HKDSE Biology Practice Paper: Ecosystems, Energy Flow And Biodiversity (12 Questions, 100 Marks, 75 Minutes, With Answers)

HKDSE Biology Practice Paper: Ecosystems, Energy Flow and Biodiversity (12 Questions, 100 Marks, 75 Minutes, with Answers)

2026-11-06

HKDSE Biology: Ecosystems, Energy Flow and Biodiversity

S6 biology practice paper, 12 questions, 100 marks, 75 minutes: net primary productivity, energy transfer, biomass pyramids, carbon and nitrogen cycles, mark-recapture, logistic growth, suc

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Section A: Energy Flow and Nutrient Cycling (24 marks)

Each question carries 6 marks; show your working or reasoning.

#	Question	Answer and marking notes
1	Producers in a grassland ecosystem fix 48,000 kJ of solar energy per square metre per year, of which 18,000 kJ is used in the producers’ own respiration. (a) Calculate the net primary productivity of the ecosystem. (b) If only 10% of the energy is transferred at each trophic level, calculate the energy available to the primary, secondary and tertiary consumers. (c) Explain why food chains rarely have more than four or five trophic levels.	(a) Net primary productivity = gross primary productivity minus respiration = 48,000 – 18,000 = 30,000 kJ per square metre per year. (2 marks) (b) Primary consumers receive 3,000 kJ, secondary consumers 300 kJ and tertiary consumers 30 kJ. (2 marks) (c) Only about one tenth of the energy passes to the next level, the rest being lost in respiration and excretion or never eaten; by the fourth or fifth level too little energy remains to support a breeding population. (2 marks)
2	A grassland food chain is grass, locusts, frogs and snakes, with standing biomass of 12,000 kg, 1,200 kg, 120 kg and 12 kg respectively. (a) Calculate the ratio of the biomass of locusts to that of snakes. (b) Explain why a biomass pyramid is pyramid-shaped. (c) State one situation in which a biomass pyramid may be inverted, and give the condition.	(a) The ratio of locust biomass to snake biomass is 1,200 kg to 12 kg, that is 100 to 1. (2 marks) (b) The standing biomass of each trophic level is about one tenth of that below it, because most energy is lost in transfer, so higher levels can support less biomass. (2 marks) (c) Where producers are tiny but multiply very fast, such as phytoplankton, their standing biomass at any moment may be less than that of the zooplankton feeding on them; the condition is that the turnover rate of the producers is much higher than that of the consumers. (2 marks)
3	A region emits 4.2 x 10^9 kg of carbon dioxide each year from burning fossil fuels, and its forests absorb 15% of this. (a) Calculate the mass of carbon dioxide absorbed by the forests each year. (b) If one hectare of forest absorbs 12,000 kg of carbon dioxide per year, calculate the forest area needed to absorb that amount, in hectares. (c) Explain two reasons why planting trees cannot fully solve the rise in atmospheric carbon dioxide.	(a) 4.2 x 10^9 x 15% = 6.3 x 10^8 kg of carbon dioxide. (2 marks) (b) 6.3 x 10^8 / 12,000 = 52,500 hectares. (2 marks) (c) The rate of absorption falls as forests age and carbon is released when they burn or are felled; suitable land is limited, so planting cannot keep pace with emissions from fossil fuels. (2 marks)
4	Nitrogen is essential for proteins and nucleic acids, but atmospheric nitrogen cannot be used directly by most organisms. (a) State two routes of nitrogen fixation. (b) Explain the roles of nitrification and denitrification in the nitrogen cycle. (c) Explain why heavy application of nitrogen fertiliser causes eutrophication of water bodies.	(a) Biological fixation by Rhizobium in root nodules and by free-living nitrogen-fixing bacteria, and industrial fixation by the Haber process; lightning fixation is also a route. (2 marks) (b) Nitrification oxidises ammonium to nitrite and then to nitrate for plant uptake; denitrification reduces nitrate back to nitrogen gas, completing the cycle. (2 marks) (c) Excess nitrate washes into water bodies and triggers algal blooms; when the algae die, decomposers use up the dissolved oxygen and fish suffocate. (2 marks)

Section B: Populations, Communities and Succession (36 marks)

Each question carries 9 marks; note the marks for each part.

#	Question	Answer and marking notes
5	Researchers estimate a pond fish population by mark-recapture. They catch 40 fish, mark them and release them; a week later they catch 60 fish, of which 8 carry marks. (a) Estimate the total fish population using the Lincoln index, showing the calculation. (b) State three assumptions of this method. (c) Explain why recapturing immediately after marking would make the estimate too low.	(a) Lincoln index N = (M x C) / R = (40 x 60) / 8 = 300 fish. (3 marks) (b) Marking does not affect movement or survival; marked and unmarked individuals mix evenly; there is no significant immigration, emigration, birth or death during the period; marks do not fall off. (3 marks) (c) Recapturing immediately after marking catches previously handled individuals more readily, so the proportion of marked fish in the second sample is too high and the estimate is too low. (3 marks)
6	A bird population on an island has a carrying capacity of 1,200 individuals and an intrinsic growth rate r of 0.4 per year; the population is currently 300. (a) Using the logistic equation dN/dt = rN(K – N)/K, calculate the population growth for that year. (b) State the population size at which the growth rate is greatest. (c) Compare logistic and exponential growth curves, and give one real example that is close to exponential growth.	(a) dN/dt = 0.4 x 300 x (1,200 – 300) / 1,200 = 90 individuals per year. (3 marks) (b) The growth rate is greatest at K/2 = 600 individuals. (3 marks) (c) The logistic curve is S-shaped, starting close to exponential growth and slowing near K, whereas exponential growth has no limit. A real example is the early growth of an invasive insect in a new environment with abundant resources and few natural enemies. (3 marks)
7	An abandoned farmland passes through secondary succession from a grass stage to a forest stage. (a) State two differences between secondary and primary succession. (b) Describe the trends in biodiversity, biomass and soil organic matter during succession. (c) Explain why species diversity is higher in later stages while some pioneer species disappear.	(a) Secondary succession starts on land where soil and a seed bank already exist, so it is faster; primary succession starts on bare ground without soil and needs pioneer species to weather rock into soil. (3 marks) (b) Biodiversity rises, biomass rises, soil organic matter and soil depth increase, and the food web becomes more complex. (3 marks) (c) Later plants provide more varied niches and resources, supporting more species; pioneer species are mostly light-demanding and tolerant of poor soil, so they are shaded out under the tree canopy. (3 marks)
8	Two paramecium species compete for the same bacteria in a test tube. Species A has a carrying capacity K1 of 200 and species B has K2 of 150; the competition coefficient alpha is 1.5 (the effect of one B individual on A). (a) Calculate the effective carrying capacity K1/alpha of species A under competition from B. (b) Hence decide whether the two species can coexist in the long term, and explain. (c) State two reasons why laboratory competition experiments cannot be applied directly to field communities.	(a) K1/alpha = 200 / 1.5 = 133.3, which is less than K2 = 150. (3 marks) (b) The two species can coexist: the carrying capacity of species B is greater than the effective carrying capacity of species A under competition, so B cannot exclude A. (3 marks) (c) Laboratory space, food and conditions are constant, while the field has seasonal change, migration and predators; only two species are used, whereas field communities involve many species and indirect effects; experiments are short and may not reflect long-term adaptation and evolution. (3 marks)

Section C: Human Impact, Global Change and Conservation (40 marks)

Each question carries 10 marks; give full explanations or evaluations.

#	Question	Answer and marking notes
9	A farm emits 1,200 kg of methane and 40 kg of nitrous oxide each year. The global warming potential of methane is 28 and that of nitrous oxide is 265 (carbon dioxide = 1). (a) Calculate the annual carbon dioxide equivalent emissions of the farm. (b) If methane emissions are halved and everything else stays the same, calculate the new carbon dioxide equivalent and the percentage reduction, to one decimal place. (c) Explain why the same mass of nitrous oxide has a much greater effect on climate than methane.	(a) 1,200 x 28 + 40 x 265 = 33,600 + 10,600 = 44,200 kg carbon dioxide equivalent. (4 marks) (b) 600 x 28 + 40 x 265 = 16,800 + 10,600 = 27,400.0 kg; the reduction is (44,200 – 27,400) / 44,200 x 100% = 38.0%. (3 marks) (c) A nitrous oxide molecule absorbs infrared radiation far more effectively than a carbon dioxide molecule and stays in the atmosphere for a long time, so each unit of mass warms the climate much more. (3 marks)
10	The forest area of an island falls from 1,000 hectares to 100 hectares. The species-area relationship can be written as S = cA^0.25. The island originally held 80 bird species. (a) Calculate the expected number of species retained after the reduction, to the nearest whole number. (b) Explain how the reduction in area affects the extinction risk of populations. (c) Suggest two management measures that reduce the impact of habitat fragmentation.	(a) S = 80 x (100 / 1,000)^0.25 = 80 x 0.5623, about 45 species. (4 marks) (b) A smaller area supports smaller, isolated populations, so genetic diversity falls and the risk of inbreeding and random extinction rises; recolonisation after disease or fire is unlikely. (3 marks) (c) Keep and connect fragmented woodland with wildlife corridors; retain strips of native vegetation at field margins to reduce pesticide and human disturbance. (3 marks)
11	A fishery targets a species with an intrinsic growth rate r of 0.5 per year and a carrying capacity K of 2,000 tonnes. (a) Calculate the maximum sustainable yield using MSY = rK/4. (b) If the catch is kept at 320 tonnes for many years, predict the change in the population and explain. (c) Explain why managers usually set catches below the MSY.	(a) MSY = rK/4 = (0.5 x 2,000) / 4 = 250 tonnes per year. (4 marks) (b) 320 tonnes exceeds the MSY, so the catch is faster than natural recruitment and the population will keep falling and may collapse. (3 marks) (c) Real populations vary with the environment and climate and are estimated with error, while the MSY calculation assumes a constant environment; setting catches lower leaves a buffer against overfishing and knock-on effects on the ecosystem. (3 marks)
12	Hong Kong protects biodiversity through country parks and marine protected areas. (a) Distinguish in-situ from ex-situ conservation and give one example of each. (b) Evaluate two strengths and one limitation of country parks in protecting biodiversity. (c) Suggest one conservation measure that uses wildlife corridors, and explain the rationale.	(a) In-situ conservation protects a species in its own habitat, for example country parks and marine protected areas; ex-situ conservation keeps the species in an artificial setting, for example botanic gardens, seed banks or zoo breeding programmes. (4 marks) (b) Strengths: large continuous habitats protect whole communities and ecological processes, and statutory controls limit development and hunting. Limitation: country parks are mostly on high ground, so lowland and coastal wetlands are under-protected and some habitats lie outside the protected network. (3 marks) (c) Build wildlife corridors between country parks and lowland reserves, for example wooded strips along river valleys; the rationale is to increase habitat connectivity so that populations can move and exchange individuals and genes, reducing the extinction risk caused by fragmentation. (3 marks)

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