Ecological Evidence & Environmental Design — Print · Life Science & Scientific Inquiry, Grade 7

1. Describe the changes numerically: cover falls by 30 percentage points, from 80% to 50%, while the herbivore count doubles from 10 to 20.

2. Give the first plausible contribution: more herbivores grazing more heavily could reduce plant cover directly.

3. Give the second: drought could limit plant growth and slow recovery, reducing cover with no change in grazing pressure.

4. Propose evidence that separates them: fenced plots excluding herbivores under the same drought, comparable sites with drought but no herbivore increase, soil moisture records, and feeding observations.

The records show plant cover falling by 30 percentage points, from 80% to 50%, while the herbivore count doubles from 10 to 20, with a drought over the same interval. Two contributions are plausible and the record supports neither over the other. More herbivores could be grazing more plant material, reducing cover directly. Drought could be limiting plant growth and slowing regrowth, reducing cover with no change in grazing at all. It is also likely that both acted together and compounded: a drought-stressed plant recovers more slowly from being grazed, so the same herbivores do more damage in a dry year, which means treating the two as rivals where only one can win may be the wrong framing from the start. Separating them requires evidence aimed at one factor at a time. Fenced plots that exclude herbivores while experiencing the same drought are the most direct test: if cover inside stays high, grazing is doing the work, and if it falls similarly, the drought is. Comparable sites that had the drought without a herbivore increase test it from the other side. Soil moisture records establish how severe the drought actually was rather than assuming it, and feeding observations establish what the herbivores ate and in what quantity. Two further limits are worth stating: the record gives no interval length, so none of these changes can be expressed as a rate, and a single site observed once cannot distinguish an unusual year from a trend.

1. Compare richness: three species detected in each sample, so richness is identical.

2. Compare total abundance: ten individuals in each, so abundance is identical too.

3. Compare the distribution: A has one species holding 8 of 10 individuals, while B spreads them 4, 3 and 3 — B is far more even.

4. State what this cannot establish: evenness at one place and time does not show which community would withstand a particular disturbance.

The two samples are identical on the first two measures and differ on the third. Richness is three species in each, and total abundance is ten individuals in each, so neither of those numbers distinguishes them at all. The difference is in distribution: sample A is dominated by one species holding 8 of the 10 individuals, with two species represented by a single individual each, while sample B spreads its individuals 4, 3 and 3 and is close to even. That difference matters for what a loss would do. In A, losing the dominant species removes 80% of the individuals and whatever function they perform, while losing either singleton removes almost nothing measurable — and a species represented by one individual in a sample may be close to absent in the community anyway. In B, no single loss removes more than 40%, and each species is common enough that the sample is reasonable evidence it is established. What these figures cannot establish is which community is more stable. Resilience depends on functional overlap — whether the remaining species could cover the roles of one that was lost — and on what disturbance is being resisted, neither of which a count at one moment reveals. A community can absorb a drought and collapse under a disease. The samples are also single equal-effort snapshots, so they say nothing about seasonal change, nothing about the genetic diversity within each species, and nothing about species present but not detected.

1. Calculate the difference: 18 L − 10 L = 8 L less runoff recorded from the treated area under the same rain.

2. Give the mechanism: vegetation and the soil structure beneath it increase infiltration and intercept rain, so less water travels across the surface.

3. Note the design strength: matched conditions and shared rainfall mean the comparison isolates the treatment better than a before-and-after measurement at one site would.

4. Qualify the water-quality claim: runoff volume was measured and water quality was not, and one trial cannot establish behaviour in heavier rain or another season.

Area B records 8 L less runoff than Area A, since 18 − 10 = 8 L under the same rainfall. A plausible pathway is straightforward: vegetation intercepts some rain on leaves, roots and organic matter improve the soil's pore space, and both increase infiltration, so a greater share of the water soaks in rather than travelling across the surface. The design is reasonably strong for a single trial, because the two areas are stated to be comparable at the start and received the same rain, which means the weather cannot explain the difference — a before-and-after measurement at one site could not have ruled that out. What the result will not support is a water-quality claim. Runoff volume is what was measured, and water quality is a different outcome requiring its own measurements: which substances are present, at what concentration, and compared with what standard or control. Less runoff plausibly carries less surface material into a watercourse, and plausibly is not evidence. The result is also one trial. It cannot say how the two areas would compare in a heavier storm, when vegetation and soil may saturate, or after a dry spell, or in winter when plants are dormant, and it cannot confirm that the areas were genuinely equivalent in slope, soil and compaction beyond the statement that they were. Repeating across several rainfall events and seasons, with more than one pair of areas, is what would turn this into a finding — and any proposal built on it should state its trade-off, since planting occupies land and requires maintenance.

1. Total the counts: 2 + 4 + 3 + 3 = 12 plants across 4 m² of quadrats.

2. Calculate density: 12 ÷ 4 = 3 plants per square metre.

3. Scale to the habitat: 3 × 20 = 60 plants as the estimated total.

4. Name the assumptions: that the quadrats represent the whole area, that every individual present in them was detected, that the species was identified correctly, and that the habitat is 20 m² as stated.

The sample density is 3 plants per square metre and the extrapolated total is 60 plants. The four quadrats contain 2 + 4 + 3 + 3 = 12 plants across 4 m², giving 12 ÷ 4 = 3 per square metre, and multiplying by the 20 m² habitat gives 3 × 20 = 60. The word estimate matters and should appear in any report of it, because four assumptions stand between the calculation and a count. First, that the four quadrats represent the whole area — which depends on how they were placed, and the figure is only defensible if positions were fixed by a procedure before the area was examined. Second, that every individual inside each quadrat was detected; small, hidden or partly buried plants are easily missed, and a method that misses them consistently biases the estimate low in a way that more quadrats will not correct. Third, that the species was identified correctly throughout, which matters where similar species grow together. Fourth, that the habitat really is 20 m² and is reasonably uniform. The scatter is reassuring here: the counts run from 2 to 4 around a mean of 3, which suggests a fairly even distribution and makes scaling reasonable. Counts of 0, 0, 1 and 11 would have given the same mean and a very different situation, where the total would depend on how many clumps the area happened to hold. Reported properly: four 1 m² quadrats at predetermined random positions, counts 2, 4, 3, 3, density 3 per m², estimated total 60 individuals in 20 m², assuming representative placement and complete detection.

1. Calculate the reductions: A gives 20 − 8 = 12 L; B gives 20 − 10 = 10 L.

2. Calculate reduction per token: A gives 12 ÷ 80 = 0.15 L per token; B gives 10 ÷ 60 = 0.167 L per token.

3. State the criterion before recommending, since the two measures disagree — maximum total reduction favours A, best value per token favours B.

4. Propose a next test aimed at the largest unknown, which the model says is maintenance.

Plan A reduces runoff by 20 − 8 = 12 L and plan B by 20 − 10 = 10 L, so A achieves 2 L more in total. On cost effectiveness the ranking reverses: A delivers 12 ÷ 80 = 0.15 litres of reduction per token and B delivers 10 ÷ 60 = 0.167 litres per token, so B buys slightly more for each token spent. Both fit the 100-token budget and both meet the stated space requirement, so neither constraint decides it, and the two performance measures point different ways — which is exactly why the criterion has to be stated rather than discovered. Under a criterion of maximum runoff reduction within the budget, plan A is the recommendation. Under a criterion of best value per token, or where the remaining 40 tokens have another worthwhile use such as maintenance or a second site, plan B is. I would recommend B under a stated criterion of value per token plus reserve for upkeep, precisely because the model says maintenance needs are unknown for both, and a plan that consumes 80 of 100 tokens leaves almost nothing to repair or adjust it. Each option's downside should be named: A leaves little budget for anything else, and B accepts 2 L less reduction. The next test should target that largest unknown rather than the easiest measurement. Build plan B on one section, leave a comparable section untreated, measure runoff from both over several rainfall events across at least one season — long enough for maintenance needs to appear — and extend, revise or switch to A depending on whether the reduction holds up once the installation is no longer new.