Biodiversity — AP Biology Study Guide

For: AP Biology candidates sitting AP Biology.

Covers: species richness, species evenness, Simpson’s Biodiversity Index calculation, island biogeography theory, categorization of ecosystem services, and assessment of human threats to biodiversity, aligned to the AP Biology CED Unit 8 Ecology.

You should already know: Basic population growth dynamics. Evolution by natural selection. Community ecological interactions.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Biology style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Biodiversity?

Biodiversity is the total variation of all living organisms and the ecological complexes they inhabit, measured across three hierarchical scales: genetic diversity within populations, species diversity within communities, and ecosystem diversity across landscapes. AP Biology primarily focuses on species diversity at the community level, the most frequently tested form of the concept. According to the AP Biology CED, this topic accounts for approximately 6-8% of the total exam score, and questions appear on both the multiple-choice (MCQ) and free-response (FRQ) sections.

Biodiversity is not just a simple count of species: it combines two core components, the number of different species (richness) and the relative abundance of each species (evenness). Higher biodiversity is consistently linked to greater ecosystem stability, higher resistance to disturbance, and higher net primary productivity, connections that AP exam questions frequently probe. Most FRQ prompts ask students to connect biodiversity to ecosystem function or human impacts, so you must be able to both quantify biodiversity and describe its real-world benefits.

2. Quantifying Biodiversity: Richness, Evenness, and Simpson's Index

Species richness ($S$) is defined as the total number of distinct species present in a community. Species evenness describes how similar the abundances of different species are; two communities can have identical richness but very different evenness, leading to very different overall biodiversity.

The most commonly tested metric for biodiversity on the AP exam is Simpson's Biodiversity Index, which accounts for both richness and evenness. The formula is: $$D=1-\sum \frac{n(n-1)}{N(N-1)}$$ Where $n$ = the number of individuals of a single species, $N$ = the total number of individuals across all species, and $\sum$ indicates the sum of the $\frac{n(n-1)}{N(N-1)}$ term for every species. The index ranges from 0 to 1: the term $\sum \frac{n(n-1)}{N(N-1)}$ equals the probability that two randomly selected individuals belong to the same species. Subtracting this probability from 1 gives the probability that the two individuals belong to different species, a direct measure of diversity. Higher $D$ values always correspond to higher biodiversity.

Worked Example

Two adjacent prairie communities are sampled one year after a wildfire. Community 1 has species counts: Big Bluestem Grass (100 individuals), Indian Grass (50), Milkweed (30), Coneflower (20). Community 2 has counts: Big Bluestem Grass (150), Indian Grass (20), Milkweed (15), Coneflower (15). Calculate Simpson's Index for both communities and identify which has higher biodiversity.

1. Calculate total $N$ for both communities: $N_1=100+50+30+20=200$, $N_2=150+20+15+15=200$.
2. Calculate $n(n-1)$ for each species in Community 1, then sum: $(100\ast 99)+(50\ast 49)+(30\ast 29)+(20\ast 19)=9900+2450+870+380=13600$.
3. Calculate $D_1$: $N(N-1)=200\ast 199=39800$, so $D_1=1-(13600/39800)\approx 1-0.342=0.658$.
4. Repeat for Community 2: Sum of $n(n-1)=(150\ast 149)+(20\ast 19)+(15\ast 14)+(15\ast 14)=22350+380+210+210=23150$. $D_2=1-(23150/39800)\approx 1-0.582=0.418$.
5. Compare values: $D_1(0.66)>D_2(0.42)$, so Community 1 has higher biodiversity.

Exam tip: On AP FRQs, always show each step of your Simpson's Index calculation; even if you make an arithmetic error, you can earn partial credit for correctly applying the formula structure.

3. Equilibrium Theory of Island Biogeography

Island biogeography theory explains how species richness on isolated habitats (called "islands" for the model) is determined by two competing ecological processes: immigration of new species from a mainland source population, and extinction of existing species on the island. The theory predicts that equilibrium species richness occurs where the rate of immigration equals the rate of extinction.

Two key factors shift this equilibrium: 1) Island size: Smaller islands support smaller population sizes for each species, which increases extinction risk from genetic drift, disturbance, and resource limitation. Smaller islands therefore have higher extinction rates, leading to lower equilibrium species richness. 2) Distance from the mainland source: Islands farther from the source have lower immigration rates, because fewer colonizing individuals are able to disperse to the distant island, leading to lower equilibrium species richness. Critically, the model applies to any isolated habitat fragment (not just oceanic islands), which is the most common context for AP exam questions.

Worked Example

A large intact forest is fragmented by a new highway, creating three separate forest fragments: Fragment A (100 km², 10 km from intact forest), Fragment B (100 km², 50 km from intact forest), Fragment C (20 km², 10 km from intact forest). Predict which fragment will have the highest equilibrium species richness, and justify your prediction.

1. First, recall the two variables that determine equilibrium richness: island size and distance from the source population (intact forest).
2. Compare size first: Fragments A and B are the same size (100 km²), which is much larger than Fragment C (20 km²). Smaller islands have higher extinction rates, so C will have lower richness than A or B and can be eliminated.
3. Compare distance from the source for A and B: A is only 10 km from the source, while B is 50 km away. Closer islands have higher immigration rates of new species.
4. Conclusion: Fragment A has the same extinction rate as B (same size) but higher immigration, so it has the highest equilibrium species richness.

Exam tip: Always remember that "island" in AP questions does not only mean oceanic islands; any isolated habitat fragment surrounded by inhospitable habitat counts as an island for this model.

4. Categorization of Ecosystem Services

Ecosystem services are the direct and indirect benefits that humans gain from intact, biodiverse ecosystems. AP Biology requires you to correctly categorize services into four main groups, and to connect biodiversity loss to reduced service provision:

1. Provisioning services: Tangible products obtained directly from ecosystems, such as food, fresh water, timber, and medicinal plants.
2. Regulating services: Benefits gained from the regulation of natural ecosystem processes, such as climate regulation, flood control, water purification, and pollination.
3. Cultural services: Non-material benefits for humans, such as recreational opportunities, aesthetic and spiritual value, and scientific discovery.
4. Supporting services: Services that are required to produce all other ecosystem services, such as nutrient cycling, soil formation, and primary production.

Intact, high-biodiversity ecosystems provide more reliable ecosystem services than degraded, low-biodiversity systems, which is a common connection tested in FRQs.

Worked Example

A developer proposes draining a coastal wetland to build a new resort. For each of the four categories of ecosystem services, name one service provided by the intact wetland that would be lost after draining.

1. Provisioning: Intact wetlands support wild-harvested shellfish and filter groundwater to provide clean fresh water for nearby coastal communities, so this provisioning service would be lost.
2. Regulating: Wetlands absorb storm surge from hurricanes and reduce inland flooding, so the regulating service of flood control would be lost.
3. Cultural: Wetlands provide habitat for migratory birds that attract recreational birdwatchers and ecotourists, so this cultural service would be lost.
4. Supporting: Wetlands cycle excess nitrogen and phosphorus from agricultural runoff, preventing eutrophication of nearshore oceans, so this nutrient cycling supporting service would be lost.

Exam tip: When asked to name an ecosystem service on the AP exam, always explicitly tie the service to a human benefit to earn full points, do not just name a process.

5. Major Threats to Biodiversity

The six most significant human-caused threats to global biodiversity are commonly summarized by the acronym HIPPCO: Habitat destruction/fragmentation, Invasive species, human population growth, Pollution, Climate change, Overexploitation. Habitat destruction and fragmentation is the number one cause of current biodiversity loss globally, followed by invasive species, overexploitation, pollution, and climate change. Threats frequently interact: for example, habitat fragmentation creates small isolated populations that are more vulnerable to extinction from climate change or invasive species.

Worked Example

Researchers studying freshwater fish in a dammed river basin find that fish in affected river stretches have a 10x higher extinction rate than fish in undammed stretches. Affected stretches also experience agricultural nutrient runoff and have non-native predatory northern pike present. Identify three distinct threats to biodiversity in this system and explain how each increases extinction risk.

1. Habitat fragmentation from damming: Damming splits continuous river habitat into isolated upstream and downstream sections, so fish populations are trapped in small isolated patches. Small populations have higher risk of inbreeding depression and cannot recolonize after local extinction, increasing extinction risk.
2. Pollution from agricultural runoff: Nutrient runoff causes eutrophication, which leads to low dissolved oxygen levels that kill native fish. Pollution reduces survival and reproduction of native fish populations, increasing extinction risk.
3. Invasive species (northern pike): Non-native pike prey on small native fish species and outcompete native predators for food, reducing native population sizes and increasing extinction risk.

Exam tip: Never answer with a vague term like "human impact" when asked to name a threat; always name the specific threat (e.g. habitat fragmentation, invasive species) to earn full credit.

6. Common Pitfalls (and how to avoid them)

• Wrong move: Interpreting a low Simpson's Index D value as indicating high biodiversity. Why: Students confuse AP's Simpson's Biodiversity Index with the opposite-structured Simpson's Dominance Index taught in some textbooks. Correct move: Memorize that for the AP formula $D=1-\sum ...$, D closer to 1 always equals higher biodiversity.
• Wrong move: Assuming two communities with the same species richness have the same biodiversity. Why: Students only count the number of species and forget that evenness is also a core component of biodiversity. Correct move: Whenever comparing biodiversity, always account for both richness and evenness, and calculate Simpson's Index if abundance data is provided.
• Wrong move: Only applying island biogeography to oceanic islands, not recognizing terrestrial habitat fragments as islands. Why: Most textbooks introduce the model with oceanic islands, so students miss the most common AP exam context. Correct move: Automatically apply island biogeography to any isolated patch of habitat surrounded by inhospitable altered habitat.
• Wrong move: Mis-categorizing supporting ecosystem services as regulating services. Why: Many services have overlapping descriptions, leading to misclassification. Correct move: If the service is a direct benefit to humans, it is not supporting; supporting services are only those required to produce all other services.
• Wrong move: Stating that climate change is the number one threat to global biodiversity. Why: Students overemphasize climate change from news coverage and misremember the ranking. Correct move: Memorize that habitat destruction and fragmentation is the leading cause of current global biodiversity loss.
• Wrong move: Adding $n$ instead of calculating $n(n-1)$ for each species in Simpson's Index. Why: Students confuse species richness (summing counts) with the Simpson's calculation step. Correct move: Follow the formula step-by-step for every species, do not skip the $n(n-1)$ multiplication.

7. Practice Questions (AP Biology Style)

Question 1 (Multiple Choice)

Researchers sampled three amphibian communities in the Pacific Northwest. The data is shown below:

• Community 1: 5 species, 100 total individuals, 20 individuals per species
• Community 2: 4 species, 100 total individuals, counts: 70, 10, 10, 10
• Community 3: 6 species, 100 total individuals, counts: 85, 3, 3, 3, 3, 3

Which of the following correctly ranks the communities from highest to lowest Simpson's Biodiversity Index? A. 1 > 2 > 3 B. 3 > 1 > 2 C. 2 > 1 > 3 D. 3 > 2 > 1

Worked Solution: First, we can use the formula to approximate D for each community. For Community 1: $\sum n(n-1)=5\ast (20\ast 19)=1900$, N(N-1) = 10099 = 9900, so D = 1 - 1900/9900 ≈ 0.81. For Community 2: $\sum n(n-1) = 7069 + 3*(109) = 5100$,soD\approx 1-5100/9900\approx \mathrm{0.48.}ForCommunity3:$\sum n(n-1) = 8584 + 5*(3*2) = 7170$, so D ≈ 1 - 7170/9900 ≈ 0.275. Ranking from highest to lowest D gives 1 > 2 > 3. The correct answer is A.

Question 2 (Free Response)

A timber company harvests trees in a large old-growth forest, leaving three separate forest stands: Stand X (150 ha, connected to intact old-growth forest), Stand Y (150 ha, 20 km from intact old-growth forest), Stand Z (20 ha, connected to intact old-growth forest). (a) Predict the relative species richness of the three stands, using the theory of island biogeography. Justify your prediction. (b) Explain why smaller stands of forest have higher extinction rates than larger stands of forest. (c) A local conservation group argues that protecting the intact forest preserves valuable ecosystem services. Identify one provisioning service and one regulating service that an intact forest provides.

Worked Solution: (a) According to island biogeography, larger islands and islands closer to the source population (intact old-growth forest) have higher equilibrium species richness. Stand X and Y are the same size (150 ha), which is much larger than Stand Z (20 ha), so X and Y have higher richness than Z. X is closer to the source population than Y, so X has higher richness than Y. Final ranking: $\text{X}>\text{Y}>\text{Z}$. (b) Smaller forest stands support smaller total population sizes of each native species. Small populations are more vulnerable to genetic drift, inbreeding depression, and local extinction from random disturbances such as drought or disease outbreaks. Smaller stands also have less niche and resource diversity, so they cannot support specialist species that require specific conditions, further increasing extinction rates. (c) Provisioning service: Intact forests provide timber for construction and edible forest products for human use. Regulating service: Intact forests absorb carbon dioxide from the atmosphere, regulating global climate, and reduce soil erosion by holding soil with tree roots.

Question 3 (Application / Real-World Style)

A city wants to restore an abandoned industrial lot to native prairie to support pollinator biodiversity for local urban agriculture. Two restoration designs are proposed for the 10 ha site:

• Design 1: 5 native plant species, 200 individuals per species
• Design 2: 8 native plant species, 800 individuals of one dominant non-native grass, 25 individuals of each of 7 native species

Calculate Simpson's Biodiversity Index for each design, and recommend which design the city should select to meet its goal of maximum pollinator biodiversity.

Worked Solution: First calculate total N for each design: $N_1=5\ast 200=1000$, $N_2=800+(7\ast 25)=975$. For Design 1: $\sum n(n-1)=5\ast (200\ast 199)=199000$, $N(N-1)=1000\ast 999=999000$, so $D_1=1-(199000/999000)\approx 0.80$. For Design 2: $\sum n(n-1)=(800\ast 799)+7\ast (25\ast 24)=639200+4200=643400$, $N(N-1)=975\ast 974=949650$, so $D_2=1-(643400/949650)\approx 0.32$. Design 1 has a much higher Simpson's Index, so the city should select Design 1. The high evenness of native plant species in Design 1 will support a more diverse community of pollinators, meeting the city's goal.

8. Quick Reference Cheatsheet

9. What's Next

Biodiversity is the foundational concept for understanding how human impacts alter global ecosystems, and it sets up the study of disruptions to ecosystems and conservation biology, the remaining core topics in Unit 8 Ecology. Without a solid understanding of how to quantify biodiversity, how island biogeography shapes species richness, and how biodiversity supports ecosystem services, you will not be able to correctly analyze conservation strategies or predict the impacts of human disturbance on ecological communities. Biodiversity also connects backward to core AP Biology concepts: it is the product of millions of years of evolution, and genetic diversity within populations (a core component of biodiversity) is required for natural selection to occur when environments change.

• Disruptions to Ecosystems
• Conservation Biology
• Community Ecology