Ecosystems, Matter & Energy — Print · Life Science & Scientific Inquiry, Grade 7

1. Trace the pathway: fewer insects means less food reaching mice along that arrow, which could reduce mouse numbers and in turn reduce food reaching foxes.

2. Identify the alternatives the web shows: mice also eat seeds, and foxes also eat rabbits, so each has another route.

3. State what the web omits: arrows show that a relationship exists, not how much of each diet it supplies or how strongly each consumer depends on it.

4. Name what would be needed: counts of seeds and rabbits over the same interval, and a measure of what proportion of each consumer's diet each source provides.

One affected pathway runs grass → insects → mice → foxes. A decline in insects reduces the food arriving at mice along that arrow, which could reduce mouse numbers, and fewer mice could in turn mean less food reaching foxes. That is a plausible chain of effects and it is as far as the web supports going. The exact change in mice cannot be determined, because mice also eat seeds, and the web does not say what proportion of a mouse's diet comes from insects rather than seeds, nor whether seeds are plentiful or scarce over the same interval. If seeds are abundant, mice may be barely affected; if insects were most of their intake, the effect could be large. The same reasoning applies one step further up: foxes also eat rabbits, so even a substantial fall in mice might be offset, and rabbit numbers are not recorded either. The general limitation is that an arrow in a food web states that a feeding relationship exists and nothing about its strength — a critical dependency and an occasional meal look identical on the page. To make a quantitative prediction you would need the proportion of each consumer's diet supplied by each source, counts of the other resources over the same interval, and some measure of how quickly each population responds, since effects move through a web with delays. Without those, the honest statement names the direction of a possible effect and the alternatives that could absorb it.

1. First transfer: 300 ÷ 2,000 = 0.15, which is 15%.

2. Second transfer: 60 ÷ 300 = 0.2, which is 20%.

3. Reject the shortcut: the two transfers differ from each other and neither is ten per cent, which is an average across systems rather than a law.

4. Account for the missing energy: 1,700 units and then 240 went to respiration as heat, to uneaten parts, to undigested waste and to decomposers — transferred elsewhere, not destroyed.

The two transfers are 15% and 20%. The first is 300 ÷ 2,000 = 0.15 and the second is 60 ÷ 300 = 0.2, and the fact that they differ is the first thing worth noticing: a single efficiency does not apply across a chain, and neither value is the ten per cent often quoted. That figure is a rough average across many studied systems, with real measurements ranging from around one per cent to over twenty, so applying it as a rule here would have predicted 200 and 20 units instead of the 300 and 60 the model states. Where a measured or stated figure exists, it should be used; where none does, a calculation should say explicitly that it is using an approximation. The second error to avoid is treating the shortfall as destruction. At the first transfer, 2,000 − 300 = 1,700 units did not reach the consumers, and every one of them went somewhere. Most was used by the producers' own respiration and released as heat as they grew, transported materials and maintained themselves. Some was in parts never eaten — roots, woody stems — and some was in plant material that died and went to decomposers. At the second transfer the same routes apply, plus food that was eaten but not digested and left as waste, and consumers that died without being eaten. Energy is conserved throughout; what the pyramid shows is energy leaving the chain, not ceasing to exist. That is also why the chain is short: at these efficiencies, a fourth or fifth level has too little arriving to sustain it.

1. Sort the pathways: rain and runoff cross inward as inputs; evaporation and outflow cross outward as outputs.

2. Total the inputs: 40 L + 20 L = 60 L entering over the interval.

3. Total the outputs: 25 L + 15 L = 40 L leaving over the interval.

4. Calculate the storage change: 60 L − 40 L = +20 L, so the pond gains 20 L.

The pond gains 20 L over the interval. Rain at 40 L and runoff at 20 L are the inputs, crossing the boundary inward and totalling 60 L. Evaporation at 25 L and outflow at 15 L are the outputs, crossing outward and totalling 40 L. The change in storage is inputs minus outputs: 60 − 40 = +20 L. That increase is entirely consistent with matter being conserved, and the two ideas are often confused. Conservation says that water is not created or destroyed, not that any particular store stays the same size. Every litre in this model is accounted for by a pathway across a stated boundary: the 20 L the pond gained came from the rain and runoff that entered and did not leave. A store grows whenever inputs exceed outputs and shrinks when they do not, and in both cases the total water in the wider system is unchanged — it has simply moved between reservoirs, exactly as carbon does when it moves from rock to atmosphere. Two limits are worth stating. The model explicitly excludes other pathways, and a real pond would exchange water with groundwater through its bed, lose some to organisms and gain some from them, and might overflow differently at different levels. And the figures are for one stated interval, so they cannot be extended to a season without knowing how rainfall and evaporation vary. The method, though, transfers directly: name the boundary, list the pathways, sort them, and subtract.

1. Calculate A's loss: 20 g − 14 g = 6 g.

2. Calculate B's loss: 20 g − 18 g = 2 g.

3. Compare: A lost 4 g more than B over the same interval, which is three times as much.

4. State the boundary limitation: mass could leave a sample by washing out, physical crumbling or removal by animals, so the measurement is of mass lost rather than of decomposition directly.

Sample A lost 20 − 14 = 6 g and sample B lost 20 − 18 = 2 g, so A lost 4 g more over the same interval — three times B's loss. Because the samples started matched at 20 g of dry material and all conditions other than moisture are stated as matched, the comparison is structured to attribute that difference to the moisture treatment, which is more than most informal comparisons manage. The supported claim is that under these conditions, over this interval, the wetter treatment was associated with three times the dry-mass loss. Using dry mass is deliberate and worth noting: a sample weighed wet could appear to gain mass simply by absorbing water, or appear to decompose by drying out, so drying to constant mass at both ends is what makes the figures comparable at all. The boundary limitation is that mass loss and decomposition are not the same quantity. Material can leave a sample by being washed out in solution, by crumbling physically, or by being carried off by an animal, and none of those is biological breakdown — yet all of them would show up as mass lost. A stronger study would state which routes out were possible and which were prevented, for instance by enclosing the samples in mesh fine enough to exclude larger animals while admitting water and microorganisms, and would ideally measure carbon dioxide released as a direct signal of decomposer respiration. Two samples run once is also a small study, so repetition would be needed before treating the threefold difference as reliable.

1. Total the additions: 12 births + 5 immigrants = 17 individuals added.

2. Total the removals: 8 deaths + 4 emigrants = 12 individuals removed.

3. Calculate the ending count: 50 + 17 − 12 = 55, a net increase of five.

4. Explain the limitation: the same net increase of five is consistent with many combinations of the four pathways, so the net cannot identify which were active.

The population ends at 55. Additions are 12 births and 5 immigrants, totalling 17; removals are 8 deaths and 4 emigrants, totalling 12; so the ending count is 50 + 17 − 12 = 55, a net increase of five over the interval. The reason a before-and-after count could not have produced this account is that it would have given only the net figure. A rise from 50 to 55 is equally consistent with 5 births and no other activity, with 40 births and 35 deaths, or with no births at all and heavy immigration from neighbouring land — and those describe completely different situations. A population with high births and high deaths is under very different pressure from one that is simply drifting upward, and only the first would be worth investigating for disease or resource shortage. Distinguishing the pathways needs evidence aimed at each one. Marking individuals separates movement from death, since a marked animal recorded elsewhere emigrated rather than died. Counting nests, eggs or young gives a birth rate directly. Surveying the surrounding area finds emigrants and identifies where immigrants came from. Each pathway needs its own measurement, and none can be inferred from the net change. Two further limits apply to this model: it stipulates that no other pathways exist, where a real study would have to argue for that, and it gives no interval length, without which none of these figures can be turned into a rate.