2023 Cambridge IGCSE Environmental Management (0680) Practice Paper with Answers

Overfishing leads to a decline in fish stocks, which causes direct job losses for fishers. This has a knock-on effect on the local economy, reducing revenue for fish processing factories, shipbuilders, and maintenance services.

Award 1 mark for each valid economic impact listed, up to a maximum of 2 marks:
- Loss of employment / jobs in fishing or processing sectors [1]
- Loss of income / revenue for fishermen or local businesses [1]
- Increased price of fish / seafood for consumers [1]
- Increased cost of importing fish to meet local demand [1]
- Reject: environmental impacts (e.g., loss of biodiversity, food chain disruption).

Bioremediation involves introducing or encouraging the growth of specific living organisms (like microbes or plants) at a contaminated mine site. These organisms absorb or metabolise toxic elements, rendering them harmless or concentrating them for safer removal.

Award 1 mark for each point:
- Use of living organisms / microbes / bacteria / plants [1]
- To absorb / break down / neutralise / remove toxic substances / heavy metals [1]
- Reject: generic "cleaning up the site" without referencing biological mechanisms.

A multipurpose dam stores water in its reservoir during periods of high rainfall. This water can be channeled to nearby agricultural fields for irrigation, ensuring crops grow year-round. Additionally, controlling the water release prevents destructive downstream flooding of agricultural land.

Award 1 mark for each point up to 2 marks:
- Provides a reliable / consistent supply of water for irrigation (especially in dry seasons) [1]
- Controls downstream water flow to prevent flooding of agricultural land / washing away crops [1]

1. Nitrates and phosphates from fertilisers enter the water, triggering rapid algal growth (algal bloom) which blocks light to submerged plants.
2. Submerged plants and algae die; aerobic bacteria decompose the dead matter and consume the dissolved oxygen, leading to anoxia and the death of fish.

Award 1 mark for any two steps in the sequence, up to a maximum of 2 marks:
- Runoff of nitrates/phosphates causes rapid growth of algae / algal bloom / blocks sunlight [1]
- Death of algae / aquatic plants [1]
- Aerobic bacteria decompose dead organic matter [1]
- Bacteria consume dissolved oxygen, causing fish and other aquatic animals to suffocate / die [1]

Habitat fragmentation isolates animal populations, leading to inbreeding and vulnerability to disasters. Wildlife corridors provide physical links between these fragments, permitting gene flow (interbreeding between populations) and safe movement across human-dominated landscapes.

Award 1 mark for each point up to 2 marks:
- Connects fragmented / isolated habitats / allows safe migration [1]
- Prevents inbreeding / increases genetic diversity / allows finding mates from other populations [1]
- Reduces roadkill / mortality from crossing human infrastructure [1]
- Helps species access new food sources / seasonal breeding grounds [1]

Rising global temperatures speed up evaporation and transpiration, drying out the soil more quickly. Concurrently, shifts in atmospheric circulation patterns alter precipitation zones, causing prolonged periods of low or no rainfall in drought-prone regions.

Award 1 mark for each valid point, up to 2 marks:
- Higher temperatures increase evaporation / evapotranspiration (drying out soils) [1]
- Shift in atmospheric/wind patterns leads to reduced rainfall in specific regions [1]
- Less predictable/irregular rainfall seasons [1]
- Accelerated melting of glaciers leading to reduced river flow during dry seasons [1]

Cholera is a water-borne disease spread through contaminated water. During floods, sewage often mixes with drinking water. Boiling or chlorinating water kills the cholera bacteria, while containment of human waste stops further contamination of floodwaters.

Award 1 mark for each logical method, up to 2 marks:
- Boiling drinking water / using chlorine tablets / purification filters [1]
- Using bottled water / clean water from tankers [1]
- Avoiding contact with floodwaters / wearing protective boots [1]
- Using temporary sealed chemical toilets / preventing sewage contamination [1]
- Handwashing with soap / hygiene education campaigns [1]
- Reject: methods strictly for vector control (e.g., using mosquito nets, as cholera is not insect-borne).

Unlike clear-cutting, which removes all trees in an area, selective logging leaves younger trees to grow and preserves the forest's vertical structure. This protects biodiversity by keeping habitats intact and reduces soil erosion because the root systems and canopy remain to buffer rainfall.

Award 1 mark for each descriptive point, up to 2 marks:
- Only specific / mature / commercial trees are harvested (leaving younger trees to grow) [1]
- Forest canopy/ecosystem/habitats remain largely intact [1]
- Root systems remain to bind the soil / canopy protects soil from rainfall erosion [1]
- Allows the forest to regenerate naturally over time [1]

Landscaping is an essential step in restoring a mine site. By filling in empty pits and grading steep slopes, developers prevent landslides and runoff-related erosion. Applying topsoil and planting native vegetation afterwards assists in returning biodiversity and reducing visual impact.

Award 1 mark for each valid description up to a maximum of 2 marks:
- filling in open pits / cavities / hollows [1]
- grading / reshaping / smoothing steep slopes to prevent landslides or soil erosion [1]
- covering the land with topsoil (to allow plant growth) [1]
- planting native trees / grasses / vegetation to restore habitats or reduce visual impact [1]

A double-hulled tanker possesses two physical barriers of steel plates. The space between the outer hull and inner hull acts as a safety buffer. If the vessel grounds or collides with another object and ruptures the outer hull, the inner hull containing the oil cargo is protected from damage, preventing an oil spill.

Award 1 mark for each point up to a maximum of 2 marks:
- presence of an inner and outer hull / double layer of steel [1]
- outer hull absorbs impact / inner hull remains intact / prevents breach of the cargo tank [1]
- oil is contained / does not leak into the sea or marine environment [1]

(a)
Decrease in percentage points = \( 45\% - 12\% = 33\% \)
Percentage decrease = \( \frac{33}{45} \times 100 = 73.3\% \) (accept \( 73\% \) or \( 73.33\% \))

(b)
A minimum landing size ensures that immature/juvenile fish that have not yet reproduced are returned to the water. This gives them the opportunity to grow to breeding age, spawn, and replenish the population biomass.

(a)
- M1: For showing correct subtraction: \( 45 - 12 = 33 \) [1]
- A1: Correct final percentage: \( 73.3\% \) (range \( 73\% \) to \( 73.33\% \)) [1]

(b)
- Max 2 marks from:
- Allows young/juvenile fish to escape capture / ensures they are not harvested [1]
- Allows fish to reach maturity/breeding age [1]
- Gives fish a chance to reproduce at least once before being caught [1]

(a) At 300 m, the dust concentration is \( 95\ \mu\text{g/m}^3 \), which is the first recorded value below thesafe limit of \( 100\ \mu\text{g/m}^3 \).

(b) Quarry operators can suppress dust emissions by spraying water onto haul roads and active excavation faces. They can also use dust extraction hoods and bag filters on drilling or crushing machinery, or establish vegetative buffer zones (tree screens) to trap wind-blown dust.

(a)
- A1: 300 m [1]

(b)
- Max 3 marks from:
- Spraying water / water misting on roads / blast areas (to bind dust particles) [1]
- Planting trees / establishing a vegetation barrier around quarry boundaries (to block/trap dust) [1]
- Covering trucks with tarpaulins when transporting material [1]
- Installing bag filters / dust collectors on crushing and drilling machinery [1]

(a) In semi-arid regions, natural rainfall is low and evaporation rates are high, meaning crops require large volumes of artificial irrigation water to survive and produce high yields.

(b) Conflict arises because reservoir managers prioritize holding back water during dry seasons to maintain high hydraulic heads for electricity generation or municipal storage. Downstream farmers, however, require steady, high-volume releases of water during these same dry periods to irrigate their crops, resulting in competing demands for a limited resource.

(a)
- A1: Low natural rainfall / high evaporation rates require artificial watering of crops [1]

(b)
- Max 3 marks from:
- Reservoir managers want to store water to maintain electricity generation capacity / urban water supply [1]
- Downstream farmers need consistent flow/water release for crop irrigation during dry periods [1]
- Reduced water downstream can lead to crop failure / economic loss for farmers [1]
- Sudden water releases for flood control/electricity spikes can cause downstream flooding/erosion [1]

(a) Point C. It has the highest BOD (18.0 mg/l) and the lowest dissolved oxygen concentration (1.2 mg/l), showing the greatest biological oxygen consumption by decomposers.

(b) It takes time for the organic waste to travel downstream and mix fully, during which the aerobic bacterial population multiplies rapidly. By the time the water reaches Point C, the decomposition rate has peaked, causing maximum consumption of dissolved oxygen.

(a)
- M1: Point C [1]
- A1: Refers to highest BOD (18.0 mg/l) and/or lowest dissolved oxygen (1.2 mg/l) [1]

(b)
- Max 2 marks from:
- Time lag for organic matter to be decomposed / bacterial population to grow [1]
- Aerobic bacteria consume oxygen rapidly as they digest the organic waste [1]
- Re-oxygenation from the atmosphere is slower than consumption at Point C [1]

(a)
Decrease in species = \( 48 - 18 = 30 \)
Percentage decrease = \( \frac{30}{48} \times 100 = 62.5\% \)

(b)
Invasive species thrive in Area B because logging creates large gaps in the canopy, allowing light to reach the forest floor. In addition, soil disturbance caused by logging machinery clears native ground cover, providing empty niches for opportunistic invasive seeds to germinate.

(a)
- M1: Correct fraction or subtraction: \( \frac{30}{48} \) or \( 48 - 18 = 30 \) [1]
- A1: 62.5% [1]

(b)
- Max 2 marks from:
- Logging creates gaps in the canopy, increasing sunlight on the forest floor [1]
- Soil disturbance by machinery clears space / creates bare soil for opportunist seeds [1]
- Logging vehicles/roads transport and introduce invasive seeds [1]
- Reduced competition from native species [1]

(a) Years 3 and 4. These years had the lowest rainfall (310 mm and 290 mm) and the lowest reservoir levels (20% and 8% respectively).

(b) Drought kills natural vegetation and crops, meaning there is less organic matter/leaf litter returning to the soil to form humus. Furthermore, the dry, bare soil is highly vulnerable to wind erosion, which blows away the nutrient-rich topsoil layer.

(a)
- M1: Years 3 and 4 [1]
- A1: Justified by lowest rainfall (310 mm and 290 mm) and lowest reservoir levels (20% and 8%) [1]

(b)
- Max 2 marks from:
- Lack of plant growth reduces organic matter/humus input to the soil [1]
- Soil dries out and is blown away by wind erosion (removing fertile topsoil) [1]
- Death of soil microorganisms that cycle nutrients [1]

(a)
Natural Increase = \( \text{Birth Rate} - \text{Death Rate} = 28 - 6 = 22 \text{ per } 1000 \)
Rate of Natural Increase (RNI) as a percentage = \( \frac{22}{1000} \times 100 = 2.2\% \)

(b)
The net migration rate is negative (-3 per 1000), which indicates net emigration (more people leaving the country than entering). This migration loss subtracts from the natural population increase, slowing down the overall annual growth rate of Country Y.

(a)
- M1: Calculation of difference per 1000: \( 28 - 6 = 22 \) [1]
- A1: Convert to percentage: \( 2.2\% \) [1]

(b)
- Max 2 marks from:
- Negative net migration indicates emigration exceeds immigration / loss of population [1]
- This reduces the overall population growth rate (below the natural increase rate of 2.2%) [1]

(a) In a calm bay, the oil spreads slowly in a thin film. Rapidly placing booms contains the slick in a small area, preventing the oil from drifting onto sensitive coastlines, sandy beaches, or fishing zones.

(b) Benefit: Breaks the oil slick into tiny droplets, accelerating natural bacterial decomposition.
Limitation: Detergents can be toxic to marine organisms, such as fish larvae and coral reefs, and they disperse the oil into the water column rather than removing it.

(a)
- Max 2 marks from:
- Prevents the oil slick from spreading wider [1]
- Protects sensitive coastal ecosystems/beaches/mangroves from contamination [1]
- Concentrates the oil, making mechanical recovery with skimmers more efficient [1]

(b)
- Max 2 marks from:
- Benefit: Speeds up natural biodegradation / disperses oil to prevent surface birds/mammals from being oiled [1]
- Limitation: Chemicals can be toxic/poisonous to marine life / does not remove the oil from the marine ecosystem [1]

(a) Calculate the decrease: \(450 - 320 = 130\) thousand tonnes.
Calculate the percentage decrease: \(\frac{130}{450} \times 100 = 28.88\%\), which rounds to 28.9\%.

(b) The increase in net mesh size from 80 mm to 100 mm allowed juvenile fish to escape through the larger gaps. This reduced the average juvenile bycatch from 22\% in 2019 to 8\% in 2020 (and further to 5\% in 2021). Consequently, more juveniles survived to breeding age, leading to a recovery in the Spawning Stock Biomass (SSB) from 320 thousand tonnes in 2020 to 380 thousand tonnes in 2021.

(a)
- Award 1 mark for correct calculation of decrease: 130 (thousand tonnes) OR for showing the correct division: \(\frac{130}{450} \times 100\).
- Award 1 mark for the correct final answer: 28.9\% (accept 28.8\% to 29.0\%).

(b)
- Award 1 mark for linking the larger mesh size to the decrease in juvenile bycatch (e.g., from 22\% to 8\%).
- Award 1 mark for explaining that allowing juveniles to escape permits them to mature and reproduce, leading to the observed increase in Spawning Stock Biomass (from 320 to 380 thousand tonnes).

(a) Cost per 1\% of vegetation cover:
- Strategy A: \(\frac{\$50,000}{15\%} = \$3,333.33\) per 1\% cover
- Strategy B: \(\frac{\$450,000}{85\%} = \$5,294.12\) per 1\% cover
- Strategy C: \(\frac{\$250,000}{70\%} = \$3,571.43\) per 1\% cover
Therefore, Strategy A is the most cost-effective because it costs the least per percent of vegetation cover achieved.

(b) Strategy C achieves a very high level of vegetation cover (70\%), which likely satisfies government compliance requirements, unlike Strategy A (only 15\%). At the same time, Strategy C is significantly cheaper than Strategy B (saving $200,000 per hectare), making it suitable for a limited budget. Additionally, bioremediation actively cleans up heavy metal toxic residues left by copper mining.

(a)
- Award 1 mark for correct calculations of cost-effectiveness for at least two strategies (e.g., $3,333 vs $3,571 or $5,294 per percent cover).
- Award 1 mark for identifying Strategy A as the most cost-effective based on calculation.

(b)
- Award 1 mark for noting that Strategy C achieves a high/satisfactory restoration success (70\% cover) while staying within a tighter budget (costs $200,000 less than B).
- Award 1 mark for identifying that bioremediation neutralizes or cleans up chemical toxins/heavy metals typical of copper mining, helping to meet strict regulations.

(a) Total capacity lost = \(1200 - 900 = 300\) million \(\text{m}^3\).
Average annual loss = \(\frac{300 \text{ million m}^3}{30 \text{ years}} = 10\) million \(\text{m}^3\) per year.

(b) Sedimentation fills up the reservoir basin, which reduces the total volume of water it can hold. With less stored water, there is less potential energy/water flow to turn the hydroelectric turbines, decreasing electricity output. During heavy rains, the reduced storage capacity means the reservoir fills up too quickly, forcing dam operators to release large volumes of water, increasing the risk of downstream flooding.

(a)
- Award 1 mark for correct calculation of total capacity lost (300 million \(\text{m}^3\)) or showing \(\frac{1200 - 900}{30}\).
- Award 1 mark for the correct final answer of 10 with appropriate units (million \(\text{m}^3\) per year).

(b)
- Award 1 mark for explaining that sediment build-up reduces the volume of water available to drive turbines, reducing electricity.
- Award 1 mark for explaining that the loss of reservoir volume reduces flood-retention capacity, forcing the release of excess water during heavy rains.

(a) As distance increases from 2 km (Point B) to 8 km (Point D), dissolved oxygen decreases significantly from 92\% to a minimum of 32\%. Beyond 8 km, between Point D and Point E (12 km), the dissolved oxygen level recovers and rises back up to 85\%.

(b) High nitrate concentrations in the agricultural runoff enter the water, acting as a nutrient source that causes rapid growth of algae (an algal bloom). When these algae die, they are decomposed by aerobic bacteria. The population of these decomposers increases rapidly, and their respiration consumes large amounts of dissolved oxygen from the water, resulting in the lowest oxygen levels at Point D.

(a)
- Award 1 mark for identifying the initial decrease/fall in dissolved oxygen downstream of the source (down to 32\%).
- Award 1 mark for identifying the subsequent recovery/increase in oxygen further downstream (up to 85\% at Point E).

(b)
- Award 1 mark for describing the rapid growth of algae/algal bloom due to high nitrate nutrients, followed by their death.
- Award 1 mark for explaining that aerobic bacteria decompose the dead algae and consume dissolved oxygen through respiration.

Protecting marine ecosystems is crucial, and both marine protected areas (MPAs) and conventional management strategies play key roles.

**Arguments in favor of marine protected areas (permanent bans):**
- They allow entire ecosystems and habitats to recover without any human disruption.
- Breeding stocks can recover fully within the reserve, leading to a "spillover effect" where fish migrate out into surrounding harvestable waters.
- They eliminate the issue of bycatch completely within the zone.
- They are often easier to police visually compared to monitoring individual catch quotas at sea.

**Arguments in favor of conventional management (quotas, mesh limits, closed seasons) / limitations of MPAs:**
- Complete bans cause immediate, severe economic hardship for local fishing communities and industries.
- Fishing quotas, when scientifically set and enforced, allow for a sustainable yield that supports human food security.
- Mesh size regulations are highly effective at allowing juvenile fish to escape and reach reproductive age before being harvested.
- Closed seasons protect species during their critical spawning periods without requiring a permanent ban.
- MPAs can shift fishing pressure to other unregulated areas, worsening the situation there.

**Conclusion:**
While permanent marine protected areas are highly effective for ecological restoration, they are most successful when integrated with conventional tools like quotas and gear restrictions outside the reserves to maintain economic viability and food security.

**Level 3 [5–6 marks]**
A coherent response is given that develops and supports the candidate’s conclusion using relevant details and examples. Indicative content and subject-specific vocabulary are used precisely and accurately. Presents a balanced evaluation of the statement, comparing marine protected areas with conventional management methods.

**Level 2 [3–4 marks]**
Development and support of the conclusion is evident, though the response may lack some coherence and/or detail. Discussion of both sides is present but may not be fully balanced.

**Level 1 [1–2 marks]**
The response may be limited in development and/or support. Contradictions or irrelevant details may be present, or the answer is presented as a simple list.

**No response or no creditable response [0 marks]**

**Indicative content:**
*Agreeing with the statement (supporting MPAs/bans):*
- Protects and restores fragile benthic habitats (e.g., coral reefs, seagrass beds).
- Provides a safe haven for target species to grow larger and produce exponentially more offspring.
- Avoids the enforcement complexities of checking mesh sizes or weighing landing catches.
- Enhances biodiversity and ecosystem resilience against climate change.

*Disagreeing with the statement (supporting conventional methods/highlighting MPA limits):*
- Fishing quotas (TACs) maintain employment and economic revenue for coastal nations.
- Net mesh size regulations specifically protect juveniles, ensuring future recruitment.
- Closed seasons protect specific life-cycle stages without closing off entire fishing grounds year-round.
- MPAs require high enforcement costs (patrol boats, satellite monitoring) to prevent illegal poaching.

The rehabilitation of open-pit mines is a highly debated topic balancing economic development with environmental stewardship.

**Arguments in agreement with the official (prioritizing economic benefits):**
- Developing nations face urgent financial demands for basic infrastructure, healthcare, and education; high restoration costs divert public or corporate funds from these needs.
- Restoring a site to its precise original ecological state is technically difficult and highly expensive.
- The mineral wealth extracted provides high employment, export revenue, and industrial growth, which are immediate national priorities.

**Arguments in disagreement with the official (prioritizing restoration):**
- Unrestored open-pit mines cause severe, long-term environmental degradation, including soil erosion, landslides, and loss of local biodiversity.
- Abandoned pits accumulate toxic water and lead to acid mine drainage, polluting local freshwater aquifers and agricultural soils.
- Spoil heaps and exposed heavy metals pose severe health risks to nearby human populations.
- Restoration strategies, such as bioremediation, re-vegetation, or converting pits into safe water reservoirs or forestry plantations, can create long-term sustainable land-use opportunities post-mining.

**Conclusion:**
While economic development is crucial for developing nations, ignoring restoration leads to externalized, long-term costs that exceed short-term mining profits. Governments should mandate basic, cost-effective rehabilitation to protect public health and resources.

**Level 3 [5–6 marks]**
A coherent response is given that develops and supports the candidate’s conclusion using relevant details and examples. Indicative content and subject-specific vocabulary are used precisely and accurately. Presents a balanced evaluation of the statement, contrasting economic priorities with environmental and social impacts.

**Level 2 [3–4 marks]**
Development and support of the conclusion is evident, though the response may lack some coherence and/or detail. Both economic and environmental aspects are discussed, but the evaluation may be unbalanced.

**Level 1 [1–2 marks]**
The response is limited in development and/or support. It may focus on only one side of the argument, contain errors, or be written as a simple list of points.

**No response or no creditable response [0 marks]**

**Indicative content:**
*Arguments for prioritizing mineral extraction over restoration (supporting the official):*
- Immediate economic growth, infrastructure development, and foreign investment.
- Creation of jobs in rural areas.
- High costs of backfilling and landscaping can make mining projects financially unviable.
- Land can sometimes be repurposed for industrial use rather than ecological restoration.

*Arguments for mandatory restoration (opposing the official):*
- Prevention of severe environmental damage, such as soil erosion, siltation of rivers, and habitat destruction.
- Mitigation of chemical pollution (acid mine drainage, heavy metal leaching) into local drinking water.
- Long-term safety risks of open pits (drowning, collapsing walls) to local communities.
- Restored land can be returned to agriculture, forestry, or ecotourism, ensuring long-term sustainability after the resource is depleted.

1. Find the difference between the actual catch and recommended TAC in 2021:
$$1050 - 800 = 250\text{ thousand tonnes}$$

2. Divide the difference by the recommended TAC:
$$\frac{250}{800} = 0.3125$$

3. Convert to a percentage:
$$0.3125 \times 100 = 31.25\%$$

(Accept 31.3%)

M1: For calculating the catch exceedance: \(1050 - 800 = 250\) (thousand tonnes)
M1: For showing a correct division step: \(\frac{250}{800} \times 100\)
A1: Correct final answer: \(31.25\%\) (accept \(31.3\%\))

At the downwind boundary (Location B), the \(PM_{10}\) level is \(74\ \mu\text{g/m}^3\), which exceeds the national limit of \(50\ \mu\text{g/m}^3\). However, at the local village (Location C), the level is \(42\ \mu\text{g/m}^3\), which is safe. The upwind background level is low (\(18\ \mu\text{g/m}^3\)). Thus, management is only partially successful as boundary pollution is too high, though residential impact is mitigated.

1 mark: Stating that air quality management is unsuccessful at the boundary (Location B) as the \(PM_{10}\) value of \(74\ \mu\text{g/m}^3\) exceeds the limit of \(50\ \mu\text{g/m}^3\).
1 mark: Stating that air quality management is successful at the village (Location C) as the \(PM_{10}\) value of \(42\ \mu\text{g/m}^3\) is within safe limits.
1 mark: For a concluding evaluative statement: Overall management is only partially successful, indicating a need for more dust suppression at source / boundary buffers.

1. Convert \(4.5\text{ billion}\) to millions:
$$4.5 \times 1000 = 4500\text{ million m}^3$$

2. Calculate the volume for agricultural irrigation (\(30\%\)):
$$4500\text{ million m}^3 \times 0.30 = 1350\text{ million m}^3$$

(Or \(1.35 \times 10^9\text{ m}^3\))

M1: Converting the total reservoir volume to millions: \(4500\text{ million m}^3\) (or showing a calculation using \(4.5 \times 10^9\))
M1: Correct percentage calculation process: \(4500 \times 0.30\) (or \(4.5 \times 10^9 \times 0.30\))
A1: Correct final numerical value: \(1350\text{ million m}^3\) (or \(1.35 \times 10^9\text{ m}^3\))

Between Station 2 and Station 3, the BOD is very high (decreasing slightly from 22 to 15 mg/l), indicating large amounts of organic pollutants. Microorganisms, such as aerobic bacteria, break down this organic matter through respiration. This biological process consumes dissolved oxygen faster than it can be replenished, causing the DO to drop to its lowest level (2 mg/l) at Station 3, forming an oxygen sag.

1 mark: Stating that high BOD at Station 2/3 represents high organic matter / waste in the water.
1 mark: Explaining that aerobic bacteria / decomposers consume oxygen during respiration while breaking down the waste.
1 mark: Linking this microbial consumption to the depletion of DO (DO drops to \(2\text{ mg/l}\) at Station 3).

1. Calculate the total area of the forest clearing:
$$\text{Total Area} = 200\text{ m} \times 100\text{ m} = 20000\text{ m}^2$$

2. Note that the quadrat size is \(1\text{ m}^2\), so the plant density is \(3.4\text{ plants/m}^2\).

3. Calculate the estimated total population:
$$\text{Total Population} = 20000\text{ m}^2 \times 3.4\text{ plants/m}^2 = 68000\text{ plants}$$

(Accept 68,000)

M1: Correct calculation of total clearing area: \(200\text{ m} \times 100\text{ m} = 20000\text{ m}^2\)
M1: Multiplying the total clearing area by the mean density per quadrat: \(20000 \times 3.4\)
A1: Correct final estimated population: \(68000\) (or \(68,000\))

1. Determine the drought threshold:
$$450\text{ mm} \times 0.50 = 225\text{ mm}$$

2. Identify years with rainfall below \(225\text{ mm}\):
* **Year 3:** \(210\text{ mm}\)
* **Year 4:** \(180\text{ mm}\)

3. Calculate the deficit for each drought year:
* **Year 3 Deficit:** \(450 - 210 = 240\text{ mm}\)
* **Year 4 Deficit:** \(450 - 180 = 270\text{ mm}\)

4. Sum the deficits:
$$\text{Combined Deficit} = 240 + 270 = 510\text{ mm}$$

1 mark: Correctly identifying Year 3 and Year 4 as the drought years (threshold is \(225\text{ mm}\)).
1 mark: Correct calculation process for individual deficits: \(450 - 210 = 240\text{ mm}\) and \(450 - 180 = 270\text{ mm}\).
1 mark: Correct final combined deficit: \(510\text{ mm}\).

1. Calculate percentage reduction:
$$\text{Reduction} = 120 - 18 = 102\text{ cases}$$
$$\text{Percentage Reduction} = \frac{102}{120} \times 100 = 85\%$$

2. Explanation:
Chlorine is added as a chemical disinfectant. It kills the pathogenic bacteria (*Vibrio cholerae*) present in water bodies, which are responsible for the spread of water-borne cholera, preventing clean water from carrying pathogens to consumers.

1 mark: Correct calculation of the percentage reduction: \(85\%\).
1 mark: Explaining that chlorine acts as a disinfectant / sanitising chemical that kills pathogens/microbes/bacteria.
1 mark: Stating that cholera is a water-borne disease caused by bacteria (*Vibrio cholerae*) and treating water interrupts transmission.

1. Calculate annual sequestration rate for the selectively logged forest:
$$\text{Change} = 145 - 120 = 25\text{ tonnes/ha}$$
$$\text{Annual Rate} = \frac{25}{10} = 2.5\text{ tonnes/ha/year}$$

2. Calculate annual sequestration rate for the degraded forest:
$$\text{Change} = 90 - 85 = 5\text{ tonnes/ha}$$
$$\text{Annual Rate} = \frac{5}{10} = 0.5\text{ tonnes/ha/year}$$

3. Calculate the difference:
$$\text{Difference} = 2.5 - 0.5 = 2.0\text{ tonnes/ha/year}$$

M1: Correct calculation of the selective logging forest's annual sequestration rate: \(2.5\text{ tonnes/ha/year}\) (or total change of \(25\text{ tonnes/ha}\)).
M1: Correct calculation of the degraded forest's annual sequestration rate: \(0.5\text{ tonnes/ha/year}\) (or total change of \(5\text{ tonnes/ha}\)).
A1: Correct final difference: \(2.0\text{ tonnes/ha/year}\) (or \(2\text{ tonnes/ha/year}\)).

To carry out random sampling:
1. Lay out two perpendicular tape measures (e.g., 20m x 20m) to form a grid area in the forest.
2. Use a random number generator (on a calculator or phone) to obtain pairs of numbers to act as X and Y coordinates.
3. Locate the point of intersection of these coordinates on the grid.
4. Place the quadrat at this exact intersection point.
5. Repeat this process until 10 quadrats have been sampled in each of the two forest zones.

Award 1 mark for each of the following points (up to 4 marks max):
- Establish a coordinate grid system using tape measures / defining boundaries [1]
- Use of a random number generator / table / drawing numbers from a hat to select coordinates [1]
- Placing the quadrat at the selected coordinate intersection [1]
- Repeating the process to obtain 10 samples in both locations [1]

1. Calculate the absolute increase in turbidity:
\(24.1\text{ NTU} - 3.2\text{ NTU} = 20.9\text{ NTU}\)
2. Calculate the percentage increase relative to the initial value (Site 1):
\(\frac{20.9}{3.2} \times 100 = 653.125\%\)
3. Rounding to 1 decimal place gives \(653.1\%\) (accept \(653\%\)).

Award marks as follows:
- Correct calculation of the difference: \(24.1 - 3.2 = 20.9\) [1]
- Correct formulation: \(\frac{20.9}{3.2} \times 100\) [1]
- Correct answer: \(653.1\%\) or \(653\%\) [1]
- Appropriate unit shown with final answer (\% or percentage) [1]
(If the final answer is correct, award full marks even if working is omitted.)

To ensure a fair test, standardizing the method is essential. Controlled variables could include:
- **Sampling duration:** Standardizing the time spent kicking (e.g., exactly 3 minutes) ensures equal sampling effort.
- **Mesh size of the kick net:** Using identical nets ensures that smaller macroinvertebrates are captured or excluded consistently at both sites.
- **Depth and speed of river flow:** Sampling in areas with similar water velocity prevents flow rate from affecting the number of organisms washed into the net.
- **Surface area of riverbed disturbed:** Keeping the sampled riverbed area constant ensures a consistent population density assessment.

Award marks as follows:
- First valid controlled variable identified (e.g., sampling time / net mesh size / riverbed area) [1]
- Corresponding explanation of why it must be controlled to maintain fairness/accuracy [1]
- Second valid controlled variable identified [1]
- Corresponding explanation of why it must be controlled [1]

1. **Reduces Selection Bias:** Opportunistic sampling at the dock only targets active fishers or people out during a specific hour, ignoring households that may have lost livelihood options entirely.
2. **Even Spatial Distribution:** Sampling every 5th house ensures residents living further away from the immediate shore or dock are represented.
3. **Representative Demographics:** It captures a wider cross-section of the population (women, children, elderly, non-fishers) rather than just those gathered at public meeting hubs.
4. **Reliability:** Standardized intervals can be repeated by other researchers to verify findings, unlike unpredictable opportunistic encounters.

Award 1 mark for each of the following points (up to 4 marks max):
- Explaining that opportunistic sampling introduces selection bias / only samples a specific subgroup [1]
- Explaining that systematic sampling ensures equal probability of selection for all household types / more representative sample [1]
- Noting that systematic sampling provides better spatial/geographic coverage of the area [1]
- Linking representativeness to increased reliability/validity of the conclusions drawn [1]

1. Calculate the mean for the terraced slope:
\(\text{Mean} = \frac{45 + 52 + 48}{3} = \frac{145}{3} = 48.33\text{ g}\)
2. Calculate the mean for the unterraced slope:
\(\text{Mean} = \frac{210 + 245 + 225}{3} = \frac{680}{3} = 226.67\text{ g}\)
3. Divide the unterraced average by the terraced average to find the ratio:
\(\text{Ratio} = \frac{226.67}{48.33} \approx 4.69\text{ times greater}\) (rounds to 4.7).

Award marks as follows:
- Correct mean calculated for the terraced slope (\(48.3\text{ g}\)) [1]
- Correct mean calculated for the unterraced slope (\(226.7\text{ g}\)) [1]
- Correct calculation setup for comparison (e.g., dividing the two averages) [1]
- Correct final calculation of the ratio: \(4.7\) times greater (accept range \(4.6\) to \(4.7\)) [1]

Common safety hazards and control measures in a managed forest include:
- **Falling debris/branches:** Overhanging deadwood can fall, especially in logged areas. Control: Wear protective hard hats.
- **Uneven, steep, or wet ground:** Risk of slips, trips, or falls. Control: Wear sturdy hiking boots with ankle support and plan paths carefully.
- **Animal or insect bites (ticks/wasps/snakes):** Control: Wear long-sleeved clothing, tuck trousers into socks, apply insect repellent, and carry a basic first-aid kit.
- **Getting lost:** Logged forests can have confusing pathways. Control: Carry a GPS, map, compass, and ensure a supervisor knows your intended route.

Award marks as follows:
- First realistic hazard identified (e.g., falling branches, tripping/slipping, bites/stings, getting lost) [1]
- Appropriate and matching control measure for the first hazard [1]
- Second realistic hazard identified [1]
- Appropriate and matching control measure for the second hazard [1]

Replication is critical in ecological fieldwork because:
1. **Reduces Anomalies:** A single survey might catch a large school of fish passing by chance, which would artificially inflate the count, or capture nothing due to temporary noise.
2. **Allows Statistics:** It enables researchers to calculate a mean (average) and assess standard deviation to see how consistent the data is.
3. **Natural Variation:** Marine animals move constantly due to tide, light, temperature, and feeding patterns; multiple replicates help smooth out this natural variation.
4. **Reliability:** Increases the statistical power and overall reliability of the study's conclusions.

Award 1 mark for each of the following points (up to 4 marks max):
- Explaining that replication helps identify or minimize the impact of anomalies/unrepresentative counts [1]
- Stating that doing multiple surveys allows for the calculation of a mean/average [1]
- Explaining that it accounts for natural environmental or behavioral variation (e.g., fish movement, tidal changes) [1]
- Linking the practice directly to improving the reliability / validity of the comparative conclusion [1]

1. **Trend:** The concentration of mosquito larvae decreases continuously as the distance from the village center increases (a negative correlation).
2. **Management Methods:**
- **Physical/Engineering:** Drain standing pools of water or fill them with sand/dirt to eliminate breeding sites. Cover open domestic water containers.
- **Biological:** Introduce larvivorous (larvae-eating) fish, such as *Gambusia* (mosquito fish), into permanent ponds to eat the larvae.
- **Chemical/Microbiological:** Apply eco-friendly targeted larvicides or *Bacillus thuringiensis israelensis* (Bti) bacteria to the pools to kill larvae without harming other wildlife.

Award marks as follows:
- Correct description of the trend (concentration of larvae decreases as distance increases) [1]
- One valid physical management strategy (e.g., draining pools, filling them in, covering water sources) [1]
- One valid biological or chemical control strategy (e.g., introducing larvae-eating fish, applying targeted larvicides) [1]
- Explaining how the chosen method directly prevents larval development or reduces survival [1]

Part (i) requires identifying the scientific purpose of sampling upstream. This serves as a control or baseline to compare the polluted downstream sites against the unpolluted river water.

Part (ii) requires a clear description of the trend using the data provided (e.g., state that the mayfly population increases from 0 to 42 per m² as you move from Site 2 to Site 5) and an explanation linking this trend to the dissolved oxygen data (organic matter is broken down, lowering oxygen at Site 2, but oxygen levels rise to 8.1 mg/dm³ at Site 5, allowing the sensitive nymphs to survive).

Part (i):
- Award 1 mark for stating that Site 1 acts as a control / determines baseline/natural conditions of the river. [1]

Part (ii):
- Award 1 mark for describing the increase in mayfly abundance from Site 2 to Site 5 (must cite values: 0 to 42 per m²). [1]
- Award 1 mark for linking this recovery to the increase in dissolved oxygen (must cite values: 1.2 to 8.1 mg/dm³). [1]
- Award 1 mark for explaining that organic waste is diluted/decomposed by microorganisms downstream, reducing oxygen demand and allowing the oxygen-sensitive species to survive. [1]

Part (i) focuses on fieldwork methodology. To prevent bias, a random sampling strategy is required, which involves using random numbers to select grid coordinates. Representativeness is achieved by repeating the process to get multiple samples.

Part (ii) tests understanding of biodiversity and conservation management. Plantation forests lack structural diversity (they are usually monocultures with heavy shade) whereas ancient forests have diverse microhabitats, stratified layers, and have had long, undisturbed periods for colonisation.

Part (i):
- Award 1 mark for setting up a grid system / random coordinates selected using a generator/computer to avoid selection bias. [1]
- Award 1 mark for repeating the sampling multiple times (e.g., 10+ quadrats per site) to obtain a representative average. [1]

Part (ii):
- Award 1 mark for suggesting that ancient forests have more varied niches / microhabitats / structural complexity (e.g., decaying wood, diverse light gaps). [1]
- Award 1 mark for explaining that plantation forests are often managed monocultures of a single species with dense canopies that heavily shade the floor, suppressing other plant growth, or suffer from soil compaction from forestry machinery. [1]

To achieve Level 3 (7-8 marks), candidates must provide a balanced evaluation showing a deep understanding of both sides of the argument. They should weigh the positive aspects of the dam (e.g., hydroelectric power generation, reliable water supply, flood control, irrigation) against the negative environmental and social impacts (e.g., habitat loss, biodiversity reduction, greenhouse gas emissions from decomposing vegetation, displacement of indigenous groups, downstream siltation). They must also reach a clear, justified conclusion.

Indicative content:
- Arguments in favour of the statement (benefits of dams):
* Generates clean, renewable hydroelectric power (HEP), decreasing dependence on fossil fuels.
* Provides a consistent domestic and industrial water supply.
* Supplies water for agricultural irrigation to enhance local food security.
* Controls downstream flooding during periods of heavy rainfall.
- Arguments against the statement (drawbacks of dams):
* Flooding reservoirs destroys natural forest habitats, risking species extinction.
* Decomposing submerged forest vegetation releases greenhouse gases like methane \( (CH_4) \) and carbon dioxide \( (CO_2) \).
* Relocating indigenous communities causes social disruption and loss of cultural heritage.
* Dam walls block the migration pathways of river fish, impacting wild fish populations.
* Siltation reduces the useful capacity of the reservoir over time and deprives downstream floodplains of natural nutrients.

Level 3 [7–8 marks]
A coherent response is given that develops and supports the candidate’s conclusion using relevant details and examples. There is a balanced evaluation of both the economic/resource benefits and the socio-environmental drawbacks.

Level 2 [4–6 marks]
Development and support of the conclusion is evident, though the response may lack balance or specific detail. Key concepts (e.g., habitat loss, renewable energy, relocation) are discussed but with limited elaboration.

Level 1 [1–3 marks]
The response is limited in development and mostly one-sided. It may consist of a simple list of benefits or drawbacks with no clear conclusion reached.

No response or no creditable response [0 marks]

To achieve Level 3 (7-8 marks), candidates should evaluate both sides of the issue. They should discuss the ecological necessity of MPAs and permanent fishing bans (such as stock recovery, protecting biodiversity, and the 'spillover effect') while also acknowledging the serious socio-economic impacts (such as job losses, reduced food security, and poverty in coastal communities). Candidates should also mention alternative sustainable options (e.g., seasonal bans, quotas, or selective gear) and conclude with a well-reasoned personal viewpoint.

Indicative content:
- Arguments supporting permanent, total bans:
* Creating 'no-take zones' allows depleted fish populations a safe environment to breed and mature.
* Helps restore marine food webs and protect fragile benthic habitats from destructive gear like bottom trawls.
* Promotes the 'spillover effect', where increased fish biomass inside the MPA moves into surrounding fishable waters, eventually benefiting commercial fisheries.
* Easier for authorities to monitor and enforce a total ban than complex regulations like size limits or quotas.
- Arguments against permanent, total bans:
* Leads to immediate loss of income and jobs for local, small-scale fishermen who rely on these waters.
* Restricts access to a cheap, vital source of protein, threatening local food security.
* Can cause fishers to displace their efforts to other, unprotected marine zones, accelerating overfishing there.
* Complete exclusion of local stakeholders from decision-making can reduce compliance and increase poaching / illegal, unreported, and unregulated (IUU) fishing.

Level 3 [7–8 marks]
A well-structured, coherent response that provides a balanced assessment of the ecological benefits of no-take zones versus the socio-economic impacts on communities. Discusses the trade-off and offers a clear, logical conclusion.

Level 2 [4–6 marks]
The candidate presents arguments for both sides, but the essay may be unbalanced or lack specific details regarding marine conservation concepts (such as the spillover effect or food web dynamics). A simple conclusion is provided.

Level 1 [1–3 marks]
Response is mostly descriptive, brief, or heavily biased towards one side with little to no evaluation of the counterarguments. No clear conclusion.

No response or no creditable response [0 marks]