Population Dynamics

Population Growth

Part 1 of 7 — Population & Community Ecology

A population is a group of individuals of the same species living in the same area and capable of interbreeding. Population ecology asks a deceptively simple question: how and why does the number of individuals change over time? The answer underlies everything from fisheries management to conservation to the spread of invasive species.

Three properties describe a population at any instant:

• Population size (N) — the total number of individuals.
• Population density — the number of individuals per unit area or volume (for example, 250 oak trees per hectare).
• Dispersion — the spatial pattern in which individuals are arranged.

Three dispersion patterns

Clumped is by far the most common pattern in nature, because resources themselves are usually patchy.

The Per-Capita Growth Rate

Populations change through four processes: births (B), deaths (D), immigration, and emigration. Ignoring migration (a closed population), the change in population size over a time interval is the difference between births and deaths. Written as a continuous rate, the overall growth rate of the population is:

$\frac{dN}{dt} = B - D$

It is more useful to express births and deaths per individual. Define the per-capita birth rate (b) and per-capita death rate (d) — the number of births or deaths per individual per unit time. Then total births $B = bN$ and total deaths $D = dN$, so:

$\frac{dN}{dt} = bN - dN = (b - d)N$

The quantity $(b - d)$ is the per-capita growth rate, symbolized $r$:

$r = b - d$

This gives the foundational equation of population growth:

$\frac{dN}{dt} = rN$

Critical distinction — growth RATE vs. per-capita rate. The per-capita rate $r$ is a property of each individual (units: per individual per unit time, e.g. $0.05\,\text{yr}^{-1}$). The population growth rate $\frac{dN}{dt}$ is the whole population's change per unit time (units: individuals per unit time). A population with a small, constant $r$ can still have a huge $\frac{dN}{dt}$ if N is large — this is the engine of exponential growth.

When $r > 0$ the population grows; when $r = 0$ it is stable (zero population growth); when $r < 0$ it declines.

Checkpoint — Density, Dispersion & Per-Capita Rates

Exponential Growth — The J-Shaped Curve

Imagine a population in an idealized environment: unlimited food, unlimited space, no predators, no disease, no competition. Under these conditions every individual reproduces at its biological maximum, so $r$ takes its largest possible value, the intrinsic rate of increase, $r_{max}$. The model becomes:

$\frac{dN}{dt} = r_{max}N$

This is exponential growth. Because the growth rate $\frac{dN}{dt}$ is proportional to N, the larger the population gets, the faster it adds individuals. Plotting N against time produces the characteristic J-shaped curve: it starts shallow, then sweeps steeply upward.

Worked example — computing the growth rate

A bacterial culture has $r_{max} = 1.5\,\text{hr}^{-1}$. Find $\frac{dN}{dt}$ at two population sizes.

At N = 100 cells: $\frac{dN}{dt} = r_{max}N = (1.5)(100) = 150$ cells per hour.

At N = 1000 cells: $\frac{dN}{dt} = (1.5)(1000) = 1500$ cells per hour.

Notice: $r_{max}$ did not change, but the population is now adding cells ten times faster simply because N is ten times larger. The per-capita rate is constant; the population rate accelerates. That accelerating slope IS the J-curve.

Why exponential growth is rarely sustained

True exponential growth occurs only in bursts: a colonizing species entering a new habitat, bacteria in fresh medium, or a population rebounding after a crash. It cannot continue indefinitely because no real environment offers unlimited resources. As N rises, food dwindles, waste accumulates, and predators and pathogens find easy targets. These pressures pull the realized $r$ below $r_{max}$, bending the J-curve toward the logistic model you will meet in Part 2.

A famous illustration: a single E. coli dividing every 20 minutes would, in 36 hours of unchecked exponential growth, produce a bacterial mass greater than the planet. Reality intervenes long before that — which is exactly why $r_{max}$ is a ceiling, not a sustained rate.

Doubling Time — How Fast Is "Fast"?

A vivid way to feel the power of exponential growth is the doubling time: how long it takes a population to double in size. Under exponential growth the doubling time depends only on the per-capita rate $r_{max}$, not on the current population size — a population of 100 and a population of 100,000 with the same $r_{max}$ take the same time to double. A higher $r_{max}$ means a shorter doubling time.

Worked example — recognizing exponential growth in data

A biologist counts yeast cells in a flask of fresh medium every 2 hours:

Step 1 — Look at the change, not just the totals. The number of cells added in each 2-hour interval keeps growing (5, 10, 20, 40), so $\frac{dN}{dt}$ is rising as N rises.

Step 2 — Check the doubling. The population doubles every 2 hours no matter whether it starts at 5,000 or 40,000 cells. A constant doubling time is the signature of a constant per-capita rate, so this is exponential growth: $\frac{dN}{dt} = r_{max}N$.

Step 3 — Predict. If the pattern continued, the flask would hold 160,000 cells at 10 hours. In a real flask, sugar runs out and waste builds up, so the counts will soon fall below this prediction and the curve will start to level off.

Why doubling time is independent of N: because $\frac{dN}{dt} = r_{max}N$ scales with N, a larger population adds more individuals per unit time, but it also needs more individuals to double. The two effects cancel, leaving the doubling time a function of $r_{max}$ alone. This is a hallmark of true exponential growth — and another reminder that it cannot persist once resources become limiting.

Checkpoint — Exponential Growth

Carrying Capacity

Part 2 of 7 — Population & Community Ecology

In Part 1 you saw that exponential growth cannot last: real environments impose limits. The carrying capacity, symbolized $K$, is the maximum population size that a particular environment can sustain indefinitely, given its available resources (food, water, space, nesting sites). As a population approaches $K$, resources per individual shrink, birth rates fall, death rates rise, and growth slows.

The logistic growth model captures this self-limiting behavior:

$\frac{dN}{dt} = r_{max}N\frac{(K-N)}{K}$

Compare it to the exponential model $\frac{dN}{dt} = r_{max}N$. The logistic model simply multiplies the exponential rate by an extra term, $\frac{(K-N)}{K}$, the fraction of unused carrying capacity (an "environmental resistance" term). This braking factor is the entire difference between the two models.

Plotting N versus time produces an S-shaped (sigmoidal) curve: slow start, a rapid middle phase, then a plateau at $K$.

Reading the Braking Term

The behavior of logistic growth lives entirely in the factor $\frac{(K-N)}{K}$. Trace what happens as N climbs from small to large:

The single most tested fact in this unit: the population GROWTH RATE $\frac{dN}{dt}$ is maximal when $N = K/2$, not when N is largest. Students who assume "bigger population = faster growth" miss this. At $N = K/2$ the product of "many reproducing individuals" and "plenty of remaining resources" is balanced for the greatest absolute growth. This $N = K/2$ point is where harvesting produces the maximum sustainable yield.

Density-dependent vs. density-independent limiting factors

The factors that enforce carrying capacity fall into two categories:

Logistic regulation toward a stable $K$ is driven by density-dependent factors, because only a density-dependent brake can grow stronger precisely when the population grows larger. Density-independent events can crash a population but do not, on their own, hold it at a particular carrying capacity.

Worked Examples — Logistic Computations

A deer population grows logistically with $r_{max} = 0.5\,\text{yr}^{-1}$ and carrying capacity $K = 1000$ deer. Compute $\frac{dN}{dt}$ at four population sizes.

Case 1 — N = 100 (small relative to K):

$\frac{dN}{dt} = (0.5)(100)\frac{(1000-100)}{1000} = (0.5)(100)(0.9) = 45$ deer per year.

The braking term is 0.9, so growth is 90% of the exponential value $(0.5)(100) = 50$. At low density the population behaves almost exponentially.

Case 2 — N = 500 (exactly K/2):

$\frac{dN}{dt} = (0.5)(500)\frac{(1000-500)}{1000} = (0.5)(500)(0.5) = 125$ deer per year.

This is the maximum growth rate of the entire trajectory. No other value of N produces a larger $\frac{dN}{dt}$.

Case 3 — N = 900 (near K):

$\frac{dN}{dt} = (0.5)(900)\frac{(1000-900)}{1000} = (0.5)(900)(0.1) = 45$ deer per year.

Even though there are far more deer than in Case 1, growth has fallen back to 45/yr because only 10% of the carrying capacity remains unused.

Case 4 — N = 1000 (at K):

$\frac{dN}{dt} = (0.5)(1000)\frac{(1000-1000)}{1000} = (0.5)(1000)(0) = 0$ deer per year.

The population is stable. Births balance deaths exactly.

The symmetry to notice: Case 1 (N = 100) and Case 3 (N = 900) both give 45 deer/yr — the growth rate is the same far below and far above K/2, and peaks in the middle at $N = K/2$. This is the hallmark of the logistic curve's S-shape.

Checkpoint — Logistic Computations

How Density-Dependent Factors Enforce K

The logistic plateau is not magic — it is the visible result of specific biological mechanisms that grow stronger as a population becomes crowded. Each one lowers the per-capita birth rate $b$, raises the per-capita death rate $d$, or both, pulling $r = b - d$ toward zero as $N \rightarrow K$.

Tracing one mechanism quantitatively

Suppose at low density a songbird has $b = 0.5$ and $d = 0.1$, giving $r = 0.4$. As the population approaches K, competition cuts the birth rate to $b = 0.2$ and disease raises the death rate to $d = 0.2$. Now $r = b - d = 0.2 - 0.2 = 0$ — zero population growth, exactly what defines carrying capacity. The population has not stopped reproducing; rather, births have fallen to match deaths because the density-dependent factors squeezed both rates together.

Why density-INDEPENDENT factors cannot set K. A flood or hard frost might kill 30% of the population whether it is sparse or crowded. Because its proportional impact does not change with density, it cannot create the negative feedback needed to home in on a specific equilibrium. Only a factor whose strength rises with crowding can balance births against deaths precisely at K — which is why the logistic plateau is fundamentally a density-dependent phenomenon.

Exponential vs. Logistic — Side by Side

Real populations and overshoot

Real populations rarely glide smoothly to a flat plateau. Many overshoot $K$ — when a population grows so fast that reproduction continues past the point where resources can support it — and then crash back down, oscillating around K. Reindeer introduced to St. Paul Island famously exploded to roughly 2,000, overshot the lichen supply, then collapsed to a few dozen. The logistic model is an idealization; the key insight it captures is that density-dependent feedback opposes growth more strongly the closer N gets to K.

Carrying capacity is not fixed. $K$ can rise or fall with climate, resource availability, and species interactions. A drought lowers K for a herbivore; a wet, productive year raises it. Treat K as the environment's current ceiling, not a permanent constant.

Checkpoint — Interpreting the Logistic Curve

Energy and Reproductive Strategies

Part 3 of 7 — Population & Community Ecology

Organisms use energy to organize, grow, reproduce, and maintain homeostasis. Because the energy an organism can acquire is limited, every species faces trade-offs: energy spent on one thing (growth, maintenance, many offspring) cannot also be spent on another (a few large, well-provisioned offspring, survival to reproduce again). An organism's life history — when it grows, when and how often it reproduces, and how much it invests in each offspring — reflects how natural selection has shaped those trade-offs in its environment.

Many textbooks sort species into "r-selected" and "K-selected" groups and classify survival graphs as Type I, II, or III. Those labels are not part of the AP Biology framework; the exam tests the energy reasoning below, and any survival graph on the exam comes with the information you need to read it.

The Energy Budget

The AP framework describes what happens when an organism's energy intake and use are out of balance:

Body temperature strategies change the budget.

• Endotherms (birds, mammals) use thermal energy generated by their own metabolism to keep a stable body temperature. This costs a lot of energy, so an endotherm needs much more food than an ectotherm of the same mass.
• Ectotherms (most fish, amphibians, reptiles, invertebrates) lack efficient internal mechanisms for holding body temperature. They regulate it behaviorally, by moving into sun or shade or by clustering together, and they spend far less energy on heat.

Size matters too. Smaller animals have a larger surface area-to-volume ratio, so they lose heat faster and typically have a higher metabolic rate per gram of body mass than larger animals.

Checkpoint — Energy Budgets

Reproductive Strategies Respond to Energy Availability

The framework states that different organisms use various reproductive strategies in response to energy availability, and that some organisms alternate between asexual and sexual reproduction depending on how much energy is available.

Why switch to sexual reproduction when times get hard? Asexual reproduction is fast and cheap but makes genetically identical offspring. Sexual reproduction makes genetically varied offspring, which raises the chance that some will survive a changing or worsening environment.

Reading Survival Data

AP questions often give survival data for a group of individuals born at the same time (a cohort). Read the data, not a label:

• Where is mortality concentrated? From age 0 to 1, 900 of 1,000 die (90%). After that, losses are much smaller. Most deaths happen early in life.
• Connect to energy: heavy early mortality is typical of species that produce many offspring and invest little in each; species that invest heavily in a few offspring usually show high survival until old age.
• Watch the axis: survival graphs are often drawn with a logarithmic y-axis. On a log scale, a straight line means a constant percentage dies in each interval, not a constant number.

Checkpoint — Strategies and Data

Community Ecology

Part 4 of 7 — Population & Community Ecology

A community is an assemblage of all the populations of different species that live and interact in a particular area. While population ecology tracks one species' numbers, community ecology asks how species affect one another. These interspecific interactions are classified by their effect on the per-capita growth of each participant, scored with a sign:

• (+) the interaction benefits that species (raises its $r$ — more food, protection, etc.).
• (−) the interaction harms that species (lowers its $r$ — death, lost resources).
• (0) the interaction has no significant effect on that species.

Reading interactions as pairs of signs is the fastest way to classify them on the AP exam. The sign convention summarizes whether each partner gains, loses, or is unaffected.

The Interspecific Interaction Sign Table

Sign-convention trap: predation, herbivory, and parasitism all share the +/− signature — they differ in mechanism, not in their effect signs. Do not assume +/− automatically means "predator eats prey"; a parasite or herbivore also fits +/−. Conversely, mutualism (+/+) and commensalism (+/0) are easy to confuse: in commensalism one partner gains while the other gets nothing (a true 0), whereas in mutualism both gain.

The interactions that involve one partner consuming another (predation, herbivory, parasitism) are sometimes grouped as exploitation: +/− interactions in which one organism's gain is the other's loss.

Competition, Niches, and Coexistence

The ecological niche

A species' ecological niche is the sum total of how it uses the biotic and abiotic resources of its environment — its "occupation," not just its "address" (which is the habitat). Ecologists distinguish two versions:

• Fundamental niche — the full range of conditions and resources a species could use in the absence of competitors.
• Realized niche — the portion of that range the species actually occupies once competitors and other limiting interactions are present. The realized niche is usually smaller than the fundamental niche.

The competitive exclusion principle

Two species cannot coexist indefinitely on exactly the same limiting resource — if their niches overlap completely, the better competitor drives the other locally extinct. This is the competitive exclusion principle (Gause). Coexistence requires that the niches differ in some way.

Resource partitioning and character displacement

Species often coexist by resource partitioning — dividing a contested resource so their realized niches diverge. Classic example: MacArthur's warblers, which feed in different vertical zones of the same spruce trees, partitioning the canopy rather than competing head-to-head. Over evolutionary time, competition can drive character displacement, in which competing species evolve differences (e.g., beak sizes in Galápagos finches) that reduce niche overlap.

Niche reasoning worked through: Suppose two barnacle species both could settle across an entire intertidal zone (overlapping fundamental niches). In practice the competitively dominant species monopolizes the lower zone, restricting the weaker species to the upper zone — its realized niche. Remove the dominant competitor, and the weaker species expands downward toward its fundamental niche. The shrinkage from fundamental to realized niche is the signature of competition (−/−) acting in the community.

Checkpoint — Interactions & Niches

Keystone Species and Trophic Cascades

Not every species affects a community equally. A keystone species has an impact on community structure far out of proportion to its abundance. Like the keystone of an arch, removing it causes the whole structure to collapse, often triggering a sharp drop in diversity.

• Pisaster sea stars (Paine's classic experiment): this predatory sea star preferentially eats mussels. When researchers removed Pisaster from rocky intertidal plots, mussels outcompeted everything else and species richness fell from about 15 species to roughly 8. The sea star's predation had been keeping a dominant competitor in check.
• Sea otters control sea urchins; without otters, urchins overgraze and destroy kelp forests.

Trophic cascades

A trophic cascade is the chain of indirect effects that ripples down through trophic levels when a top consumer is added or removed. The reintroduction of gray wolves to Yellowstone is the canonical case:

Notice the pattern: the predator's effect "cascades" downward to organisms it never directly touches. A keystone predator can thus increase plant abundance two links away by suppressing herbivores — an indirect, top-down effect.

Distinguish keystone from dominant. A dominant species exerts its influence through sheer biomass or abundance (e.g., the most common tree in a forest). A keystone species is influential despite being relatively rare — its role, not its numbers, makes it pivotal.

Checkpoint — Keystones & Cascades

Biodiversity

Part 5 of 7 — Population & Community Ecology

Biodiversity describes the variety of life in a community. At the community level it has two distinct components, and a strong AP answer keeps them separate:

• Species richness — the number of different species present.
• Species evenness — how equally individuals are distributed among those species (their relative abundances).

These are independent. Two communities can have identical richness yet very different diversity:

Both have 4 species (equal richness), but Community A is far more diverse because no single species dominates. A useful diversity index must reward both high richness and high evenness — which is exactly what Simpson's index does.

Simpson's Diversity Index

Simpson's Diversity Index combines richness and evenness into one number. One common form is:

$D = 1 - \sum (p_i)^2$

where $p_i$ is the proportion of the community made up by species $i$ (that is, species $i$'s count divided by the total number of individuals), and the sum $\sum$ runs over all species. The term $\sum (p_i)^2$ is the probability that two randomly drawn individuals belong to the same species; subtracting from 1 gives the probability they belong to different species. Higher $D$ means greater diversity, on a scale from 0 (one species only) up toward 1.

Worked computation

Consider a meadow sample of 100 insects:

Step 1 — find each $p_i$: divide each count by 100 (column 3).

Step 2 — square each proportion (column 4).

Step 3 — sum the squares: $\sum (p_i)^2 = 0.25 + 0.09 + 0.0225 + 0.0025 = 0.365$.

Step 4 — subtract from 1: $D = 1 - 0.365 = 0.635$.

So $D \approx 0.64$. Interpretation: there is about a 64% chance that two randomly chosen insects are different species.

Evenness check: if instead all four species had 25 individuals each, every $p_i = 0.25$, so $\sum (p_i)^2 = 4 \times (0.25)^2 = 4 \times 0.0625 = 0.25$, giving $D = 1 - 0.25 = 0.75$. Same richness (4 species), but higher evenness pushes $D$ from 0.64 up to 0.75 — confirming that the index rewards evenness, not just species count.

Checkpoint — Diversity & Indices

Diversity Within a Species: Genetic Diversity

Biodiversity is not only the number of species. It also includes genetic diversity within each species, and the AP framework treats it as a key factor in how populations respond to change:

• The ability of a population to respond to changes in the environment is influenced by its genetic diversity. Species and populations with little genetic diversity are at risk of decline or extinction.
• Genetically diverse populations are more resilient to environmental perturbation, because they are more likely to contain individuals that can withstand the new pressure.
• An allele that is adaptive in one environment may be harmful in another, because selective pressures differ.

Connection: genetic drift and bottlenecks (Natural Selection unit) are the usual reasons small populations lose genetic diversity.

Diversity, Stability, and Resilience

Why does biodiversity matter beyond cataloging species? Greater diversity tends to confer:

• Stability — resistance to change. A diverse community is buffered because different species respond differently to perturbations; if one declines, others can fill its functional role.
• Resilience — the capacity to recover after a disturbance.
• Productivity — diverse communities often use resources more completely (complementary niches), supporting higher overall productivity.

The mechanism is sometimes called the insurance or redundancy effect: when several species perform similar ecological functions, the loss or failure of one is compensated by the others, so ecosystem processes (pollination, decomposition, nutrient cycling) continue. A community dominated by a single species is far more vulnerable — a pest or pathogen specific to that species can devastate the whole system.

Common AP misconception to avoid: high biodiversity does not make a community immune to disturbance. Diverse systems can still be damaged by fire, drought, or invasive species. The accurate claim is statistical and relative: greater diversity tends to make a community more resistant and more able to recover than an otherwise similar low-diversity community — not invulnerable. State the relationship as a tendency, not a guarantee.

Checkpoint — Genetic Diversity & Stability

Problem-Solving Workshop

Part 6 of 7 — Population & Community Ecology

The AP Biology exam rewards students who can compute, not just define. This workshop walks through three fully worked quantitative problem types that appear repeatedly: mark-recapture population estimation, logistic growth prediction, and diversity index calculation. Work each problem yourself before reading the solution, then check the reasoning at every step.

The three formulas you will use:

Problem 1 — Mark-Recapture (Lincoln-Petersen)

You cannot line up and count every fish in a lake, so ecologists estimate population size by capturing, marking, releasing, and recapturing. The Lincoln-Petersen estimator is:

$N = \frac{MC}{R}$

where:

• $M$ = number of individuals marked in the first capture (then released),
• $C$ = total number captured in the second sample,
• $R$ = number in the second sample that are recaptures (already marked),
• $N$ = estimated total population size.

The logic: the proportion of marked individuals in the second sample $\frac{R}{C}$ should equal the proportion of marked individuals in the whole population $\frac{M}{N}$. Solving $\frac{R}{C} = \frac{M}{N}$ for $N$ gives $N = \frac{MC}{R}$.

Worked solution

A researcher captures, tags, and releases 60 trout ($M = 60$). A week later she captures 80 trout ($C = 80$), of which 24 are tagged ($R = 24$). Estimate the population.

$N = \frac{MC}{R} = \frac{(60)(80)}{24} = \frac{4800}{24} = 200 \text{ trout}$

The lake holds an estimated 200 trout.

Sanity check and assumptions. The recapture fraction is $\frac{24}{80} = 0.30$, so marked fish are 30% of the sample; if 60 marked fish are 30% of the population, the population is $\frac{60}{0.30} = 200$ — consistent. This estimate assumes the marked individuals mixed randomly back into the population, no births/deaths/migration occurred between samples, and marks were not lost. A common error is dividing by $C$ instead of $R$ — remember $R$ (recaptures) sits in the denominator.

Problem 2 — Logistic Growth Prediction

A wildlife biologist models a reintroduced bison herd with logistic growth: $r_{max} = 0.3\,\text{yr}^{-1}$ and carrying capacity $K = 600$ bison. The current population is $N = 200$.

Part (a): What is the current growth rate $\frac{dN}{dt}$?

$\frac{dN}{dt} = r_{max}N\frac{(K-N)}{K} = (0.3)(200)\frac{(600-200)}{600}$

$= (0.3)(200)\frac{400}{600} = (0.3)(200)(0.6667) = 40 \text{ bison per year}$

Part (b): At what population size will the herd grow fastest, and what is that maximum rate?

Logistic growth is fastest at $N = K/2 = 600/2 = 300$ bison. At that point:

$\frac{dN}{dt} = (0.3)(300)\frac{(600-300)}{600} = (0.3)(300)(0.5) = 45 \text{ bison per year}$

So the herd's growth peaks at 45 bison/year when N = 300. Note that the current rate (40/yr at N = 200) is slightly below this peak — the herd has not yet reached its fastest-growing size.

Part (c): What happens to $\frac{dN}{dt}$ as $N \rightarrow 600$?

As N approaches K = 600, the term $\frac{(K-N)}{K} \rightarrow 0$, so $\frac{dN}{dt} \rightarrow 0$. For example at $N = 580$:

$\frac{dN}{dt} = (0.3)(580)\frac{(600-580)}{600} = (0.3)(580)(0.0333) \approx 5.8 \text{ bison per year}$

Growth has nearly stopped — the herd is leveling off at carrying capacity, the plateau of the S-curve.

Pattern recap: the growth rate climbs from N small, peaks at $N = K/2$ (300 here), then falls back toward 0 as N approaches K (600). The maximum growth rate is not at the maximum population size — a point the exam tests relentlessly.

Problem 3 — Diversity Index

A biologist surveys a 1 $m^{2}$ quadrat of tide-pool invertebrates and records the counts below. Compute Simpson's Diversity Index, $D = 1 - \sum (p_i)^2$.

Step 1 — proportions: divide each count by the total of 100 to get each $p_i$.

Step 2 — square each $p_i$ (rightmost column).

Step 3 — sum the squares: $\sum (p_i)^2 = 0.16 + 0.16 + 0.0225 + 0.0025 = 0.345$.

Step 4 — subtract from 1: $D = 1 - 0.345 = 0.655$.

The tide pool's diversity is $D \approx 0.66$. About 66% of the time, two randomly chosen invertebrates will be different species.

Comparison practice: if a nearby disturbed pool held 90 mussels, 5 barnacles, 3 snails, and 2 anemones, then $\sum (p_i)^2 = 0.81 + 0.0025 + 0.0009 + 0.0004 \approx 0.814$ and $D \approx 0.19$. Same four species (equal richness) but heavy dominance by mussels collapses the diversity from 0.66 to 0.19 — a vivid reminder that evenness, not just richness, drives the index.

Problem 4 — A Combined, Multi-Step Problem

Real AP free-response items often chain methods together. Here is one that uses mark-recapture to find N, then logistic growth to predict the trajectory.

Setup. A conservation team studies a reintroduced tortoise population on an island with estimated carrying capacity $K = 500$ and $r_{max} = 0.25\,\text{yr}^{-1}$. To find the current size, they tag 40 tortoises ($M = 40$), release them, and later capture 50 tortoises ($C = 50$), of which 8 are tagged ($R = 8$).

Part (a): Estimate the current population.

$N = \frac{MC}{R} = \frac{(40)(50)}{8} = \frac{2000}{8} = 250 \text{ tortoises}$

Part (b): Compute the current growth rate. With $N = 250$ and $K = 500$, note that $N = K/2$ exactly:

$\frac{dN}{dt} = (0.25)(250)\frac{(500-250)}{500} = (0.25)(250)(0.5) = 31.25 \text{ tortoises per year}$

Because $N = K/2$, this is the maximum growth rate the population will ever reach.

Part (c): Predict what happens next. Since the population sits right at $N = K/2$, any further increase moves it past the inflection point, so the growth rate will begin to slow even as the population keeps rising — the curve bends from its steepest tangent toward the plateau at $K = 500$. Management can expect the fastest absolute gains now and diminishing gains as N climbs toward 500.

Why chaining is testable: part (a)'s answer (250) is the input to part (b), and recognizing that 250 = K/2 is what makes part (c)'s prediction precise. A single arithmetic slip in the mark-recapture step would cascade — which is why writing each formula and checking units at every stage protects your score.

Workshop Problem Set

Workshop Problem Set — Assumptions & Combined Methods

AP Review

Part 7 of 7 — Population & Community Ecology

This final part synthesizes the unit and drills the specific traps that cost students points on the AP exam. Use the master summary, then the trap table, then the application questions to confirm you can apply each idea under exam conditions.

Master equation summary

The AP Trap Table

Each row is a place students reliably lose points. Memorize the correction.

Putting It Together — A Synthesis Walkthrough

Consider a single invasive beetle introduced to a new forest, and trace the unit end to end.

1. Colonization (Part 1). Resources are unlimited and N is tiny, so the population grows nearly exponentially: $\frac{dN}{dt} \approx r_{max}N$, a J-curve. With abundant food, each beetle takes in more energy than it uses, and that net energy gain goes into growth and high reproductive output (Part 3): many offspring, little parental care, and heavy early mortality.

2. Approaching limits (Part 2). As N rises, density-dependent factors kick in — competition for host trees, predators and parasites that find the now-abundant beetle, disease spreading through the dense population. Growth bends from a J toward an S: logistic, $\frac{dN}{dt} = r_{max}N\frac{(K-N)}{K}$, fastest at $N = K/2$, leveling at K.

3. Community effects (Part 4). The beetle is an herbivore (+/−) on native trees. If it suppresses a previously dominant tree, it may act almost like a keystone agent, shifting which species coexist and triggering a small trophic cascade. Native insects competing for the same trees may be driven out by competitive exclusion, or persist via resource partitioning.

4. Biodiversity outcome (Part 5). If the beetle homogenizes the forest by favoring one tree species, evenness drops and Simpson's $D$ falls. A less diverse forest is also less resilient: a later drought or pathogen could cause much greater losses than in the original, more diverse forest.

5. Quantifying it (Part 6). A manager estimates beetle numbers by mark-recapture ($N = \frac{MC}{R}$) and uses the logistic model to predict whether the population is below, at, or past $K/2$ — which determines whether control efforts should expect the fastest growth (near K/2) ahead.

This single scenario touches every concept in the unit, which is exactly how AP free-response questions are built: one ecological story, many linked principles.

AP-Style Application Questions

Free-Response Strategy for Ecology Prompts

AP ecology free-response questions usually combine a graph or data table with several short tasks: describe a trend, calculate a value, justify a claim, and predict an outcome. A reliable approach:

1. Identify the model first. Is the curve J-shaped (exponential) or S-shaped (logistic)? Naming the model tells the reader which equation governs your reasoning and frames every later part.
2. Show the formula, then the numbers. For any calculation, write the equation ($\frac{dN}{dt} = r_{max}N\frac{(K-N)}{K}$, $N = \frac{MC}{R}$, or $D = 1 - \sum (p_i)^2$), substitute, and report units. Graders award points for the setup even if arithmetic slips.
3. Justify with a mechanism, not a restatement. "Growth slows near K because density-dependent factors such as competition and disease intensify as crowding increases" earns the point; "growth slows because N is high" does not.
4. Match the scale of your claim to the data. Do not invoke ecosystem-level conclusions from a single-population graph, and keep per-capita rate ($r$) distinct from population rate ($\frac{dN}{dt}$).

Reading a logistic graph at a glance

One-line reminders for the exam: maximum growth at $N = K/2$ (not at K); read survival graphs carefully (log axes); +/− covers predation, herbivory, AND parasitism; diversity needs richness AND evenness; K is not a fixed constant.

Final Synthesis Checkpoint