Full module assessment covering all three inquiry questions: nutrient acquisition, environmental factors, and species effects on ecosystems.
Which group of organisms converts inorganic nutrients into organic compounds through photosynthesis or chemosynthesis?
The “10% rule” in ecology describes:
Which abiotic factor is most directly responsible for determining the alpine treeline in Australia?
In a mark-recapture study using the Lincoln-Petersen formula, what effect does increasing the number of marked recaptures (R) have on the population estimate (N)?
Which marine ecosystem has the highest biodiversity per unit area?
Removing dingoes from Australian grazing land leads to increased kangaroo populations, overgrazing, and soil erosion. This chain of effects is best described as:
In primary succession, facilitation refers to:
What is the single largest driver of global biodiversity loss?
Which of the following is an example of ex-situ conservation?
A student claims that decomposers recycle both matter and energy back to producers in an ecosystem. Which statement best evaluates this claim?
The following food chain occurs in an Australian grassland: grass → grasshopper → frog → snake → eagle.
(a) If the grass contains 50,000 kJ m² yr¹, calculate the energy available to the frog assuming 10% trophic efficiency at each transfer. Show your working. 2 MARKS
(b) Explain why pyramids of energy are always upright, regardless of the ecosystem. 2 MARKS
(c) Predict and explain what would happen to the grass and eagle populations if all snakes were removed from this food chain. 3 MARKS
(a) Grass (T1) = 50,000 kJ. Grasshopper (T2) = 50,000 × 0.10 = 5,000 kJ. Frog (T3) = 5,000 × 0.10 = 500 kJ [1 mark for method, 1 mark for correct answer].
(b) Pyramids of energy are always upright because energy is lost at each trophic level, primarily as heat through cellular respiration [1 mark]. The second law of thermodynamics dictates that energy transfers are never 100% efficient; some energy is always dissipated as heat that cannot be recaptured by organisms [1 mark].
(c) If snakes were removed, the frog population would increase due to reduced predation pressure [1 mark]. More frogs would consume more grasshoppers, reducing grasshopper grazing pressure on grass, so grass biomass would increase [1 mark]. Eagles, which prey on snakes, would lose their food source and their population would decline unless they could switch to alternative prey [1 mark].
Compare quadrat sampling and mark-recapture as methods for estimating population size.
(a) Describe when each method is appropriate and explain why. 2 MARKS
(b) State the Lincoln-Petersen formula and define each variable. 2 MARKS
(c) Explain one source of error specific to each method. 1 MARK
(a) Quadrat sampling is appropriate for sessile or slow-moving organisms such as plants, barnacles, and corals because they remain within the quadrat long enough to be counted [1 mark]. Mark-recapture is appropriate for mobile animals such as mammals, birds, and fish that cannot be counted in quadrats because they move too quickly [1 mark].
(b) N = (M × C) / R [1 mark]. N = total population estimate; M = number marked and released in first sample; C = total number captured in second sample; R = number of marked individuals recaptured in second sample [1 mark].
(c) Quadrat error: observer bias (researchers place quadrats where organisms look abundant) or non-random distribution (clumping means too few quadrats miss patches) [0.5 marks]. Mark-recapture error: open population (births, deaths, migration between samples) or trap shyness (marked animals learn to avoid traps) [0.5 marks].
Evaluate whether Australia should prioritise creating new national parks (in-situ conservation) or establishing more captive breeding programs (ex-situ conservation) for its threatened mammals. Your answer should include at least two advantages and two disadvantages of each approach, and make a justified recommendation that considers ecological, economic, and ethical dimensions.
6 MARKSNational parks (in-situ) — advantages: Protects entire ecosystems and hundreds of species simultaneously at relatively low cost per species. Preserves evolutionary processes, ecological relationships, and ecosystem services [1 mark].
National parks — disadvantages: Cannot protect species if specific habitat is destroyed or if threats operate inside park boundaries. Many parks are in marginal land, not the most biodiverse areas [1 mark].
Captive breeding (ex-situ) — advantages: Can prevent imminent extinction when no safe habitat exists. Maintains genetic material for future reintroduction and raises public awareness [1 mark].
Captive breeding — disadvantages: Extremely expensive per individual. Risks adaptation to captivity and loss of wild behaviours. Does not preserve ecosystem function [1 mark].
Recommendation: Australia should prioritise in-situ conservation as the foundation because it protects ecosystems and is cost-effective. However, ex-situ captive breeding should be maintained as a safety net for critically endangered species [1 mark]. The two approaches are complementary — the eastern barred bandicoot recovered because captive breeding provided insurance while predator-proof fences prepared habitat for reintroduction [1 mark].
Mark module quiz as complete