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Checkpoint 2 — IQ2: How does the environment affect distribution and abundance?

Covering Lessons 06–12: abiotic factors, population growth, competition, symbiotic relationships, ecological sampling, ecosystem comparison, and multi-factor prediction.

~30 min 10 MC · 3 Short Answer Lessons 06–12

What is Covered

L06
Abiotic Factors
  • Temperature, water, salinity, pH, soil, light
  • Shelford’s Law of Tolerance
  • Liebig’s Law of the Minimum
L07
Population Growth
  • Exponential vs logistic growth
  • Carrying capacity (K)
  • Density-dependent and density-independent factors
L08
Interspecific and Intraspecific Competition
  • Competitive exclusion principle
  • Resource partitioning
  • Niche differentiation
L09
Symbiotic Relationships
  • Mutualism, commensalism, parasitism
  • Mycorrhizae, coral-zooxanthellae
  • Predation vs parasitism
L10
Ecological Sampling
  • Quadrat sampling and density calculation
  • Line and belt transects
  • Mark-recapture (Lincoln-Petersen)
L11
Comparing Ecosystems
  • Terrestrial vs aquatic ecosystems
  • Thermocline and upwelling
  • Diversity-productivity paradox
L12
Multi-Factor Synthesis
  • Multi-factor prediction framework
  • Case studies (mangroves, alpine, coral, starlings)
  • Common misconceptions

Section A — Multiple Choice (10 questions)

Question 1

Which abiotic factor is the primary determinant of the alpine treeline for snow gums in the Australian Alps?

A Soil pH, because acidic soils prevent root development at high altitude
B Mean growing season temperature dropping below the threshold for cambial growth
C Wind speed, because strong winds physically uproot all seedlings above 1,500 m
D UV radiation, because high UV damages DNA in tree cells
Question 2

A population of European rabbits is introduced to an island with abundant food and no predators. For the first three years, the population doubles every six months. Which growth curve does this represent?

A Exponential (J-curve) growth, because resources are unlimited and there are no density-dependent constraints
B Logistic (S-curve) growth, because all populations eventually reach carrying capacity
C Linear growth, because rabbit reproduction is constant over time
D Cyclical growth, because predator populations will eventually arrive on the island
Question 3

Two species of honeyeater feed on nectar from the same eucalypt trees. One species feeds in the morning, the other in the afternoon. What ecological principle does this demonstrate?

A Competitive exclusion, because they cannot coexist indefinitely
B Mutualism, because they cooperate to pollinate the trees
C Resource partitioning, because they use the same resource at different times
D Commensalism, because one species benefits and the other is unaffected
Question 4

The relationship between mycorrhizal fungi and eucalyptus tree roots is best described as:

A Mutualism, because the fungi absorb water and minerals for the tree while the tree provides sugars to the fungi
B Parasitism, because the fungi extract resources from the tree roots
C Commensalism, because the fungi benefit but the tree is unaffected
D Predation, because the fungi consume the tree roots
Question 5

A wildlife biologist needs to estimate the population of feral pigs in a national park. The pigs are nocturnal, wide-ranging, and difficult to observe directly. Which sampling method is most appropriate?

A Quadrat sampling with randomly placed 1 m×1 m squares
B Belt transect across the park from north to south
C Line transect recording all species touched by the line
D Mark-recapture using ear tags and remote camera traps
Question 6

In a mark-recapture study, researchers tag and release 30 agile wallabies. In the second capture of 50 wallabies, 10 are tagged. What is the estimated population size (N)?

A 80
B 150
C 300
D 1,500
Question 7

Which marine ecosystem has the highest biodiversity per unit area?

A The open ocean surface waters of the Southern Ocean
B The deep-sea benthic zone
C A tropical coral reef
D A temperate estuary
Question 8

Why are tropical open ocean surface waters typically nutrient-poor despite high sunlight?

A Phytoplankton consume nutrients faster than rivers can replace them
B High salinity prevents nutrient dissolution
C Nutrients are locked in the bodies of large marine mammals
D The thermocline prevents vertical mixing of nutrients from deep water
Question 9

A student states: “The carrying capacity of a grassland is fixed at 500 kangaroos because that is the maximum number the land can support.” Which statement best evaluates this claim?

A The claim is correct because carrying capacity is determined by soil type, which does not change
B The claim is incorrect because K changes with resource availability, climate, predation, and disturbance
C The claim is partially correct: K is fixed for herbivores but changes for carnivores
D The claim is incorrect because carrying capacity applies only to logistic growth curves
Question 10

Climate models predict a 2°C warming in the Australian Alps. Which additional factor could prevent the alpine treeline from shifting upward as predicted?

A More frequent bushfires killing adult snow gums before seedlings can establish higher up
B Increased rainfall improving soil moisture for tree seedlings
C Reduced wind exposure making high-altitude conditions more favourable
D Expansion of grassland habitat into former woodland areas

Section B — Short Answer (3 questions)

Question 11

A researcher places twelve 1 m×1 m quadrats randomly in a 3,000 m² grassland to estimate the population of kangaroo grass (Themeda triandra). The counts are: 5, 8, 12, 6, 9, 7, 11, 4, 10, 8, 6, 14.

(a) Calculate the mean density of kangaroo grass per square metre. Show your working. 2 MARKS

(b) Estimate the total kangaroo grass population in the grassland. 1 MARK

(c) Explain two sources of error that could make this estimate inaccurate, and describe how the researcher could reduce each error. 4 MARKS

Show Model Answer

(a) Total individuals = 5 + 8 + 12 + 6 + 9 + 7 + 11 + 4 + 10 + 8 + 6 + 14 = 100 [1 mark]. Total quadrat area = 12 × 1 = 12 m². Mean density = 100 / 12 = 8.33 plants per m² (accept 8.3 or 8) [1 mark].

(b) Population estimate = 8.33 × 3,000 = 25,000 plants (accept 24,900–25,000) [1 mark].

(c) Any two of the following, with matching solutions [2 marks each]:

  • Observer bias: The researcher might subconsciously place quadrats where grass looks abundant. Solution: Use random number coordinates to determine quadrat positions objectively [2 marks].
  • Non-random distribution / clumping: Kangaroo grass may grow in patches, so 12 quadrats might over-sample or under-sample clumps. Solution: Increase the number of quadrats or use stratified random sampling across habitat zones [2 marks].
  • Edge effects: Plants on the quadrat boundary may be counted inconsistently. Solution: Establish a clear rule (e.g., only count individuals whose centre lies inside the quadrat) [2 marks].
  • Size mismatch: A 1 m×1 m quadrat may be inappropriate if plants are very large or very small. Solution: Match quadrat size to organism size [2 marks].
Question 12

Compare line transects and belt transects as methods for investigating species distribution along an environmental gradient.

(a) Describe the data produced by each method. 2 MARKS

(b) State one situation where a line transect would be preferred and one where a belt transect would be preferred. Justify your choices. 3 MARKS

Show Model Answer

(a) A line transect produces qualitative or semi-quantitative data: presence/absence and relative abundance of species along the gradient, recorded where each species touches the line [1 mark]. A belt transect produces quantitative data: actual counts, density, or percentage cover of all individuals within a defined strip on either side of the line [1 mark].

(b) A line transect is preferred when time is limited and the research question only requires knowing which species occur where along the gradient — for example, a rapid survey of rocky shore zonation to document species presence with tide height [1 mark]. A belt transect is preferred when the research question requires comparing population density or biomass between zones — for example, measuring how barnacle density changes along an intertidal gradient, where actual counts are needed for statistical analysis [1 mark]. The belt transect is more time-consuming but produces data suitable for quantitative comparison [1 mark]. Total: 5 marks.

Question 13

Use the multi-factor framework to predict what would happen to mangrove distribution in a NSW estuary if sea level rises by 0.5 m over the next 50 years and existing seawalls prevent landward migration. In your answer, integrate at least one abiotic factor, one biotic factor, and one population dynamic concept.

6 MARKS
Show Model Answer

Abiotic factor: Mangroves require a specific salinity range (approximately 20–40 ppt) and regular tidal inundation. A 0.5 m sea level rise increases the frequency and duration of tidal inundation at current mangrove sites [1 mark]. At the seaward edge, prolonged submersion may exceed the tolerance limit of some mangrove species, causing stress and reduced growth [1 mark].

Biotic factor: Where mangroves are prevented from shifting landward by seawalls, they are squeezed into a narrower band. This increases intraspecific competition for space and light among mangrove seedlings [1 mark]. At the landward edge, salt marsh species that previously competed with mangroves will be outcompeted as salinity increases, but the seawall prevents the mangroves from occupying that space, leaving it as degraded habitat [1 mark].

Population dynamic: The effective carrying capacity for mangroves in the estuary decreases because the total suitable habitat area shrinks [1 mark]. With reduced recruitment space, the mangrove population may stabilise at a lower density or decline if adult mortality exceeds seedling establishment. This population decline would cascade to species that depend on mangroves for habitat, such as juvenile fish and crab species [1 mark]. Total: 6 marks.

Self-Assessment Score Tracker

Section A — Multiple Choice
Question 11
Question 12
Question 13
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