Biology by Bradford
Unit 8: Ecology. Practice questions with scoring guidelines.
215 original AP-style questions on Unit 8, written to the CED learning objectives: 150 multiple-choice, 65 free-response. Below is a 22-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 31 questions
- 8.2 Energy Flow Through Ecosystems 56 questions
- 8.3 Population Ecology 33 questions
- 8.4 Effect of Density on Populations 30 questions
- 8.5 Community Ecology 42 questions
- 8.6 Biodiversity 29 questions
- 8.7 Disruptions in Ecosystems 34 questions
In the bank for Unit 8
- 150 multiple-choice
- 65 free-response
Every question is original and tagged to a learning objective and science practice.
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The practice set
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Two forest sites each contain 100 individual invertebrates belonging to four species. The table shows how the individuals are distributed at each site.
Suppose a disease sharply reduces the beetle population at both sites. Based on the relationship between diversity and resilience, which site's community is more likely to maintain its overall function, and why?
| Species | Site A (individuals) | Site B (individuals) |
|---|---|---|
| Beetle | 25 | 85 |
| Ant | 25 | 5 |
| Spider | 25 | 5 |
| Fly | 25 | 5 |
- Site B, as low diversity increases stability
- Both equally, as diversity has no effect
- Site B, as its dominant species regrows fast
- Site A, as greater diversity brings resilience
In 2025, ecologists repeated a survey of the butterflies on a mountain that was first made in 1925, using the same methods. The summit of the mountain is at 2,900 m, and the mean summer temperature on the mountain rose by about 1.5 °C between the two surveys. The table shows the lowest and highest elevations at which four species were found in each survey.
If summer temperatures on the mountain keep rising, which species is most likely to be lost from the mountain first, and why?
| Species | Range in 1925 (m) | Range in 2025 (m) |
|---|---|---|
| P | 600 to 1,500 | 750 to 1,700 |
| Q | 1,100 to 2,200 | 1,350 to 2,450 |
| R | 1,800 to 2,600 | 2,100 to 2,900 |
| S | 2,300 to 2,900 | 2,650 to 2,900 |
- Species S, because its lower limit keeps rising while its upper limit cannot rise above the summit
- Species R, because its range moved up by 300 m, farther than the ranges of species P and Q moved
- Species Q, because its range moved up by 250 m at both ends, so it is shifting the fastest
- Species P, because it lives lowest on the mountain, where the summer temperatures are highest
After a glacier retreats, bare rock is first colonized by lichens and mosses; over decades these are replaced by grasses, then shrubs, and eventually a forest supporting many animal species. Which statement best describes what is happening to the community over this time?
- Only the number of individuals changes, while the set of species present stays the same
- Community composition and diversity change as new species replace earlier ones
- Species diversity necessarily decreases as succession proceeds
- Community structure is fixed from the moment the glacier retreats and does not change
At deep-sea hydrothermal vents where no sunlight penetrates, bacteria oxidize hydrogen sulfide (H₂S) to fix carbon into organic molecules, and tube worms and other animals depend on these bacteria. Which statement best describes how energy enters this community?
- The animals photosynthesize using faint light emitted by the vents
- Chemoautotrophs convert chemical energy into biomass that heterotrophs consume
- Energy is created by the tube worms and passed downward to the bacteria
- The community needs no energy input because it is completely isolated
Ecologists compared aboveground plant biomass across four grazing intensities on the same grassland. Sixteen 2-hectare plots on the same soil type were used, with four plots randomly assigned to each of no grazing, light grazing, moderate grazing, and heavy grazing by cattle; stocking rates were set at the start of the season and held constant. All sixteen plots received the same rainfall and were harvested on the same date. Eight 1 m² quadrats were clipped in each plot and the dried biomass weighed, and each plot mean contributed one value to the mean for its grazing intensity. Means are reported with standard error, and the error bars in a plot of these data represent plus or minus 2 SE of the mean.
A rancher concludes that increasing grazing intensity always increases aboveground biomass. Which statement best evaluates this conclusion?
| Grazing intensity | Mean aboveground biomass (Mean ± SE, g / m²) |
|---|---|
| None | 420 ± 12 |
| Light | 455 ± 14 |
| Moderate | 470 ± 10 |
| Heavy | 305 ± 11 |
- It is supported, because the moderately grazed plots have the highest mean biomass of the four treatments
- It is supported, because all three grazing treatments exceed the ungrazed plots in mean aboveground biomass
- It is not supported, because the heavily grazed plots are significantly lower than the ungrazed plots
- It cannot be evaluated, because only one plot was assigned to each grazing intensity in the study
Researchers grew young trees of one deciduous species in growth chambers held at a constant 20 °C. Trees given a day length that shortened each week, as in autumn, dropped their leaves after about 8 weeks. Trees kept on a constant 16-hour day kept their leaves for the whole 8 weeks. Which conclusion about this physiological response is best supported?
- Shortening day length alone can trigger leaf drop in this species
- Leaf drop occurs only when day length and temperature both fall
- Leaf drop happens at a fixed age, whatever the day length
- Falling temperature is the cue that triggers leaf drop in this species
A small lizard lives on an offshore island. No lizards reach the island from elsewhere and none leave it. Feral cats, which eat the lizards, lived on the island until conservation workers removed every cat at the end of Year 3. Each year ecologists estimated the number of lizards at the start of the year, the per capita birth rate (young hatched per lizard per year) and the per capita death rate (deaths per lizard per year).
| Year | Lizards at start of year (N) | Per capita birth rate, b (per year) | Per capita death rate, d (per year) |
|---|---|---|---|
| 1 | 800 | 0.40 | 0.45 |
| 2 | 760 | 0.39 | 0.44 |
| 3 | 722 | 0.40 | 0.45 |
| 4 | 686 | 0.40 | 0.25 |
| 5 | 789 | 0.41 | 0.26 |
| 6 | 907 | 0.39 | 0.24 |
At the edge of a woodland, ecologists chose 24 young saplings of one deciduous tree species of similar height. In spring they covered 12 saplings, chosen at random, with netting that kept out insect-eating birds but let insects pass through freely and did not measurably change the light reaching the leaves. The other 12 saplings were left open to birds. At the end of the summer they made three measurements on each sapling. Each value in the table is the mean of 12 saplings ± 2SE.
| Measurement | Open to birds (mean ± 2SE) | Birds excluded (mean ± 2SE) |
|---|---|---|
| Caterpillars per 100 leaves | 1.8 ± 0.6 | 4.6 ± 1.1 |
| Leaf area eaten by insects (percent) | 6 ± 2 | 15 ± 4 |
| Height growth over the summer (cm) | 31 ± 5 | 27 ± 6 |
Female moths of many species attract mates by releasing a sex pheromone, a chemical signal that drifts away from the female in the air. Males of the same species detect the pheromone, even at very low concentrations, with receptors on their antennae. In some vineyards, growers control a moth whose caterpillars feed inside grape berries by hanging hundreds of small dispensers on the vines. Each dispenser slowly releases a synthetic copy of the female moth's pheromone, so that the air throughout the vineyard carries it.
A simple model of population growth states that dN/dt = (b − d)N, where N is the population size, b is the per-capita birth rate, and d is the per-capita death rate. The difference (b − d) is the per-capita growth rate, r. The table applies the model to snail populations in four ponds, each holding N = 1,000 snails when measured. The ponds differ in the amount of algae, the snails’ food, available per snail.
| Pond | Algae available per snail (relative units) | Per-capita birth rate, b (yr⁻¹) | Per-capita death rate, d (yr⁻¹) | Growth rate, dN/dt (snails·yr⁻¹) |
|---|---|---|---|---|
| S | 4.0 | 0.60 | 0.35 | 250 |
| P | 3.0 | 0.50 | 0.30 | 200 |
| Q | 2.0 | 0.40 | 0.40 | 0 |
| R | 1.0 | 0.30 | 0.45 | −150 |
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.
- D: at Site A the beetle is one of four equally common species, so the ant, spider and fly (75 of 100 individuals) can compensate; at Site B the beetle is 85 of 100 individuals, so losing it removes most of the community. The greater diversity (here, evenness) of Site A makes it more resilient.
- A: all four ranges have moved upslope as the mountain warmed, but species S already reached the summit in 1925, so only its lower limit could move; its range has shrunk from 600 m to 250 m wide and will shrink further. A is wrong because species R, though it now reaches the summit, still has a range 800 m wide; B is wrong because the range of species P has simply moved up and widened slightly; C is wrong because species Q kept its full 1,100 m range, and the lower limits of R and S rose farther.
- B: succession changes both which species are present and the diversity of the community as pioneers give way to later-arriving species.
- B: chemoautotrophs convert chemical energy in H₂S into biomass, and heterotrophs obtain that energy by consuming them, enabling energy flow with no light.
- C: heavy grazing gives 305 ± 22, or 283 to 327 g / m², well below the ungrazed 396 to 444 g / m² with no overlap, so biomass peaks at moderate grazing and then falls rather than rising throughout.
- A: temperature was the same for both groups and only the trees on shortening days dropped their leaves, so photoperiod, an external cue, can trigger leaf drop on its own; A and D are ruled out by the constant temperature and C by the trees that kept their leaves.
- The per capita birth rate stays about the same (about 0.40 per year) in both periods, but the per capita death rate falls from about 0.45 to about 0.25 per year
- Births per lizard do not change (0.39 to 0.41 throughout), whereas deaths per lizard drop by about 0.2 per year after Year 3
Total for part (A): 1 point
- Year 3: 0.40 − 0.45 = −0.05 per year; Year 4: 0.40 − 0.25 = 0.15 per year
- r = −0.05 per year in Year 3 and +0.15 per year in Year 4
Total for part (B): 1 point
- Not supported: more young hatched in Year 6 only because there were more lizards (907 against 722); the per capita birth rate was about the same (0.39 against 0.40), so each lizard did not produce more young, and the population grew because the per capita death rate fell from about 0.45 to about 0.25
- The claim confuses the total number of young with the young per lizard: b stayed at about 0.40 per year before and after the cats were removed, so the larger total in Year 6 comes from the larger population, not from more young per lizard
- Not supported: 354 is about 0.39 × 907 and 289 is about 0.40 × 722, so the extra young are explained by the extra lizards, not by a higher birth rate per lizard; the lizards became more numerous because fewer died once the cats were gone
Total for part (C): 1 point
- Measure the death rate over the same years on a similar nearby island that had the same winters but where the cats were not removed; if the death rate fell after Year 3 on the island from which the cats were removed but not on the island that kept its cats, the removal is supported
- Compare with a control lizard population that still has cats and had the same weather: no fall in d there, but a fall in d on the island where the cats were removed, would support the removal as the cause
- Compare death rates in mild and harsh winters on islands with cats and islands without cats; a lower death rate without cats in winters of the same kind would support the removal as the cause
Total for part (D): 1 point
Total for question 7: 4 points
- Both increased when birds were excluded: caterpillars rose from 1.8 to 4.6 per 100 leaves and leaf area eaten from 6 to 15 percent
- Excluding birds roughly doubled to tripled (about 2.5 times) both the number of caterpillars and the percentage of leaf area eaten
- Saplings without birds had more caterpillars and lost more leaf area to insects than saplings open to birds
Total for part (A): 1 point
- No significant difference: the open interval (26 to 36 cm) and the birds-excluded interval (21 to 33 cm) overlap, so the null hypothesis is not rejected
- Not significantly different, even though open saplings had the larger mean (31 against 27 cm), because the ±2SE intervals overlap between 26 and 33 cm
- The intervals overlap, so the 4 cm difference in mean height growth could be due to chance (sampling variation) and the data do not support rejecting the null hypothesis
Total for part (B): 1 point
- The birds prey on the caterpillars, which are herbivores (primary consumers); with birds present there are fewer caterpillars, so less leaf tissue of the producer is eaten
- Predation by birds (secondary consumers) keeps the herbivore population low, which indirectly protects the producers from herbivory (a trophic cascade / top-down control)
- Birds eat the caterpillars that eat the leaves, so the effect of the birds on the saplings is indirect, passing through the caterpillar population
Total for part (C): 1 point
- The netted saplings would fall further behind in growth, and the difference could become statistically significant, because losing more leaf area each year leaves less leaf area to capture light energy, so less energy is stored as new biomass for growth
- The difference would increase, because heavier herbivory reduces the net primary productivity of each netted sapling year after year, so these saplings have less energy and organic matter available to grow
- Netted saplings would become shorter than open saplings by a growing margin, because the extra leaf tissue eaten each summer is energy and biomass the sapling loses instead of using for growth
Total for part (D): 1 point
Total for question 8: 4 points
- Males detect the pheromone and move toward higher concentrations (upwind) to find its source, so a female that releases it is found by males and mates, and her eggs are fertilized
- The signal brings males to the female from a distance, so she mates and produces offspring; a female that did not signal might not be found by a male
Total for part (A): 1 point
- With pheromone everywhere, males cannot follow a concentration gradient to a real female (or follow the dispensers instead), so fewer females are found and mated; unmated females lay no fertilized eggs, so fewer caterpillars hatch
- The synthetic pheromone masks the signal of the real females and disrupts communication between the sexes, so fewer matings occur and fewer offspring are produced
Total for part (B): 1 point
- Vineyard 1, where the population is sparse
- The treatment will be more effective where the moths are at low density
Total for part (C): 1 point
- In a sparse population, males and females are far apart, so a male can find a female only by following her pheromone over a long distance, which the dispensers prevent; in a dense population, males and females are often close together and meet by chance or find each other at short range (by sight, or by the strong pheromone right next to a female), so many still mate
- The chance of a male meeting a female without following the long-distance chemical signal increases with population density, so disrupting the signal reduces mating less when the moths are dense
- In a dense population, many real females release pheromone, so real sources make up a larger share of all the sources in the air; a male searching among the dispensers is more likely to come upon a real female's plume, so a larger percentage of females still mate
Total for part (D): 1 point
Total for question 9: 4 points
- As food per snail decreases, b − d decreases: 0.25 in Pond S, 0.20 in P, 0 in Q and −0.15 in R
- More food per snail gives a higher per-capita growth rate; with the least food (Pond R) deaths exceed births and r is negative
Total for part (A): 1 point
- About 850 snails (1,000 + (−150))
- About 860 snails (1,000 × e^(−0.15), treating growth as continuous)
Total for part (B): 1 point
- The population will decline
- It will shrink, because dN/dt becomes negative
Total for part (C): 1 point
- r = b − d = 0.50 − 0.55 = −0.05 per year, so dN/dt = −0.05N is negative and the population decreases (by about 50 snails per year at N = 1,000)
- Deaths per individual now exceed births per individual, so (b − d) is negative and the model gives a negative dN/dt
Total for part (D): 1 point
Total for question 10: 4 points
Other AP Biology units
- Unit 1 Chemistry of Life 196
- Unit 2 Cell Structure and Function 260
- Unit 3 Cellular Energetics 195
- Unit 4 Cell Communication and Cell Cycle 187
- Unit 5 Heredity 203
- Unit 6 Gene Expression and Regulation 219
- Unit 7 Natural Selection 316
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