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IB Environmental Systems and Societies SL Paper 2 Data-Response Practice Paper (10 Questions, 100 Marks, 75 Minutes)
IB ESS SL Paper 2
IB ESS SL Paper 2 data-response practice paper: dependency ratios, net primary production, carbon fluxes, water use, soil erosion, Simpson index, oxygen sag.
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Paper details Level: IB Diploma | Subject: Environmental Systems and Societies SL | Paper: 2 (Data response) | Time: 75 minutes | Marks: 100 | Questions: 10
Section A Population, Production and Energy Flow (40 marks)
Questions 1-2
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | A country has a population (thousands) of 3600 aged 0-14, 7800 aged 15-64 and 1200 aged 65 and over. (a) Calculate the youth dependency ratio, the old-age dependency ratio and the total dependency ratio (%). (4 marks) (b) Define the dependency ratio and explain why a high ratio does not automatically mean a heavy economic burden. (2 marks) (c) Give two specific areas of government spending that rise as a population ages. (2 marks) |
(a) Youth dependency = 3600 / 7800 x 100 = 46.2%; old-age dependency = 1200 / 7800 x 100 = 15.4%; total = (3600 + 1200) / 7800 x 100 = 61.5%, which equals 46.2 + 15.4. (4 marks: 1 each, 1 for consistency) (b) The dependency ratio is the number of people outside working age (0-14 and 65+) divided by the number of working age (15-64), multiplied by 100: how many dependants each 100 working-age people support. A high ratio is not necessarily a heavy burden because the real burden depends on employment and participation rates, productivity, savings and the design of social security; if unemployment is high, or many older people still work, the ratio and the actual transfers diverge. (2 marks) (c) Health care and long-term care (hospital, residential and home care); pensions and social security; also acceptable: elderly housing, rehabilitation transport, community care facilities. (2 marks) Marking notes: numerator and denominator must not be mixed up; a vague “children plus old people” without a numeric definition earns half credit. |
| 2 | In a forest, gross primary production (GPP) is 42000 kJ m-2 yr-1 and producer respiration is 18900 kJ m-2 yr-1. Herbivores ingest 2310 kJ m-2 yr-1 and respire 1848 kJ m-2 yr-1. (a) Calculate net primary production (NPP) and NPP as a percentage of GPP. (4 marks) (b) Calculate the net production of primary consumers and the efficiency of transfer from producer NPP to that net production. (4 marks) (c) Explain why only a small fraction of energy passes between trophic levels. (2 marks) |
(a) NPP = GPP – R = 42000 – 18900 = 23100 kJ m-2 yr-1; NPP/GPP = 23100 / 42000 x 100 = 55%. (4 marks: 3 for NPP, 1 for the percentage) (b) Net production of primary consumers = 2310 – 1848 = 462 kJ m-2 yr-1; ecological efficiency from NPP = 462 / 23100 x 100 = 2.0%. (4 marks: 2 for net production, 1 for the working, 1 for the answer) (c) Most energy is lost between levels: much plant material is not eaten (roots, wood, fibre); of the material ingested, some is indigestible and lost in faeces; most assimilated energy is used in respiration for warmth, movement and metabolism and leaves as heat that cannot be reused; only what remains is stored as growth and reproduction, so roughly 10% passes on. (2 marks) Marking notes: NPP subtracts respiration only, never excretion; transfer efficiency must use NPP as the base, using GPP gives 5.5% and is wrong. |
Questions 3-4
| # | Question | Answer and marking notes |
|---|---|---|
| 3 | Energy in a grassland food chain (kJ m-2 yr-1) is: producers 100000, primary consumers 12000, secondary consumers 1500, tertiary consumers 180. (a) Calculate the three transfer efficiencies (%) and identify the highest. (6 marks) (b) Explain why a pyramid of biomass can be inverted in an aquatic system but a pyramid of energy cannot. (2 marks) (c) Calculate the energy lost in the transfer from secondary to tertiary consumers (kJ m-2 yr-1) and the percentage lost. (2 marks) |
(a) 100000 to 12000: 12.0%; 12000 to 1500: 12.5%; 1500 to 180: 12.0%. The highest is the second transfer at 12.5%. (6 marks: 2 per transfer) (b) In aquatic systems the producers are phytoplankton: they are tiny and turn over very fast, being eaten as quickly as they multiply, so standing biomass is low even though production is high, and consumer biomass can exceed producer biomass at any moment. A pyramid of energy records energy flow per unit time, and the second law of thermodynamics requires a loss at every step, so it can never be inverted. (2 marks) (c) Loss = 1500 – 180 = 1320 kJ m-2 yr-1; percentage lost = 1320 / 1500 x 100 = 88%. (2 marks) Marking notes: efficiencies are always divided by the energy of the level below; adding the three percentages together is a common error. |
| 4 | Global carbon fluxes (GtC yr-1): fossil fuel combustion 10.2, land-use change 1.6; ocean uptake 2.6, land sink 3.1. Each 1 ppm of atmospheric carbon dioxide corresponds to about 2.13 GtC. (a) Calculate total anthropogenic carbon emissions. (2 marks) (b) Calculate the carbon that remains in the atmosphere each year and its percentage of total emissions. (3 marks) (c) Calculate the annual rise in atmospheric carbon dioxide in ppm (two decimal places). (3 marks) (d) Suggest two reasons why the ocean sink may weaken in future. (4 marks) |
(a) Total = 10.2 + 1.6 = 11.8 GtC yr-1. (2 marks) (b) Sinks = 2.6 + 3.1 = 5.7; residual in the atmosphere = 11.8 – 5.7 = 6.1 GtC yr-1; share = 6.1 / 11.8 x 100 = 51.7%. (3 marks: 1 for sinks, 1 for the residual, 1 for the percentage) (c) 6.1 x (1 / 2.13) = 2.86 ppm yr-1. (3 marks: 2 for the working, 1 for the answer) (d) Warming reduces the solubility of carbon dioxide in seawater; acidification lowers carbonate ion concentration and weakens the buffering capacity; also acceptable: stronger stratification slows deep mixing, falling plankton productivity weakens the biological pump, and droughts and fires reduce the land sink. (4 marks: 2 each) Marking notes: sinks are added, not subtracted; the ppm conversion must use 2.13 GtC per ppm, and quoting GtC alone scores nothing. |
Section B Water, Soil and Biodiversity (30 marks)
Questions 5-7
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | A catchment receives 2400 mm of rain a year over 1200 km2, with a runoff coefficient of 0.40 and evaporation of 55% of rainfall; the population is 2.4 million. (a) Calculate the annual volume of rainfall and the annual runoff (m3). (3 marks) (b) Calculate the water available per person (m3 per person per year). (3 marks) (c) If absolute water scarcity is defined as below 500 m3 per person per year, classify this place and suggest one way to increase available water. (2 marks) |
(a) Rainfall volume = 2400 mm / 1000 x 1200 km2 x 10^6 = 2.88e+09 m3 (2.88 x 10^9 m3); runoff = 1.15e+09 x 0.40 = 1.152e+09 m3 (1.152 x 10^9 m3). (3 marks: 2 for rainfall, 1 for runoff) (b) Per person = 1.152e+09 m3 / 2.4 x 10^6 = 480 m3 per person per year. (3 marks) (c) 480 m3 is below 500 m3, so the area is in absolute water scarcity. Ways to increase supply: desalination, treated greywater reuse, rainwater harvesting, drip irrigation and mulching to cut evaporation, inter-basin transfer; reducing demand through quotas and leak control is also acceptable. (2 marks) Marking notes: mm and km2 must be converted to metres first; counting evaporation as usable water is a conceptual error. |
| 6 | Farmland loses 12 t ha-1 of topsoil a year while soil forms at 0.8 t ha-1 yr-1; soil organic matter starts at 2.4% and falls by 0.05 percentage points a year. (a) Calculate the net loss rate (t ha-1 yr-1). (2 marks) (b) Calculate the cumulative net loss over 40 years (t ha-1). (3 marks) (c) Calculate the organic matter percentage after 20 years and the relative fall (%). (3 marks) (d) Suggest two cultivation practices that reduce the loss. (2 marks) |
(a) Net loss = 12 – 0.8 = 11.2 t ha-1 yr-1. (2 marks) (b) Over 40 years = 11.2 x 40 = 448 t ha-1. (3 marks: 2 for the working, 1 for the answer) (c) After 20 years = 2.4 – 0.05 x 20 = 1.4%; relative fall = (2.4 – 1.4) / 2.4 x 100 = 41.7%. (3 marks) (d) Contour ploughing or terracing, no-till or reduced tillage, cover crops, mulch and stubble retention, windbreaks, rotation with organic manure; each needs a reason such as slowing runoff or raising infiltration. (2 marks: 1 each) Marking notes: the net rate must subtract soil formation; a fall of 1 percentage point is not the same as a 41.7% fall. |
| 7 | One hundred beetles are sampled at random in each of two woodlands. Woodland A: 50, 30, 20 individuals per species; woodland B: 90, 6, 4. Simpson index D = 1 – sum(n/N)^2. (a) Calculate D for woodland A (three decimal places). (4 marks) (b) Calculate D for woodland B (three decimal places). (4 marks) (c) Which woodland is more diverse, and why is counting species alone not enough? (4 marks) |
(a) sum(n/N)^2 = 0.50^2 + 0.30^2 + 0.20^2 = 0.25 + 0.09 + 0.04 = 0.38; D = 1 – 0.38 = 0.620. (4 marks: 1 per square, 1 for the sum, 1 for D) (b) sum(n/N)^2 = 0.90^2 + 0.06^2 + 0.04^2 = 0.81 + 0.0036 + 0.0016 = 0.8152; D = 1 – 0.8152 = 0.185. (4 marks) (c) Woodland A is more diverse, 0.620 against 0.185. Both have three species, but in B 90%% of individuals belong to one species, so evenness is very low and the community is fragile: a single disease of the dominant species would collapse it. The Simpson index combines richness and evenness, so richness alone cannot distinguish them. (4 marks: 1 for the choice, 2 for evenness, 1 for what the index measures) Marking notes: proportions, not raw counts, are squared; equal species numbers do not give equal D. |
Section C Pollution, Energy and Sustainability (30 marks)
Questions 8-10
| # | Question | Answer and marking notes |
|---|---|---|
| 8 | Dissolved oxygen (DO) upstream and downstream of a discharge is 9.2 and 3.4 mg/L, and BOD5 is 1.5 and 11.0 mg/L; the saturation concentration at that temperature is 9.2 mg/L. (a) Calculate the percentage change in DO downstream and the percentage saturation. (3 marks) (b) Calculate the increase in BOD5 and the ratio between the two sites. (2 marks) (c) Explain how organic pollution produces an oxygen sag curve and give one limitation of a single monitoring measurement. (3 marks) |
(a) Change = (3.4 – 9.2) / 9.2 x 100 = -63.0%, a fall of 63.0%; saturation = 3.4 / 9.2 x 100 = 37.0%. (3 marks: 2 for the change, 1 for saturation) (b) Increase = 11.0 – 1.5 = 9.5 mg/L; ratio = 11.0 / 1.5 = 7.3 times. (2 marks) (c) After the discharge, decomposers multiply and respire while breaking down the organic matter, using up dissolved oxygen, so DO falls sharply below the outfall and reaches a minimum in the oxygen sag; as the organic load is broken down and turbulence and algal photosynthesis re-aerate the water, DO recovers. Limitation: a single DO reading depends on time of day, temperature and flow and shows only that moment, while BOD5 needs five days of incubation and gives no immediate warning. (3 marks: 2 for the mechanism, 1 for the limitation) Marking notes: a percentage change needs a sign or an explicit fall; percentage saturation is not the same as concentration. |
| 9 | A region generates 40000 GWh a year: coal 45%, natural gas 30%, nuclear 15%, renewables 10%. Emission factors are 0.9 and 0.4 tCO2 per MWh for coal and gas, with nuclear and renewables counted as zero. (a) Calculate the electricity from coal and from gas (GWh) and the emissions of each (Mt CO2). (4 marks) (b) Calculate total emissions (Mt CO2) and the average emission intensity (g CO2 per kWh). (4 marks) (c) If 10% of generation switches from coal to wind, calculate the emissions avoided (Mt CO2) and the percentage of present total emissions. (2 marks) |
(a) Coal = 40000 x 0.45 = 18000 GWh, emitting 18000 x 10^3 MWh x 0.9 / 10^6 = 16.2 Mt; gas = 40000 x 0.30 = 12000 GWh, emitting 12000 x 10^3 x 0.4 / 10^6 = 4.8 Mt. (4 marks: 2 each) (b) Total = 16.2 + 4.8 = 21.0 Mt; intensity = 21.0 x 10^12 / (40000 x 10^6) = 525 g CO2 per kWh. (4 marks: 2 for the total, 2 for the intensity) (c) Electricity switched = 4000 GWh; avoided = 4000 x 10^3 x 0.9 / 10^6 = 3.6 Mt; share = 3.6 / 21.0 x 100 = 17.1%. (2 marks) Marking notes: GWh to MWh multiplies by 10^3 and tonnes to Mt divides by 10^6; intensity uses total generation in the denominator. |
| 10 | A country has an ecological footprint of 4.8 gha per person and a biocapacity of 1.9 gha per person; the world average footprint is 2.7 gha per person. (a) Calculate the country ecological deficit (gha per person). (3 marks) (b) Calculate the footprint as a multiple of biocapacity (two decimal places) and explain what overshoot means. (2 marks) (c) Name three sustainability indicators and explain why one indicator alone cannot assess sustainability. (4 marks) (d) Suggest one policy to reduce the deficit and say which data would be used to monitor it. (3 marks) |
(a) Deficit = 4.8 – 1.9 = 2.9 gha per person. (3 marks: 2 for the working, 1 for the answer) (b) Multiple = 4.8 / 1.9 = 2.53 times. Overshoot means demand for resources and for waste absorption exceeds what the ecosystems of that area can regenerate, so the gap is met by importing resources, drawing down natural capital such as over-felling or over-abstraction of groundwater, or exporting pollution, which cannot continue. (2 marks) (c) Examples: ecological footprint, Human Development Index, carbon dioxide emissions per person, Gini coefficient, access to clean water, forest cover, share of renewable energy. One indicator is not enough because sustainability has environmental, social and economic dimensions: a high HDI can come with a high footprint, and low emissions can come with poverty, so they must be read together. (4 marks: 3 for the three indicators, 1 for the explanation) (d) A carbon tax, public transport and rail investment, higher building energy standards, cutting food waste and meat consumption. Monitor with footprint per person (gha) and carbon dioxide emissions per person (t) over time, alongside changes in biocapacity. (3 marks: 2 for the policy, 1 for the monitoring data) Marking notes: the deficit is footprint minus biocapacity, never the reverse; overshoot is about regenerative capacity, not just pollution. |
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