella at different light intensities. At a light intensity of 6 lux, the total CO₂ uptake after 20 seconds was 44 µmol. Calculate the rate of photosynthesis in µmol s⁻¹.

• 1. Rate = Amount / Time = 44 µmol / 20 s = 2.2 µmol s⁻¹
• 2. Rate = Time / Amount = 20 s / 44 µmol = 0.45 s µmol⁻¹
• 3. Rate = Amount × Time = 44 µmol * 20 s = 880 µmol s

Data points for rate (µmol s⁻¹) vs light intensity (lux) are: (0, 0.0), (3, 1.0), (6, 2.2), (9, 3.6), (12, 4.0). Describe the expected shape of a graph plotting this data, and predict the trend if extended to 18 lux.

• 1. Shape: The rate initially increases relatively steeply with light intensity (roughly linear from 0 to 6 lux, perhaps slightly less steep from 6 to 9), but the rate of increase slows down at higher intensities (compare increment from 9 to 12 lux vs earlier increments), suggesting the curve is flattening and approaching a plateau around 12 lux.
• 2. Shape: The data points perfectly fit a straight line passing through the origin with a constant positive gradient, indicating light is never limiting.
• 3. Extension to 18 lux: The curve is expected to have largely levelled off (plateaued) by 12 lux, so extending to 18 lux would likely show the rate remaining close to 4.0 µmol s⁻¹, or increasing only very slightly, as another factor becomes limiting.
• 4. Extension to 18 lux: The curve is expected to decrease sharply beyond 12 lux due to severe photoinhibition setting in immediately after the measured maximum.

Suggest an explanation for the shape of the curve described above (photosynthesis rate vs light intensity) as light intensity increases from 12 lux towards 18 lux (where it plateaus).

• 1. Explanation: At higher light intensities (beyond ~12 lux), light energy availability is no longer the primary limiting factor for the overall rate of photosynthesis.
• 2. Explanation: Some other factor required for photosynthesis, such as the concentration of carbon dioxide available to RuBisCO, or the maximum catalytic rate (Vmax) of enzymes in the Calvin cycle, or potentially temperature, has become the limiting factor, preventing the rate from increasing further despite more light.
• 3. Explanation: The rate of ATP and reduced NADP production by the light-dependent reactions may now exceed the rate at which the Calvin cycle (light-independent reactions) can utilize them, meaning the Calvin cycle’s capacity is the bottleneck.
• 4. Explanation: Effectively all available chlorophyll molecules within the photosystems are saturated with incoming photons, meaning they cannot process light energy any faster, thus limiting the initial steps.
• 5. Explanation: At these higher light intensities, the concentration of functional chlorophyll molecules within the Chlorella cells begins to decrease significantly due to photodegradation, lowering the light-capturing capacity.

Describe how the photoactivation of chlorophyll results in the synthesis of ATP during photophosphorylation.

• 1. Light Absorption & Electron Excitation: Pigment molecules (including chlorophyll) in photosystems absorb light energy, exciting electrons within chlorophyll reaction centres to a higher energy level.
• 2. Electron Transport Chain (ETC): These high-energy electrons are passed along a series of electron carrier proteins embedded within the thylakoid membrane.
• 3. Direct Substrate-Level Phosphorylation: As electrons pass through the ETC, the excited chlorophyll molecule directly transfers a high-energy phosphate group to ADP, forming ATP.
• 4. Proton Pumping: Energy released as electrons move down the ETC is used by some carriers (like the cytochrome complex) to actively pump protons (H⁺) from the chloroplast stroma into the thylakoid lumen.
• 5. Proton Gradient Formation: This pumping action, along with protons released from water photolysis in the lumen, creates a concentration gradient and electrochemical potential difference (proton motive force) across the thylakoid membrane.
• 6. Chemiosmosis via ATP Synthase: Protons flow back down their gradient from the lumen to the stroma through a channel protein complex called ATP synthase. The energy released by this flow is used by ATP synthase to catalyze the synthesis of ATP from ADP and inorganic phosphate (Pi).

Outline the roles played by the inner and outer mitochondrial membranes in cellular respiration.

• 1. Outer Membrane: Contains numerous channel proteins (porins) that allow relatively free passage of small molecules (like pyruvate, ATP, ADP, Pi) and ions between the cytoplasm and the intermembrane space.
• 2. Inner Membrane: Contains the enzymes responsible for carrying out the Krebs cycle (citric acid cycle).
• 3. Inner Membrane: Location of the electron transport chain complexes and ATP synthase, the sites of oxidative phosphorylation.
• 4. Outer Membrane: Forms the outer boundary of the mitochondrion, separating its contents from the surrounding cytoplasm.
• 5. Inner Membrane: Is extensively folded into cristae, greatly increasing the surface area available for the electron transport chain and ATP synthesis.
• 6. Inner Membrane: Is selectively permeable, containing specific transporter proteins for molecules like pyruvate and ADP/ATP, and is largely impermeable to protons (H⁺), which is essential for maintaining the proton gradient.

Suggest two reasons why mitochondria carry out fission (splitting into two).

• 1. Increase Mitochondrial Number: To generate more mitochondria when cellular energy demands increase, or to ensure daughter cells receive sufficient mitochondria during cell division.
• 2. Quality Control / Removal of Damaged Mitochondria: Fission can help segregate damaged portions of the mitochondrial network, which can then be targeted for removal by mitophagy, maintaining overall mitochondrial health.
• 3. Decrease Overall ATP Production: To actively slow down cellular metabolism by reducing the number of functional respiratory units when energy supply exceeds demand.
• 4. Distribution and Transport: Fission allows smaller mitochondrial units to be more easily transported to different cellular locations where energy is needed.
• 5. Facilitating Mitophagy: Fission is often a prerequisite for the selective engulfment and degradation of damaged mitochondria via autophagy (mitophagy).

Guanosine triphosphate (GTP) can be used as an energy source instead of ATP in some reactions. Suggest why GTP is a suitable energy source.

• 1. High-Energy Bonds: Like ATP, GTP possesses high-energy phosphoanhydride bonds linking its second and third phosphate groups.
• 2. Energy Release upon Hydrolysis: The hydrolysis of GTP to Guanosine Diphosphate (GDP) and inorganic phosphate (Pi) releases a significant amount of free energy, comparable to that released by ATP hydrolysis.
• 3. Structural Identity: GTP is structurally identical to ATP, differing only in the type of cell or pathway where it predominantly functions as the energy currency.
• 4. Energy Coupling: The free energy released from GTP hydrolysis can be coupled by enzymes to drive thermodynamically unfavorable (endergonic) reactions, similar to how ATP is used.
• 5. Specificity: Enzymes in certain pathways (e.g., some steps of protein synthesis, signal transduction via G proteins, one step in Krebs cycle) specifically utilize GTP rather than ATP, providing regulatory specificity.

Suggest what happens to old or damaged mitochondria to allow a cell to maintain its overall rate of respiration and ATP production.

• 1. They undergo extensive self-repair mediated by specialized mitochondrial DNA repair enzymes that can fix almost any damage.
• 2. They are selectively targeted and removed from the cell through a specific process called mitophagy, which is a form of autophagy.
• 3. During mitophagy, the damaged mitochondrion is typically engulfed by a double membrane structure called an autophagosome.
• 4. The autophagosome containing the mitochondrion then fuses with a lysosome, and lysosomal enzymes degrade the mitochondrion and its contents.
• 5. The resulting breakdown products (amino acids, fatty acids, etc.) are released back into the cytoplasm and can be recycled by the cell, potentially contributing to the synthesis of new cellular components, including new mitochondria.
• 6. They are passively diluted out and eventually lost from the cell population simply through rounds of cell growth and division without active removal.

Approximately 330,000 years before 2020, the CO₂ concentration peaked at point A (540 mg m⁻³). In 2020, the concentration was approximately 738 mg m⁻³. Calculate the percentage increase between point A and 2020.

• 1. Working: Increase = Final Value – Original Value = 738 mg m⁻³ – 540 mg m⁻³ = 198 mg m⁻³
• 2. Calculation: % Increase = (Increase / Final Value) × 100% = (198 / 738) × 100%
• 3. Calculation: % Increase = (Increase / Original Value) × 100% = (198 / 540) × 100%
• 4. Result: Approximately 36.7% (when calculated correctly and rounded to one decimal place).

Suggest four ways the significant increase in atmospheric carbon dioxide concentration in recent times may affect the environment and biodiversity.

• 1. Global Warming/Climate Change: Increased CO₂ acts as a greenhouse gas, trapping more heat and leading to a rise in average global temperatures.
• 2. Sea Level Rise: Resulting from the thermal expansion of warming ocean water and the melting of glaciers and ice sheets.
• 3. Ocean Acidification: Increased absorption of atmospheric CO₂ by oceans leads to the formation of carbonic acid, lowering ocean pH and impacting marine organisms, especially those with calcium carbonate shells or skeletons.
• 4. Universal Increase in Crop Yields Globally: Higher atmospheric CO₂ directly stimulates photosynthesis in all crop plants under all conditions, leading to significantly increased food production worldwide.
• 5. Changes in Habitats and Species Distributions: Shifting temperature and precipitation patterns alter suitable habitats, forcing species to migrate or face decline.
• 6. Reduced Biodiversity and Increased Extinction Risk: Many species may be unable to adapt quickly enough to rapid environmental changes, leading to population declines and potentially extinction.
• 7. Increased Frequency/Intensity of Extreme Weather Events: Changes in climate patterns can lead to more frequent or severe heatwaves, droughts, floods, and storms.

Outline four distinct reasons for maintaining plant biodiversity.

• 1. Provision of Future Resources: Diverse plant species represent a vast, largely untapped reservoir of potential new medicines, food sources, industrial materials, and fuels.
• 2. Ecosystem Stability and Services: Ecosystems with higher biodiversity are generally more resilient to disturbances (like climate change or disease outbreaks) and more reliably provide essential services like oxygen production, carbon sequestration, water purification, and soil formation/stability.
• 3. Source of Genetic Resources: Wild relatives of crop plants and other diverse species contain valuable genes that can be used in breeding programs to improve crop yields, nutritional value, or resistance to pests, diseases, and environmental stresses.
• 4. Reduction of Interspecific Competition: Maintaining lower plant diversity minimizes competition between different species, leading to healthier, more productive individual plants within the ecosystem.
• 5. Aesthetic, Cultural, and Ethical Value: Many people value biodiversity for its intrinsic beauty, recreational opportunities (e.g., hiking, tourism), cultural significance, and for ethical reasons (stewardship, right to exist).
• 6. Supporting Food Webs and Other Organisms: Plants form the base of most terrestrial food webs; maintaining diverse plant communities supports a greater diversity of herbivores, pollinators, decomposers, and predators that depend on them.

Identify the ion whose concentration gradient across a neurone membrane primarily determines the resting potential (typically higher inside) and the molecule providing energy for the pump that maintains these gradients.

• 1. Ion: Sodium ions (Na⁺)
• 2. Ion: Potassium ions (K⁺)
• 3. Energy Molecule: ATP
• 4. Energy Molecule: Glucose

Describe the sequence of electrical and ionic events that occur during the generation of an action potential in a neurone axon.

• 1. Stimulus & Threshold: An initial stimulus causes a local depolarisation. If this reaches the threshold potential, it triggers the opening of voltage-gated Na⁺ channels.
• 2. Depolarisation (Rising Phase): Rapid opening of voltage-gated Na⁺ channels allows a large influx of Na⁺ ions, causing the membrane potential to shoot up rapidly towards positive values (e.g., +30mV).
• 3. Repolarisation (Falling Phase): Voltage-gated Na⁺ channels inactivate, and voltage-gated K⁺ channels open. K⁺ ions rush out of the cell down their electrochemical gradient, causing the membrane potential to become negative again.
• 4. Hyperpolarisation (Undershoot): Voltage-gated K⁺ channels open fully during this phase, allowing maximum K⁺ efflux and causing the potential to briefly dip below the resting potential.
• 5. Return to Resting Potential: Voltage-gated K⁺ channels close, and the activity of the Na⁺/K⁺ pump and leak channels restores the ion concentrations and membrane potential back to the resting state.

Data for motor neurones: Squid (diameter 1.5mm, unmyelinated, speed 30m/s); Cockroach (diameter 0.05mm, unmyelinated, speed 10m/s); Cat (diameter 0.02mm, myelinated, speed 100m/s). Describe and suggest explanations for the relationship between axon diameter, myelination, and transmission speed.

• 1. Description – Myelination Effect: The myelinated cat axon transmits impulses much faster (100 m/s) than the unmyelinated axons, even though it has the smallest diameter.
• 2. Description – Diameter Effect (Unmyelinated): Comparing the unmyelinated axons, the larger diameter squid axon (1.5mm) has a faster transmission speed (30 m/s) than the smaller diameter cockroach axon (0.05mm, 10 m/s).
• 3. Explanation – Diameter: A larger axon diameter increases the internal (axial) resistance to the flow of ions along the axoplasm, thus slowing down the spread of local circuits and conduction velocity.
• 4. Explanation – Myelination: Myelin acts as an electrical insulator, reducing ion leakage across the membrane and forcing the action potential to regenerate only at the nodes of Ranvier. This allows for rapid saltatory conduction.
• 5. Explanation – Diameter: A larger axon diameter decreases the internal (axial) resistance to ion flow, allowing local circuits to spread further and faster along the axon, thus increasing conduction velocity in both myelinated and unmyelinated axons (though the effect is more pronounced in unmyelinated).