Open with the Gros Michel banana collapse and the looming threat to today's Cavendish. Nearly every banana sold in the U.S. is a genetically identical clone. In the 1950s the dominant Gros Michel cultivar was wiped out globally by Panama disease (Fusarium wilt Tropical Race 1); the industry replaced it with the Cavendish, also a clone, which is now being destroyed by TR4. Ask students: 'You just ate a banana this week that is genetically identical to every other banana in the store. Why should that scare an ecologist?' Take 2–3 quick student responses, then pivot: today we define biodiversity at three scales and learn to measure it. Targets SP 1 Concept Explanation and SP 7 Environmental Solutions.

1. 7m

   Three nested scales of biodiversity

   Content

   Biodiversity is not one number — it exists at three nested scales that fit inside one another. Genetic diversity is the variety of alleles in a population's gene pool (for example, the ~1,000+ potato varieties Andean farmers cultivate versus the handful of clones in Irish fields in 1845). Species diversity is the number and relative abundance of species in a community (a coral reef with 500 fish species versus a parking-lot lot with 3 weed species). Ecosystem or habitat diversity is the variety of habitat types across a landscape (a watershed with wetlands, riparian forest, meadow, and rocky outcrops versus a monoculture cornfield). The scales are nested: destroy habitat diversity and you lose species; drive species toward small isolated populations and you lose genetic diversity. All three matter because each scale contributes independently to resilience.

   Delivery

   Emphasize the nesting — the slide shows genetic inside species inside ecosystem as concentric rings. Ask: 'If two national parks each have 40 mammal species, are they equally biodiverse?' Push students to name what else could differ (habitat types, genetic variation within species). Pre-empt the misconception that biodiversity = species count by asking, 'A cornfield in Iowa and a cornfield in Nebraska planted with the same hybrid — what's the genetic diversity between the two fields?' (Essentially zero.) Targets SP 2 Visual Representations.

2. 7m

   Richness vs. evenness — why counting species isn't enough

   Content

   Species diversity has two components. Species richness is simply how many species are present. Species evenness describes how equally individuals are distributed among those species. Two communities can have identical richness but very different diversity. Consider Community A with 100 individuals split 25/25/25/25 across four species, and Community B with 100 individuals split 97/1/1/1. Both have richness = 4, but A is highly even and B is dominated by one species. Ecologists treat A as more diverse because losing any one species in B (other than the dominant) has almost no functional effect, while the community is essentially a monoculture ecologically. This is why Simpson's index — which incorporates proportions — gives A a much higher score than B.

   Delivery

   The slide shows two pie charts with identical richness but opposite evenness. Ask: 'Which pie looks more diverse to your eye? Why?' Then formalize: evenness is the missing variable when someone quotes only species counts. Head off the misconception that identical richness means identical diversity by explicitly walking through the 25/25/25/25 vs 97/1/1/1 case. Note that a lawn with dandelions is technically 'richness = 5' but functionally a grass monoculture. Targets SP 2 and SP 5.

3. 8m

   Simpson's diversity index — the calculation

   Content

   Simpson's diversity index quantifies both richness and evenness in a single number. We use the form D = 1 − Σ(pᵢ)², where pᵢ = (number of individuals of species i) ÷ (total individuals). D ranges from 0 (no diversity) to nearly 1 (very high diversity). Worked example — Community A (25/25/25/25, total 100): each pᵢ = 0.25, so Σ(pᵢ)² = 4 × (0.25)² = 4 × 0.0625 = 0.25, and D = 1 − 0.25 = 0.75. Community B (97/1/1/1, total 100): Σ(pᵢ)² = (0.97)² + 3 × (0.01)² = 0.9409 + 0.0003 = 0.9412, so D = 1 − 0.9412 = 0.0588. Community A (D ≈ 0.75) is about 13× more diverse by this index than Community B (D ≈ 0.06), despite identical richness. Units: none — D is dimensionless.

   Delivery

   Do the calculation live, one step per line, calling on students for each pᵢ. The slide shows the formula with substitutions. Emphasize: (1) proportions must sum to 1, (2) square each proportion BEFORE summing, (3) subtract from 1 at the end. Common student error: forgetting to square, or squaring the sum instead of summing the squares. Ask students to predict what D approaches as one species comes to dominate (→ 0) and as species become perfectly even with many species (→ 1). Targets SP 6 Mathematical Routines.

1. 25m

   Bead Quadrat Lab: Calculating & Comparing Simpson's DLab

   Setup (before class): Prepare bead cups for each pair. Quadrat 1 (rich but uneven, mimics disturbed community): 60 red, 15 blue, 12 yellow, 8 green, 5 white (total 100). Quadrat 2 (rich and even, mimics mature community): 22 red, 21 blue, 20 yellow, 19 green, 18 white (total 100). Both have richness = 5. Students should NOT know the counts in advance. Run: Pairs sort each cup by color, tally counts, compute D for each, then answer the analysis questions. Circulate to check that students are squaring proportions BEFORE summing. Targets SP 5 Data Analysis and SP 6 Mathematical Routines and SP 7 Environmental Solutions. Student handout — Bead Quadrat Lab Scenario: You are an ecologist sampling two forest quadrats. Each bead color represents a tree species. You will determine which quadrat is more biodiverse. Part 1 — Data collection - Pour Quadrat 1 onto your tray. Sort by color. Record counts in the table. - Repeat for Quadrat 2. Quadrat 1 counts: - Red (Species R): ______ - Blue (Species B): ______ - Yellow (Species Y): ______ - Green (Species G): ______ - White (Species W): ______ - Total N₁ = ______ Quadrat 2 counts: - Red: ______ - Blue: ______ - Yellow: ______ - Green: ______ - White: ______ - Total N₂ = ______ Part 2 — Calculate Simpson's D for each quadrat Formula: D = 1 − Σ(pᵢ)², where pᵢ = nᵢ / N Quadrat 1: - p_R = ______, p_B = ______, p_Y = ______, p_G = ______, p_W = ______ - Σ(pᵢ)² = ______ - D₁ = ______ Quadrat 2: - p_R = ______, p_B = ______, p_Y = ______, p_G = ______, p_W = ______ - Σ(pᵢ)² = ______ - D₂ = ______ Square each proportion before you sum. Do not square the sum. Part 3 — Analysis (answer in complete sentences) 1. Both quadrats contain 5 species. Explain, using your D values, why they are not equally diverse. Reference richness and evenness explicitly. 2. A bark beetle arrives that attacks only Species R (red). Predict the percent of individuals lost in each quadrat and describe the likely effect on community structure in each case. 3. Which quadrat is more resilient to the beetle outbreak, and why? Argue from your data. 4. Propose one land-management action that would raise the Simpson's D of Quadrat 1 over 20 years. Justify with reference to evenness.

   Materials

   • Two opaque cups labeled 'Quadrat 1' and 'Quadrat 2' per pair
   • Colored beads (or beans) — 5 colors representing species: red, blue, yellow, green, white
   • Small trays or paper plates for sorting
   • Calculators
   • Student handout (printed from description below)

   Example outputs

   • Quadrat 1: pᵢ = 0.60, 0.15, 0.12, 0.08, 0.05; Σ(pᵢ)² = 0.36 + 0.0225 + 0.0144 + 0.0064 + 0.0025 = 0.4058; D₁ = 1 − 0.4058 ≈ 0.59
   • Quadrat 2: pᵢ = 0.22, 0.21, 0.20, 0.19, 0.18; Σ(pᵢ)² = 0.0484 + 0.0441 + 0.0400 + 0.0361 + 0.0324 = 0.2010; D₂ = 1 − 0.201 ≈ 0.80
   • Analysis Q2 exemplar: Quadrat 1 loses 60% of individuals because red dominates; Quadrat 2 loses only 22%. Quadrat 1's canopy would collapse; Quadrat 2 retains structure because remaining species can fill the niche.
   • Analysis Q4 exemplar: Selectively thin red-species (Species R) individuals and replant with underrepresented species over time to raise evenness; D rises even though richness stays at 5.

   No-equipment fallback

   Provide the two count sets on paper (Quadrat 1: 60/15/12/8/5; Quadrat 2: 22/21/20/19/18) and have students complete Parts 2 and 3 as a paper calculation exercise.

1. Two forest plots each contain 4 tree species. Plot X has 40, 40, 40, 40 individuals. Plot Y has 150, 5, 3, 2 individuals. Which statement is best supported? A) X and Y have equal diversity because richness is equal. B) Y has higher diversity because it has more individuals. C) X has higher diversity because it has greater evenness. D) Diversity cannot be compared without knowing the species identities.

   multiple choiceC. Richness is identical (4 species each), so the difference is evenness. Plot X is perfectly even (each pᵢ = 0.25), while Plot Y is dominated by one species (p ≈ 0.94). Simpson's D confirms: D_X = 0.75, D_Y ≈ 0.12. Targets SP 5.

2. A pond community contains: 80 mosquitofish, 15 sunfish, 5 bass (total N = 100). Calculate Simpson's diversity index D = 1 − Σ(pᵢ)². Show work and report D to two decimal places.

   calculationpᵢ = 0.80, 0.15, 0.05 Σ(pᵢ)² = (0.80)² + (0.15)² + (0.05)² = 0.6400 + 0.0225 + 0.0025 = 0.6650 D = 1 − 0.6650 = 0.34 (or 0.335) Targets SP 6.

3. The Cavendish banana is grown worldwide as a genetic clone. Explain how low genetic diversity within this cultivated 'population' increases its extinction risk when a novel pathogen (Fusarium TR4) appears. Reference the gene pool and connect to the concept of a monoculture.

   short answerBecause every Cavendish plant shares essentially the same genome, the gene pool contains no alleles conferring resistance to TR4 — if the pathogen defeats one plant's defenses, it defeats all of them. A monoculture eliminates the variation that natural selection normally acts on, so no resistant individuals survive to repopulate. Contrast with a wild banana population containing hundreds of genotypes, where some individuals would likely carry resistance alleles and persist. Targets SP 1 and SP 7.

• ☐Colored beads or beans in 5 colors (approximately 100 of each color per class)
• ☐Opaque cups (2 per pair) pre-loaded with the Quadrat 1 and Quadrat 2 counts
• ☐Small sorting trays or paper plates (2 per pair)
• ☐Calculators (or student phones with calculator)
• ☐Printed student handout (one per student)
• ☐Projector for the auto-generated slide deck