Energy flow in ecosystems

Evidence: the food web shows arrows from grass to the grasshopper, the mouse, and the rabbit. The passage explains that "Each arrow points from an organism to the organism that eats it."

Step by step:

1. Grass makes its own food by photosynthesis, so it is the producer.
2. A primary consumer eats the producer. The mouse eats grass (arrow from grass → mouse).
3. The frog eats grasshoppers (a primary consumer), so the frog is a secondary consumer.
4. The hawk eats mice, rabbits, songbirds, and snakes, so it is a higher-level consumer, not primary.

The answer is D. Mouse.

Tip: follow the arrows. An arrow means "is eaten by," so count how many steps an organism is from the producer.

Evidence: in the food web, arrows go from the frog → snake and the mouse → snake, and from the snake → hawk.

Step by step:

1. Snakes eat frogs and mice. With fewer snakes, fewer frogs are eaten.
2. So the frog population is likely to increase. (Mice would likely increase too, but that is not offered as a choice.)
3. Check the others: B is backwards (mice lose a predator, so they increase, not decrease). C is also backwards: more frogs eat more grasshoppers, so grasshoppers would likely decrease. D is wrong: hawks lose one food source (snakes), so hawks would not be expected to increase.

The answer is A.

Tip: for "what if a species disappears" questions, look one arrow away in both directions: its prey usually increase; its predators lose food.

Evidence: in the web, arrows run from algae and pondweed to snail, and from snail to frog. Pondweed has only one arrow leaving it, to the snail.

Step by step:

1. The snail is the only organism shown that eats pondweed.
2. With most snails gone, almost nothing is eating the pondweed, so the pondweed is likely to increase.
3. Check the others: frogs lose a food source, so they are more likely to decrease (C is backward). Zooplankton would have more algae, not less (B is backward). Herons do not eat snails, and fewer frogs would mean less food for herons, not more (D).

The answer is A.

Tip: when one species disappears, look first at what it ate (that food may increase) and at what ate it (those predators may decrease).

Evidence: the table shows 2,500 kJ/m² in primary consumers (including mice) and only about 25 kJ/m² in tertiary consumers (hawks). The passage says that at each step "the rest is used for life processes and lost as heat."

Step by step:

1. Each step up the food chain keeps only about 10% of the energy.
2. From mice (primary) to hawks (tertiary) is two steps, so hawks get about 2{,}500 \times 0.1 \times 0.1 = 25 kJ/m².
3. With 1% as much energy, the grassland can support far fewer hawks than mice.
4. Eliminate B (large predators need more energy, not less), C (mice are consumers; grasses and wildflowers are the producers), and D (energy is not recycled; it is lost as heat).

The answer is A.

Tip: fewer organisms at the top of a food chain is a result of energy loss at every step, not of predators being stronger or bigger.

Evidence: the passage says the rest of the energy "is used by organisms for life processes such as cellular respiration, movement, and keeping warm, and much of it is lost as heat. Some energy also stays in parts that are not eaten or digested."

Step by step:

1. Each organism uses most of the energy it takes in to stay alive, and that energy leaves the food chain as heat.
2. Only the energy stored in its body, about 10%, can be passed on when it is eaten.
3. Eliminate A: energy is never destroyed (law of conservation of energy); it changes form, mostly to heat. Eliminate C: body size has nothing to do with it; seals are much larger than phytoplankton. Eliminate D: consumers get their energy from food, not directly from sunlight; only producers capture sunlight.

The answer is B.

Tip: energy flows one way through an ecosystem and is lost as heat at every step. It is not recycled, and it is never destroyed.

Evidence: rabbits peak at 1,300 in year 3 and foxes peak at 80 in year 4. Rabbits rise again to 1,250 in year 9, and foxes rise to 75 in year 10. The passage warns that "rabbits are read on the left axis, and foxes are read on the right axis."

Step by step:

1. Rabbit peak (year 3) → fox peak one year later (year 4). Rabbit peak (year 9) → fox numbers still climbing in year 10.
2. Why the lag? More rabbits mean more food for foxes, so more fox pups survive, but that takes time.
3. Eliminate B: the peaks happen in different years. Eliminate C: the largest fox count is 80; the largest rabbit count is 1,300; reading both lines on one axis causes this error. Eliminate A: foxes fall from 80 to 25 between years 4 and 7.

The answer is D.

Tip: with two vertical axes, match each line to its axis using the legend and axis colors before comparing values.

Evidence: rabbits fell from 1,300 (year 3) to 400 (year 6). Foxes were near their peak, 65 (year 3) and 80 (year 4), as the decline began. The passage says "foxes are the main predator of rabbits."

Step by step:

1. A large predator population eats more prey.
2. Foxes had grown to their peak just as rabbits began to fall, so predation was greatest then.
3. Eliminate B: foxes were increasing from year 3 to 4, not decreasing. Eliminate A: rabbits eat grass; they are consumers and do not need sunlight directly. Eliminate D: foxes were present the whole time.

The answer is C. (A shortage of grass from so many rabbits could also contribute, but of the choices, only C fits the data.)

Tip: in predator-prey graphs, prey usually fall when predators are near their peak, and predators fall after prey have become scarce.

Evidence: the passage says "The island's grasses are the rabbits' only food."

Step by step:

1. Carrying capacity is set by limiting factors: food, water, space, shelter, predators, and disease.
2. Since grass is the rabbits' only food, more grass means the island can feed more rabbits over time, so carrying capacity rises.
3. Eliminate A: more foxes means more predation, which lowers the number of rabbits the island can support. Eliminate C: disease is a limiting factor that lowers the population. Eliminate D: a higher birth rate makes the population grow faster, but it does not change how many rabbits the island's resources can support.

The answer is B.

Tip: carrying capacity is about the environment (resources and limits), not about how fast a population reproduces.

Evidence: hares peaked at 80 thousand in year 3 and 70 thousand in year 9. Lynx peaked at 9 hundred in year 4 and 8 hundred in year 10.

Step by step:

1. Hare peaks: years 3 and 9. Lynx peaks: years 4 and 10.
2. Each lynx peak came about 1 year after a hare peak. With many hares to eat, more lynx survived and raised young; then, as hares became scarce, lynx numbers fell.
3. Eliminate B (lynx peaks came after, not before, hare peaks), C (the units differ: 80 thousand hares vs 9 hundred lynx, so hares always far outnumbered lynx), and D (the peaks were a year apart, not at the same time).

The answer is A.

Tip: in predator-prey data, the prey usually peaks first and the predator follows. Always check the units before comparing two populations.

Evidence: the graph shows the population growing from 50 to 760 by year 12, then staying near 800 (800, 790, 810, 800) from year 14 to year 20. The passage defines carrying capacity as "the largest population of a species that it can support over time."

Step by step:

1. Find where the curve flattens: from year 14 on.
2. The values bounce a little around 800 but stay there.
3. So the plateau can support about 800 deer over time.
4. Eliminate A (50 is the starting number), B (300 was the population at year 6, while it was still growing fast), and D (the data show no rise above about 810, so 1,000 predicts beyond the data).

The answer is C. About 800 deer.

Tip: on an S-shaped growth curve, carrying capacity is the flat top, not the starting value, the steep middle, or a guess beyond the data.

Evidence: in the web, arrows run from zooplankton and mayfly larva to sunfish, and from sunfish to bass and heron. Algae has arrows to both zooplankton and mayfly larva.

Step by step:

1. Sunfish eat zooplankton and mayfly larvae, and both of those eat algae.
2. If algae sharply decline, zooplankton and mayfly larvae lose food and decline. Then sunfish lose both of their foods, so sunfish likely decrease.
3. Eliminate B and C: herons and bass eat sunfish, so fewer of them would likely let sunfish increase. Eliminate D: more zooplankton means more food for sunfish.

The answer is A.

Tip: a change at the bottom of a food web travels upward. Follow the arrows from the producer to the organism in the question.

Evidence: the graph shows growth of 160 deer from year 8 to 10 (480 → 640) but only 40 from year 12 to 14 (760 → 800), then almost none. The passage says limiting factors "such as the amount of food, water, and space, and the spread of disease, keep a population from growing forever."

Step by step:

1. Early on, food and space were plentiful, so the population grew quickly.
2. As the number of deer neared about 800, each deer had less food and space. Limiting factors slowed births and raised deaths.
3. When births and deaths balance, the population levels off at the carrying capacity.
4. Eliminate B (a steady population means births about equal deaths, not that births stopped), C (the passage says there were very few predators and mentions none being added), and D (nothing suggests the plateau grew; a bigger plateau would allow more growth).

The answer is A.

Tip: slowing growth near the top of an S-curve is explained by limiting factors. Reject answers that add events the passage never mentions.

Evidence: the table shows 0.6 ppm in walleye and 3.0 ppm in loons, but only 0.000002 ppm in lake water. The passage says "Organisms take in methylmercury faster than their bodies get rid of it" and "A predator eats many prey animals over its lifetime, so the methylmercury in its body becomes more concentrated than in its prey."

Step by step:

1. Loons are at the top of this food chain; they eat walleye.
2. Each walleye carries methylmercury, and a loon eats many walleye over its life.
3. Because the loon's body does not get rid of methylmercury quickly, it builds up. Its level becomes 3.0 \div 0.6 = 5 times the walleye's.
4. Eliminate B (lake water has a tiny amount, 0.000002 ppm, so drinking water cannot explain 3.0 ppm), C (energy decreases up the chain; mercury builds up because it is stored, not because of energy), and D (mercury comes from burning coal; organisms do not make it).

The answer is A.

Tip: biomagnification happens when a substance is stored in the body instead of being broken down or passed out. Predators collect it from every prey they eat.

Evidence: Flask 1 (A alone) shows 2, 40, 110, 124, 125 cells/mL on days 0, 4, 8, 12, and 16.

Step by step:

1. Growth is fast at first (2 → 40 → 110), then slows (110 → 124), then nearly stops (124 → 125).
2. A population that levels off has reached the carrying capacity of its environment. For A alone, that is about 125 cells/mL.
3. Eliminate B: 100 is Species A grown with B (Flask 3), a different condition. Eliminate C: 80 is Species B alone. Eliminate A: 2 is the starting population, not the limit.

The answer is D. About 125 cells/mL.

Tip: tables with several similar columns are a trap. Put your finger on the column named in the question before you read any numbers.

Evidence: the passage says "only about 10% of the energy at one trophic level becomes part of the next level." So each level holds about 10 times the energy of the level above it.

Step by step:

1. Tertiary consumers: 40 kJ/m².
2. Secondary consumers: 40 \times 10 = 400 kJ/m².
3. Primary consumers: 400 \times 10 = 4{,}000 kJ/m².
4. Producers: 4{,}000 \times 10 = 40{,}000 kJ/m².
5. The traps are stopping too early (400 or 4,000) or going one step too far (400,000).

The answer is D. 40,000 kJ/m².

Check: going back up: 40{,}000 \to 4{,}000 \to 400 \to 40 ✓, the same pattern as the table (25{,}000 \to 25).

Tip: count the steps between the two levels before calculating. Producers to tertiary consumers is three steps, so the factor is 10 \times 10 \times 10 = 1{,}000.

Evidence: the table shows energy dropping from 25,000 to 2,500 to 250 kJ/m², and the 10% rule gives about 25 for hawks. The passage says only about 10% passes to the next level, and "the rest is used for life processes and lost as heat."

Step by step:

1. A fifth level (animals that eat hawks) would get about 25 \times 0.10 = 2.5 kJ/m², just 0.01% of the producers' energy.
2. That is too little energy to support a population of large, active predators.
3. So the loss of about 90% at each step limits how long a food chain can be.
4. Eliminate B (energy decreases up the chain, it does not increase), C (decomposers break down dead matter at every level; they do not "eat the fifth level"), and D (the number of levels depends on energy, not on a fixed rule about kinds of animals).

The answer is A.

Tip: many "why" questions about ecosystems come back to the 10% rule. Run the numbers one more step to see the effect.

Evidence: the passage says lynx "main food is snowshoe hares." In the table, hares fell from 25 thousand (year 5) to 10 thousand (year 6) while lynx were still at 7 hundred in year 5. Hares eat "shrubs and young trees."

Step by step:

1. Lynx eat hares, so lynx were one cause of the hare decline.
2. With most lynx removed, far fewer hares would be eaten.
3. So hares would most likely drop less and start recovering sooner than they did in the real data.
4. Eliminate B (removing a predator does not make its prey decline faster), C (the passage links the two species directly), and D (hares eat shrubs and young trees, a limited food supply, so they cannot grow without limit).

The answer is A.

Tip: removing a predator usually helps its prey in the short run. But no population grows forever: food is still a limiting factor.

Evidence: the table shows 0.15 ppm in yellow perch and 0.6 ppm in walleye. The passage says "a health guideline advises people to limit fish that contain more than 0.3 ppm."

Step by step:

1. Yellow perch: 0.15 ppm, which is below 0.3 ppm.
2. Walleye: 0.6 ppm, which is above 0.3 ppm (twice the guideline).
3. So the family should choose yellow perch and limit walleye. This fits biomagnification: walleye eat perch and are one level higher, so they have more methylmercury.
4. Eliminate B (walleye has the higher level), C (perch is under the guideline), and D (the low level in the water does not matter; the fish have built up much more).

The answer is A.

Tip: for questions that apply a limit, write each value next to the limit and mark it above or below before choosing.

Evidence: fine-mesh bags "let bacteria and fungi in but kept out earthworms and insects"; coarse-mesh bags "let earthworms and insects in too." Average mass left: fine 15.1 g (4.9 g lost); coarse 9.9 g (10.1 g lost).

Step by step:

1. The only planned difference between the bag types was whether earthworms and insects could get in.
2. Coarse bags lost about twice as much mass: 10.1 \div 4.9 \approx 2.1.
3. So letting earthworms and insects in roughly doubled how much leaf matter was broken down in 6 months.
4. Eliminate B (fine bags still lost 4.9 g, so bacteria and fungi did decompose leaves), C (coarse bags lost more, not less), and D (one forest over 6 months cannot show what happens in all forests).

The answer is A.

Tip: link the result to the one difference between groups, and keep the conclusion as narrow as the experiment.

Evidence: on day 16, Species A reached 125 alone but only 100 with B. Species B reached 80 alone but fell to 4 with A (after peaking at 20 on day 8). The passage says "Both species feed on the same kind of bacteria."

Step by step:

1. Both species did worse together than alone, which is the sign of competition for a shared, limited resource.
2. Species B declined steadily after day 8 (20 → 10 → 4), so it was being pushed toward elimination; A was the stronger competitor.
3. Eliminate B (predation): if A ate B, A should gain extra food and do better than when alone, but A did worse (100 vs. 125). Eliminate C (mutualism): neither species benefited. Eliminate D: B was falling, not rising, at the end; predicting B would win contradicts the trend.

The answer is A.

Tip: to identify an interaction, compare each species with and without the other. Both lower = competition; both higher = mutualism; one higher and one lower = predation or parasitism.

Evidence: P alone: 60 g for 20 plants; Q alone: 50 g for 20 plants; mixed pots: 42 g of P and 9 g of Q, with "10 plants of each species together." The passage notes that "Species P grows taller than Species Q" and both need light, water, and nutrients.

Step by step:

1. Mass per plant, Species P: alone 60 \div 20 = 3.0 g; mixed 42 \div 10 = 4.2 g. P plants grew more each when mixed with Q.
2. Mass per plant, Species Q: alone 50 \div 20 = 2.5 g; mixed 9 \div 10 = 0.9 g. Q plants grew much less when mixed with P.
3. So competition with P greatly reduced Q's growth, while P did at least as well beside Q. Being taller, P probably shaded Q.
4. Eliminate B (Q plants were smaller, 0.9 vs 2.5 g), C (P's pot total fell from 60 to 42 g only because there were half as many P plants; per plant P grew more), and D (Q's sharp drop shows the species did compete).

The answer is A.

Tip: before comparing groups, make sure you are comparing the same unit. Here that means grams per plant.

Evidence: paragraph 1: decomposers "use the energy stored in this dead matter for their own life processes, and that energy is eventually lost as heat. At the same time, decomposers release nutrients such as nitrogen and phosphorus into the soil, where plants can take them up and use them again."

Step by step:

1. Nitrogen is matter. Decomposers return it to the soil, and plants can absorb it again. Matter cycles.
2. Energy is used by the decomposers to live and is eventually lost as heat. Plants cannot take it back up; they get new energy from sunlight. Energy flows one way.
3. Eliminate B (energy is not returned to plants), C (nitrogen is not destroyed; it goes back into the soil), and D (it reverses the two: nitrogen is recycled and energy is lost).

The answer is A.

Tip: remember the rule: matter (nutrients) cycles through an ecosystem; energy flows through it once and leaves as heat.

Evidence: paragraph on oxpeckers: they "eat ticks off their skin," but "researchers have also seen oxpeckers peck at open wounds and drink blood, which can keep the wounds from healing."

Step by step:

1. The oxpecker benefits (food) in both behaviors: +.
2. The rhino gains from tick removal (+), but it is harmed when wounds are kept open (−).
3. +/+ is mutualism, and +/− is parasitism. The evidence supports parts of both, so the relationship is not clearly mutualism; it may depend on how much each behavior happens.
4. Eliminate B (it ignores the wound-feeding evidence), C (the rhino is affected, both helped and harmed, so it is not +/0), and D (oxpeckers do not kill and eat rhinos).

The answer is A.

Tip: the GED likes evidence that does not fit neatly. The best answer accounts for all the evidence, not just the part that supports the first idea.