Open with a real-world provocation: in 1859, Thomas Austin released 24 European rabbits (Oryctolagus cuniculus) onto his estate in Victoria, Australia, for sport hunting. Within 50 years there were an estimated 600+ million rabbits stripping vegetation across the continent. Ask students: 'If those 24 rabbits reproduced at a constant 30% annual growth rate, would that be enough to explain 600 million by 1910?' Take a quick show-of-hands prediction (yes / no / not sure), then tell students: 'By the end of this class you will be able to answer that with a calculator and a shortcut called the rule of 70.' Do NOT solve it yet — hold the tension. This hook targets SP 7 (framing an environmental problem) and previews SP 6 (mathematical routines).

1. 6m

   The growth rate equation — it's not just births

   Content

   Population growth rate (r) is the net percent change in a population per unit time. The full equation is r = [(births + immigration) − (deaths + emigration)] ÷ population × 100. All four terms matter: a country can have a high birth rate but shrink if emigration and deaths outpace births (see Bulgaria, r ≈ −0.7%). Worked example: a town of 10,000 has 200 births, 100 deaths, 50 immigrants, 30 emigrants in one year. Net change = (200 + 50) − (100 + 30) = 120. r = 120 ÷ 10,000 × 100 = 1.2% per year. Note this is a PERCENT — the whole-number form (1.2), which we will need for the rule of 70.

   Delivery

   Emphasize that students routinely forget immigration and emigration and treat 'growth rate' as birth rate minus death rate only — this is the #4 misconception on the standard list. Ask a cold-call: 'If births = deaths, can a population still grow?' (Yes — via net immigration.) Have students annotate the equation in their notes with the four inputs. Confirm units: r is in percent per year, NOT a decimal. Head off the decimal-vs-percent trap now because it will bite them in the rule of 70 beat.

2. 6m

   Exponential growth is COMPOUNDING, not adding

   Content

   A constant percent growth rate does NOT mean the same number of individuals added each year — it means each year's increase is a percent of the NEW, larger total. That's compounding, and it's why exponential curves bend upward into a J-curve. Worked example: start with 1,000 organisms at r = 10%/yr. Year 1: +100 → 1,100. Year 2: +110 → 1,210. Year 3: +121 → 1,331. Year 10: ~2,594. The absolute additions grow every year even though the percent is fixed. Contrast this with LINEAR growth (adding a constant 100/yr), which would only reach 2,000 by year 10. The formula N_t = N₀(1 + r)^t (with r as a decimal here) captures the compounding.

   Delivery

   This is misconception #2 — students think 5%/yr means 'add the same 5 each year.' Walk the 1,000 → 1,100 → 1,210 sequence out loud and ask students to predict year 4 before you reveal it (1,464). The J-curve on the slide shows the visual signature: shallow at first, then rocketing upward. Point out that the curve LOOKS like it has a lag phase early on, but it doesn't — that apparent flatness is just small absolute numbers before compounding takes off. This distinction matters for the S-curve beat next.

3. 5m

   Logistic growth: when resources push back

   Content

   In the real world, resources (food, water, space, nesting sites) are finite. As population density rises, per-capita resources fall, death rates climb, and birth rates drop — the growth rate itself decreases. The population traces an S-curve: a slow lag phase, a near-exponential middle, and a plateau at carrying capacity (K). Mathematically, dN/dt = rN(1 − N/K): when N is small, (1 − N/K) ≈ 1 and growth is nearly exponential; when N approaches K, (1 − N/K) → 0 and growth halts. Classic real example: reindeer introduced to St. Paul Island, Alaska in 1911 (26 individuals) grew to ~2,000 by 1938, then crashed as lichen was depleted — an overshoot of K, not a clean S.

   Delivery

   This is misconception #1 — students expect J-curves to keep climbing even when resources are capped. Anchor the difference: J-curve = unlimited resources (theoretical or short-term); S-curve = real-world with limits. Ask: 'Which curve better describes bacteria in a fresh petri dish in the first 6 hours? After 24 hours?' (Exponential early; logistic once nutrients deplete.) Preview that overshoot-and-crash is a variation on logistic they'll see in unit 3.7–3.8.

4. 6m

   The rule of 70 — doubling time made easy

   Content

   For any quantity growing exponentially at r percent per period, the doubling time ≈ 70 ÷ r, where r is the WHOLE-number percent. Derivation shortcut: solving 2 = (1 + r)^t gives t = ln(2)/ln(1+r) ≈ 0.693/r (with r as a decimal), and multiplying numerator and denominator by 100 gives 69.3/(r as percent) ≈ 70/r%. Worked examples: (a) r = 2%/yr → doubling time ≈ 70/2 = 35 years. (b) r = 7%/yr → 10 years. (c) r = 0.5%/yr → 140 years. Trap: if a student plugs in 0.02 instead of 2, they get 70/0.02 = 3,500 years — off by a factor of 100. Rabbit callback: at r = 30%, doubling time ≈ 70/30 ≈ 2.33 years. Starting from 24 rabbits, doubling every ~2.3 years for 50 years gives ~21 doublings → 24 × 2²¹ ≈ 50 million (and real growth rates were even higher).

   Delivery

   Drill the units. Say it three times: 'Use the whole-number percent — 2, not 0.02.' This is misconception #3 and the single most common AP FRQ error on this topic. Have students silently compute the r = 7% case, then thumbs-up/thumbs-down check. Then release the rabbit hook: work through 24 × 2²¹ on the calculator and let them react.

1. 25m

   Yeast Population Lab + Rule-of-70 AnalysisLab

   Students obtain cell-count data from six yeast cultures at increasing incubation times, graph the data on linear axes, identify whether the curve is J or S, calculate r and doubling time from the exponential phase, and use the rule of 70 to predict when the population would double again — and whether the prediction holds. Targets SP 4 (scientific experiments), SP 5 (data analysis), SP 6 (mathematical routines), and SP 2 (visual representations). Setup: Six flasks are pre-labeled 0h, 2h, 4h, 8h, 12h, 24h. Groups of 3–4 rotate through microscope stations. Each group loads a hemocytometer with 10 µL of culture, counts yeast cells in the four corner squares (each 1 mm × 1 mm × 0.1 mm = 0.1 µL), averages, and multiplies by 10⁴ to get cells/mL. If hemocytometers are limited, provide the pre-counted values in the fallback data table below and have all groups analyze them. Student handout: Part 1 — Data collection (or use provided table) Count yeast cells in the four large corner squares of the hemocytometer. Average the four counts, then multiply by 10⁴ to get cells per mL. Cells/mL = (average count per square) × 10⁴ - 0 h: ______ cells/mL - 2 h: ______ cells/mL - 4 h: ______ cells/mL - 8 h: ______ cells/mL - 12 h: ______ cells/mL - 24 h: ______ cells/mL Part 2 — Graph Plot time (x-axis, hours) vs. cell density (y-axis, cells/mL) on linear axes. 1. Is the shape a J-curve or an S-curve? ______ 2. Identify and label on your graph: the lag phase, the exponential (log) phase, and the stationary phase (if present). 3. Estimate carrying capacity K from the plateau: K ≈ ______ cells/mL Part 3 — Calculate the growth rate during the exponential phase Using your two exponential-phase data points (typically 2 h and 8 h), compute the per-hour percent growth rate. - N at time 1 = ______ cells/mL (t₁ = ____ h) - N at time 2 = ______ cells/mL (t₂ = ____ h) - Number of doublings between them = log₂(N₂ / N₁) = ______ - Time per doubling (observed) = (t₂ − t₁) ÷ doublings = ______ hours Part 4 — Apply the rule of 70 From your observed doubling time, back-calculate r: r (% per hour) ≈ 70 ÷ (observed doubling time in hours) = ______ %/hr Remember: r is a whole-number percent, not a decimal. Using this r, predict the population 3 hours after your last exponential-phase point: Predicted N = ______ cells/mL Part 5 — Evaluate (SP 4, SP 7) 1. Compare your predicted N (Part 4) to the actual value at 12 h or 24 h. Is the prediction accurate? Why or why not? ______ 2. Name TWO resources in the flask that likely became limiting between the exponential and stationary phases: ______ 3. If you transferred 1 mL of the 24-h culture into a fresh 100 mL flask of sugar-water, sketch what the new growth curve would look like on the back of the handout. Justify the shape in one sentence. Walk around and check: (1) students are using the whole-number percent in the rule of 70, (2) they identify the S-curve shape, (3) they can name at least one limiting resource (glucose, dissolved O₂, space, ethanol toxicity buildup).

   Materials

   • Pre-prepared yeast cultures (Saccharomyces cerevisiae) in sugar-water broth, started 0h, 2h, 4h, 8h, 12h, and 24h before class (one flask per time point, labeled)
   • Hemocytometers (or pre-counted data table if hemocytometers unavailable)
   • Compound microscopes (400× total magnification)
   • Micropipettes or transfer pipettes
   • Slides and coverslips
   • Calculators
   • Graph paper (linear and semi-log, or graphing software)
   • Student handout (below)

   Example outputs

   • Graph shows classic S-curve: near-flat 0–2 h (lag), steep rise 2–12 h (exponential), leveling 12–24 h (stationary near K ≈ 7.5 × 10⁶ cells/mL). Observed doubling time ≈ 2 hr in the exponential phase; r ≈ 70/2 = 35%/hr; predicted N at 11 h from an 8-h value of 1.6 × 10⁶ ≈ 4.5 × 10⁶ — matches the 12-h data reasonably well. Limiting resources named: glucose depletion and ethanol accumulation.
   • Student who used r = 0.35 instead of 35 in the rule of 70 gets doubling time = 70/0.35 = 200 hr — flagged during walk-around as the classic decimal error and corrected. Corrected value 2 hr matches observation.
   • Extension sketch: fresh flask transfer curve shows a new lag phase followed by another S-curve leveling at similar K — justified by 'resources reset but the carrying capacity of the flask is the same.'

   No-equipment fallback

   Provide the pre-counted data table so students skip Parts 1 and go straight to graphing and calculation: - 0 h: 1.0 × 10⁵ cells/mL - 2 h: 2.0 × 10⁵ cells/mL - 4 h: 4.1 × 10⁵ cells/mL - 8 h: 1.6 × 10⁶ cells/mL - 12 h: 4.8 × 10⁶ cells/mL - 24 h: 7.5 × 10⁶ cells/mL (near K) Expected observed doubling time in exponential phase ≈ 2 h, so r ≈ 70/2 = 35%/hr.

1. A country of 50,000,000 people records in one year: 900,000 births, 400,000 deaths, 150,000 immigrants, and 50,000 emigrants. (a) Calculate the percent annual growth rate. (b) Using the rule of 70, calculate the doubling time. (c) Assuming exponential growth continues, what will the population be after two doubling times? Targets SP 6 (Mathematical Routines).

   calculation(a) Net change = (900,000 + 150,000) − (400,000 + 50,000) = 600,000. r = 600,000 ÷ 50,000,000 × 100 = 1.2%/yr. (b) Doubling time ≈ 70 ÷ 1.2 ≈ 58.3 years. (c) After 2 doubling times: 50,000,000 × 2² = 200,000,000 people.

2. A student claims: 'A population growing at 3% per year adds the same number of individuals every year.' Explain why this claim is incorrect, using a specific numerical example from a starting population of 10,000. Targets SP 1 (Concept Explanation) and SP 6.

   short answerThe claim confuses percent growth with linear (constant-number) growth. At 3%, each year's addition is 3% of the NEW total, so additions compound. Example: Year 1 adds 300 (→10,300); Year 2 adds 309 (→10,609); Year 3 adds ~318 (→10,927). The absolute number added grows each year, producing a J-curve, not a straight line.

3. Which of the following best describes the difference between a J-curve and an S-curve? A) A J-curve represents a smaller population than an S-curve. B) A J-curve occurs under unlimited resources; an S-curve levels off at carrying capacity when resources become limiting. C) A J-curve is logistic growth; an S-curve is exponential growth. D) Both curves eventually reach carrying capacity at the same time. Targets SP 2 (Visual Representations).

   multiple choiceB. A J-curve represents exponential growth under unlimited resources and keeps rising; an S-curve represents logistic growth that slows and plateaus at K as resources become limiting. A and D are unrelated to shape; C reverses the definitions.

4. A student calculating doubling time for a country with r = 2%/yr writes: doubling time ≈ 70 ÷ 0.02 = 3,500 years. Identify the error and give the correct doubling time. Targets SP 6.

   short answerError: the student used the decimal form (0.02) instead of the whole-number percent (2) in the rule of 70. Correct calculation: 70 ÷ 2 = 35 years. The rule of 70 requires r expressed as a whole-number percent.

• ☐Pre-prepared yeast (Saccharomyces cerevisiae) cultures in 5% sugar-water broth, incubated to reach 0h, 2h, 4h, 8h, 12h, and 24h time points at class time
• ☐Hemocytometers (1 per group of 3–4)
• ☐Compound microscopes with 400× capability
• ☐Micropipettes or transfer pipettes and tips
• ☐Microscope slides and coverslips
• ☐70% ethanol or disinfectant wipes for cleanup
• ☐Nitrile gloves and safety goggles
• ☐Calculators
• ☐Linear graph paper (semi-log optional)
• ☐Printed student handouts (yeast lab + rule of 70)
• ☐Whiteboard/marker for worked examples