2025 Cambridge IGCSE Environmental Management (0680) Practice Paper with Answers

1. Floating booms are deployed on the surface of the water to surround and contain the spread of the spilled oil.
2. Skimmers (oil recovery vessels or pumps) are then used to vacuum or scrape the contained oil layer off the surface of the water.

Award 1 mark for each correct physical containment or recovery method identified, up to a maximum of 2 marks.
- Acceptable answers include: floating booms / barriers, skimmers, sorbents / absorbent pads.
- Reject: chemical dispersants, burning, or bioremediation (as these are chemical, thermal, or biological treatment processes rather than physical containment or recovery).

1. Social disadvantage: Local communities and indigenous populations must be relocated (displaced) when the reservoir area is flooded behind the dam.
2. Environmental/Economic disadvantage: Large areas of fertile agricultural land, forests, and natural habitats are permanently submerged upstream.

Award 1 mark for each valid social or environmental disadvantage, up to a maximum of 2 marks.
- Accept: displacement / relocation of local populations, flooding of agricultural land / villages, disruption of fish migration routes, silting up of the reservoir, loss of nutrients downstream, increased risk of water-borne diseases.

1. Draining puddles, marshes, and uncovered water containers removes stagnant breeding sites (physical method).
2. Introducing biological control agents, such as larvae-eating fish (e.g., Gambusia or guppies), to open water bodies naturally reduces the mosquito population before they mature (biological method).

Award 1 mark for each valid physical or biological method, up to a maximum of 2 marks.
- Acceptable physical methods: draining pools of standing water, covering water storage containers, clearing vegetation around ponds.
- Acceptable biological methods: introducing predator fish (larvivorous fish), using bacterial pathogens (e.g., Bacillus thuringiensis israelensis).
- Reject: spraying chemical insecticides (e.g., DDT) or using bed nets (as these target adult mosquitoes, not their breeding process, and are chemical/mechanical barriers).

During rock formation, newly deposited sediment layers are buried by subsequent layers. This weight squeezes the sediments together in a process called compaction (Process A). Minerals dissolved in the groundwater then precipitate out and bind the sediment grains together in a process called cementation (Process B).

Award 1 mark for Process A and 1 mark for Process B.
- Process A: Compaction (accept: compression / consolidation).
- Process B: Cementation (accept: binding / crystallization of mineral glues).
- Do not accept answers in reverse order.

1. Seed banks prevent the extinction of rare plant species and crop wild relatives, conserving genetic resources that could otherwise be permanently lost.
2. The stored seeds serve as a genetic library, allowing plant breeders to retrieve genes for developing new crop varieties that can resist emerging diseases, pests, or withstand severe drought and heat caused by climate change.

Award 1 mark for each clear, distinct reason explained, up to a maximum of 2 marks.
- Accept: source of genes for pest/disease resistance, source of genes for climate change adaptation (e.g., drought resistance), safeguarding genetic diversity against natural disasters/wars, conserving wild ancestors of modern crops for future selective breeding.

1. Backfilling: The large open pits are filled with the stored overburden (waste rock and soil removed during initial extraction) and graded to match the surrounding landscape.
2. Revegetation: Saved topsoil is spread back over the area, and native grass, trees, or crops are planted to prevent erosion and restore soil ecosystems.

Award 1 mark for each valid land restoration method described, up to a maximum of 2 marks.
- Acceptable methods: backfilling the pit with overburden, replacing stored topsoil, revegetation / tree planting (reforestation), bioremediation / treatment of contaminated soils, converting the pit into a reservoir / recreational lake / landfill site.

1. Afforestation (planting trees in the upper river catchment) increases water interception and transpiration, reducing surface runoff.
2. Floodplain zoning restricts high-value development (like housing or industry) on land adjacent to the river that is natural flood territory, reducing potential economic damage.

Award 1 mark for each correct soft engineering strategy identified, up to a maximum of 2 marks.
- Accept: afforestation, floodplain zoning, river restoration (allowing the river to take its natural course), wetland creation/preservation, riparian buffer strips.
- Reject hard engineering methods: dams, levees, embankments, flood walls, channelization, dredging.

Decrease in crop yield = \(2.5 \text{ tonnes/ha} - 1.1 \text{ tonnes/ha} = 1.4 \text{ tonnes/ha}\). Percentage decrease = \(\frac{1.4}{2.5} \times 100 = 56\%\).

1 mark for correct working to find the yield reduction of 1.4 tonnes/ha or setting up the correct fraction \(\frac{2.5 - 1.1}{2.5}\). 1 mark for the correct final percentage of 56%.

First, identify the renewable energy resources: Hydroelectric (15%) and Solar and Wind (8%). Total renewable percentage = \(15\% + 8\% = 23\%\). Calculate the TWh: \(23\%\) of \(250 \text{ TWh} = 0.23 \times 250 = 57.5 \text{ TWh}\).

1 mark for correctly identifying the total percentage of renewables as 23% or showing the calculation \(0.23 \times 250\). 1 mark for the correct final answer of 57.5 (TWh).

Overall rate of change per 1000 = \(\text{Birth Rate} - \text{Death Rate} + \text{Net Migration Rate} = 18 - 6 + (-2) = 10\) per 1000. Percentage growth rate = \(\frac{10}{1000} \times 100\% = 1\%\).

1 mark for calculating the net change of 10 per 1000 (or showing the correct equation of \(18 - 6 - 2\)). 1 mark for converting this to 1% or 1.0%.

1. Microorganisms (such as bacteria or fungi) that can feed on hydrocarbons are introduced to the affected area, or indigenous populations are stimulated. 2. Nitrogen and phosphorus fertilizers are applied to the shoreline to accelerate microbial growth and activity (bio-stimulation). 3. The microorganisms metabolize the complex hydrocarbons in the oil. 4. The hydrocarbons are broken down into non-toxic end-products, primarily carbon dioxide, water, and biomass.

Award marks as follows: - Identify the role of oil-degrading bacteria/microorganisms [1 mark] - Use of fertilizers (nitrogen/phosphorus) to stimulate microbial growth [1 mark] - Breakdown/metabolism of complex hydrocarbons [1 mark] - Production of harmless end-products such as \(CO_2\) and water [0.5 mark]

1. An infected person excretes the Vibrio cholerae bacteria in their feces. 2. In areas with inadequate sanitation, sewage mixes with drinking water sources or crops are irrigated with contaminated water. 3. Uninfected people ingest the contaminated water or food. 4. The bacteria colonize the small intestine, releasing toxins that lead to severe diarrhea, continuing the cycle.

Award marks as follows: - Feces from an infected person contaminate the water supply/soil due to lack of sanitation [1 mark] - Uninfected individuals ingest this contaminated water or food (fecal-oral route) [1 mark] - Pathogen multiplies in the small intestine/releases toxins [1 mark] - Mention of diarrhea/vomiting leading to further environmental contamination and cycle perpetuation [0.5 mark]

1. Siltation: The dam blocks the river flow, dramatically reducing water velocity. This loss of kinetic energy causes suspended sediments to settle to the bottom of the reservoir. 2. Downstream erosion: The water discharged through the dam turbines or spillways is mostly sediment-free (known as 'hungry water'). 3. Because it lacks a sediment load, the discharged water has higher kinetic energy and capacity to erode downstream. 4. The river compensates by actively eroding its own bed and banks downstream.

Award marks as follows: - Reduction in river velocity as it enters reservoir leads to deposition of sediment/siltation [1 mark] - Water released downstream is sediment-free/clear [1 mark] - Clear water has higher kinetic energy/erosive power ('hungry water') [1 mark] - Resulting in severe erosion of the downstream river channel/banks [0.5 mark]

1. Mining activities expose previously buried rocks containing metal sulfides, such as pyrite (\(FeS_2\)), to the atmosphere. 2. Oxygen and rainwater react with these exposed sulfide minerals. 3. This chemical oxidation reaction produces sulfuric acid (\(H_2SO_4\)) and dissolved iron. 4. Rainwater transports this acidic solution out of the mine site into local water bodies, significantly lowering their pH.

Award marks as follows: - Exposure of sulfide-rich minerals (e.g., pyrite) to oxygen and water/rain [1 mark] - Chemical oxidation reaction occurs [1 mark] - Production of sulfuric acid and dissolved metals [1 mark] - Transport of acidic runoff into surrounding aquatic ecosystems via rainwater leaching [0.5 mark]

1. Agricultural fertilizers (nitrates and phosphates) are washed off farmland by rain (runoff) and enter coastal marine ecosystems. 2. The high nutrient levels trigger rapid growth of phytoplankton and algae, forming an algal bloom. 3. The algae die and sink to the bottom, where they are decomposed by aerobic bacteria. 4. The bacteria multiply rapidly and consume most of the dissolved oxygen in the water, causing hypoxia ('dead zones') which suffocates fish and other marine life.

Award marks as follows: - Runoff of fertilizers (nitrates/phosphates) from land to coastal waters [1 mark] - Nutrients stimulate rapid growth of algae/algal bloom [1 mark] - Aerobic bacteria decompose dead algae, consuming dissolved oxygen [1 mark] - Hypoxia/depletion of oxygen leads to death of marine life ('dead zones') [0.5 mark]

1. Meteorological drought starts with a prolonged lack of precipitation and high rates of evapotranspiration. 2. This leads to a depletion of soil moisture, causing agricultural drought where plants wilt. 3. Without rainfall, there is no percolation to recharge aquifers, causing the water table to drop. 4. Consequently, groundwater baseflow to rivers stops, leading to significantly reduced streamflow, dried-up reservoirs, and low lake levels (hydrological drought).

Award marks as follows: - Low precipitation and high evapotranspiration deplete soil moisture (meteorological/agricultural phase) [1 mark] - Lack of infiltration stops groundwater/aquifer recharge [1 mark] - Water table falls [1 mark] - River discharge, streamflow, and reservoir levels drop significantly (hydrological drought) [0.5 mark]

1. Natural soil and vegetation are replaced with impermeable surfaces (concrete, tarmac, roofs). 2. This prevents rainwater infiltration into the soil and percolation into groundwater. 3. Surface runoff increases dramatically and is channeled rapidly via storm drains directly into river systems. 4. The lag time (difference between peak rainfall and peak discharge) is shortened, resulting in a rapid surge in river discharge (high peak discharge) that overflows banks.

Award marks as follows: - Replacement of natural ground with impermeable surfaces, preventing infiltration [1 mark] - Increase in surface runoff and rapid transport of water via artificial drains/sewers [1 mark] - Shortened lag time on the river hydrograph [1 mark] - Significant increase in peak discharge causing the river to exceed its channel capacity [0.5 mark]

1. Identification and isolation: The gene of interest (e.g., the Bt toxin gene for pest resistance) is identified and cut from the donor organism's DNA using restriction enzymes. 2. Vector insertion: The gene is inserted into a vector, such as a bacterial plasmid or coated onto micro-particles for a gene gun. 3. Transformation: The vector is used to introduce the gene into the DNA of host crop plant cells. 4. Selection and tissue culture: Cells that successfully integrated the gene are grown into whole plants using plant tissue culture, resulting in a GM crop.

Award marks as follows: - Identify and isolate the desired gene from a donor organism using restriction enzymes [1 mark] - Insert the gene into a vector (e.g., plasmid or gene gun) [1 mark] - Transfer the gene/vector into host plant cells (transformation) [1 mark] - Grow transformed plant cells into full mature plants expressing the new trait [0.5 mark]

Bioremediation is an oil spill cleanup process that utilizes living organisms, primarily bacteria and fungi, to degrade hazardous hydrocarbons into non-toxic substances. The process begins when native oil-degrading microorganisms already present in the environment consume hydrocarbons as a carbon and energy source. To accelerate this slow natural process, nutrients containing nitrogen and phosphorus are applied to the shoreline (a process called biostimulation) to stimulate rapid microbial growth and activity. In some cases, specific, highly active oil-degrading microbes are introduced to the site (known as bioaugmentation). The microbes enzymatically break down the complex hydrocarbons into harmless end products, specifically carbon dioxide (\(CO_2\)), water (\(H_2O\)), and harmless organic biomass, effectively restoring the contaminated shoreline.

Award marks as follows: [1 mark] for explaining that microorganisms (bacteria or fungi) break down/biodegrade hydrocarbons into simpler substances. [1 mark] for detailing the addition of nutrients/fertilizers (nitrogen and phosphorus) to stimulate microbial growth (biostimulation). [1 mark] for identifying the harmless end products of the process, specifically carbon dioxide (\(CO_2\)) and water (\(H_2O\)). [0.5 marks] for mentioning bioaugmentation (adding specialized non-native microbes) or the requirement of oxygen (aerobic conditions) to sustain the process. Maximum 3.5 marks.

Schistosomiasis transmission is a multi-step cyclic process dependent on freshwater environments and specific hosts. First, an infected human releases parasite eggs into fresh water through their urine or faeces. Once in the water, these eggs hatch into free-swimming larvae called miracidia. The miracidia actively search for and infect specific species of freshwater snails, which serve as the intermediate host. Inside the snail, the parasite undergoes asexual reproduction and develops into a different larval stage called cercariae. The snails release thousands of these free-swimming cercariae into the water. Finally, when another human comes into contact with the infested water (during swimming, washing, or agricultural activities), the cercariae penetrate the unbroken human skin, migrate through the bloodstream to mature, and complete the transmission cycle.

Award marks as follows: [1 mark] for stating that parasite eggs enter fresh water via human faeces or urine. [1 mark] for explaining that hatched larvae (miracidia) must infect freshwater snails (intermediate hosts) to develop. [1 mark] for describing how the snails release free-swimming larval forms (cercariae) into the water. [0.5 marks] for explaining that transmission to a new human occurs when these cercariae penetrate the skin during water contact (swimming/washing/wading). Maximum 3.5 marks.

An effective evaluation of these three methods includes:

1. Containment Booms:
- How they work: Floating physical barriers used to restrict the spread of oil and keep it concentrated.
- Advantages: Highly effective in calm water; prevents oil from reaching sensitive ecosystems like salt marshes and beaches; allows for physical removal using skimmers without adding chemical pollutants to the water.
- Limitations: Ineffective in rough seas, high winds, or strong currents, as oil can wash over or flow under the booms.

2. Chemical Dispersants:
- How they work: Chemicals sprayed onto the spill that break the oil down into tiny droplets, allowing it to disperse into the water column.
- Advantages: Quickly disperses large slicks, reducing immediate threats to seabirds and marine mammals on the water surface; promotes natural biodegradation by increasing the surface area of the oil.
- Limitations: Does not remove oil from the environment; dispersant chemicals can themselves be toxic to corals, fish, and other marine organisms; moves the pollution from the surface into the deeper water column.

3. Bioremediation:
- How they work: The introduction of microorganisms (bacteria) or nutrients (like nitrogen and phosphorus) to accelerate the natural breakdown of hydrocarbons into harmless water and carbon dioxide.
- Advantages: Extremely environmentally friendly, natural process with no toxic chemical residues; useful for long-term clean-up of shorelines and hard-to-reach coastal areas.
- Limitations: Very slow process compared to physical or chemical methods; not suitable as an immediate emergency response to prevent oil from spreading; performance depends heavily on water temperature, oxygen levels, and nutrient availability.

Conclusion:
No single method is entirely sufficient on its own. Immediate containment using booms is the most ideal first response where weather permits. When containment is impossible, dispersants are effective offshore to protect surface wildlife, while bioremediation is the best long-term strategy for restoring contaminated coastlines.

Level 3 (5-6 marks):
- Evaluates all three methods (booms, dispersants, bioremediation) with clear explanations of how they work, their advantages, and their limitations.
- Information is organized, clear, and coherent.
- Provides a reasoned conclusion/judgment on their overall combined or relative effectiveness.

Level 2 (3-4 marks):
- Describes at least two of the methods, outlining some advantages and/or limitations.
- Structure is mostly logical but may lack depth in evaluation or fail to provide a clear concluding judgment.

Level 1 (1-2 marks):
- Identifies or briefly describes one or two methods with limited detail.
- The response is largely descriptive rather than evaluative.

Level 0 (0 marks):
- No creditable response.

To calculate the sector angles for a pie chart, multiply each percentage by 3.6 (since 360 degrees / 100% = 3.6 degrees per percent). Off-shore drilling: \(5 \times 3.6 = 18^\circ\). Natural seeps: \(45 \times 3.6 = 162^\circ\). Land-based runoff: \(35 \times 3.6 = 126^\circ\). Marine transportation: \(15 \times 3.6 = 54^\circ\). For a bar chart: (i) The independent variable is the source of oil pollution, which is plotted on the horizontal x-axis. (ii) The dependent variable is the percentage of oil pollution, which is plotted on the vertical y-axis.

[Part a: 2.0 marks] Award 0.5 marks for each correct angle: Off-shore drilling = 18 degrees, Natural seeps = 162 degrees, Land-based runoff = 126 degrees, Marine transportation = 54 degrees. [Part b: 2.5 marks] Award 1.0 mark for correctly identifying the independent variable (Source of oil pollution) and its axis (x-axis). Award 1.0 mark for correctly identifying the dependent variable (Percentage of oil pollution) and its axis (y-axis). Award 0.5 marks for clear structure and terminology.

(a) Mean calculation: \((12 + 15 + 8 + 18 + 7) / 5 = 60 / 5 = 12\) bluebells per square meter. (b) X-axis (independent variable) must be labeled 'Light intensity / lux' with a linear scale from 0 to at least 500. Y-axis (dependent variable) must be labeled 'Bluebell abundance' or 'Number of plants' with a linear scale from 0 to at least 25 (e.g., up to 30). The trend shows a strong negative correlation: as light intensity increases, the abundance of bluebells decreases.

[Part a: 1.5 marks] Award 1.0 mark for showing the correct working \((12+15+8+18+7)/5\) and 0.5 marks for the correct final answer of 12. [Part b: 3.0 marks] Award 1.0 mark for correctly stating the x-axis label with units (Light intensity / lux) and scale. Award 1.0 mark for correctly stating the y-axis label (Bluebell abundance / number of plants) and scale. Award 1.0 mark for identifying the negative trend (abundance decreases as light intensity increases).

(a) The highest yield is 5.5 tonnes/ha, which occurs at a fertilizer application rate of 60 kg/ha. (b) The independent variable, 'Fertilizer applied (kg/ha)', is plotted on the x-axis. The dependent variable, 'Crop yield (tonnes/ha)', is plotted on the y-axis. A suitable scale for the y-axis goes from 0 to 6.0 with intervals of 1.0 or 0.5. A curve of best fit is preferred because biological growth and response to nutrients change smoothly; straight line segments would imply unrealistic, abrupt changes in the rate of response at each exact measurement point.

[Part a: 1.5 marks] Award 1.0 mark for identifying the optimum fertilizer rate of 60 kg/ha and 0.5 marks for stating the yield of 5.5 tonnes/ha. [Part b: 3.0 marks] Award 1.0 mark for correctly assigning the variables to the correct axes (Fertilizer on x-axis, Yield on y-axis). Award 1.0 mark for suggesting a suitable linear scale for the crop yield axis (e.g. 0 to 6 tonnes/ha). Award 1.0 mark for explaining that biological response is continuous/smooth, making a curve of best fit more representative than rigid point-to-point lines.

(a) Total normal rainfall = \(45 + 50 + 40 + 30 + 15 + 10 = 190\text{ mm}\). Total actual rainfall = \(40 + 35 + 15 + 5 + 0 + 0 = 95\text{ mm}\). Deficit = \(190 - 95 = 95\text{ mm}\). (b) (i) The horizontal x-axis should be labeled 'Month'. The vertical y-axis should be labeled 'Rainfall / mm' or 'Rainfall (mm)'. (ii) For each month, two bars (one for Normal, one for Actual) should be drawn side-by-side touching each other, with a gap between different months. The bars must have different colors/shading, and a key must be included to define which bar represents normal and actual rainfall.

[Part a: 1.5 marks] Award 1.0 mark for correct calculation of both totals (190 mm and 95 mm) or showing the subtraction step. Award 0.5 marks for the correct deficit of 95 mm. [Part b: 3.0 marks] Award 1.0 mark for correct axes labels including unit for rainfall (mm). Award 1.0 mark for describing the side-by-side/clustered bar configuration. Award 1.0 mark for explaining how to distinguish the bars (using distinct colors/shading and a key).

1. Set up a coordinate grid in each field using two long tape measures placed at right angles.
2. Use a random number generator to generate pairs of coordinates to determine where to place the quadrats, which avoids sampler bias.
3. Place a standard size quadrat (e.g., \(1\,\text{m} \times 1\,\text{m}\)) at each selected coordinate.
4. Estimate the percentage cover of dandelions in each quadrat (e.g., using a grid quadrat or estimating visual cover).
5. Repeat this process at least 10 times in each field to calculate a reliable mean percentage cover.

Award 1 mark for each valid point up to a maximum of 2.2 marks:
- Use of coordinate grid and random number generator to select quadrat positions (prevents bias) [1 mark]
- Placing a quadrat of standard size and estimating percentage cover [1 mark]
- Repeating the procedure (at least 5-10 times) in each field to calculate an average / mean [1 mark]

To support biological data on sewage pollution, the student can measure:
1. Dissolved oxygen (DO): measured using an electronic DO probe or a chemical test kit (Winkler method).
2. Water temperature: measured using a standard thermometer or temperature probe.
3. pH: measured using a pH probe or universal indicator paper.

Example measurement method: The electronic dissolved oxygen probe is calibrated, lowered into the water at each sampling site, and the reading is recorded after stabilizing.

Award marks as follows (up to 2.2 marks):
- Identifying two relevant abiotic variables (e.g., dissolved oxygen, temperature, pH, nitrate concentration, phosphate concentration, turbidity) [1 mark for both, or 0.5 marks for one]
- Stating a correct method/instrument for measuring one of these variables (e.g., electronic probe, thermometer, pH meter, Secchi disc) [1.2 marks]
Reject: 'pollution' or 'cleanliness' as variables.

To ensure representative and reliable sampling:
- Use multiple transect lines (at least 3) radiating in different directions from the factory to account for wind direction variations.
- Take replicate soil samples (e.g., 3-5 samples) at the same distance at each sampling point, then calculate an average or combine them to create a composite sample.
- Ensure all samples are taken from the same depth (e.g., top 10 cm of soil) to avoid variations in pH caused by different soil horizons.

Award 1 mark for each valid methodology point up to 2.2 marks:
- Sampling along multiple transect lines in different directions from the factory [1 mark]
- Taking replicate samples at each distance and calculating an average [1 mark]
- Keeping sampling depth constant across all locations [1 mark]

To make this a fair test, the independent variable is the type of natural sorbent. All other factors that could influence absorption must be kept constant (controlled variables):
1. The volume or mass of motor oil added to the beaker (e.g., exactly 20 mL of oil).
2. The volume of water in each container (e.g., 200 mL).
3. The mass of each sorbent used (e.g., exactly 5.0 g of coconut coir, sawdust, or cotton wool).
4. The contact time allowed for absorption (e.g., exactly 5 minutes before removal).

Award 1.1 marks for each correct control variable up to a maximum of 2.2 marks:
- Volume / mass of oil added [1.1 marks]
- Mass of the sorbent used [1.1 marks]
- Volume of water used [1.1 marks]
- Duration/time the sorbents are left in contact with the oil [1.1 marks]
Do not accept: 'amount' of oil / sorbent (must specify mass or volume).

1. Measure the initial mass of the collected leaf with the dust on it using a high-precision balance (\(M_1\)).
2. Carefully wash the dust off the leaf using distilled water, then dry the leaf completely using paper towels.
3. Measure the final mass of the clean, dry leaf (\(M_2\)).
4. Calculate the mass of the dust: \(\text{Mass of dust} = M_1 - M_2\).
5. Determine the surface area of the leaf by tracing its outline on graph paper and counting the number of \(1\,\text{mm}^2\) or \(1\,\text{cm}^2\) squares (\(A\)). Since dust is deposited on the top surface, use the area of one side.
6. Calculate dust deposition: \(\text{Dust deposition} = \frac{M_1 - M_2}{A}\).

Award marks up to 2.2 marks:
- Explain how to determine the mass of dust (weigh dusty leaf, wash/dry, weigh clean leaf, find the difference) [1.1 marks]
- Explain how to determine the surface area of the leaf (using graph paper / grid squares) [1.1 marks]
- Correct calculation formula (mass of dust divided by surface area) [1 mark, max 2.2 marks total]

To avoid bias in selection, researchers can:
1. Use systematic sampling: Select every \(n\)-th house along a street or path (e.g., every 5th or 10th household) instead of choosing houses that look more accessible or wealthier.
2. Use random sampling: Use a village map to assign numbers to all households, and then use a random number generator to select which households to interview.
3. Avoid convenience sampling (e.g., do not only interview houses closest to the main road or clinic).

Award 1.1 marks for each valid suggestion (up to 2.2 marks):
- Use systematic sampling (e.g., selecting every N-th house) [1.1 marks]
- Use random sampling with a map and a random number generator [1.1 marks]
- Ensure sampling covers different areas of the village/district (e.g., stratified sampling based on distance to water source) [1.1 marks]
Reject: 'asking random people on the street' (this is convenience sampling and highly biased).

1. Manipulation of the independent variable: Setup identical soil trays, but vary the ground cover. For example, one tray is left bare (control), one has grass cover, and one is covered with leaf litter.
2. Measurement of the dependent variable: Place splash collector cups or paper cards around the perimeter of each tray. After a set period of simulated rainfall, collect the soil particles splashed out of the tray into these cups, dry the soil in an oven, and weigh the dry mass of splashed soil using a balance.

Award marks as follows (up to 2.2 marks):
- Manipulation of independent variable: planting different vegetation types/densities or comparing bare soil with planted soil in separate trays [1.1 marks]
- Measurement of dependent variable: collecting the soil particles splashed outside the tray and weighing their dry mass [1.1 marks]

To standardise the trees used for lichen sampling:
- Select only trees of the same species (as bark texture and acidity affect lichen growth).
- Select trees of a similar trunk diameter/girth (which indicates similar age and exposure time).
- Sample from the same side of the tree trunk (e.g., always the north-facing side) at a fixed height from the ground (e.g., between 1.0 m and 1.5 m) to ensure microclimatic conditions like sunlight and moisture are consistent.

Award 1.1 marks for each valid standardisation method (up to 2.2 marks):
- Choose trees of the same species [1.1 marks]
- Choose trees of similar trunk diameter / girth / age [1.1 marks]
- Sample at the same height from the ground [1.1 marks]
- Sample from the same aspect / compass direction on the trunk [1.1 marks]

The student needs to lay a tape measure from the forest edge 50 m straight into the forest. Quadrats should be placed at fixed, systematic intervals (for example, every 5 m) along this line. Within each quadrat, the student must record either the number of wild flowers or the percentage cover. To ensure reliability, this systematic sampling should be repeated along multiple parallel transect lines to find a mean.

Award marks for any two valid methodological steps, up to 2.2 marks total:
- Lay out a tape measure from the forest edge 50 m inside / use systematic sampling along a line [1.1]
- Place quadrats at regular intervals (e.g., every 5m) along the tape [1.1]
- Count the number of flowers / estimate percentage cover in each quadrat [1.1]
- Repeat with multiple transect lines / calculate an average to ensure reliability [1.1]

To compare soil drainage, the student must control key variables. They should place equal volumes of dry sandy soil and dry clay soil into identical funnels, each fitted with a piece of filter paper. A measuring cylinder is placed below each funnel. The student then pours the same volume of water into both funnels simultaneously. After a timed interval of 5 minutes, the volume of water collected in each measuring cylinder is recorded and compared.

Award marks for any two valid points, up to 2.2 marks total:
- Use equal masses/volumes of dry soil AND equal volumes of water (control variables) [1.1]
- Record the volume of water collected after a timed interval (e.g., 5 minutes) [1.1]
- Use funnels, filter paper, and measuring cylinders to measure the drainage [1.1]

To use a Secchi disc, the researcher lowers it into the water until the black and white segments disappear from view. They note the depth using markings on the line. Then, they lower it slightly further and slowly pull it up until the disk is just visible again, noting this second depth. The average of these two depths is calculated as the Secchi depth. To make it a fair comparison between the two ponds, measurements should be taken at the same time of day, on the same side of the boat (avoiding direct glare), and without sunglasses.

Award marks for any two valid points, up to 2.2 marks total:
- Lower disc until it disappears and record depth, then raise until it reappears and record depth, finding the average [1.1]
- Keep conditions constant (e.g., same time of day, same observer, avoid direct sunlight/glare) [1.1]

When using lichens as bioindicators on tree trunks, physical differences in the substrate can affect lichen growth independently of air pollution. The student must select trees of the same species because different tree species have bark with different pH, moisture retention, and roughness. Additionally, the student must sample lichens from the same aspect (direction) of the trunk (e.g., north-facing side) because different sides of a tree receive different amounts of sunlight, wind, and rain, which affects microclimatic humidity and lichen survival.

Award 1.1 marks for each correct control variable up to a maximum of 2.2 marks:
- Select trees of the same species / same bark characteristics [1.1]
- Sample from the same height on the trunk [1.1]
- Sample from the same aspect/compass direction on the trunk [1.1]
- Select trees of similar age/girth [1.1]

First, calculate the actual decrease in oil spills:
\(358 - 63 = 295\) spills.

Next, calculate the percentage decrease based on the original decade's total:
\(\text{Percentage decrease} = \left( \frac{295}{358} \right) \times 100 = 82.4022...\%\)

Rounded to one decimal place, the percentage decrease is \(82.4\%\).

* [1 mark] for calculating the correct absolute difference of 295.
* [1 mark] for dividing the difference by 358 and multiplying by 100.
* [0.5 marks] for the correct final answer of 82.4% (accept 82.4; reject 82 or 82.40).

1. Identify shallow earthquakes (depth < 70 km): Earthquake 1 (12 km) and Earthquake 2 (45 km).
Calculate mean depth: \(\frac{12 + 45}{2} = 28.5\text{ km}\).

2. Identify deep earthquakes (depth > 300 km): Earthquake 4 (310 km) and Earthquake 5 (620 km). This represents 2 out of 5 total earthquakes.
Calculate percentage: \(\left( \frac{2}{5} \right) \times 100 = 40\%\).

* [1 mark] for correctly calculating the mean depth of shallow earthquakes as 28.5 km (allow working: (12+45)/2).
* [1 mark] for identifying that 2 out of 5 earthquakes are deep and setting up the calculation: \(\frac{2}{5} \times 100\).
* [0.5 marks] for the correct final percentage of 40%.

First, calculate the squared proportion \(\left( \frac{n}{N} \right)^2\) for each species:
* Species 1: \(\left( \frac{15}{40} \right)^2 = 0.140625\)
* Species 2: \(\left( \frac{14}{40} \right)^2 = 0.1225\)
* Species 3: \(\left( \frac{11}{40} \right)^2 = 0.075625\)

Sum of these values:
\(0.140625 + 0.1225 + 0.075625 = 0.33875\)

Calculate D:
\(D = 1 - 0.33875 = 0.66125\)

Rounding to two decimal places gives \(0.66\).

* [1 mark] for calculating individual proportions squared and summing them up to obtain 0.33875 (or 542/1600).
* [1 mark] for subtracting this sum from 1 to obtain 0.66125.
* [0.5 marks] for rounding the final answer correctly to 0.66.

1. Calculate percentage increase from 0 to 100 kg/ha:
Yield at 0 kg/ha = 2.4 tonnes/ha.
Yield at 100 kg/ha = 5.2 tonnes/ha.
\(\text{Increase} = 5.2 - 2.4 = 2.8\text{ tonnes/ha}\).
\(\text{Percentage increase} = \left( \frac{2.8}{2.4} \right) \times 100 = 116.67\%\) (or 116.7%).

2. Identify optimum level:
The maximum mean yield is 5.8 tonnes/ha, which is achieved at an application rate of 150 kg/ha. After this, yield declines to 5.6 tonnes/ha.

* [1 mark] for correct percentage increase formula setup: \(\frac{5.2 - 2.4}{2.4} \times 100\).
* [0.5 marks] for the correct percentage calculation of 116.7% (accept range 116.6% to 117.0%).
* [1 mark] for identifying the optimum fertilizer level as 150 kg/ha.

1. Lag time is the time difference between peak rainfall and peak discharge:
19:30 hours minus 14:00 hours = 5 hours and 30 minutes.

2. Storm runoff discharge = Peak discharge \( - \) Baseflow = \(180\text{ m}^3\text{/s} - 20\text{ m}^3\text{/s} = 160\text{ m}^3\text{/s}\).
As a percentage of peak discharge:
\(\left( \frac{160}{180} \right) \times 100 = 88.88...\%\), which rounds to 88.9%.

* [1 mark] for the correct lag time of 5 hours 30 minutes (or 5.5 hours).
* [1 mark] for correctly calculating the runoff discharge of 160 and setting up the fraction: \(\frac{160}{180} \times 100\).
* [0.5 marks] for the correct final percentage of 88.9% (accept 88.8% to 89.0%).

1. Drought frequency:
Number of drought years (\(n\)) = 11.
Total years (\(N\)) = 40.
\(\text{Frequency} = \left( \frac{11}{40} \right) \times 100 = 27.5\%\).

2. Recurrence interval:
\(T = \frac{40}{11} = 3.636...\text{ years}\).
Rounded to one decimal place, this is 3.6 years.

* [0.5 marks] for correctly noting 11 drought years.
* [1 mark] for calculating the correct frequency of 27.5%.
* [1 mark] for calculating the correct recurrence interval of 3.6 years (accept range 3.6 to 3.64).

First, calculate the total capacity loss from 1980 to 2020:
\(450\text{ million m}^3 - 344\text{ million m}^3 = 106\text{ million m}^3\).

Next, divide this total loss by the 40-year period to find the annual rate:
\(\text{Annual rate of loss} = \frac{106\text{ million m}^3}{40\text{ years}} = 2.65\text{ million m}^3\text{ per year}\).

* [1 mark] for calculating the total volume loss of 106 million \(\text{m}^3\).
* [1 mark] for setting up the division of the total loss by 40 years.
* [0.5 marks] for the final correct rate of 2.65 million \(\text{m}^3\)/year.

First, sum the percentages of underfished and maximally sustainably fished stocks:
\(7.2\% + 55.4\% = 62.6\%\).

Next, calculate this proportion of the total assessed biomass of 120 million tonnes:
\(120\text{ million tonnes} \times 0.626 = 75.12\text{ million tonnes}\).

* [1 mark] for correctly adding the percentage of underfished and sustainably fished stocks to get 62.6%.
* [1 mark] for multiplying the combined percentage by 120 million tonnes.
* [0.5 marks] for the correct final answer of 75.12 million tonnes (accept 75.1).

1. Larger mesh size: Enables young, immature fish to swim through the net. This allows them to survive, grow, and reach sexual maturity so they can reproduce at least once before being harvested.
2. Seasonal closures: Prohibits fishing in specific spawning zones during the breeding season. This protects spawning adults from capture and prevents disruption to eggs and larvae, facilitating successful recruitment into the population.

Award 1 mark for each correct mechanism explained (up to 2.8 marks maximum):
- Larger mesh size allows juvenile / young / undersized fish to escape (1)
- This allows fish to reach sexual maturity / reproduce before being caught (1)
- Seasonal closures protect spawning/breeding adult fish during their reproductive period (1)
- This increases the survival rate of eggs/larvae and boosts population replenishment (1)

Booms and skimmers are physical recovery methods that actually extract the oil from the marine ecosystem. On the other hand, chemical dispersants do not remove the oil; they only break it up into tiny droplets that sink. In shallow coastal waters, these dispersed oil droplets settle onto mangrove pneumatophores (breathing roots), causing suffocation, and the chemical agents themselves are highly toxic to sensitive marine organisms in coastal nurseries.

Award marks for the following points (up to 2.8 marks):
- Booms and skimmers physically remove the oil from the aquatic environment entirely (1)
- Chemical dispersants do not remove oil, but instead disperse/sink it into the water column (1)
- Dispersed oil droplets easily coat and suffocate mangrove breathing roots (pneumatophores) (1)
- Chemical dispersants are highly toxic to sensitive coastal organisms and nursery species (1)

In a wet FGD system, flue gases containing acidic sulphur dioxide (\(\text{SO}_2\)) are scrubbed with an alkaline slurry of sprayed limestone (\(\text{CaCO}_3\)). The \(\text{SO}_2\) reacts with the calcium carbonate and is oxidised by added air to produce calcium sulphate (\(\text{CaSO}_4 \cdot 2\text{H}_2\text{O}\)), known as synthetic gypsum. This gypsum is a valuable industrial by-product used to manufacture plasterboard and cement.

Award marks for the following points (up to 2.8 marks):
- Sulphur dioxide reacts with limestone/calcium carbonate to neutralise the acidic gas (1)
- Oxygen/air is added to oxidise the mixture to produce calcium sulphate / gypsum (1)
- Gypsum is a highly useful industrial by-product used in building and construction (e.g. plasterboard, cement) (1)
- This commercial use reduces the environmental impact of mining natural gypsum (1)

1. Secondary treatment: Wastewater is aerated so that aerobic bacteria can decompose dissolved organic matter, which greatly lowers the Biological Oxygen Demand (BOD) and prevents oxygen depletion in the lake.
2. Tertiary treatment: This advanced stage removes dissolved mineral plant nutrients, specifically nitrates and phosphates, which directly prevents eutrophication (algal blooms) in the lake. It also includes disinfection (via chlorination, UV light, or ozone) to kill any remaining pathogens.

Award marks for the following points (up to 2.8 marks):
- Secondary treatment uses aerobic bacteria/microorganisms to decompose dissolved organic matter (1)
- This reduces biological oxygen demand (BOD) and prevents oxygen depletion in the lake (1)
- Tertiary treatment removes plant nutrients (nitrates and phosphates) to prevent eutrophication/algal blooms (1)
- Tertiary treatment includes disinfection (chlorination, UV, or ozone) to kill waterborne pathogens (1)

1. Cholera control: Provide clean, treated drinking water. This can be achieved by distributing chlorine tablets (water purification tablets) or teaching communities to boil all drinking water. This kills the pathogen (Vibrio cholerae) and halts transmission.
2. Malaria vector control: Eliminate stagnant breeding sites for the Anopheles mosquito vector. This involves draining pools of standing floodwater, using chemical larvicides, or introducing biological control agents (such as fish that eat mosquito larvae) to disrupt their life cycle.

Award marks for the following points (up to 2.8 marks):
- Cholera strategy: provide safe drinking water (e.g. through chlorination, boiling, or filtration) (1)
- Cholera explanation: kills Vibrio cholerae bacteria, stopping transmission via ingestion (1)
- Malaria strategy: eliminate breeding pools (draining standing water) OR apply larvicides OR use biological control (e.g. larvivorous fish) (1)
- Malaria explanation: disrupts the life cycle of the Anopheles mosquito vector, reducing vector population size (1)

Improving female education acts as a highly effective population management tool through several key mechanisms:
1. Delaying marriage: Girls who stay in school longer tend to marry later in life, delaying their first pregnancy and shortening their total reproductive lifespan.
2. Career opportunities: Education enables women to seek formal employment and financial independence, which raises the opportunity cost of raising children and encourages smaller family sizes.
3. Family planning knowledge: Educated women have better awareness of health services, reproductive rights, and are more likely to successfully adopt modern contraceptive methods.

Award marks for the following points (up to 2.8 marks):
- Staying in school delays the age of marriage and first pregnancy (1)
- Shortens the overall reproductive lifespan of women (1)
- Provides career/employment opportunities, increasing the economic cost of having children / changing aspirations (1)
- Enhances knowledge, access, and effective use of family planning and contraceptives (1)

1. Contour ploughing: Ploughing across the slope (following elevation contours) rather than up and down creates horizontal ridges and furrows. These act as miniature dams that slow down the speed of surface runoff and give water more time to infiltrate the soil, minimizing water-driven sheet and rill erosion.
2. Windbreaks: Planting dense rows of trees or shrubs around fields acts as physical barriers to block and redirect strong winds. This reduces wind velocity at the soil surface, preventing dry topsoil particles from being lifted and blown away, while the tree roots help bind the soil structure together on the slopes.

Award marks for the following points (up to 2.8 marks):
- Contour ploughing creates ridges across/perpendicular to the slope (1)
- This slows down surface runoff and increases water infiltration, preventing water erosion (1)
- Windbreaks consist of rows of trees/shrubs that block and reduce wind speed at ground level (1)
- This prevents wind from blowing away loose topsoil, and tree roots help bind and stabilise the soil on slopes (1)

1. Landform reconstruction (regrading): The company must grade and shape the steep quarry walls and fill in deep pits to make the landscape stable, safe, and prevent landslides.
2. Soil and waste management: Toxic tailings or acidic waste rock must be capped (e.g., with clay) to prevent chemical leaching and acid mine drainage. The original topsoil (which was stripped and stored during initial excavation) must then be spread back over the graded land.
3. Revegetation: Native grasses, shrubs, or pioneer trees must be planted. Their roots bind the soil to prevent erosion, and as they decompose, they restore nutrients and organic matter, facilitating ecological succession.

Award marks for the following points (up to 2.8 marks):
- Regrading / reshaping of steep quarry walls and filling pits to ensure physical stability (1)
- Capping toxic tailings/waste with clay or sealing layers to prevent acid mine drainage/water pollution (1)
- Spreading stored topsoil back over the area to provide a nutrient-rich growing medium (1)
- Planting native/pioneer species to bind the soil, control erosion, and restore biodiversity (1)

Sustainable harvesting can be achieved using several methods: 1. Fishing quotas: Limiting the maximum mass of blue-lined snapper that can be landed prevents overfishing. 2. Mesh size restrictions: Using larger net meshes allows juvenile fish to escape, grow, and reproduce. 3. Closed seasons: Banning fishing during the spawning season protects breeding adults and allows population recovery. Other valid methods include establishing marine protected areas (MPAs) and licensing vessels.

Award 1 mark for each valid management strategy suggested, up to a maximum of 3 points (scaled to 2.8 marks). Acceptable points: - Introduce fishing quotas / limits on catch size. - Restrict net mesh sizes / mandate larger mesh size. - Establish closed seasons / ban fishing during breeding seasons. - Create marine protected areas (MPAs) / no-take zones. - Restrict the size, number, or type of fishing vessels. - Issue fishing licenses to limit the number of fishers. Reject: - Complete fishing ban (as excluded by the question). - 'Education' without context.

To manage domestic organic pollution: 1. Sewage treatment plants: Directing domestic waste to treatment facilities ensures organic matter is broken down before effluent is released. 2. Regulations and monitoring: Implementing and enforcing strict fines prevents illegal dumping of untreated waste. 3. Riparian buffer zones: Planting vegetation along riverbanks slows runoff and naturally filters out solid organic waste. Other options include public awareness campaigns to encourage proper household waste disposal.

Award 1 mark for each distinct, realistic management action, up to a maximum of 3 points (scaled to 2.8 marks): - Construction or upgrading of sewage/wastewater treatment plants. - Enforcing strict regulations, laws, or fines against illegal waste dumping. - Planting riparian buffer zones / vegetated borders to filter runoff. - Improving municipal waste collection and disposal systems. - Educating citizens about correct domestic waste disposal practices. Reject: - General 'clean up the river' without a specific method.

Non-chemical vector control targets the life cycle and transmission of the Anopheles mosquito: 1. Eliminating breeding sites: Draining small, stagnant pools of water prevents female mosquitoes from laying eggs. 2. Biological control: Introducing predator fish to reservoirs consumes larvae, reducing the adult population. 3. Physical barriers: Using bed nets and window screens physically prevents mosquitoes from biting humans, disrupting the pathogen transmission cycle.

Award 1 mark for each valid non-chemical vector control method, up to a maximum of 3 points (scaled to 2.8 marks): - Draining stagnant water / puddles / filling in hollows to remove breeding sites. - Introducing biological predators (e.g., mosquito fish / Gambusia) to eat larvae. - Using physical barriers (e.g., window screens, bed nets). - Clearing tall grass/vegetation around dwellings where adult mosquitoes rest. - Using bacterial larvicides (e.g., Bti) which target larvae specifically and are biological, not synthetic chemical insecticides. Reject: - Chemical spraying / DDT / insecticide-treated nets (as question specifies 'without relying on chemical insecticides').

Slope soil erosion can be managed by: 1. Terracing: Cutting flat steps into the hillside slows the downward flow of water and encourages infiltration rather than runoff. 2. Contour ploughing: Ploughing horizontally across the slope creates ridges that trap water and soil. 3. Cover cropping and mulching: Maintaining vegetative cover or crop residues protects bare soil from the kinetic energy of heavy raindrops and binds soil particles with roots. Other valid methods include agroforestry and windbreaks.

Award 1 mark for each correct soil management practice appropriate for steep slopes, up to 3 points (scaled to 2.8 marks): - Terracing / building terraces. - Contour ploughing / planting along contours. - Planting cover crops / maintaining vegetative cover. - Mulching / applying organic residue to soil surface. - Agroforestry / planting trees along with crops to bind soil. - Strip cropping / planting alternating strips of close-growing plants. Reject: - General crop rotation on its own without explanation of erosion control. - Fertiliser application (does not directly prevent erosion).