Ecology, Populations & Biodiversity — Chapter review · Physics & Environmental Science, Grade 12

Grass = producer (trophic level 1); grasshopper = primary consumer (level 2); mouse = secondary consumer (level 3); hawk = tertiary consumer (level 4).

The producer that makes its own food is trophic level 1 (grass). Each step up is one level higher: the grasshopper eats the producer (primary consumer, level 2), the mouse eats the grasshopper (secondary consumer, level 3), and the hawk eats the mouse (tertiary consumer, level 4).

GPP is the total energy producers capture through photosynthesis; NPP is what remains after producers' own respiration: NPP = GPP - respiration.

GPP measures all the energy fixed by photosynthesis. But producers must spend energy on their own life processes (respiration), so the energy actually stored as new biomass and available to consumers is the net amount: NPP = GPP - respiration.

Primary consumers: 9,000 kcal/m²/yr; secondary consumers: 900 kcal/m²/yr; tertiary consumers: 90 kcal/m²/yr.

Apply the 10% rule (× 0.10) at each step. Primary consumers: 90,000 × 0.10 = 9,000. Secondary consumers: 9,000 × 0.10 = 900. Tertiary consumers: 900 × 0.10 = 90 kcal/m²/yr. Energy drops by about 90% at every level.

(a) genetic diversity, (b) species diversity, (c) ecosystem diversity.

(a) Variation within a single population is genetic diversity. (b) The count and balance of species in one community is species diversity (richness plus evenness). (c) The variety of whole ecosystems across a region is ecosystem diversity.

D = 0.75.

Each proportion is 25/100 = 0.25, and (0.25)² = 0.0625. With four equally common species, Σ(n/N)² = 4 × 0.0625 = 0.25. Then D = 1 - 0.25 = 0.75 — high because the community is both rich and perfectly even.

(a) regulating, (b) provisioning, (c) supporting, (d) cultural.

(a) Pollination stabilizes and maintains crop production — a regulating service. (b) Lumber is a tangible good the ecosystem supplies — provisioning. (c) Soil formation and nutrient cycling underpin all the other services — supporting. (d) Recreation and beauty are cultural services.

The diverse forest is likely to recover better, because its variety of species and genes provides redundancy and resistance, giving it greater resilience than the uniform plantation.

In the single-species plantation, every tree is vulnerable to the same pest, so the outbreak can devastate the whole stand at once, and there are no other tree species to fill the gap — low biodiversity means low resilience. The diverse forest is buffered on two levels: high genetic diversity means some individuals may resist the pest, and high species diversity means that even if one tree species is hit hard, others continue performing similar ecological roles. This redundancy keeps the forest functioning and lets it recover, illustrating why biodiversity underpins ecosystem stability and resilience.

160 individuals/year at that instant.

.4×500×(1−500/2500)=160. The future population and rate change, so this is not an exact finite-year increment.

Growth rate = 3% per year; doubling time ≈ 23 years.

Net change = 320 - 80 = 240. Growth rate = 240/8,000 = 0.03 = 3% per year. Rule of 70: doubling time ≈ 70 ÷ 3 ≈ 23.3 years.

A density-dependent factor (e.g., disease or food competition) intensifies as the herd crowds, pulling the population back toward K; a density-independent factor (e.g., a severe winter storm) reduces the population regardless of how dense it is.

Density-dependent factors get stronger as density rises: as deer crowd near K, food competition sharpens and disease spreads more easily, raising deaths until the population eases back toward carrying capacity. Density-independent factors, like a blizzard, drought, or wildfire, kill a set fraction of deer whether the herd is large or small, so they can knock the population down at any density. Real populations are shaped by both, which is why they fluctuate around K rather than sitting exactly on it.

Temperate deciduous forest.

Moderate temperatures with distinct seasons and moderate-to-high precipitation (around 100 cm/yr) support broadleaf trees that shed their leaves to survive cold winters. On a temperature-versus-precipitation diagram, this middle-temperature, fairly wet climate is the temperate deciduous forest.

The open ocean is nutrient-poor at the sunlit surface because nutrients sink into the dark depths, so productivity is low; estuaries and reefs are shallow, sunlit, and nutrient-rich, so productivity is high.

Photosynthesis needs both light and nutrients. In the open ocean, light is available only in the shallow photic zone, but nutrients continually sink below it into the lightless deep, leaving surface waters nutrient-starved — productivity per square meter is low. Estuaries receive a steady load of nutrients from rivers in shallow, well-lit water, and coral reefs combine sunlit shallows with efficient nutrient recycling, so both are among the most productive ecosystems on Earth.

First 1000, second 100 units. State that the given efficiency is modeled and energy is not recycled like matter.

Response criteria

• State a scientifically accurate conclusion.
• Show the calculation or mechanism that supports it.
• Include units, conditions, or limits where relevant.

NPP=7000 kJ/m²/yr. Efficiency=840/7000=12%. The fraction uses NPP, not GPP, and does not imply every intermediate process has12% efficiency.

NPP=7000 kJ/m²/yr.

Efficiency=840/7000=12%.

The fraction uses NPP, not GPP, and does not imply every intermediate process has12% efficiency.

Response criteria

• Use the stated measurements or conditions, with correct units.
• Connect the result to the physical or environmental mechanism or evidence relationship.
• Identify the assumption, model limit or further evidence requested.

D=.5. State this version of the index and the sampling effort before comparing sites.

Response criteria

• State a scientifically accurate conclusion.
• Show the calculation or mechanism that supports it.
• Include units, conditions, or limits where relevant.

Richness=3. Fractions0.6,0.3,0.1 give D=1−(0.36+0.09+0.01)=0.54. The sample may miss rare or hard-to-detect species; matched effort and replication matter.

Richness=3.

Fractions0.6,0.3,0.1 give D=1−(0.36+0.09+0.01)=0.54.

The sample may miss rare or hard-to-detect species; matched effort and replication matter.

Response criteria

• Use the stated measurements or conditions, with correct units.
• Connect the result to the physical or environmental mechanism or evidence relationship.
• Identify the assumption, model limit or further evidence requested.

dN/dt=.4×500×(1−.2)=160individuals/year at that moment; future N changes the rate.

Response criteria

• State a scientifically accurate conclusion.
• Show the calculation or mechanism that supports it.
• Include units, conditions, or limits where relevant.

At200:0.5×200×0.75=75 individuals/year. At400:0.5×400×0.5=100/year. Total growth rises even while per-capita growth falls.

At200:0.5×200×0.75=75 individuals/year.

At400:0.5×400×0.5=100/year.

Total growth rises even while per-capita growth falls.

Response criteria

• Use the stated measurements or conditions, with correct units.
• Connect the result to the physical or environmental mechanism or evidence relationship.
• Identify the assumption, model limit or further evidence requested.

Low local nutrients can limit per-area production, but its enormous total area matters for aggregate output.

Response criteria

• State a scientifically accurate conclusion.
• Show the calculation or mechanism that supports it.
• Include units, conditions, or limits where relevant.

Mass change=−300 mg/day; concentration change=−0.30 mg/L/day. The result assumes the stated constant volume, uniform mixing and continuing fluxes; real fluxes can respond to changing conditions.

Mass change=−300 mg/day; concentration change=−0.30 mg/L/day.

The result assumes the stated constant volume, uniform mixing and continuing fluxes; real fluxes can respond to changing conditions.

Response criteria

• Use the stated measurements or conditions, with correct units.
• Connect the result to the physical or environmental mechanism or evidence relationship.
• Identify the assumption, model limit or further evidence requested.

Write the relationship first; substitute with consistent units.

0.32

1−(.04+.64)=.32.

Area, effort, detectability, locations, and timing determine its interpretation.

Response criteria

• State a scientifically accurate conclusion.
• Show the calculation or mechanism that supports it.
• Include units, conditions, or limits where relevant.

Total16 over2.0 m² gives8 individuals/m². Extrapolated abundance≈800. Spatial variation, finite sampling and imperfect detection make it an estimate rather than a census.

Total16 over2.0 m² gives8 individuals/m².

Extrapolated abundance≈800.

Spatial variation, finite sampling and imperfect detection make it an estimate rather than a census.

Response criteria

• Use the stated measurements or conditions, with correct units.
• Connect the result to the physical or environmental mechanism or evidence relationship.
• Identify the assumption, model limit or further evidence requested.