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S6 Biology Paper: Energy Flow, Biomass, Nutrient Cycles, Biodiversity and Succession (100 marks)
S6 Biology Practice Paper
S6 Biology: 6 questions, 24 parts, 100 marks, 60 minutes on energy flow, biomass, nutrient cycles, biodiversity, succession and mark-recapture.
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Paper details S6 biology practice paper with 6 questions and 24 parts on energy flow, biomass, nutrient cycles and conservation.
Section A: Energy Flow in Ecosystems (16 marks)
Researchers measured energy data for a grassland ecosystem over one year (unit: kJ m-2 yr-1): producers fixed 87,000 in total; producer respiration used 39,000; primary consumers ingested 12,000, of which 3,600 was egested in faeces; primary consumer respiration used 5,400; secondary consumers assimilated 1,680.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | Calculate the net primary production (NPP) of the producers. | Answer: NPP = 87,000 – 39,000 = 48,000 kJ m-2 yr-1. (Marking: 2 marks for the method, 1 mark for the value, 1 mark for the unit.) |
| 2 | Calculate the assimilation of the primary consumers. | Answer: assimilation = ingested – egested = 12,000 – 3,600 = 8,400 kJ m-2 yr-1. Energy in faeces was never assimilated. (Marking: 3 marks for the subtraction, 1 mark for the explanation.) |
| 3 | Calculate the ecological efficiency of energy transfer from primary to secondary consumers. | Answer: efficiency = 1,680 divided by 8,400, times 100 = 20.0 per cent. (Marking: 2 marks for the formula, 1 mark for the value, 1 mark for the unit.) |
| 4 | Explain why an energy pyramid rarely has more than four or five levels. | Answer: at each trophic level only about one tenth of the energy is built into growth and passed on; the rest is lost in respiration, excretion and uneaten parts. After four or five levels the energy left cannot support another viable population. (Marking: 2 marks for energy loss at each level, 2 marks for insufficient energy.) |
Section B: Biomass and Production (16 marks)
A student measured biomass in a grassland using quadrats of 0.5 m by 0.5 m. The first harvest gave a dry mass of 18.0 g. One year later, a quadrat on another plot with the same conditions gave a dry mass of 24.5 g. The growing season lasted 150 days.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | Calculate the standing biomass per square metre. | Answer: area = 0.5 x 0.5 = 0.25 m2; biomass = 18.0 divided by 0.25 = 72.0 g m-2. (Marking: 2 marks for the area, 2 marks for the value with unit.) |
| 6 | Calculate the increase in biomass per square metre over the year. | Answer: increase = (24.5 – 18.0) divided by 0.25 = 26.0 g m-2 yr-1. (Marking: 2 marks for the difference, 2 marks for dividing by the area.) |
| 7 | Calculate the mean daily production during the growing season. | Answer: daily production = 26.0 divided by 150 = 0.173 g m-2 per day. (Marking: 2 marks for the division, 1 mark for the value, 1 mark for the unit.) |
| 8 | Explain why measuring only the above-ground parts underestimates net primary production. | Answer: net primary production includes root growth, material eaten by herbivores, leaf litter and exudates. Harvesting only shoots leaves out all below-ground production and the parts already consumed, so the value is always underestimated. (Marking: 2 marks for below-ground parts, 2 marks for consumed and shed parts.) |
Section C: Carbon and Nitrogen Cycles (16 marks)
A woodland produces 1,000 g of leaf litter each year, containing 45 per cent carbon by dry mass. Decomposers release 60 per cent of that carbon as carbon dioxide; the rest remains in soil organic matter. (Relative atomic masses: C = 12, O = 16.)
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | Calculate the mass of carbon in the annual leaf litter. | Answer: carbon = 1,000 x 45 per cent = 450 g. (Marking: 3 marks for the working, 1 mark for the value with unit.) |
| 10 | Calculate the mass of carbon released as carbon dioxide each year. | Answer: carbon released = 450 x 60 per cent = 270 g. (Marking: 3 marks for the working, 1 mark for the value with unit.) |
| 11 | Convert that released carbon into the mass of carbon dioxide. | Answer: mass of carbon dioxide = 270 x 44 divided by 12 = 990 g. (Relative molecular mass of CO2 is 12 + 16 x 2 = 44.) (Marking: 2 marks for the molar ratio, 2 marks for the calculation.) |
| 12 | State the difference between nitrogen fixation and nitrification. | Answer: nitrogen fixation reduces atmospheric nitrogen to ammonia or ammonium ions, carried out by nitrogen-fixing bacteria; nitrification oxidises ammonium ions to nitrite and then to nitrate by nitrifying bacteria, and nitrate is the form plants absorb most readily. (Marking: 2 marks for defining fixation, 2 marks for nitrification and its product.) |
Section D: Biodiversity and Conservation (16 marks)
Two islands were surveyed. Island A has an area of 250 km2 and lies 40 km from the mainland, with 180 native species. Island B has an area of 60 km2 and lies 12 km from the mainland, with 95 native species. An endangered bird numbers 320 on Island A and 80 on Island B. Assume heterozygosity falls by 1 divided by (2Ne) each generation, with Ne = 50 on both islands.
| # | Question | Answer and marking notes |
|---|---|---|
| 13 | Calculate the number of species per square kilometre on each island. | Answer: Island A 180 divided by 250 = 0.72 species per km2; Island B 95 divided by 60 = 1.58 species per km2. (Marking: 2 marks for each island, working and value.) |
| 14 | Decide which island carries the higher extinction risk, using area, population size and isolation. | Answer: Island B is at higher risk. Although it is closer to the mainland, its area is only about a quarter of Island A, so it supports fewer habitats and niches; with only 80 birds, genetic drift and inbreeding act more strongly, and a disease outbreak or typhoon could remove the whole population. (Marking: 1 mark for the decision, 1 mark for area, 1 mark for population size, 1 mark for comparing isolation.) |
| 15 | Calculate the percentage of heterozygosity retained after 10 generations. | Answer: fraction retained per generation = 1 – 1 divided by (2 x 50) = 0.99; after 10 generations = 0.99 to the power 10, times 100 = 90.4 per cent, so about 9.6 percentage points are lost. (Marking: 2 marks for the per-generation fraction, 1 mark for raising to the power 10, 1 mark for the value.) |
| 16 | Suggest two in-situ conservation measures and explain how each lowers extinction risk. | Answer: first, designate protected areas and restrict development to keep large, continuous habitats, which keeps population and effective population size high and slows the loss of genetic diversity. Second, build ecological corridors between fragmented woodlands so individuals can move and breed between populations, reducing inbreeding and allowing recolonisation after local disasters. (Marking: 2 marks for each measure with its explanation.) |
Section E: Ecological Succession (16 marks)
Researchers recorded species numbers on cooled volcanic lava: 6 species in year 1, 14 species in year 5, 27 species in year 20 and 31 species in year 80.
| # | Question | Answer and marking notes |
|---|---|---|
| 17 | Calculate the mean number of new species added each year from year 1 to year 20. | Answer: mean gain per year = (27 – 6) divided by 19 = 1.11 species per year. (Marking: 2 marks for the difference, 1 mark for dividing by 19 years, 1 mark for the value.) |
| 18 | Calculate the percentage increase in species number from year 20 to year 80. | Answer: percentage increase = (31 – 27) divided by 27, times 100 = 14.8 per cent. (Marking: 2 marks for the formula, 1 mark for the value, 1 mark for the unit.) |
| 19 | Decide whether this is primary or secondary succession, and explain. | Answer: this is primary succession. The lava has no soil and no existing seeds, roots or spores, so pioneer species must arrive from elsewhere and build soil from nothing. (Marking: 2 marks for the decision, 1 mark each for absent soil and absent propagules.) |
| 20 | Explain why species number levels off in the later stages of succession. | Answer: a climax community forms, with stable biomass and layered structure. Resources and niches are fully occupied, so new species cannot easily invade, while competitive exclusion removes some early species, and gains roughly balance losses. The climate and soil can support only a limited number of species. (Marking: 2 marks for the climax community, 2 marks for competition and niche saturation.) |
Section F: Ecological Survey and Data Analysis (20 marks)
Students used the mark-recapture method to estimate a snail population: 60 snails were caught, marked and released; a second sample of 80 snails contained 12 marked individuals. They also counted snails in five 1 m2 quadrats and obtained 4, 6, 5, 7 and 3.
| # | Question | Answer and marking notes |
|---|---|---|
| 21 | Estimate the population size of the snails. | Answer: population size = (60 x 80) divided by 12 = 400 snails. (Marking: 3 marks for the formula, 2 marks for the value with unit.) |
| 22 | Calculate the percentage of marked individuals in the second sample. | Answer: percentage = 12 divided by 80, times 100 = 15.0 per cent. (Marking: 3 marks for the working, 2 marks for the value and unit.) |
| 23 | State two assumptions of the mark-recapture method. | Answer: first, the population is closed, with no immigration, emigration, births or deaths during the study; second, marking does not affect survival or the chance of being caught, and marked individuals mix evenly so the second sample is representative. (Marking: 2.5 marks for each assumption.) |
| 24 | Calculate the mean snail density in the five quadrats. | Answer: total = 4 + 6 + 5 + 7 + 3 = 25 snails; mean density = 25 divided by 5 = 5.0 snails per m2. (Marking: 2 marks for the total, 2 marks for the mean, 1 mark for the unit.) |
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