A field of grass soaks up sunlight all summer. A grasshopper eats the grass, a shrew eats the grasshopper, a hawk eats the shrew. Here's the puzzle that runs through all of environmental science: why can't that field support very many hawks? Why is the world full of grass and grasshoppers but you have to search for a top predator? The answer is not about space or meanness — it's about energy, and how much of it is lost at every hand-off in a food chain. Master that single idea and you've unlocked a huge fraction of Unit 1. Grab a pencil; this lesson is where APES becomes a numbers game.
Almost every ecosystem on Earth runs on solar energy. Producers (autotrophs) — plants, algae, cyanobacteria — capture sunlight through photosynthesis and convert it into chemical energy stored in glucose:
6 CO₂ + 6 H₂O + light energy → C₆H₁₂O₆ + 6 O₂
A small number of ecosystems (deep-sea hydrothermal vents) instead run on chemosynthesis, where bacteria oxidize inorganic molecules like hydrogen sulfide. But for the exam, assume solar energy and photosynthesis unless a vent is explicitly named.
Energy passes from organism to organism in a food chain, and each feeding position is a trophic level:
Real ecosystems are not simple chains but food webs — interconnected chains where most organisms eat, and are eaten by, several species.
[DIAGRAM: Food web. Grass → grasshopper and rabbit. Grasshopper → shrew and sparrow. Rabbit → hawk and fox. Shrew → hawk. Sparrow → hawk. Fox and hawk at top. Arrows point in the direction energy flows (from eaten to eater). Decomposer box (fungi/bacteria) receives arrows from every organism.]
Here is the master concept: only about 10% of the energy at one trophic level is passed up to the next. The other ~90% is lost, mostly as heat from cellular respiration, plus energy spent moving, energy in undigested waste, and energy in parts not eaten.
[DIAGRAM: Energy pyramid, 4 levels. Base = producers 10,000 kcal/m²/yr; primary consumers 1,000; secondary consumers 100; tertiary consumers 10. "×10% transfer" arrow between each level. Width of each bar proportional to energy.]
This is why pyramids narrow at the top, and it answers the warm-up: there simply isn't enough energy left at the top to support many large predators. It's also why food chains rarely exceed 4–5 links — beyond that, there's almost nothing left to eat.
The 10% figure is an average; real ecological efficiency ranges roughly 5–20%. On the exam, use 10% unless told otherwise.
Primary productivity is the rate at which producers capture and store energy.
NPP = GPP − respiration (R)
NPP is what matters ecologically, because it's the energy actually available to the rest of the food web. Units are usually energy per area per time (kcal/m²/yr) or biomass per area per time (g/m²/yr).
The most productive ecosystems per unit area are tropical rainforests, estuaries, swamps, and coral reefs; the least productive are deserts, open ocean, and tundra. (The open ocean has low productivity per square meter but, because it's so vast, contributes the largest total NPP on Earth — a classic trap.)
Every energy question on the exam — pyramids, biomagnification of toxins up food chains, why eating lower on the food chain feeds more people — traces back to the 10% rule and the GPP/NPP distinction. Lock these in now.
Producers in a meadow store 45,000 kcal/m²/yr. Using the 10% rule, how much energy reaches the secondary consumers?
Strategy: Multiply by 10% (×0.10) for each level climbed. Producers → primary → secondary is two steps.
Solution:
45,000 kcal × 0.10 = 4,500 kcal (primary consumers)
4,500 kcal × 0.10 = 450 kcal (secondary consumers)
Answer: 450 kcal/m²/yr.
Interpretation: Two trophic steps means two factors of 10 — energy dropped by 100×.
In a forest, gross primary productivity is 12,000 kcal/m²/yr and plant respiration consumes 6,500 kcal/m²/yr. Find NPP.
Strategy: NPP = GPP − R.
Solution:
NPP = 12,000 − 6,500 = 5,500 kcal/m²/yr
Answer: 5,500 kcal/m²/yr is available to consumers and for plant growth.
Interpretation: Less than half the captured energy is left after the plants "pay their own bills." Respiration is a real energy sink, not a rounding error.
A population of hawks requires 30 kcal/m²/yr and sits at the 4th trophic level. Assuming the 10% rule, how much energy must the producers capture (NPP)?
Strategy: Going down the pyramid, divide by 0.10 (multiply by 10) per level. From 4th to 1st is three steps.
Solution:
Tertiary (hawks): 30 kcal
Secondary: 30 ÷ 0.10 = 300 kcal
Primary: 300 ÷ 0.10 = 3,000 kcal
Producers: 3,000 ÷ 0.10 = 30,000 kcal
Answer: 30,000 kcal/m²/yr.
Interpretation: Supporting a top predator takes a thousand times more energy at the base — the ecological reason apex predators need huge territories.
A lab finds only 8% of energy transfers from producers (5,000 kcal) to primary consumers. How much reaches primary consumers, and how does this compare to the 10% assumption?
Solution:
5,000 × 0.08 = 400 kcal (at 8%)
5,000 × 0.10 = 500 kcal (at 10%)
Answer: 400 kcal, which is 100 kcal less than the 10% estimate would predict.
Interpretation: Read the problem — if it gives an efficiency, use it. The 10% rule is only the default.
80,000 × 0.10 = 8,000 kcal. (A) divides by 100; (C) by 1,000; (D) multiplies.NPP = 9,000 − 4,000 = 5,000. (A) adds; (C) is respiration; (D) is nonsense.15 → 150 → 1,500 → 15,000. (A)/(B) too few steps; (D) is one step.(C) Earthworm = detritivore. (A) herbivore = primary consumer; (B) fungi = decomposer; (D) algae = producer.
(a) 50,000 × 0.10 = 5,000 kcal/m²/yr to primary consumers. (b) primary→secondary→tertiary = two more steps: 5,000 × 0.10 = 500, 500 × 0.10 = 50 kcal/m²/yr to tertiary consumers.
(a) R = GPP − NPP = 20,000 − 7,000 = 13,000 kcal/m²/yr. (b) A low NPP:GPP ratio means the plants spend most of their captured energy on their own respiration, leaving little for growth or for consumers — typical of stressed or mature ecosystems.
FRQ rubric (10 pts):
- (a) 1 pt independent variable (amount/concentration of fertilizer added); 1 pt dependent variable (NPP / algae growth / dissolved O₂). (2)
- (b) 1 pt multiple ponds/tanks with varying fertilizer levels; 1 pt a control group receiving no fertilizer; 1 pt measuring NPP the same way in all groups over the same period. (3)
- (c) 1 pt names a valid constant (temperature, light, algae species, water volume, pH); 1 pt explains it could otherwise confound the result (be an alternative cause of NPP change). (2)
- (d) 1 pt NPP = 3,000 − 1,200 = 1,800 mg O₂/L/day. (1)
- (e) 1 pt identifies cultural eutrophication / algal bloom; 1 pt explains that decomposition of dead algae depletes dissolved oxygen, creating a hypoxic dead zone that kills fish. (2)
Ecologists want to test the hypothesis that fertilizer runoff increases the net primary productivity of pond algae.
(a) Identify the independent and dependent variables. (2 pts) (b) Describe a controlled experimental design to test the hypothesis, including a control group. (3 pts) (c) Identify one variable that must be held constant and explain why. (2 pts) (d) The algae's GPP is measured at 3,000 mg O₂/L/day and respiration at 1,200 mg O₂/L/day. Calculate NPP. (1 pt) (e) Describe one way fertilizer runoff reaching a natural pond could ultimately harm the ecosystem despite raising productivity. (2 pts)
MC:
1. (B) Producers — the base of the pyramid holds the most energy; every level above loses ~90%. (A) and (D) are higher up with less; (C) decomposers get scraps.
2. (C) Heat from cellular respiration is the dominant loss. (A) light is the input; (B) undigested biomass is only part of the loss; (D) is not a loss mechanism.
3. (B) 80,000 × 0.10 = 8,000 kcal. (A) divides by 100; (C) by 1,000; (D) multiplies.
4. (B) NPP = 9,000 − 4,000 = 5,000. (A) adds; (C) is respiration; (D) is nonsense.
5. (C) Too little energy remains after repeated 90% losses. Others are not the limiting reason.
6. (C) Tropical rainforest — high light, water, warmth. Open ocean, desert, tundra are all low per m².
7. (C) Three steps down: 15 → 150 → 1,500 → 15,000. (A)/(B) too few steps; (D) is one step.
8. (B) grass → mouse → snake → hawk. Only (B) runs producer to apex.
9. (B) Hydrothermal vents. The rest are photosynthetic.
10. (C) Earthworm = detritivore. (A) herbivore = primary consumer; (B) fungi = decomposer; (D) algae = producer.
(a) 50,000 × 0.10 = 5,000 kcal/m²/yr to primary consumers. (b) primary→secondary→tertiary = two more steps: 5,000 × 0.10 = 500, 500 × 0.10 = 50 kcal/m²/yr to tertiary consumers.
(a) R = GPP − NPP = 20,000 − 7,000 = 13,000 kcal/m²/yr. (b) A low NPP:GPP ratio means the plants spend most of their captured energy on their own respiration, leaving little for growth or for consumers — typical of stressed or mature ecosystems.
FRQ rubric (10 pts):
- (a) 1 pt independent variable (amount/concentration of fertilizer added); 1 pt dependent variable (NPP / algae growth / dissolved O₂). (2)
- (b) 1 pt multiple ponds/tanks with varying fertilizer levels; 1 pt a control group receiving no fertilizer; 1 pt measuring NPP the same way in all groups over the same period. (3)
- (c) 1 pt names a valid constant (temperature, light, algae species, water volume, pH); 1 pt explains it could otherwise confound the result (be an alternative cause of NPP change). (2)
- (d) 1 pt NPP = 3,000 − 1,200 = 1,800 mg O₂/L/day. (1)
- (e) 1 pt identifies cultural eutrophication / algal bloom; 1 pt explains that decomposition of dead algae depletes dissolved oxygen, creating a hypoxic dead zone that kills fish. (2)
⭐ Exam strategy: Whenever a question gives you energy at one trophic level, immediately sketch a 4-bar pyramid and label which direction you're moving. Up = ×0.10, down = ×10. This one habit prevents the most common APES math error.
Content pending external review.