Biology by Bradford
Unit 8: Ecology. Practice questions with scoring guidelines.
131 original AP-style questions on Unit 8, written to the CED learning objectives: 103 multiple-choice, 28 free-response. Below is a 20-mark practice set built from them, ready to assign as a unit check or homework, or to sit yourself and score against the guidelines. Print it, project it, or build a fresh one.
Topics in this unit
- 8.1 Responses to the Environment 16 questions
- 8.2 Energy Flow Through Ecosystems 40 questions
- 8.3 Population Ecology 24 questions
- 8.4 Effect of Density of Populations 23 questions
- 8.5 Community Ecology 22 questions
- 8.6 Biodiversity 8 questions
- 8.7 Disruptions to Ecosystems 19 questions
In the bank for Unit 8
- 103 multiple-choice
- 28 free-response
Every question is original and tagged to a learning objective and science practice.
Make your own
The practice set
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Two meadows each contain 100 plants. Meadow P has 90 plants of one species and 10 of a second. Meadow Q has 25 plants of each of four species. Which statement about their diversity is correct?
- Q has both greater richness and greater evenness than P
- P has greater richness but lower evenness than Q
- The two meadows have equal diversity because both contain 100 plants
- Richness cannot be compared because the species differ
The model shows part of the food web of a rocky shore. Arrows point from a food source to the organism that eats it. Sea stars prefer to eat mussels, which are the strongest competitors for space on the rock.
Which organisms in the model are primary consumers?
- Seaweed and phytoplankton
- Limpet, mussel and barnacle
- Sea star and dog whelk
- Dog whelk, sea star and gull
Experiment 1: students investigated whether terrestrial isopods respond to moisture. A two-chamber choice apparatus was lined with damp filter paper on one side and dry filter paper of the same type on the other; both sides had the same substrate, the same dim light, and the same temperature. Sixty isopods were released at the center junction and their positions recorded after 10 minutes. Three trials were run with different groups of 60 isopods. Experiment 2: the same students then tested whether the isopods respond to light by covering one side of the apparatus with an opaque lid while leaving the other side lit, with dry filter paper on both sides.
Identify a control group missing from Experiment 2.
| Trial | Isopods on damp side | Isopods on dry side | Total |
|---|---|---|---|
| 1 | 44 | 16 | 60 |
| 2 | 41 | 19 | 60 |
| 3 | 47 | 13 | 60 |
| All trials combined | 132 | 48 | 180 |
- A trial in which one side is damp and the covered side is dry
- A trial in which both sides of the apparatus are equally lit and otherwise identical
- A trial in which the isopods are released at one end rather than at the center
- A trial run with a different species of isopod in the same apparatus
Two forest sites each contain 100 individual invertebrates belonging to four species. The table shows how the individuals are distributed at each site.
Two forest sites each contain 100 individual invertebrates distributed among the same four species, as shown in the table. Which site has the greater species diversity, and why?
| Species | Site A (individuals) | Site B (individuals) |
|---|---|---|
| Beetle | 25 | 85 |
| Ant | 25 | 5 |
| Spider | 25 | 5 |
| Fly | 25 | 5 |
- Site B, because one species is far more abundant than the others and dominates the community
- They are identical in diversity, because both contain exactly four species and 100 individuals
- Neither, because diversity depends only on the total number of individuals, which is equal
- Site A, because its individuals are distributed more evenly among the four species
Experiment 1: students investigated whether terrestrial isopods respond to moisture. A two-chamber choice apparatus was lined with damp filter paper on one side and dry filter paper of the same type on the other; both sides had the same substrate, the same dim light, and the same temperature. Sixty isopods were released at the center junction and their positions recorded after 10 minutes. Three trials were run with different groups of 60 isopods. Experiment 2: the same students then tested whether the isopods respond to light by covering one side of the apparatus with an opaque lid while leaving the other side lit, with dry filter paper on both sides.
Which statement is the null hypothesis for Experiment 1?
| Trial | Isopods on damp side | Isopods on dry side | Total |
|---|---|---|---|
| 1 | 44 | 16 | 60 |
| 2 | 41 | 19 | 60 |
| 3 | 47 | 13 | 60 |
| All trials combined | 132 | 48 | 180 |
- Isopods will gather on the damp side because moisture reduces their rate of water loss
- Isopods move toward damp filter paper more often than toward dry filter paper in every trial
- Moisture is the environmental cue that isopods use to select a microhabitat in the chamber
- Moisture has no effect on how the isopods distribute themselves between the two sides
Five warbler species feed on insects in the same spruce trees, but each species forages mainly in a different vertical zone of the tree — some near the top, others in the middle or lower branches. How does this pattern most likely allow the species to coexist in one community?
- By partitioning the resource so that competition between the species is reduced, allowing coexistence
- By eliminating all competition through the warblers preying on one another in the different zones
- Because the warblers in different zones consume different insects and so are not really competitors
- Because stronger interspecific competition always increases the number of coexisting species
Why are ecosystems with high species diversity generally more resistant to disturbance than those with low diversity?
- High-diversity ecosystems contain fewer predators, so prey populations remain stable when conditions change
- Every species in a diverse ecosystem is equally abundant, so no single species can be lost during a disturbance
- If one species declines, others with similar roles can maintain ecosystem functions such as decomposition and pollination
- Diverse ecosystems have simpler food webs with fewer links, so a disturbance to one species affects fewer others
The table shows the growth rate (dN/dt) predicted by the logistic model dN/dt = rₘₐₓN[(K−N)/K] for a population with maximum per-capita growth rate rₘₐₓ = 0.4 yr⁻¹ and carrying capacity K = 800 individuals, evaluated at four population sizes.
The table gives the growth rate (dN/dt) predicted by the logistic model dN/dt = rₘₐₓN[(K−N)/K] for a population with rₘₐₓ = 0.4 yr⁻¹ and carrying capacity K = 800. At approximately which population size is the population growing fastest?
| Population size, N | Growth rate, dN/dt (individuals·yr⁻¹) |
|---|---|
| 100 | 35 |
| 400 | 80 |
| 600 | 60 |
| 800 | 0 |
- The growth rate is the same at every population size
- N ≈ 400, near half the carrying capacity
- N ≈ 100, when the population is smallest
- N ≈ 800, at the carrying capacity
A prolonged drought sharply reduces net primary productivity in a grassland for several years. Which outcome is the most likely direct consequence of this decrease in available energy at the base of the food web?
- Decomposer activity increases enough to replace the lost producer energy.
- Predator populations increase because prey become easier to catch permanently.
- Populations of herbivores and their predators tend to decline.
- The number of trophic levels the system can support increases.
A powerful hurricane fells large areas of a coastal forest, opening gaps in the canopy. In the following years, sun-loving pioneer species colonize the gaps before shade-tolerant trees gradually return. This naturally occurring event most directly affects the ecosystem by which mechanism?
- A disturbance initiating secondary succession
- It permanently prevents any forest regrowth
- It raises energy at the highest levels first
- It immediately restores the mature forest
An unseasonal hard frost kills roughly the same large fraction of an insect population whether the population is sparse or dense. This type of limiting factor is best classified as which of the following?
- An example of exponential population growth
- Density-independent, as density does not matter
- A carrying-capacity effect from competition
- Density-dependent, as crowding causes deaths
A large forest is split by roads and fields into several small, isolated patches. Interior-dwelling songbirds decline sharply even though the total forested area lost is modest. Which mechanism best explains this decline?
- Fragmentation raises edge effects and isolates birds.
- Fragmentation has no effect if some forest remains.
- Fragmentation increases the energy available.
- Splitting the forest raises interior humidity.
Researchers monitored a population of voles (prey) and the owls (predators) that feed on them in a grassland over six years. The table shows the estimated size of each population.
| Year | Vole population | Owl population |
|---|---|---|
| 1 | 100 | 20 |
| 2 | 250 | 30 |
| 3 | 400 | 55 |
| 4 | 150 | 60 |
| 5 | 80 | 25 |
| 6 | 200 | 22 |
The exponential growth model describes population growth when resources are effectively unlimited: dN/dt = rₘₐₓN, where N is the population size and rₘₐₓ is the maximum per-capita growth rate. The table gives the growth rate (dN/dt) this model predicts for a population with rₘₐₓ = 0.2 yr⁻¹ at four population sizes.
| Population size, N | Growth rate, dN/dt (individuals·yr⁻¹) |
|---|---|
| 100 | 20 |
| 500 | 100 |
| 1000 | 200 |
| 2000 | 400 |
Original practice questions © Biology by Bradford · CC BY-NC-SA 4.0 · AP® is a trademark registered by the College Board, which was not involved in and does not endorse this site.
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Each point is worth 1 point and is credited independently. Accept any one of the listed alternatives per point; ( ) marks optional wording and / separates interchangeable wording.
- Richness counts species: four beats two. Evenness asks how equally individuals are spread across them: 25 each is as even as it gets, while 90 to 10 is heavily skewed. Q wins on both.
- B — Primary consumers feed directly on producers: the limpet grazes seaweed, and the mussel and barnacle filter phytoplankton.
- B — with both sides identical, any uneven distribution reveals a bias built into the apparatus or the release method, so without it a side preference could be mistaken for a response to light.
- D — the two sites have equal richness (4 species), so the more even distribution of individuals in Site A gives it the higher diversity.
- D — the null hypothesis states that the treatment has no effect, which here means the isopods should be distributed 50:50 regardless of moisture; the other options are predictions or mechanistic claims.
- A — dividing the foraging space is resource partitioning, which lowers interspecific competition and lets the species share the community.
- C — Functional redundancy buffers ecosystem processes.
- B — logistic growth is fastest near N = K/2; here dN/dt peaks at 80 individuals·yr⁻¹ when N = 400.
- C — reduced primary productivity means less energy passes to consumers, so herbivore and predator populations tend to shrink.
- A — a meteorological disturbance that damages an existing community but leaves soil intact sets off secondary succession as pioneers recolonize.
- B — a frost's proportional effect is unrelated to how crowded the population is, making it a density-independent limiting factor.
- A — dividing habitat raises the proportion of edge, shrinks interior habitat, and isolates small populations, all of which lower the abundance of interior specialists.
- The owl population tracks the vole population but lags behind it — owls peak (Year 4) shortly after the voles peak (Year 3), and owls decline after the voles crash
- As voles rise, owls rise afterward; when voles fall, owls fall after a delay — a lagged predator–prey oscillation
Total for part (A): 1 point
- 300% increase ((400 − 100) / 100 × 100)
- +300%
- 300 percent
Total for part (B): 1 point
- The owl population declines
- Owl numbers fall following the drop in prey
Total for part (C): 1 point
- With fewer voles, less food/energy is available to the owls, so owl births fall and deaths rise and the predator population declines after its prey — a density-dependent response
- Predator numbers depend on prey abundance; when prey (voles) collapse, the reduced food supply lowers owl survival and reproduction, so owls decline with a lag
Total for part (D): 1 point
Total for question 13: 4 points
- dN/dt increases in direct proportion to N — doubling N doubles the growth rate
- The growth rate rises linearly with population size (each individual adds the same amount, so larger N gives faster growth)
Total for part (A): 1 point
- 300 individuals per year (0.2 × 1,500)
- dN/dt = 300 yr⁻¹
Total for part (B): 1 point
- The growth rate would rise, then peak near N = K/2, and fall back toward 0 as N approached K, rather than climbing without limit
- dN/dt would stop increasing and eventually decline to zero at carrying capacity
Total for part (C): 1 point
- As N approaches K, (K−N)/K shrinks toward 0, which reduces dN/dt and brings growth to zero at carrying capacity, unlike the exponential model that lacks this term
- The factor (K−N)/K acts as a brake that gets stronger as N nears K, so growth slows and stops rather than accelerating with N
Total for part (D): 1 point
Total for question 14: 4 points
Other AP Biology units
- Unit 1 Chemistry of Life 96
- Unit 2 Cell Structure and Function 105
- Unit 3 Cellular Energetics 112
- Unit 4 Cell Communication and Cell Cycle 104
- Unit 5 Heredity 130
- Unit 6 Gene Expression and Regulation 116
- Unit 7 Natural Selection 149
All eight units → · IB Biology subtopics → · Open the exam maker →