Skill Align SACE Stage 2 Biology - Free Online Pack 0
An independently authored e-exam-style simulation with worked answers, marking guidance and diagnostic review prompts.
- Paper
- Electronic Examination Printable Simulation
- Reading
- Included in total time
- Writing
- 130 minutes total
- Assessment
- 120 marks
An approved calculator may be used. Notes and other unauthorised materials are not permitted. This printable resource simulates the content demands of an electronic examination; it does not reproduce SACE e-exam software.
Section 1 - Multiple-choice questions
Questions 1-10 are multiple choice. Select the best answer for each question. Each question is worth 1 mark.
Question 1
1 mark- Eukaryotic and prokaryotic DNA are both usually circular and lack associated proteins.
- Eukaryotic nuclear DNA is associated with histones in linear chromosomes, whereas prokaryotic DNA is usually unbound and circular.
- Prokaryotic DNA forms linear chromosomes inside a nucleus, whereas eukaryotic nuclear DNA is circular.
- Eukaryotic nuclear DNA is usually one circular chromosome, whereas prokaryotic DNA forms histone-bound linear chromosomes.
Question 2
1 mark- Heating separates the complementary DNA strands by disrupting the weak bonds between them.
- Heating allows primers to bind to their complementary target sequences.
- Heating provides the optimal conditions for polymerase to extend the bound primers.
- Heating covalently joins the amplified target fragments into a plasmid.
Question 3
1 mark- active transport through an energy-coupled carrier protein.
- simple diffusion directly through the phospholipid bilayer.
- osmosis of water through a selective membrane channel.
- facilitated diffusion.
Question 4
1 mark- provide the main site of glycolysis and pyruvate formation.
- shorten the diffusion distance between nuclear DNA and cytosolic ribosomes.
- provide a large surface area for membrane-associated reactions in aerobic respiration.
- store respiratory substrates separately from mitochondrial enzymes.
Question 5
1 mark- More water is reabsorbed from the nephron, producing a smaller volume of concentrated urine.
- Less water is reabsorbed from the nephron, producing a larger volume of dilute urine.
- More glucose is reabsorbed from the nephron, producing a smaller volume of dilute urine.
- Less sodium is filtered at the kidney, while urine volume remains unchanged.
Question 6
1 mark- neurotransmitter is destroyed before any molecule can diffuse back across the cleft.
- neurotransmitter is released from the presynaptic ending and receptors are located on the postsynaptic membrane.
- action potentials can travel only from dendrites to axons in both connected neurons.
- the postsynaptic membrane releases transmitter that binds to presynaptic vesicles.
Question 7
1 mark- Mitochondria have extensively folded inner membranes used in aerobic respiration.
- Mitochondria are inherited within the cytoplasm of many eukaryotic cells.
- Mitochondria supply ATP to many metabolically active eukaryotic cells.
- Mitochondria contain circular DNA and ribosomes resembling those of prokaryotes.
Question 8
1 mark- directional selection acting against the neutral allele.
- gene flow caused by emigration of the surviving insects.
- genetic drift following a population bottleneck.
- a founder effect caused by colonisation of a new island.
Question 9
1 mark- The template strand read in the 5' to 3' direction
- The coding strand read in the 5' to 3' direction
- The coding strand after every base has been replaced by its complement
- Both DNA strands after introns have been removed from each strand
Question 10
1 mark- Replication helps estimate random variation and increases confidence in the mean response.
- Replication eliminates all random variation from the measurements.
- Replication identifies and corrects any calibration error in the equipment.
- Replication proves that pH caused the response without requiring controlled variables.
Section 2 - Response questions
Questions 11-16 contain several parts. Answer every part and use the supplied biological evidence where required.
Question 11
18 marksA coastal banksia population contains a heat-response gene. The reference coding DNA triplet is 5'-CAC-3'; a variant triplet is 5'-CTC-3'. After a 40 °C treatment, mean gene mRNA abundance is 2.4 relative units in reference plants and 6.0 relative units in variant plants. Leaf survival is 58% and 81% respectively. A codon excerpt is supplied.
Question 12
17 marksEqual yeast suspensions containing excess glucose are incubated for 12 minutes. Mean carbon dioxide production is 8, 19, 31, 24 and 4 mL at 10, 20, 30, 40 and 50 °C respectively. Each mean comes from four independent flasks. At 50 °C, membrane leakage is also greater than at 30 °C.
Question 13
19 marksDuring a supervised desert hike, volunteers lose water through sweating. Mean plasma solute concentration rises from 286 to 298 mOsm kg⁻¹, mean urine flow falls from 1.8 to 0.45 mL min⁻¹, and plasma ADH rises fourfold. After drinking water, all three measures move toward baseline over 90 minutes.
Question 14
18 marksThree island snail populations, North, Central and South, originated from one mainland population. Mitochondrial sequence differences are: North-Central 1.2%, Central-South 4.8%, and North-South 5.1%. North and Central snails mate in laboratory trials and produce fertile offspring. Crosses involving South snails produce embryos that rarely hatch. Shell colour overlaps among all populations.
Question 15
20 marksHuman fibroblasts in culture contain a mutation in a gene product that normally stops the cell cycle when DNA is damaged. Mutant cultures continue dividing after ultraviolet exposure; reference cultures pause division. Researchers repair the mutation with CRISPR-Cas9. In repaired cultures, 74% of cells pause after ultraviolet exposure, compared with 18% of untreated mutant cells and 79% of reference cells.
Question 16
18 marksA drained saltmarsh is reconnected to tidal flow. Plant richness, bare-ground cover and soil organic matter are recorded after 0, 2, 5 and 12 years. Richness changes from 3 to 7, 12 and 15 species; bare ground changes from 82% to 51%, 24% and 16%; organic matter changes from 1.1% to 1.5%, 2.3% and 3.2%. A nearby undrained reference marsh has 17 species and 3.5% organic matter.
Worked Solutions And Marking Guide
General marking principles
- Award each listed mark independently when the required biological evidence is demonstrated.
- Accept scientifically accurate equivalent wording, representations and logically equivalent calculations.
- Credit a justified conclusion only to the extent supported by the supplied evidence and its limitations.
Section 1 Question 1
Answer: Eukaryotic nuclear DNA is associated with histones in linear chromosomes, whereas prokaryotic DNA is usually unbound and circular.
The current Stage 2 outline distinguishes histone-bound linear nuclear chromosomes from the usually unbound circular chromosome of prokaryotes.
Section 1 Question 2
Answer: Heating separates the complementary DNA strands by disrupting the weak bonds between them.
Denaturation disrupts hydrogen bonding between complementary bases so that each DNA strand can act as a template in later steps.
Section 1 Question 3
Answer: facilitated diffusion.
Facilitated diffusion uses a selective membrane protein but moves material down its gradient without direct energy input.
Section 1 Question 4
Answer: provide a large surface area for membrane-associated reactions in aerobic respiration.
Cristae increase the membrane area available for respiratory electron-transfer components and ATP-producing processes.
Section 1 Question 5
Answer: More water is reabsorbed from the nephron, producing a smaller volume of concentrated urine.
ADH increases water permeability in later nephron regions, conserving water and helping restore body-fluid concentration.
Section 1 Question 6
Answer: neurotransmitter is released from the presynaptic ending and receptors are located on the postsynaptic membrane.
The asymmetric locations of vesicle release and receptor binding give a chemical synapse its usual direction of information transfer.
Section 1 Question 7
Answer: Mitochondria contain circular DNA and ribosomes resembling those of prokaryotes.
Bacteria-like circular DNA and ribosomes are consistent with mitochondria descending from prokaryotic cells incorporated by an ancestral host.
Section 1 Question 8
Answer: genetic drift following a population bottleneck.
Chance survival during a severe population reduction can alter allele frequencies independently of the fitness effects of the alleles.
Section 1 Question 9
Answer: The coding strand read in the 5' to 3' direction
The mRNA sequence corresponds to the coding strand, while RNA polymerase uses the complementary template strand to produce it.
Section 1 Question 10
Answer: Replication helps estimate random variation and increases confidence in the mean response.
Independent repeats reveal measurement variability and make a mean less dependent on one anomalous observation.
Section 2 Question 11
(a) - The reference coding triplet produces the mRNA codon 5'-CAC-3'. - CAC specifies histidine in the supplied codon excerpt. - The variant produces 5'-CUC-3', which specifies leucine.
(b) - The DNA change is a single-nucleotide substitution from A to T in the coding strand. - The altered codon makes this a missense mutation rather than a silent or stop mutation. - Histidine is replaced by leucine in the primary structure of the encoded polypeptide.
(c) - Variant plants have 2.5 times the heat-treated mRNA abundance of reference plants. - Variant plants also show a 23 percentage-point increase in leaf survival. - The paired pattern is consistent with the variant or a linked difference contributing to heat tolerance. - The association does not establish that increased transcript abundance caused the survival difference.
(d) - Replacing one amino acid changes the protein's primary structure. - Different side-chain properties can alter local bonding and polypeptide folding. - A changed tertiary shape could modify an active site, binding site or protein stability. - Functional testing is required because a substitution outside a critical region may have little effect.
(e) - Create matched plant lines that differ only at the specified nucleotide, using an appropriate gene-editing method. - Include edited reference, edited variant and sham-edited controls from the same genetic background. - Apply identical heat treatments with random allocation and sufficient independent plant replicates. - Measure transcript abundance, protein activity and survival, then compare effect sizes with uncertainty.
Mark allocation
Part a (3 marks)
- 1 mark: The reference coding triplet produces the mRNA codon 5'-CAC-3'.
- 1 mark: CAC specifies histidine in the supplied codon excerpt.
- 1 mark: The variant produces 5'-CUC-3', which specifies leucine.
Part b (3 marks)
- 1 mark: The DNA change is a single-nucleotide substitution from A to T in the coding strand.
- 1 mark: The altered codon makes this a missense mutation rather than a silent or stop mutation.
- 1 mark: Histidine is replaced by leucine in the primary structure of the encoded polypeptide.
Part c (4 marks)
- 1 mark: Variant plants have 2.5 times the heat-treated mRNA abundance of reference plants.
- 1 mark: Variant plants also show a 23 percentage-point increase in leaf survival.
- 1 mark: The paired pattern is consistent with the variant or a linked difference contributing to heat tolerance.
- 1 mark: The association does not establish that increased transcript abundance caused the survival difference.
Part d (4 marks)
- 1 mark: Replacing one amino acid changes the protein's primary structure.
- 1 mark: Different side-chain properties can alter local bonding and polypeptide folding.
- 1 mark: A changed tertiary shape could modify an active site, binding site or protein stability.
- 1 mark: Functional testing is required because a substitution outside a critical region may have little effect.
Part e (4 marks)
- 1 mark: Create matched plant lines that differ only at the specified nucleotide, using an appropriate gene-editing method.
- 1 mark: Include edited reference, edited variant and sham-edited controls from the same genetic background.
- 1 mark: Apply identical heat treatments with random allocation and sufficient independent plant replicates.
- 1 mark: Measure transcript abundance, protein activity and survival, then compare effect sizes with uncertainty.
Section 2 Question 12
(a) - Carbon dioxide production rises from 8 mL at 10 °C to a maximum of 31 mL at 30 °C. - Production then falls to 24 mL at 40 °C and 4 mL at 50 °C. - The relationship is non-linear and shows an optimum near 30 °C under these conditions.
(b) - Higher temperature increases molecular movement and the frequency of productive enzyme-substrate collisions. - Respiratory enzymes therefore catalyse pathway reactions more rapidly up to the observed optimum. - Faster glucose breakdown produces carbon dioxide more rapidly during the fixed interval.
(c) - High temperature can disrupt bonds that maintain enzyme tertiary structure. - Altered active sites reduce catalysis in the respiratory pathway. - Heat also disrupts membrane structure, increasing uncontrolled movement of cell contents.
(d) - Glucose contains chemical energy that yeast can release through cellular reactions. - Some released energy is captured when ADP and inorganic phosphate form ATP. - ATP hydrolysis then supplies energy for cellular work. - Fermentation releases less usable energy per glucose molecule than aerobic respiration.
(e) - Four independent flasks provide some estimate of random variation, but no spread or uncertainty values are reported. - Carbon dioxide volume is an indirect rate measure and may be influenced by gas solubility or leakage. - Use calibrated gas sensors or sealed respirometers and report variation for every temperature. - Maintain equal yeast density, glucose concentration, pH and actual suspension temperature across treatments.
Mark allocation
Part a (3 marks)
- 1 mark: Carbon dioxide production rises from 8 mL at 10 °C to a maximum of 31 mL at 30 °C.
- 1 mark: Production then falls to 24 mL at 40 °C and 4 mL at 50 °C.
- 1 mark: The relationship is non-linear and shows an optimum near 30 °C under these conditions.
Part b (3 marks)
- 1 mark: Higher temperature increases molecular movement and the frequency of productive enzyme-substrate collisions.
- 1 mark: Respiratory enzymes therefore catalyse pathway reactions more rapidly up to the observed optimum.
- 1 mark: Faster glucose breakdown produces carbon dioxide more rapidly during the fixed interval.
Part c (3 marks)
- 1 mark: High temperature can disrupt bonds that maintain enzyme tertiary structure.
- 1 mark: Altered active sites reduce catalysis in the respiratory pathway.
- 1 mark: Heat also disrupts membrane structure, increasing uncontrolled movement of cell contents.
Part d (4 marks)
- 1 mark: Glucose contains chemical energy that yeast can release through cellular reactions.
- 1 mark: Some released energy is captured when ADP and inorganic phosphate form ATP.
- 1 mark: ATP hydrolysis then supplies energy for cellular work.
- 1 mark: Fermentation releases less usable energy per glucose molecule than aerobic respiration.
Part e (4 marks)
- 1 mark: Four independent flasks provide some estimate of random variation, but no spread or uncertainty values are reported.
- 1 mark: Carbon dioxide volume is an indirect rate measure and may be influenced by gas solubility or leakage.
- 1 mark: Use calibrated gas sensors or sealed respirometers and report variation for every temperature.
- 1 mark: Maintain equal yeast density, glucose concentration, pH and actual suspension temperature across treatments.
Section 2 Question 13
(a) - Sweating removes proportionally more water than dissolved solute from body fluids. - Plasma solute concentration therefore rises above its usual regulated range. - Cells may lose water by osmosis if the extracellular fluid remains more concentrated.
(b) - Osmoreceptors detect the rise in body-fluid solute concentration. - Neural signalling promotes increased release of ADH into the blood. - ADH acts on target regions of nephrons to increase water permeability. - Greater water reabsorption reduces the original concentration change and inhibits the initiating stimulus.
(c) - ADH rises at the same time that urine flow falls from 1.8 to 0.45 mL min⁻¹. - The urine-flow reduction is 75% relative to baseline. - Lower urine flow is consistent with more filtered water returning to the blood. - Movement toward baseline after drinking is consistent with reduced need for water conservation.
(d) - Sensory receptors and neurons detect and rapidly transmit information about the internal change. - ADH is a chemical messenger transported in the bloodstream. - Neural signals are rapid and directed, whereas hormonal effects are generally slower and more sustained. - The two systems act together because neural detection controls an endocrine output affecting the kidneys.
(e) - The time association and known target response support an ADH contribution. - The volunteers were not randomly assigned to different ADH conditions, so the evidence is observational. - Changes in blood pressure, kidney filtration or other hormones could also affect urine flow. - A controlled physiological study with appropriate ethical safeguards and direct water-reabsorption measures would be needed for stronger causal inference.
Mark allocation
Part a (3 marks)
- 1 mark: Sweating removes proportionally more water than dissolved solute from body fluids.
- 1 mark: Plasma solute concentration therefore rises above its usual regulated range.
- 1 mark: Cells may lose water by osmosis if the extracellular fluid remains more concentrated.
Part b (4 marks)
- 1 mark: Osmoreceptors detect the rise in body-fluid solute concentration.
- 1 mark: Neural signalling promotes increased release of ADH into the blood.
- 1 mark: ADH acts on target regions of nephrons to increase water permeability.
- 1 mark: Greater water reabsorption reduces the original concentration change and inhibits the initiating stimulus.
Part c (4 marks)
- 1 mark: ADH rises at the same time that urine flow falls from 1.8 to 0.45 mL min⁻¹.
- 1 mark: The urine-flow reduction is 75% relative to baseline.
- 1 mark: Lower urine flow is consistent with more filtered water returning to the blood.
- 1 mark: Movement toward baseline after drinking is consistent with reduced need for water conservation.
Part d (4 marks)
- 1 mark: Sensory receptors and neurons detect and rapidly transmit information about the internal change.
- 1 mark: ADH is a chemical messenger transported in the bloodstream.
- 1 mark: Neural signals are rapid and directed, whereas hormonal effects are generally slower and more sustained.
- 1 mark: The two systems act together because neural detection controls an endocrine output affecting the kidneys.
Part e (4 marks)
- 1 mark: The time association and known target response support an ADH contribution.
- 1 mark: The volunteers were not randomly assigned to different ADH conditions, so the evidence is observational.
- 1 mark: Changes in blood pressure, kidney filtration or other hormones could also affect urine flow.
- 1 mark: A controlled physiological study with appropriate ethical safeguards and direct water-reabsorption measures would be needed for stronger causal inference.
Section 2 Question 14
(a) - North and Central populations have the smallest sequence difference at 1.2%. - Fewer accumulated sequence differences are consistent with a more recent common ancestor. - South is more distantly related to both North and Central in this dataset.
(b) - Fertile North-Central offspring indicate that substantial gene flow is biologically possible between those populations. - Rare hatching in crosses with South indicates a post-zygotic barrier. - The barrier occurs after fertilisation because embryos form before most fail. - Laboratory crosses do not show how often mating occurs in the wild.
(c) - Geographic separation reduced gene flow between South and the other islands. - Mutation introduced new variation and genetic drift could alter allele frequencies in the isolated population. - Different island selection pressures could favour different heritable traits. - Accumulated genetic differences eventually reduced hybrid viability, contributing to reproductive isolation.
(d) - Overlapping shell colour does not exclude species separation because morphology may remain similar. - A mitochondrial sequence represents one inherited genetic history rather than the whole genome. - Concordant nuclear sequences and wild mating observations would strengthen the lineage inference. - Species status should integrate reproductive, molecular, morphological and ecological evidence.
(e) - A molecular-clock calculation requires an independently calibrated mutation rate. - Rates can differ among lineages, genes and environmental conditions. - Sequence sampling and ancestral variation add uncertainty to a single distance estimate.
Mark allocation
Part a (3 marks)
- 1 mark: North and Central populations have the smallest sequence difference at 1.2%.
- 1 mark: Fewer accumulated sequence differences are consistent with a more recent common ancestor.
- 1 mark: South is more distantly related to both North and Central in this dataset.
Part b (4 marks)
- 1 mark: Fertile North-Central offspring indicate that substantial gene flow is biologically possible between those populations.
- 1 mark: Rare hatching in crosses with South indicates a post-zygotic barrier.
- 1 mark: The barrier occurs after fertilisation because embryos form before most fail.
- 1 mark: Laboratory crosses do not show how often mating occurs in the wild.
Part c (4 marks)
- 1 mark: Geographic separation reduced gene flow between South and the other islands.
- 1 mark: Mutation introduced new variation and genetic drift could alter allele frequencies in the isolated population.
- 1 mark: Different island selection pressures could favour different heritable traits.
- 1 mark: Accumulated genetic differences eventually reduced hybrid viability, contributing to reproductive isolation.
Part d (4 marks)
- 1 mark: Overlapping shell colour does not exclude species separation because morphology may remain similar.
- 1 mark: A mitochondrial sequence represents one inherited genetic history rather than the whole genome.
- 1 mark: Concordant nuclear sequences and wild mating observations would strengthen the lineage inference.
- 1 mark: Species status should integrate reproductive, molecular, morphological and ecological evidence.
Part e (3 marks)
- 1 mark: A molecular-clock calculation requires an independently calibrated mutation rate.
- 1 mark: Rates can differ among lineages, genes and environmental conditions.
- 1 mark: Sequence sampling and ancestral variation add uncertainty to a single distance estimate.
Section 2 Question 15
(a) - A checkpoint monitors whether conditions such as DNA integrity are suitable for division. - Checkpoint proteins can delay progression when DNA damage is detected. - The delay provides time for repair or prevents damaged DNA from being copied into daughter cells. - Loss of checkpoint control can allow mutations to accumulate through continued division.
(b) - Only 18% of untreated mutant cells pause, showing the defective checkpoint phenotype. - Repair raises pausing to 74%, a 56 percentage-point increase over untreated mutant cells. - The repaired value is close to the 79% reference value. - The pattern supports substantial restoration of checkpoint function after editing.
(c) - The gene's nucleotide sequence determines the amino acid sequence of its protein product. - A sequence change can alter protein folding, stability or binding interactions. - A dysfunctional regulatory protein may fail to transmit a stop signal at the checkpoint. - The altered molecular interaction produces a measurable change in division behaviour.
(d) - The comparison needs sham-edited mutant cells to control for the editing procedure. - Off-target changes could affect pausing independently of repair at the intended site. - A percentage alone does not report replicate cultures or uncertainty. - Sequencing the target and predicted off-target sites plus independent edited clones would strengthen attribution.
(e) - Repairing a harmful somatic mutation could restore normal regulation in a patient's affected tissue. - Somatic treatment would not intentionally transmit the edit to future generations. - Unintended edits or abnormal growth create safety risks that require long-term evidence. - Consent, equitable access and the distinction between treatment and enhancement affect responsible use.
Mark allocation
Part a (4 marks)
- 1 mark: A checkpoint monitors whether conditions such as DNA integrity are suitable for division.
- 1 mark: Checkpoint proteins can delay progression when DNA damage is detected.
- 1 mark: The delay provides time for repair or prevents damaged DNA from being copied into daughter cells.
- 1 mark: Loss of checkpoint control can allow mutations to accumulate through continued division.
Part b (4 marks)
- 1 mark: Only 18% of untreated mutant cells pause, showing the defective checkpoint phenotype.
- 1 mark: Repair raises pausing to 74%, a 56 percentage-point increase over untreated mutant cells.
- 1 mark: The repaired value is close to the 79% reference value.
- 1 mark: The pattern supports substantial restoration of checkpoint function after editing.
Part c (4 marks)
- 1 mark: The gene's nucleotide sequence determines the amino acid sequence of its protein product.
- 1 mark: A sequence change can alter protein folding, stability or binding interactions.
- 1 mark: A dysfunctional regulatory protein may fail to transmit a stop signal at the checkpoint.
- 1 mark: The altered molecular interaction produces a measurable change in division behaviour.
Part d (4 marks)
- 1 mark: The comparison needs sham-edited mutant cells to control for the editing procedure.
- 1 mark: Off-target changes could affect pausing independently of repair at the intended site.
- 1 mark: A percentage alone does not report replicate cultures or uncertainty.
- 1 mark: Sequencing the target and predicted off-target sites plus independent edited clones would strengthen attribution.
Part e (4 marks)
- 1 mark: Repairing a harmful somatic mutation could restore normal regulation in a patient's affected tissue.
- 1 mark: Somatic treatment would not intentionally transmit the edit to future generations.
- 1 mark: Unintended edits or abnormal growth create safety risks that require long-term evidence.
- 1 mark: Consent, equitable access and the distinction between treatment and enhancement affect responsible use.
Section 2 Question 16
(a) - Plant richness increases from 3 to 15 species over 12 years. - Bare-ground cover falls by 66 percentage points during the same period. - Soil organic matter rises from 1.1% to 3.2%.
(b) - Early colonising plants tolerate exposed, disturbed conditions. - Plant growth and decomposition add organic matter to the substrate. - Improved substrate structure and resource retention allow additional species to establish. - Increasing vegetation cover leaves less ground bare and modifies later species interactions.
(c) - The restored site approaches the reference in richness but remains two recorded species lower. - Its organic matter remains 0.3 percentage points below the reference value. - Similarity in two measures does not establish equivalent species composition or ecosystem function. - The data support partial recovery rather than proof of complete restoration.
(d) - Repeated observations show temporal change at the restored site. - One reference marsh provides limited representation of natural spatial variability. - Measurements before reconnection and at multiple untreated marshes would strengthen causal attribution. - Consistent seasonal timing, plot locations and sampling effort are needed for valid comparisons.
(e) - Greater genetic and species diversity can provide varied responses to future selection pressures. - Diverse communities can retain more ecological interactions and potential recovery pathways. - Conservation decisions have intergenerational consequences because extinction permanently removes evolutionary lineages.
Mark allocation
Part a (3 marks)
- 1 mark: Plant richness increases from 3 to 15 species over 12 years.
- 1 mark: Bare-ground cover falls by 66 percentage points during the same period.
- 1 mark: Soil organic matter rises from 1.1% to 3.2%.
Part b (4 marks)
- 1 mark: Early colonising plants tolerate exposed, disturbed conditions.
- 1 mark: Plant growth and decomposition add organic matter to the substrate.
- 1 mark: Improved substrate structure and resource retention allow additional species to establish.
- 1 mark: Increasing vegetation cover leaves less ground bare and modifies later species interactions.
Part c (4 marks)
- 1 mark: The restored site approaches the reference in richness but remains two recorded species lower.
- 1 mark: Its organic matter remains 0.3 percentage points below the reference value.
- 1 mark: Similarity in two measures does not establish equivalent species composition or ecosystem function.
- 1 mark: The data support partial recovery rather than proof of complete restoration.
Part d (4 marks)
- 1 mark: Repeated observations show temporal change at the restored site.
- 1 mark: One reference marsh provides limited representation of natural spatial variability.
- 1 mark: Measurements before reconnection and at multiple untreated marshes would strengthen causal attribution.
- 1 mark: Consistent seasonal timing, plot locations and sampling effort are needed for valid comparisons.
Part e (3 marks)
- 1 mark: Greater genetic and species diversity can provide varied responses to future selection pressures.
- 1 mark: Diverse communities can retain more ecological interactions and potential recovery pathways.
- 1 mark: Conservation decisions have intergenerational consequences because extinction permanently removes evolutionary lineages.
Diagnostic Checklist
| Topic | Questions | Marks | Marks Lost | Action |
|---|---|---|---|---|
| Stage 2 science inquiry skills | Q10 | 1 | ___ | Review data representation, uncertainty, experimental design, evidence limits and justified biological conclusions. |
| Topic 1 - DNA, genes and protein synthesis | Q1, Q9 | 2 | ___ | Review DNA structure and replication, gene expression, protein synthesis and protein structure-function relationships. |
| Topic 1 - Mutations, biotechnology and molecular data | Q2, Q11 | 19 | ___ | Review mutation consequences, PCR, electrophoresis, profiling, gene transfer, CRISPR and responsible use of genetic information. |
| Topic 2 - Cells as the basis of life | Q3, Q4, Q12, Q15 | 39 | ___ | Review cell organisation, membranes, transport, energy transformations, metabolism, division and cell-cycle regulation. |
| Topic 3 - Homeostasis | Q5, Q6, Q13 | 21 | ___ | Review stimulus-response and negative-feedback models, nervous control, endocrine control and regulated internal conditions. |
| Topic 4 - Evolution and speciation | Q7, Q8, Q14 | 20 | ___ | Review origins of cells, molecular evidence, reproductive isolation, variation, selection, drift and speciation. |
| Topic 4 - Population and environmental change | Q16 | 18 | ___ | Review succession, genetic diversity, extinction risk, human selection pressures and biodiversity obligations. |