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Methods used to monitor fishing activity include: 1. Vessel Monitoring Systems (VMS) or satellite tracking, which track vessel movements. 2. Deploying official observers on board vessels to verify catches. Other methods include port-side catch inspections and auditing logbooks.

Award 1 mark for each correct monitoring method described, up to a maximum of 2 marks. Acceptable answers: satellite tracking/VMS, on-board observers, inspections at ports, logbook audits, aerial patrols. Reject: net mesh size regulations (this is a control measure, not a monitoring method).

Restoration of open-cast mines involves: 1. Landfilling or backfilling the excavated pit with waste rock/overburden, followed by laying topsoil and planting native vegetation (revegetation). 2. Flooding the deep pits to create artificial lakes or wetlands, which can serve as wildlife habitats or recreational areas.

Award 1 mark for each valid restoration method, up to a maximum of 2 marks. Accept: backfilling/landfilling with overburden, landscaping/grading the land, re-spreading topsoil, planting trees/crops/revegetation, flooding the pit to make a lake/wetland. Reject: simply 'stopping mining'.

Constructing a dam has several environmental impacts: 1. The creation of a reservoir floods vast upstream areas, destroying terrestrial habitats and reducing local biodiversity. 2. The physical barrier blocks the migration of aquatic species like salmon and prevents natural nutrient-rich sediment from flowing downstream, degrading downstream ecosystems.

Award 1 mark for each explained environmental disadvantage, up to a maximum of 2 marks. Accept: upstream habitat loss/flooding, disruption of fish migration routes, reduced nutrient/sediment downstream, increased risk of greenhouse gas emissions from decomposing vegetation in reservoir. Reject: economic or social costs (e.g., displacement of people, cost of building).

1. Runoff containing agricultural fertilizers (nitrates and phosphates) causes rapid growth of algae, leading to an algal bloom. 2. As the algae die, they are decomposed by aerobic bacteria. 3. These bacteria multiply rapidly and consume large amounts of dissolved oxygen in the water during respiration, which results in a high Biochemical Oxygen Demand (BOD).

Award 1 mark for mentioning the rapid growth of algae / algal bloom caused by the nutrients. Award 1 mark for explaining that the decomposition of dead algae by aerobic bacteria consumes dissolved oxygen, raising the BOD. Max 2 marks.

To ensure validity and safety: 1. A raised cover (such as a flat stone or piece of wood on sticks) should be placed over the trap to prevent rainwater from entering and drowning the insects, and to shield them from predators. 2. Traps must be checked frequently (at least every 24 hours) to prevent captured animals from escaping, dying of exposure, or preying on one another.

Award 1 mark for each valid precaution stated, up to a maximum of 2 marks. Accept: placing a raised lid/cover to block rain/predators, checking traps frequently (daily) to prevent escape/predation, placing the rim flush with the soil, using dry traps/moist paper to prevent drowning. Reject: general personal safety rules (e.g., wearing gloves).

Rainwater harvesting captures and stores precipitation during wet periods, creating a reserve supply of fresh water to use when standard water supplies are low. Greywater reuse recycles domestic wastewater (from showers or sinks) for non-potable uses like crop irrigation and toilet flushing, thereby conserving the limited supply of clean drinking water.

Award 1 mark for explaining that rainwater harvesting stores water during wet periods to be used as a reserve during drought. Award 1 mark for explaining that greywater reuse recycles domestic wastewater for non-drinking/irrigation uses, conserving clean potable water. Max 2 marks.

Cholera is a water-borne disease spread via the fecal-oral route. 1. Providing treated or chlorinated piped water ensures that residents do not drink contaminated water. 2. Building sealed pit latrines that are located downhill and far away from water wells prevents sewage from contaminating the groundwater supply.

Award 1 mark for each valid sanitation or infrastructure improvement, up to a maximum of 2 marks. Accept: chlorinated/treated water supply, water purification stations, sealed pit latrines/toilets, positioning toilets away from water sources, proper sewage pipes/treatment systems, handwashing stations with clean water and soap. Reject: 'washing hands' on its own (must be framed as providing facilities).

Agroforestry involves growing agricultural crops and trees together. For farmers, this provides multiple sources of income and food (such as timber, fruits, and arable crops) from the same land. For the environment, tree roots anchor the soil to prevent erosion, leaf litter restores organic matter and nutrients to the soil, and the tree canopy provides habitats that support biodiversity.

Award 1 mark for a benefit to farmers (e.g., diversified income/food supply, shade for crops). Award 1 mark for an environmental benefit (e.g., reduced soil erosion, increased soil fertility/nutrients, biodiversity preservation, carbon sequestration). Max 2 marks.

Applying regulations that mandate a minimum mesh size ensures that the holes in fishing nets are large enough for smaller, immature fish to swim through and escape.

This is beneficial because:
1. It reduces bycatch of juvenile fish that have not yet reproduced.
2. It allows these young fish to survive to breeding age, ensuring they can spawn and contribute to the long-term survival and recruitment of the fish stock.

Award 1 mark for each point up to a maximum of 2 marks:
- Explaining that larger mesh sizes allow small/young/juvenile fish to escape/pass through the net [1 mark].
- Linking this to allowing fish to reach reproductive maturity / spawn / replenish the population [1 mark].

Once extraction is complete, open-cast mines leave large voids and disrupted landscapes. Rehabilitation can be achieved by:
- **Backfilling and Landscaping:** Filling the excavated hole with the original overburden and waste rock, grading the slopes to make them stable and safe, and covering with topsoil.
- **Revegetation:** Planting grasses, shrubs, or trees to prevent wind and water erosion, improve soil structure, and restore local ecosystems.

Award 1 mark for each valid method described, up to a maximum of 2 marks:
- Backfilling / filling the pit/crater with overburden or waste rock [1 mark].
- Spreading topsoil over the area to allow plant growth [1 mark].
- Revegetation / replanting trees or grasses (to stabilize soil/prevent erosion) [1 mark].
- Creating a reservoir / flooding the pit to make a lake for wildlife or recreation [1 mark].
- Bioremediation / adding lime or microbes to neutralize toxic/acidic mine soils [1 mark].
- Landscaping to reduce slope angles / prevent landslides [1 mark].

(a) Percentage decrease = \(\frac{120000 - 105000}{120000} \times 100 = \frac{15000}{120000} \times 100 = 12.5\)%. (b) Quotas set a limit on the weight of fish that can be caught; this prevents overexploitation; and ensures a sufficient breeding population is left behind to sustain the stock.

(a) [2 marks] 1 mark for correct working: \(\frac{15000}{120000} \times 100\). 1 mark for correct answer: 12.5%. (b) [2 marks] 1 mark for explaining that quotas limit catch size to prevent overfishing. 1 mark for explaining that this allows fish stocks to reproduce and sustain the population.

(a) Total cost = $1 000 000. Revegetation cost = $180 000. Percentage = \(\frac{180000}{1000000} \times 100 = 18\)%. (b) Restoring mining sites returns the land to a useful state (e.g., agriculture or recreation); replanting vegetation restores habitats and biodiversity; it stabilizes the soil to prevent erosion; and improves the visual appearance of the landscape.

(a) [2 marks] 1 mark for correct working: \(\frac{180000}{1000000} \times 100\). 1 mark for correct answer: 18%. (b) [2 marks] Award 1 mark for each valid benefit up to 2: returns land to productive use; prevents soil erosion; restores wildlife habitats/biodiversity; improves landscape aesthetics/reduces dust.

(a) As the percentage of reservoir capacity filled increases, the amount of electricity generated also increases (a positive relationship). (b) Benefits include flood prevention, domestic/industrial water supply, irrigation, and recreational opportunities. (c) Negative impacts include displacement of local human populations, loss of agricultural land, or destruction of local terrestrial ecosystems.

(a) [1 mark] State that there is a positive correlation / as capacity increases, electricity generated increases. (b) [2 marks] Award 1 mark for each of two valid benefits: flood control, irrigation, water supply, tourism. (c) [1 mark] Award 1 mark for a valid disadvantage: flooding of habitats, displacement of local communities, disruption to fish migration.

(a) The river is under the greatest oxygen stress at 500 m downstream, where dissolved oxygen is at its lowest level (2.1 mg/L). (b) Raw sewage contains organic waste. Bacteria and other decomposers feed on this organic matter, causing their population to grow rapidly. These bacteria perform aerobic respiration, consuming the dissolved oxygen in the river water, which drastically reduces oxygen levels.

(a) [1 mark] Correctly identifies 500 m. (b) [3 marks] 1 mark for noting sewage contains organic matter. 1 mark for stating that bacteria decompose/break down this organic matter. 1 mark for explaining that bacteria consume oxygen during aerobic respiration, reducing dissolved oxygen levels.

(a) Tree density in Core Area = \(\frac{420}{2000} = 0.21\) trees per square meter. (b) The Core Area has the highest species richness (18 species compared to 11 in the Buffer Zone). (c) Management strategies include: legal prohibition of logging or mining, strict patrols/monitoring by wildlife rangers, restriction of human access to scientific research only, and prohibiting development of permanent structures.

(a) [1 mark] Correct calculation: 420 / 2000 = 0.21 trees per square meter. (b) [1 mark] Core Area. (c) [2 marks] Award 1 mark for each of two valid strategies: strict restriction of human access; banning commercial extraction (logging/hunting); regular ranger patrols; allowing only scientific/monitoring activities.

(a) Deficit = \(320 \text{ mm} - 128 \text{ mm} = 192 \text{ mm}\). Percentage deficit = \(\frac{192}{320} \times 100 = 60\)%. (b) Prolonged drought leads to widespread crop failure, causing food shortages or famine. It also causes death of livestock due to water and feed shortages, and forces communities to migrate in search of resources, leading to social disruption.

(a) [2 marks] 1 mark for correct working: \(\frac{320 - 128}{320} \times 100\). 1 mark for correct answer: 60%. (b) [2 marks] Award 1 mark for each of two valid impacts: crop failure/food insecurity/famine; loss of livestock; water shortages for drinking/sanitation; migration/refugees.

(a) Case decrease = \(12500 - 1400 = 11100\). Percentage decrease = \(\frac{11100}{12500} \times 100 = 88.8\)% (accept 89%). (b) Improved sanitation ensures that human waste is safely contained and treated, which prevents raw sewage containing Vibrio cholerae from contaminating drinking water sources. This breaks the fecal-oral transmission cycle.

(a) [2 marks] 1 mark for correct working: \(\frac{11100}{12500} \times 100\). 1 mark for correct answer: 88.8% (or 89%). (b) [2 marks] 1 mark for explaining that sewage containment/treatment prevents waste contaminating drinking water sources. 1 mark for explaining that this blocks pathogens (such as Vibrio cholerae) from being ingested by humans.

(a) Difference = \(4.2\% - 0.8\% = 3.4\)% (or 3.4 percentage points). (b) Selective logging only harvests specific, mature trees, leaving younger trees and the forest canopy mostly intact. This minimizes soil erosion, preserves forest ecosystems, and allows the forest to naturally regenerate. Replanting ensures that timber resources are replaced, maintaining long-term sustainability.

(a) [1 mark] 3.4%. (b) [3 marks] 1 mark for explaining that only selected trees are cut down, leaving the forest canopy/structure intact. 1 mark for stating that replanting replaces harvested trees to maintain resources. 1 mark for explaining that this prevents soil erosion / preserves biodiversity / maintains carbon sinks compared to clear-cutting.

(a) Decrease is 120 - 65 = 55 thousand tonnes. Percentage decrease is (55 / 120) * 100 which equals 45.83%. Accept 45.8% or 46%. (b) Closed seasons prevent fishing during breeding or spawning periods. This allows fish to reproduce without disturbance, protecting breeding adults and increasing juvenile survival rates.

Part (a): 1 mark for correct working showing (120 - 65) / 120 * 100. 1 mark for correct calculation of 45.8% or 46%. Part (b): 1 mark for identifying that closed seasons protect fish during breeding/spawning seasons. 1 mark for explaining that this allows reproduction to occur / helps recover population numbers.

(a) There is a positive correlation where both tree canopy cover and soil organic matter increase as the number of years post-restoration increases. For example, canopy cover increases from 5% to 75%, and soil organic matter increases from 1.2% to 3.8% over 12 years. (b) Tree roots bind soil particles together, stabilizing the soil structure. The canopy intercepts rainfall, reducing the impact of raindrops, while organic matter improves water retention, reducing surface runoff.

Part (a): 1 mark for identifying that both canopy cover and organic matter increase over time (positive correlation). 1 mark for using data from the table to support this trend (comparing year 2 to year 12). Part (b): 1 mark for explaining how tree roots bind soil or canopy intercepts rain. 1 mark for explaining how soil organic matter improves soil structure or water retention to reduce runoff.

(a) Combined volume of crop irrigation and domestic supply is 180 + 60 = 240 million cubic meters. Percentage of total is (240 / 300) * 100 = 80%. (b) Environmental disadvantages include: Flooding of upstream terrestrial habitats causing biodiversity loss, and blockages of migratory routes for aquatic species like river fish.

Part (a): 1 mark for correct working showing (180 + 60) / 300 * 100. 1 mark for correct calculation of 80%. Part (b): 1 mark each for any two valid environmental disadvantages (max 2 marks): flooding of habitats/loss of land species; blocking fish migration; reduction in downstream sediment/siltation behind dam; altered downstream flow patterns.

(a) There is an inverse relationship between BOD and DO. As BOD increases, DO decreases. For example, at Site B, BOD is at its highest (18.5 mg/L) while DO is at its lowest (2.1 mg/L). Further downstream at Site E, BOD decreases to 1.5 mg/L and DO recovers to 9.2 mg/L. (b) Sewage contains high levels of organic matter. Aerobic bacteria multiply rapidly to decompose this organic waste, and their respiration consumes large amounts of dissolved oxygen from the water, creating a high oxygen demand.

Part (a): 1 mark for identifying the inverse/negative relationship. 1 mark for using comparative data from the table to support the trend. Part (b): 1 mark for stating that sewage contains organic matter decomposed by bacteria. 1 mark for explaining that these bacteria respire and consume dissolved oxygen, raising the oxygen demand.

Using a combination of net mesh size regulations and marine protected areas (MPAs) provides a multi-faceted approach to fishery management:

1. **Net Mesh Size Regulations**:
- **Benefits**: Specifying larger mesh sizes allows juvenile, immature fish and smaller non-target species (bycatch) to escape. This ensures that younger fish survive to reproductive age, preserving the breeding population and supporting long-term stock recovery.
- **Limitations**: It is difficult and expensive to monitor and enforce mesh size regulations on the open ocean. Some fishers may use illegal net liners to capture smaller species, undermining the policy.

2. **Marine Protected Areas (MPAs)**:
- **Benefits**: Creating 'no-take' zones protects critical habitats (such as coral reefs, mangrove nurseries, or spawning grounds) from physical damage caused by destructive fishing methods like bottom trawling. It allows adult fish populations to grow larger and produce exponentially more offspring. This leads to a 'spillover effect' where fish migrate out of the MPA, replenishing surrounding fishing grounds.
- **Limitations**: MPAs can cause short-term economic hardship and loss of livelihood for local fishing communities who lose access to traditional grounds. Furthermore, MPAs are fixed geographical areas and offer little protection to highly migratory species (e.g., tuna) that swim across international boundaries.

**Conclusion**:
Combining both methods is highly effective because MPAs protect breeding hotspots and overall ecosystem biodiversity, while net mesh regulations ensure that sustainable harvesting practices are maintained in the open fishing zones outside the MPAs.

Award up to 6 marks for balanced points addressing benefits and limitations of both strategies, and a concluding evaluation.

**Net Mesh Size Regulations (Max 2 marks):**
- Larger mesh sizes allow juvenile fish/non-target species to escape, ensuring they reach reproductive maturity. [1]
- Limitation: High cost/difficulty of monitoring and enforcement at sea (e.g., use of illegal inner liners). [1]

**Marine Protected Areas (MPAs) (Max 2 marks):**
- Protects habitats (spawning/nursery grounds) from physical damage and allows populations to recover, leading to a 'spillover effect' into fished areas. [1]
- Limitation: Creates short-term loss of income/livelihoods for local fishers OR does not protect highly migratory species once they leave the boundary. [1]

**Synthesis/Evaluation (Max 2 marks):**
- Combining spatial protection (MPAs) with technical gear regulations (mesh size) is more effective than either alone because it protects both habitat structure and fish population age-structures simultaneously. [1]
- Successful management requires strong international cooperation, satellite tracking (VMS), and engagement with local communities to ensure compliance. [1]

Restoring an open-cast bauxite mine requires a systematic approach to reclaim the land and re-establish a functioning ecosystem:

1. **Land profiling / Geotechnical stability**:
- Deep open pits must be backfilled with the overburden (rock and soil removed during the initial extraction process) and graded to match the surrounding natural landscape. This reduces the slope angle, making the land stable and preventing future landslides, rockfalls, or severe soil erosion.

2. **Soil restoration**:
- During the initial mining phase, topsoil should have been removed and stored. This preserved topsoil must be spread back over the graded overburden. Topsoil contains essential organic matter, micro-organisms, and seed banks that are critical for supporting plant life.

3. **Revegetation / Bioremediation**:
- Fast-growing pioneer plant species should be planted first to bind the loose soil with their root systems, minimizing wind and water erosion. These plants can be followed by native trees and shrubs to rebuild the natural forest or grassland ecosystem, which helps in sequestering carbon and restoring the nitrogen cycle.

4. **Water and Wetland Management**:
- Deep depressions that cannot be fully backfilled can be converted into artificial lakes or wetlands. This provides freshwater habitats, attracts waterfowl and aquatic organisms, increases local biodiversity, and can serve as a recreational resource for local communities.

5. **Monitoring and Management**:
- The restored area must be monitored to control invasive species, check for acidic or heavy-metal toxic runoff (such as acid mine drainage), and ensure that the newly planted vegetation becomes self-sustaining.

Award 1 mark for each valid suggestion with its corresponding environmental/ecological explanation, up to a maximum of 6 marks.

- **Backfilling and grading [1]**: Fills in pits with overburden and levels steep slopes to prevent soil erosion and landslides/make the area physically safe. [1]
- **Replacing stored topsoil [1]**: Reintroduces organic matter, nutrients, and soil microbes necessary to support plant growth. [1]
- **Revegetation/replanting native species [1]**: Roots bind the soil to prevent wind/water erosion AND plants recreate habitats/food sources for wildlife. [1]
- **Creating artificial lakes/wetlands [1]**: Fills deep hollows with water to create new aquatic ecosystems, boosting local biodiversity. [1]
- **Treatment of toxic runoff/bioremediation [1]**: Uses wetlands or chemical lime to neutralize acid mine drainage before it contaminates groundwater/rivers. [1]
- **Long-term monitoring/pest control [1]**: Ensures invasive species do not overrun the developing ecosystem and monitors soil health/vegetation success. [1]

First, calculate the actual decrease in tonnes:
\(15800 - 11200 = 4600\text{ tonnes}\)

Next, calculate the percentage decrease based on the 2021 value:
\(\frac{4600}{15800} \times 100 = 29.1139...\%\)

Rounding to one decimal place gives \(29.1\%\).

- 1 mark for calculating the difference in catch: \(15800 - 11200 = 4600\text{ tonnes}\) (or showing this step in calculation).
- 1 mark for correct formula set-up: \(\frac{4600}{15800} \times 100\).
- 1 mark for correct final answer rounded to one decimal place: \(29.1\%\) (accept 29.1 without the % sign).

1. Mean soil pH of reclaimed site:
\(\frac{5.2 + 5.4 + 5.0}{3} = \frac{15.6}{3} = 5.2\)

2. Mean soil pH of natural woodland site:
\(\frac{6.5 + 6.3 + 6.4}{3} = \frac{19.2}{3} = 6.4\)

3. Reason for difference: The reclamation process involves disturbed overburden and spoil heaps. These minerals, particularly pyrite (iron sulfide), react with air and rainwater to form sulfuric acid, which lowers the pH of the soil relative to the undisturbed natural forest soil.

- 1 mark for calculating both means correctly: Reclaimed = 5.2 AND Natural = 6.4.
- 1 mark for stating that the reclaimed site is more acidic / has a lower pH.
- 1 mark for a valid scientific reason: e.g., oxidation of iron sulfides / pyrite in mining waste; production of acid mine drainage; leaching of essential alkaline bases due to lack of organic matter/canopy cover.

1. Calculate total sediment accumulation over 25 years:
\(25\text{ years} \times 1.8\text{ million m}^3/\text{year} = 45\text{ million m}^3\)

2. Calculate the percentage reduction compared to the initial capacity:
\(\frac{45}{450} \times 100 = 10\%\)

- 1 mark for calculating the total volume of silt after 25 years: \(45\text{ million m}^3\).
- 1 mark for the correct fraction/percentage methodology: \(\frac{45}{450} \times 100\).
- 1 mark for the correct final calculation: \(10\%\) (accept 10).

At 0m to 50m, organic sewage enters the river. This provides a food source for aerobic decomposers (bacteria). These bacteria multiply rapidly and consume dissolved oxygen for aerobic respiration, creating a high Biochemical Oxygen Demand (BOD) and causing the oxygen dip. Further downstream (200m to 500m), the organic matter is mostly broken down, bacterial populations decline, and oxygen is restored through turbulence/re-aeration from the atmosphere and photosynthesis by recovering river plants and algae.

- 1 mark for explaining that the initial decrease in oxygen is due to aerobic bacteria/decomposers breaking down the organic sewage.
- 1 mark for linking this decomposition to the rapid uptake of oxygen via aerobic respiration (high BOD).
- 1 mark for explaining that oxygen increases further downstream because organic waste runs out (bacterial activity decreases) and oxygen is replenished by dissolution from the air/water movement or photosynthesis by aquatic plants.

Systematic sampling is ideal here. The students should lay a line transect (using a tape measure) from the high-tide mark straight inland across the dunes. They should then place quadrats at fixed, regular intervals (e.g., every 5 meters) along the transect and record the percentage cover or abundance of Marram grass. Random sampling would fail to show the gradient/zonation pattern because sample sites would be scattered arbitrarily without a continuous, structured relation to distance from the sea.

- 1 mark for describing the transect setup: Use a tape measure/line transect placed perpendicular to the shoreline / stretching inland from the high-tide mark.
- 1 mark for systematic quadrat placement: Place quadrats at fixed/regular intervals (e.g., every 5 or 10 meters) along the line to count Marram grass.
- 1 mark for justifying why random sampling is unsuitable: Random sampling does not target/display the environmental gradient or changes in vegetation zonation over distance.

1. Identify the drought: 2021 had the lowest rainfall (\(210\text{ mm}\)) and led to the lowest maize yield (\(0.4\text{ tonnes/ha}\)).
2. Mitigation strategies: Farmers can adopt water-conserving practices such as:
- Drip irrigation (reduces evaporation losses by delivering water directly to roots).
- Rainwater harvesting (storing wet-season runoff in small reservoirs/tanks for dry-season use).
- Planting drought-tolerant crop varieties (genetically adapted crops that require less water).
- Mulching (retains soil moisture by reducing surface evaporation).

- 1 mark for identifying 2021 as the most severe drought year AND stating that it dramatically reduced crop yield to its lowest value of \(0.4\text{ tonnes/ha}\).
- 1 mark for suggesting one valid water management strategy (e.g., drip irrigation, rainwater harvesting/storage reservoirs).
- 1 mark for suggesting a second valid strategy (e.g., mulching to conserve soil moisture, planting drought-resistant crop varieties, or using conservation tillage).

1. Calculate percentage reduction:
\(\text{Reduction} = 120 - 15 = 105\text{ cases}\)
\(\text{Percentage Reduction} = \frac{105}{120} \times 100 = 87.5\%\)

2. Reasons for remaining cases:
- Contaminated food (e.g., washed in untreated river water or handled by infected individuals).
- Poor sanitation and personal hygiene (lack of handwashing with soap allows person-to-person transmission).
- Use of other contaminated water sources (farmers/laborers working in fields away from the chlorine-dispensing wells may drink directly from streams).

- 1 mark for calculating the correct percentage reduction: \(87.5\%\) (must show working or correct final value).
- 1 mark for suggesting one valid reason for ongoing cases (e.g., cholera transmission via contaminated food / flies; poor household hand hygiene/sanitation).
- 1 mark for suggesting a second valid reason (e.g., residents using untreated river/pond water when working away from the village well; improper storage of water at home leading to re-contamination).

The data shows that selective logging maintains a substantial canopy cover of \(70\%\), whereas clear-felling leaves almost no canopy (\(5\%\)). This high canopy cover protects the soil underneath from the kinetic energy of falling rain, which prevents soil detachment. As a result, the soil erosion rate in selectively logged zones is very low (\(1.5\text{ tonnes/ha/year}\)) compared to the massive soil loss of \(24.8\text{ tonnes/ha/year}\) under clear-felling. This preserves nutrients, keeps topsoil intact for future forest regeneration, and prevents sediment from choking local waterways, making selective logging far more sustainable.

- 1 mark for comparing canopy cover data: Selective logging retains significant canopy cover (\(70\%\)) compared to clear-felling which removes almost all of it (only \(5\%\) remains).
- 1 mark for linking canopy cover to soil erosion data: The intact canopy in selective logging shields soil from rain impact, resulting in an extremely low erosion rate of \(1.5\text{ tonnes/ha/year}\) compared to \(24.8\text{ tonnes/ha/year}\) in clear-felled zones.
- 1 mark for explaining the sustainability benefit: Low soil erosion preserves soil fertility/nutrients for future tree growth, prevents siltation of rivers, and maintains some habitat structure for biodiversity.

1. Place the transect line perpendicular to the footpath starting from the edge. 2. Position a quadrat at regular, predetermined intervals (e.g., every 2 meters) along the line to achieve systematic sampling. 3. Identify and count the number of different plant species within each quadrat (species richness) and estimate the percentage area covered by plants. 4. Repeat the entire process using multiple parallel transect lines in the same area to increase reliability and calculate an average.

Award 1 mark for each of the following points, up to a maximum of 4 marks: - Place a quadrat at regular/fixed intervals (e.g., every 2m) along the transect line [1] - Identify the different plant species present in each quadrat (to measure species richness) [1] - Estimate the percentage cover of vegetation or count individual plants within each quadrat [1] - Repeat the transect line in other locations or parallel lines to obtain a representative mean [1]. Reject random sampling as the question specifies systematic sampling.

Control variables when using a Secchi disk include ensuring the same person takes the reading to avoid observer bias, taking the reading from the shaded side of the boat or platform to avoid glare, and using the same disk. Taking measurements on the same day is crucial because rainfall or weather changes can cause soil runoff, increasing flow rate and naturally altering turbidity, which would invalidate comparison between sites.

Award up to 4 marks: - Control variable 1: Same observer to judge when the disk disappears / same side of the boat (e.g., shaded side) to prevent glare / same Secchi disk used [1] - Control variable 2: Second valid control variable from the list above [1] - Explanation for same day: Rainfall or weather events can change water levels/stream flow rate [1] - Rainfall/runoff would wash sediment into the river, changing turbidity naturally and making the comparison unfair [1]

To collect quantitative data, the student must weigh each dry glass slide before applying the petroleum jelly and placing it. After a fixed exposure period (e.g., one week), the slides are collected, carefully dried if damp, and weighed again to find the mass of dust deposited (change in mass). Alternatively, they can place a clear grid over the slide and count the number of dust particles under a microscope in a set area. A key limitation is that weather conditions like wind direction changes or heavy rain can wash dust off the slides, underestimating the deposition.

Award up to 4 marks: - Quantitative measurement method: Weigh slides before and after exposure / count dust particles using a grid under a microscope [1] - Standardisation: Leave slides out for the exact same duration of time at all distances [1] - Calculation: Calculate mass gain per square centimetre or percentage cover of dust [1] - Limitation: Rain can wash dust off / wind direction changes mean dust might not blow toward the slides / cannot distinguish quarry dust from background dust [1]

Closed questions require fixed response options (e.g., Yes/No or Likert scale). Question 1: 'Do you support the 5 km fishing ban?' (Options: Yes / No / Undecided). Question 2: 'How has the fishing ban affected your daily catch or income?' (Options: Decreased / Stayed the same / Increased). To select a representative sample, the student should use systematic sampling (e.g., interviewing every 10th person at the harbor) or stratified sampling (ensuring a proportional representation of fishermen, business owners, and local residents) rather than convenience sampling.

Award up to 4 marks: - Question 1: Relevant closed question with clear options (e.g. Yes/No, or scale) [1] - Question 2: Second relevant closed question focusing on attitudes/impact with clear options [1] - Sampling method name: Identify stratified sampling or systematic sampling [1] - Sampling description: Explain how to execute it (e.g., interview every nth person at the harbor, or select proportional numbers from different stakeholder groups like fishermen, merchants, and tourists) [1]. Note: Reject open-ended questions for part a. Reject simple 'ask anyone they see' for part b.

The student ensured a fair test by keeping the independent variable (slope angle) as the only factor being changed, while holding the soil type, initial soil volume, rainfall intensity, and rainfall duration constant across all plots. Two additional variables they must control are the compaction/bulk density of the soil (how firmly it is packed into the plots) and the presence/absence of vegetation or organic matter on the soil surface, as roots and cover significantly reduce erosion rates.

Award up to 4 marks: - Explanation of fair test: Soil type, volume, rainfall intensity, and duration were kept constant so that only the slope angle affects the mass of eroded soil [1] - Control variable 1: Soil compaction/packing density must be identical in all three plots [1] - Control variable 2: Vegetation cover / crop residue (all plots must be completely bare or have identical cover) [1] - Control variable 3: Height of the simulated rainfall source / droplet size of the water [1] (Accept any two of the control variables for 1 mark each, max 2 marks for controls)

A suitable table must include clear column headers with units where appropriate. Columns should be: Household ID/Number, Type of standing water container (e.g., open drums, discarded tyres, puddles), Presence/Absence of mosquito larvae, and Volume of standing water (estimated). Practical physical management actions the villagers can take include: covering all water storage containers with tight-fitting lids or mosquito nets, draining any puddles or low-lying areas of standing water, and adding larvivorous fish (like Guppies) to permanent water ponds.

Award up to 4 marks: - Table design 1: Columns/Rows clearly labeled for household identifier AND types of breeding sites / container types [1] - Table design 2: Column for presence/absence of larvae OR number of larvae observed [1] - Action 1: Cover water storage containers with tight-fitting lids or fine mesh [1] - Action 2: Drain standing water pools / empty discarded tires / clear blocked gutters to prevent accumulation [1]. (Accept: Biological control like introducing larvivorous fish for 1 mark. Reject chemical sprays unless linked to standing water treatment).

First, calculate the mean time taken by the cork: Mean time = (12.5 + 14.1 + 13.4) / 3 = 40.0 / 3 = 13.33 s. Next, calculate the mean velocity: Velocity = Distance / Time = 10 m / 13.33 s = 0.75 m/s (or 0.750 m/s). To improve accuracy, the students could use a digital flow meter which measures velocity at various depths, or use an object that is mostly submerged (like an orange) so that wind does not blow it faster or slower than the actual water current.

Award up to 4 marks: - Method to calculate mean time: 13.33 s shown [1] - Calculation of velocity: 10 / 13.33 = 0.75 m/s (accept 0.75 or 0.75 to 0.76) [1] - Accuracy improvement: Use a flow meter / impeller (which is more precise than floating objects) [1] - Alternative improvement: Use a weighted/submerged float (like an orange) to reduce the effect of wind / repeat across different parts of the river width and average [1]

To collect reliable data, the students must take light meter readings at ground level (where the understory grows) at the exact same time of day (e.g., midday) and under similar weather conditions (e.g., clear sky) in both forests to control external variables. They should take multiple systematic readings (e.g., at each quadrat) to calculate a representative mean. To show the relationship between light intensity and understory plant cover, they should present the data using a scatter graph, plotting light intensity on the horizontal (x) axis and percentage plant cover on the vertical (y) axis, and draw a line of best fit to show the correlation.

Award up to 4 marks: - Light meter usage: Hold the sensor at a standard height (ground level / understory level) [1] - Control of conditions: Take readings at the same time of day / under same cloud/weather conditions [1] - Replication: Take multiple readings in both forest types to obtain an average [1] - Data presentation: Draw a scatter graph / plot light intensity on x-axis and % plant cover on y-axis, showing a line of best fit to determine correlation [1]

(a)
Difference in beetles caught = \(80 - 16 = 64\)
Percentage decrease = \(\frac{64}{80} \times 100 = 80\%\)

(b)
1. Wear appropriate clothing (e.g. sturdy boots to prevent slipping, long trousers/sleeves to protect against biting insects or venomous species).
2. Use navigation equipment (GPS/map) or work in groups to prevent getting lost in dense forest.

(a)
1 mark for correct working: (80 - 16) / 80 * 100 [or equivalent fraction, e.g., 64/80]
1 mark for correct calculated value: 80[%]

(b)
1 mark for each sensible safety precaution up to a maximum of 2 marks:
- Wear long trousers/long sleeves / high-topped boots / protective footwear (to protect against insects/snakes/plants)
- Carry insect repellent / first-aid kit
- Use a map / GPS / compass / stay with a partner / do not wander off alone (to avoid getting lost)
- Inform a supervisor / school of planned location and return time
- Check weather forecast before entering forest / avoid entering during storm or high winds (danger of falling branches)

(a) There is an inverse relationship/negative correlation: as water temperature increases, dissolved oxygen concentration decreases.

(b) Fish rely on dissolved oxygen in water to breathe/respire. When levels are low, fish cannot get enough oxygen, leading to physiological stress, suffocation, death, or forcing them to migrate away, reducing the population.

(c) The industrial plant could direct the hot water into cooling towers or holding/cooling ponds to allow it to cool down naturally via evaporation and heat transfer before releasing it into the river system.

(a)
1 mark for describing the inverse/negative relationship: as temperature increases, dissolved oxygen decreases (or vice versa).

(b)
1 mark for identifying that fish need dissolved oxygen for respiration / breathing.
1 mark for explaining the impact: low oxygen causes suffocation / death / physiological stress / migration / reduced reproduction.

(c)
1 mark for suggesting a valid cooling method:
- cooling towers
- cooling ponds / holding ponds / retention ponds
- heat exchangers / redirecting heat for industrial use
- recycling/reusing the water within the plant

An effective evaluation compares the environmental and socio-economic outcomes of both strategies. Strategy A (no-take MPA) is highly effective at conserving biodiversity and protecting the coral habitat from physical damage. It allows fish to grow larger and produce more eggs, leading to a 'spillover effect' where adults and larvae migrate into the remaining 60% of fishing areas. It also enhances ecotourism, which brings economic benefits. However, it concentrates all fishing pressure onto the remaining 60% of the reef, which could lead to rapid overexploitation of those areas if they remain unregulated. It also causes immediate economic loss for fishers who lose access to 40% of their traditional grounds. Strategy B (gear restrictions and ITQs) addresses fishing pressure directly across the whole ecosystem. Banning gillnets reduces bycatch of non-target species, and the 10 cm mesh size allows juvenile fish to escape and reproduce before capture. ITQs limit the total catch, helping to prevent the collapse of fish stocks while giving fishers a secure share of the catch. However, monitoring and enforcing ITQs and mesh sizes in small-scale fisheries is technically challenging and highly expensive, and fishers face high upfront costs to replace their gear. Combining both strategies is the most sustainable approach. The MPA provides a safe ecological haven, while the gear regulations and quotas on the remaining 60% ensure that spillover fish are harvested sustainably without causing localized collapse.

Level 3 (7-8 marks): Answers show a comprehensive evaluation of both Strategy A and Strategy B. Arguments cover both environmental benefits (such as biodiversity, juvenile protection, and habitat conservation) and socio-economic impacts (such as fisher incomes, tourism, and gear costs). A clear, well-reasoned recommendation is made (e.g., choosing one or arguing for a combined approach). Level 2 (4-6 marks): Answers show a balanced evaluation of both strategies but may lack detail on some aspects (e.g., focusing only on ecological benefits and omitting economic costs). A recommendation is made with some supporting reasons. Level 1 (1-3 marks): Answers are descriptive rather than evaluative, listing simple facts about MPAs or fishing gear with little or no analysis. The recommendation is absent or lacks logical justification. Key points to credit include: Benefits of Strategy A: allows fish stocks to recover, increases biodiversity, spillover effect, supports ecotourism. Limitations of Strategy A: displaces fishing pressure, initial loss of income/food source, enforcement difficulties. Benefits of Strategy B: protects juvenile fish (mesh size), reduces bycatch (gillnet ban), limits total catch (ITQs). Limitations of Strategy B: high cost of monitoring and enforcement, financial burden of new gear, potential for illegal fishing. Synthesis/Combination: explaining how they complement each other to achieve both conservation and socio-economic goals.

An effective evaluation compares the ecological and socio-economic benefits and drawbacks of both restoration plans. Plan X (rewilding and wetland) has major environmental benefits. Creating a wetland increases local biodiversity by providing habitats for birds, insects, and amphibians. Planting native vegetation stabilizes slopes, reducing soil erosion and preventing sediment from polluting the nearby river. Restricting human access ensures wildlife remains undisturbed. However, this plan offers minimal economic benefit to the local town, which may have suffered job losses from the mine closure, and untreated mine water in the pit could become acidic or contaminated, posing a hazard to local wildlife. Plan Y (agriculture and recreation) focuses on socio-economic recovery. Completely backfilling the pit eliminates the hazard of a deep water-filled pit. Treating toxic leachate protects groundwater and the river. Converting the land to agriculture provides long-term employment and food security, while the park improves community well-being and can attract tourism. However, Plan Y is extremely expensive and logistically difficult. Backfilling a large open-cast pit requires massive volumes of material, and imported topsoil often lacks the structure and nutrients needed for crops, requiring heavy fertilizer use which may cause eutrophication in the nearby river. In conclusion, while Plan X is cheaper and better for pure biodiversity, Plan Y provides a better balance of socio-economic development and environmental remediation, making it more sustainable for a community transitioning away from mining.

Level 3 (7-8 marks): A detailed evaluation of both Plan X and Plan Y is provided, weighing environmental benefits (biodiversity, erosion control, water quality) against socio-economic needs (employment, safety, food security). A clear and well-justified recommendation is made. Level 2 (4-6 marks): Both plans are evaluated, but the response may be unbalanced (e.g., focusing heavily on environmental aspects of Plan X and ignoring the socio-economic challenges of Plan Y). A recommendation is made with some supporting reasons. Level 1 (1-3 marks): Basic descriptive points are made about one or both plans with little to no comparative evaluation. The recommendation is missing or lacks clear logical support. Key points to credit include: Benefits of Plan X: low cost, high biodiversity, natural erosion control, river protection. Limitations of Plan X: no economic return/jobs for the community, potential hazard of acidic water body. Benefits of Plan Y: economic recovery, safe elimination of the open pit, community facilities, active leachate treatment. Limitations of Plan Y: extremely high cost, potential poor soil quality for farming, risk of agricultural runoff (fertilizers) polluting the river.