Population Biology种群生物学
A population is a group of individuals of the same species living in the same area at the same time. This guide covers how populations are described (density, dispersion, age structure), how they grow under unlimited conditions (exponential growth, $\tfrac{dN}{dt}=rN$), how carrying capacity puts the brakes on growth (logistic model), what factors regulate population size, the r vs K life-history trade-off, human population growth and the demographic transition, and how conservation biology responds to biodiversity loss. Worked examples and quiz questions use real data throughout.种群（population，种群）是同一时间、同一地点、同一物种个体的集合。本指南涵盖种群的描述方式（种群密度、分布格局、年龄结构）、无限条件下的增长方式（指数增长，$\tfrac{dN}{dt}=rN$）、环境容纳量如何制约增长（逻辑斯蒂模型）、种群大小的调节因素、r 对策与 K 对策的生活史权衡，以及人口增长与人口转变理论，最后探讨保护生物学如何应对生物多样性丧失。全部例题与测验均以真实数据为据。
How to use this guide如何使用本指南
Population biology sits at the intersection of ecology and evolution: it asks how many individuals there are, why those numbers change, and what happens when they change too fast or too slow. The four curricula agree on the core qualitative scope: describing populations, exponential vs logistic growth, carrying capacity, and density-dependent regulation. They diverge on mathematical depth. US NGSS (HS-LS2-1) treats carrying capacity quantitatively but explicitly does not require deriving growth equations. Ontario SBI4U Strand F F2.2 requires calculating population growth using exponential, sigmoid (logistic), and sinusoidal models — this is the Honors SBI4U track. Alberta Biology 30 Unit D GO3 also requires the quantitative models (gr = ΔN/Δt, logistic S-curve, r/K strategies) — flagged Honors Bio 30. BC Life Sciences 11 treats population change at the conceptual level under microevolution. The table below locates each section in your curriculum.种群生物学处于生态学与进化论的交汇点：它追问种群数量的多少、变化的原因，以及变化过快或过慢时会发生什么。四套大纲在定性核心范围上高度一致：描述种群、指数增长与逻辑斯蒂增长、环境容纳量、密度制约调节。分歧主要在数学深度上。US NGSS（HS-LS2-1）定量处理环境容纳量，但明确不要求推导增长方程。安大略 SBI4U Strand F F2.2 要求使用指数、S 型（逻辑斯蒂）和正弦模型计算种群增长——这是 荣誉 SBI4U 轨道。阿尔伯塔 Biology 30 Unit D GO3 同样要求定量模型（gr = ΔN/Δt、逻辑斯蒂 S 型曲线、r/K 对策）——标注 荣誉 Bio 30。BC Life Sciences 11 在微进化框架下以概念层面处理种群变化。下表定位各节在你大纲中的位置。
| If you are in…如果你在… | Focus on these sections重点学习 | Defer / lighter可推迟 / 减负 | Source依据 |
|---|---|---|---|
| 🇺🇸 US NGSS HS Life Sciences美国 NGSS 生命科学 | §1 (population metrics), §3 (carrying capacity — HS-LS2-1), §4 (regulation, ecosystem stability — HS-LS2-6), §7 (biodiversity loss — HS-LS2-2)§1（种群指标）、§3（环境容纳量—HS-LS2-1）、§4（调节、生态系统稳定性—HS-LS2-6）、§7（生物多样性丧失—HS-LS2-2） | Deriving growth equations: NGSS Assessment Boundary for HS-LS2-1 explicitly excludes "deriving mathematical equations." Read §2 and §3 math for conceptual understanding推导增长方程：NGSS HS-LS2-1 评估边界明确排除"推导数学方程"，阅读 §2、§3 数学内容以获得概念性理解即可 | NGSS HS Life Science — HS-LS2-1, HS-LS2-2, HS-LS2-6 PEs and Assessment Boundaries— HS-LS2-1、HS-LS2-2、HS-LS2-6 表现期望及评估边界 |
| 🇨🇦 ON Grade 12 — SBI4U Honors安大略 12 年级 — SBI4U 荣誉 | All 7 sections fully, including mathematical growth models (§2 and §3 — F2.2 exponential, sigmoid, sinusoidal), population characteristics (§1 — F3.2), regulation (§4 — F3.3), and r vs K life-history (§5)全部 7 节完整学习，含数学增长模型（§2 和 §3—F2.2 指数、S 型、正弦）、种群特征（§1—F3.2）、调节（§4—F3.3）和 r/K 生活史（§5） | Nothing — SBI4U Strand F Population Dynamics is directly assessed无 — SBI4U F 单元种群动态直接列入考核 | Ontario SBI3U/4U Biology — SBI4U Strand F: F2.2, F3.2, F3.3, F3.4, F3.5— SBI4U F 单元：F2.2、F3.2、F3.3、F3.4、F3.5 |
| 🇨🇦 BC Life Sciences 11BC Life Sciences 11 | §1, §4, §5, §7 are core: Big Idea 2 ("Evolution occurs at the population level") and Content bullet "microevolution: change within a species … population genetics; natural selection"§1、§4、§5、§7 为核心：大概念 2（"进化发生在种群层面"）及内容条目"微进化：种群内变化……种群遗传学；自然选择" | Quantitative growth models (§2 and §3 math): BC Life Sciences 11 treats population change conceptually; the logistic equation is not required定量增长模型（§2 和 §3 数学）：BC Life Sciences 11 以概念层面处理种群变化，不要求逻辑斯蒂方程 | BC Life Sciences 11 / Anatomy 12 — Life Sciences 11 Big Idea 2 + microevolution Content bullet— Life Sciences 11 大概念 2 + 微进化内容条目 |
| 🇨🇦 AB Biology 20 & 30阿尔伯塔 Biology 20 & 30 | Bio 20 Unit B GO1 (§1 vocabulary) and GO2 (§4, §5 framing). Bio 30 Unit D GO3 (§2, §3 quantitative models; §5 r/K) is Grade-12 and requires full logistic and exponential growth calculations Honors Bio 30Bio 20 Unit B GO1（§1 词汇）和 GO2（§4、§5 框架）。Bio 30 Unit D GO3（§2、§3 定量模型；§5 r/K 对策）为 12 年级内容，要求完整的逻辑斯蒂和指数增长计算 荣誉 Bio 30 | Hardy-Weinberg (§1 note): Bio 30 D1.3k requires quantitative HW calculations; this guide covers only conceptual population genetics哈温方程（§1 注）：Bio 30 D1.3k 要求定量 HW 计算；本指南仅涵盖概念性种群遗传学 | Alberta Biology 20/30 — Biology 20 Unit B GO1–GO2; Biology 30 Unit D GO3— Biology 20 Unit B GO1–GO2；Biology 30 Unit D GO3 |
Once you have located your row, use the two cards below for the approach that fits your timeline.找到所在行后，用下面两张卡片选择适合你时间安排的方式。
Know the three population characteristics (density, dispersion, age structure); the shape difference between a J-curve (exponential) and an S-curve (logistic); what carrying capacity ($K$) means; and four density-dependent factors (food, predation, disease, competition). Read every cram-cheat box. Skip the going-deeper sections on demographic transition math and the logistic differential equation derivation.掌握三种种群特征（种群密度、分布格局、年龄结构）；J 型曲线（指数增长）与 S 型曲线（逻辑斯蒂增长）的形状差异；环境容纳量（$K$）的含义；以及四种密度制约因素（食物、捕食、疾病、竞争）。读每个速记框，跳过人口转变数学和逻辑斯蒂微分方程推导的深入内容。
Be precise about the logistic growth equation $\tfrac{dN}{dt} = rN\!\left(1-\tfrac{N}{K}\right)$ and what each term means. Explain why growth rate peaks at $N = K/2$. Distinguish density-dependent from density-independent factors with real examples. For SBI4U/Bio 30, be ready to calculate $\Delta N$, growth rate $gr = \Delta N / \Delta t$, and per-capita growth rate. Explain why a positive population growth rate does not mean the population is healthy (see §6 and §7).精准掌握逻辑斯蒂增长方程 $\tfrac{dN}{dt} = rN\!\left(1-\tfrac{N}{K}\right)$ 及各项的含义。解释为何增长率在 $N = K/2$ 时达到峰值。用实例区分密度制约与非密度制约因素。SBI4U/Bio 30 轨道需能计算 $\Delta N$、增长率 $gr = \Delta N / \Delta t$ 和人均增长率。解释为何正的种群增长率并不意味着种群健康（见 §6 和 §7）。
Population Characteristics: Density, Dispersion, and Age Structure种群特征：种群密度、分布格局与年龄结构
- Population density:种群密度： number of individuals per unit area or volume. $D_p = N/A$. High density increases competition and disease transmission; low density risks extinction (Allee effect).单位面积或体积内的个体数目。$D_p = N/A$。高密度加剧竞争与疾病传播；低密度有灭绝风险（阿利效应）。
- Population dispersion:分布格局：
- Clumped — individuals cluster near resources or social groups (e.g. wolves in a pack, trees near water). Most common in nature.集群分布 — 个体聚集于资源或社会群体附近（如狼群、水边树木）。自然界中最常见。
- Uniform — individuals are evenly spaced due to competition or territoriality (e.g. nesting penguins, creosote bushes).均匀分布 — 因竞争或领地行为个体均匀分布（如筑巢企鹅、木馏油灌木）。
- Random — positions are unpredictable; resources are uniformly available and individuals do not interact (e.g. dandelion seeds blown by wind). Rarest in nature.随机分布 — 位置无法预测；资源均匀分布且个体间无互动（如风吹散的蒲公英种子）。自然界中最罕见。
- Age structure:年龄结构： proportion of individuals in pre-reproductive, reproductive, and post-reproductive age classes. A broad base (many young) predicts future growth; a narrow base predicts decline. Tracked via age-structure pyramids.处于繁殖前、繁殖中和繁殖后年龄组的个体比例。宽基底（幼年个体多）预示未来增长；窄基底预示种群衰退。通过年龄结构金字塔追踪。
A 4 km² forest contains 120 white-tailed deer. (a) Calculate population density. (b) A survey finds 60 fawns (pre-reproductive), 40 adults (reproductive), and 20 elderly deer (post-reproductive). Predict whether this population will grow, stay stable, or decline.一片 4 km² 的森林中有 120 头白尾鹿。(a) 计算种群密度。(b) 调查发现 60 头幼鹿（繁殖前期）、40 头成年鹿（繁殖期）和 20 头老年鹿（繁殖后期）。预测该种群将增长、稳定还是衰退。
(a) $$ D_p = \frac{N}{A} = \frac{120}{4} = 30 \text{ deer/km}^2 $$
(b) Growing population.(b) 增长型种群。 The pre-reproductive class (60 = 50%) outnumbers both reproductive (40 = 33%) and post-reproductive (20 = 17%) classes. The large cohort of fawns will enter the reproductive class in future years, increasing the birth rate. This broad-base pyramid is characteristic of a growing population.繁殖前期个体（60 只，占 50%）多于繁殖期（40 只，33%）和繁殖后期（20 只，17%）。大量幼鹿将在未来几年进入繁殖期，提高出生率。这种宽基底金字塔是增长型种群的典型特征。
Going deeper — mark-recapture method for estimating population size (Lincoln-Petersen)深入 — 标记重捕法估算种群大小（林肯-彼得森法）
When direct counting is impractical, ecologists use mark-recapture. Catch a sample ($n_1$), mark and release. Later catch a second sample ($n_2$); count how many are marked ($m$). The Lincoln-Petersen estimate is: $$ N \approx \frac{n_1 \times n_2}{m} $$ Assumptions: the population is closed (no births/deaths/migration between samples), marks do not affect survival, and marks do not wash off. SBI4U F2 and Biology 30 D3 both expect students to apply this formula and state its assumptions.当直接计数不可行时，生态学家使用标记重捕法。捕获一个样本（$n_1$），标记后放回。之后再捕第二个样本（$n_2$），统计其中有标记的个体数（$m$）。林肯-彼得森估算公式为：$$ N \approx \frac{n_1 \times n_2}{m} $$ 假设条件：两次取样之间种群封闭（无出生/死亡/迁移），标记不影响生存，标记不会脱落。SBI4U F2 和 Biology 30 D3 均要求学生运用该公式并陈述其假设条件。
Exponential Growth: The J-Curve指数增长：J 型曲线
- Biotic potential ($r$): the maximum per-capita rate of increase a population can achieve under ideal conditions. $r = \text{birth rate} - \text{death rate}$ (per individual per unit time).种群增长潜力（$r$）：在理想条件下种群能达到的最大人均增长率。$r = \text{birth rate} - \text{death rate}$（出生率减去死亡率，每个体每单位时间）。
- Exponential growth equation: $\dfrac{dN}{dt} = rN$. Each individual produces $r$ offspring per unit time, so growth rate accelerates as $N$ increases. The graph is a J-shaped curve.指数增长方程：$\dfrac{dN}{dt} = rN$。每个个体每单位时间产生 $r$ 个后代，因此增长率随 $N$ 增大而加速。图形呈 J 型曲线。
- Discrete-time form (for calculations): $\Delta N = r \cdot N \cdot \Delta t$, so $N_{t} = N_0 e^{rt}$. A population with $r = 0$ is stable; $r > 0$ grows; $r < 0$ declines.离散时间形式（用于计算）：$\Delta N = r \cdot N \cdot \Delta t$，即 $N_{t} = N_0 e^{rt}$。$r = 0$ 时种群稳定；$r > 0$ 时增长；$r < 0$ 时衰退。
- Why it can't last: no real population has unlimited resources. Exponential growth is only a starting model; logistic growth (§3) adds the realistic brake.为何无法持续：现实中没有资源无限的种群。指数增长只是起点模型；逻辑斯蒂增长（§3）加入了现实的制约。
$N$ = population size $r$ = per-capita rate of increase $t$ = time. Growth rate $\Delta N / \Delta t$ increases with $N$ — a hallmark of exponential growth.$N$ = 种群大小 $r$ = 人均增长率 $t$ = 时间。增长速率 $\Delta N / \Delta t$ 随 $N$ 增大而增加 — 这是指数增长的标志。
A bacterial colony starts with $N_0 = 500$ cells. The per-capita growth rate is $r = 0.4$ per hour. (a) Calculate the growth rate $\Delta N / \Delta t$ at this moment. (b) Predict $N$ after 1 hour using $\Delta N = rN\Delta t$.一个细菌菌落初始 $N_0 = 500$ 个细胞，人均增长率 $r = 0.4$/小时。(a) 计算此时的增长率 $\Delta N / \Delta t$。(b) 使用 $\Delta N = rN\Delta t$ 预测 1 小时后的 $N$。
(a) $$ \frac{dN}{dt} = rN = 0.4 \times 500 = 200 \text{ cells/hour} $$
(b) $$ \Delta N = r \cdot N \cdot \Delta t = 0.4 \times 500 \times 1 = 200 $$ $$ N_1 = 500 + 200 = 700 \text{ cells} $$
Note: if $r$ stays constant and $N$ keeps growing, $\Delta N / \Delta t$ grows too — this is the accelerating characteristic of the J-curve.注意：若 $r$ 保持不变而 $N$ 持续增长，$\Delta N / \Delta t$ 也会增大 — 这正是 J 型曲线加速特征的体现。
Going deeper — doubling time and the rule of 70深入 — 倍增时间与 70 法则
For a continuously growing population, the doubling time $t_d \approx \ln 2 / r \approx 0.693 / r$. A quick approximation is the rule of 70: $t_d \approx 70 / (\% \text{growth rate})$. For example, a human population growing at 2% per year would double in approximately 35 years. This relationship is used in both human demography (§6) and microbiology (bacterial doubling time). Alberta Biology 30 D3 and Ontario SBI4U F2 both require students to interpret growth rates quantitatively.对于连续增长的种群，倍增时间 $t_d \approx \ln 2 / r \approx 0.693 / r$。快速近似法为 70 法则：$t_d \approx 70 / (\% \text{growth rate})$（% 增长率）。例如，年增长率为 2% 的人口大约 35 年翻倍。这一关系既用于人口统计学（§6），也用于微生物学（细菌倍增时间）。阿尔伯塔 Biology 30 D3 和安大略 SBI4U F2 均要求学生对增长率进行定量解读。
Logistic Growth and Carrying Capacity ($K$)逻辑斯蒂增长与环境容纳量（$K$）
- Carrying capacity ($K$):环境容纳量（$K$）： the maximum population size an environment can sustain indefinitely, given available food, water, space, and other resources. $K$ is not fixed — it changes when the environment changes.在可用食物、水、空间等资源条件下，环境能无限期维持的最大种群数量。$K$ 不是固定的 — 环境改变时 $K$ 也会变化。
- Logistic growth equation:逻辑斯蒂增长方程： $$ \frac{dN}{dt} = rN\!\left(1 - \frac{N}{K}\right) $$ The term $(1 - N/K)$ is the environmental resistance. When $N \ll K$, this term $\approx 1$ and growth is nearly exponential. When $N \to K$, this term $\to 0$ and growth stops. When $N = K$, $dN/dt = 0$.$(1 - N/K)$ 项为环境阻力。当 $N \ll K$ 时，该项 $\approx 1$，增长近似指数增长。当 $N \to K$ 时，该项 $\to 0$，增长停止。当 $N = K$ 时，$dN/dt = 0$。
- Maximum growth rate occurs at $N = K/2$.最大增长率出现在 $N = K/2$ 时。 This is the inflection point of the S-curve. Fisheries managers aim to keep populations at $K/2$ for maximum sustainable yield.这是 S 型曲线的拐点。渔业管理者的目标是将种群维持在 $K/2$ 以获得最大可持续产量。
- J-curve vs S-curve:J 型曲线 vs S 型曲线： exponential growth produces a J-curve (accelerating, no ceiling). Logistic growth produces an S-curve (sigmoid): slow growth at low $N$, rapid growth near $K/2$, then slowing to zero at $K$.指数增长产生 J 型曲线（加速，无上限）。逻辑斯蒂增长产生 S 型曲线（乙状曲线）：低 $N$ 时增长缓慢，接近 $K/2$ 时增长最快，然后趋近 $K$ 时增长为零。
$N$ = current population size $K$ = carrying capacity $r$ = biotic potential. Maximum growth rate at $N = K/2$.$N$ = 当前种群大小 $K$ = 环境容纳量 $r$ = 种群增长潜力。最大增长率出现在 $N = K/2$ 时。
A deer population has $r = 0.5$ per year and $K = 1000$. Calculate $dN/dt$ when (a) $N = 100$ and (b) $N = 500$. Which gives a higher growth rate?一个鹿群 $r = 0.5$/年，$K = 1000$。计算 (a) $N = 100$ 和 (b) $N = 500$ 时的 $dN/dt$。哪种情况增长率更高？
(a) $N = 100$: $$ \frac{dN}{dt} = 0.5 \times 100 \times \left(1 - \frac{100}{1000}\right) = 50 \times 0.9 = 45 \text{ deer/year} $$
(b) $N = 500$ ($= K/2$): $$ \frac{dN}{dt} = 0.5 \times 500 \times \left(1 - \frac{500}{1000}\right) = 250 \times 0.5 = 125 \text{ deer/year} $$
Conclusion:结论： $N = K/2 = 500$ gives the higher growth rate (125 vs 45 deer/year). This confirms that the maximum instantaneous growth rate occurs at half the carrying capacity.$N = K/2 = 500$ 时增长率更高（125 vs 45 只/年）。这证实了最大瞬时增长率出现在环境容纳量一半时。
Going deeper — what determines $K$? Changing carrying capacity with NGSS HS-LS2-1深入 — 什么决定 $K$？环境容纳量的变化（NGSS HS-LS2-1）
Carrying capacity is determined by the interplay of food availability, water, shelter/space, light (for autotrophs), and waste-disposal capacity. NGSS HS-LS2-1 emphasises that $K$ varies across scales: local habitat patch, watershed, and continent-scale biome. Human activities can raise $K$ (irrigation, fertilisers) or lower it (deforestation, pollution, climate change). A population overshoot — where $N$ temporarily exceeds $K$ — is followed by a die-off as resources are depleted. This overshoot-crash pattern is documented in reindeer introduced to isolated islands and in classic predator-prey cycles.环境容纳量由食物可用性、水、庇护所/空间、光照（对自养生物而言）以及废物处理能力的相互作用决定。NGSS HS-LS2-1 强调 $K$ 在不同尺度上各有不同：局部栖息地斑块、流域和大陆尺度生物群落。人类活动可以提高 $K$（灌溉、施肥）或降低 $K$（砍伐森林、污染、气候变化）。种群超调 — $N$ 暂时超过 $K$ — 之后随资源耗尽而发生种群崩溃。这种超调-崩溃模式在引入孤立岛屿的驯鹿种群和经典捕食者-猎物循环中均有记录。
Density-Dependent vs Density-Independent Regulation密度制约与非密度制约调节
- Density-dependent factors:密度制约因素： their effect intensifies as population density increases. They provide negative feedback that pushes $N$ back toward $K$. Key examples:
- Intraspecific competition for food, water, territory — intensifies at high density, increasing mortality and reducing fecundity.种内竞争（食物、水、领地）— 高密度时加剧，增加死亡率，降低生育率。
- Predation — predator populations often grow in response to prey abundance, increasing predation pressure at high prey density.捕食 — 捕食者种群通常随猎物丰富度增长，在猎物高密度时增加捕食压力。
- Disease and parasitism — spread more rapidly through dense populations (greater contact rate).疾病与寄生 — 在密集种群中传播更快（接触率更高）。
- Stress and emigration — crowding increases stress hormones, reducing reproductive success; some individuals emigrate.压力与迁出 — 拥挤增加应激激素，降低繁殖成功率；部分个体向外迁移。
- Density-independent factors:非密度制约因素： affect population regardless of density. Examples: temperature extremes, storms, floods, drought, volcanic eruption, fire. These can cause sudden population crashes to well below $K$. After a density-independent crash, the population typically rebounds via exponential growth if survivors remain.无论种群密度如何都会产生影响。例如：极端温度、风暴、洪水、干旱、火山喷发、火灾。这些因素可导致种群急剧崩溃至远低于 $K$ 的水平。非密度制约性崩溃后，若有幸存者，种群通常通过指数增长反弹。
| Feature特征 | Density-dependent密度制约 | Density-independent非密度制约 |
|---|---|---|
| Effect increases with density?效果随密度增加？ | Yes是 | No否 |
| Acts as negative feedback to $K$?对 $K$ 提供负反馈？ | Yes是 | No否 |
| Examples例子 | Competition, predation, disease竞争、捕食、疾病 | Drought, frost, fire, storm干旱、霜冻、火灾、风暴 |
| Typical curve effect典型曲线效果 | Smooth S-curve sigmoid平滑 S 型曲线 | Sudden crash then rebound突然崩溃后反弹 |
Life-History Strategies: r vs K Selection生活史策略：r 对策与 K 对策
- r-selected species:r 对策生物： maximise $r$, the intrinsic growth rate. Traits: short lifespan, early maturity, many small offspring, low parental investment, high mortality. Populations boom rapidly and often overshoot $K$. Common in unstable/disturbed environments. Examples: insects, mice, annual plants, bacteria.最大化内禀增长率 $r$。特征：寿命短、成熟早、后代多而小、亲本投资少、死亡率高。种群快速爆发，常超过 $K$。多见于不稳定/受扰动的环境。例如：昆虫、小鼠、一年生植物、细菌。
- K-selected species:K 对策生物： populations maintained near $K$. Traits: long lifespan, late maturity, few large offspring, high parental investment, low mortality. Stable populations close to carrying capacity. Common in stable, competitive environments. Examples: elephants, whales, humans, oak trees.种群维持在接近 $K$ 的水平。特征：寿命长、成熟晚、后代少而大、亲本投资多、死亡率低。种群稳定，接近环境容纳量。多见于稳定、竞争激烈的环境。例如：大象、鲸、人类、橡树。
- Key insight:关键认识： r vs K is a spectrum, not a binary. Most species lie somewhere in between. The concept explains why invasive species (often r-strategists) can rapidly outcompete native K-strategists, and why large K-selected animals are more vulnerable to extinction.r 对策与 K 对策是一个连续谱，而非二元对立。大多数物种介于两者之间。这一概念解释了为何入侵物种（通常为 r 对策者）能迅速超越本土 K 对策者，以及为何大型 K 对策动物更易灭绝。
| Feature特征 | r-selectedr 对策 | K-selectedK 对策 |
|---|---|---|
| Lifespan寿命 | Short短 | Long长 |
| Time to maturity成熟时间 | Early早 | Late晚 |
| Offspring number后代数量 | Many (small)多（小） | Few (large)少（大） |
| Parental care亲本照顾 | Little or none很少或无 | Extensive充分 |
| Juvenile mortality幼年死亡率 | High高 | Low低 |
| Typical population curve典型种群曲线 | J-curve (boom-bust)J 型曲线（爆发-崩溃） | S-curve (stable near $K$)S 型曲线（稳定接近 $K$） |
Human Population Growth and the Demographic Transition人口增长与人口转变
- Scale: global human population reached 1 billion around 1800, doubled to 2 billion by 1928, and reached 8 billion in 2022. Growth slowed from ~2% in the 1960s to ~0.9% today, but absolute numbers added per year remain large.规模：全球人口约在 1800 年达到 10 亿，1928 年翻倍至 20 亿，2022 年达到 80 亿。年增长率从 20 世纪 60 年代的约 2% 降至今日的约 0.9%，但每年新增的绝对人数仍然巨大。
- Demographic transition model:人口转变模型：
- Stage 1 — High birth rate, high death rate. Population stable at low level (pre-industrial).第 1 阶段 — 高出生率、高死亡率。种群稳定在低水平（前工业化）。
- Stage 2 — Death rate falls (improved sanitation, medicine). Birth rate remains high. Rapid population growth.第 2 阶段 — 死亡率下降（卫生、医疗改善）。出生率维持高位。种群快速增长。
- Stage 3 — Birth rate falls (urbanization, female education, contraception). Growth slows.第 3 阶段 — 出生率下降（城镇化、女性教育、避孕）。增长放缓。
- Stage 4 — Low birth rate, low death rate. Population stabilizes again at high level (post-industrial).第 4 阶段 — 低出生率、低死亡率。种群再次稳定在高水平（后工业化）。
- Ecological footprint: human population growth increases resource consumption and waste production, raising questions about Earth's carrying capacity for humans. SBI4U F1.1 expects students to analyse this relationship (Ontario) and NGSS HS-LS2-7 asks students to design solutions to reduce human impacts.生态足迹：人口增长增加资源消耗和废物产生，引发关于地球对人类的环境容纳量的问题。安大略 SBI4U F1.1 要求学生分析这一关系，NGSS HS-LS2-7 要求学生设计减少人类影响的方案。
Conservation Biology and Biodiversity Loss保护生物学与生物多样性丧失
- Biodiversity loss:生物多样性丧失： current extinction rate is estimated at 100–1000 times the natural background rate. NGSS HS-LS4-5 and HS-LS2-7 and SBI4U F1 all frame biodiversity loss as driven by habitat destruction, overexploitation, invasive species, pollution, and climate change (acronym: HIPCO).目前灭绝速率估计为自然背景速率的 100 至 1000 倍。NGSS HS-LS4-5、HS-LS2-7 和 SBI4U F1 均将生物多样性丧失的驱动因素概括为栖息地破坏、过度开发、入侵物种、污染和气候变化（缩写：HIPCO）。
- Minimum viable population (MVP):最小可存活种群： the smallest population size that can persist long-term despite demographic and environmental stochasticity. Very small populations suffer inbreeding depression, reduced genetic diversity, and Allee effects (cooperation breaks down below a threshold density). SBI4U F3.2 names "minimum viable size" explicitly.在人口统计和环境随机性下能长期存续的最小种群数量。极小种群会遭受近交衰退、遗传多样性降低和阿利效应（密度低于阈值时协作关系崩溃）。SBI4U F3.2 明确提及"最小可存活种群数量"。
- Conservation strategies:保护策略：
- Habitat protection and corridors — reserves reduce habitat loss; wildlife corridors connect fragments to enable gene flow and recolonization.栖息地保护与廊道 — 自然保护区减少栖息地丧失；野生动物廊道连接破碎地块，实现基因流动和再定殖。
- Captive breeding and reintroduction — used for species with very small wild populations (e.g. California condor, black-footed ferret).圈养繁殖与再引入 — 用于野外种群极小的物种（如加州神鹫、黑足雪貂）。
- Reducing human ecological footprint — sustainable resource use, reduced carbon emissions, controlling invasive species. Linked to SBI4U F1.1 and NGSS HS-LS2-7.减少人类生态足迹 — 可持续资源利用、减少碳排放、控制入侵物种。与 SBI4U F1.1 和 NGSS HS-LS2-7 相关联。
- Ecosystem stability and resilience (NGSS HS-LS2-6):生态系统稳定性与韧性（NGSS HS-LS2-6）： complex interactions in ecosystems maintain relatively consistent numbers and types of organisms in stable conditions; changing conditions (modest or extreme) may result in a new ecosystem state. Biodiversity increases resilience by providing functional redundancy.生态系统中的复杂互动在稳定条件下维持相对一致的生物数量和类型；变化条件（温和或极端）可能导致新的生态系统状态。生物多样性通过提供功能冗余来增强韧性。
Population biology tools — growth models, carrying capacity, age-structure analysis, density-dependent regulation — are the quantitative foundation of conservation biology. A wildlife manager who does not understand logistic growth cannot set a sustainable harvest quota. A conservation geneticist who does not understand minimum viable population cannot design an effective reserve. NGSS HS-LS2-1 and HS-LS2-2 frame this explicitly: mathematical representations of carrying capacity and population factors are the tools students should use to evaluate biodiversity claims.种群生物学工具 — 增长模型、环境容纳量、年龄结构分析、密度制约调节 — 是保护生物学的定量基础。不了解逻辑斯蒂增长的野生动物管理者无法制定可持续的捕获配额。不了解最小可存活种群的保护遗传学家无法设计有效的自然保护区。NGSS HS-LS2-1 和 HS-LS2-2 明确将此框架化：环境容纳量和种群因素的数学表示是学生用于评估生物多样性论断的工具。
Exam Strategy and Common Pitfalls考试策略与常见陷阱
- Identify curve type first: J = exponential, S = logistic.首先识别曲线类型：J 型 = 指数增长，S 型 = 逻辑斯蒂增长。 Then read off where $K$ is (the plateau of the S-curve), and where $K/2$ is (the inflection point — steepest slope).然后读取 $K$ 的位置（S 曲线的平台期）以及 $K/2$ 的位置（拐点——最陡斜率处）。
- Growth rate $\Delta N / \Delta t$ is NOT the same as $N$.增长率 $\Delta N / \Delta t$ 与种群大小 $N$ 不同。 A population can be large AND have a low growth rate (near $K$). Always read the question: it may ask for growth rate, not population size.种群可以很大但增长率很低（接近 $K$ 时）。始终仔细阅读题目：它可能询问的是增长率，而非种群大小。
- Show the formula, substitution, and units at every step.每一步都写出公式、代入过程和单位。 For exponential growth: write $dN/dt = rN$, substitute, calculate. For logistic: write $dN/dt = rN(1-N/K)$, substitute, calculate. Partial credit for correct formula even if arithmetic is wrong.指数增长：写出 $dN/dt = rN$，代入，计算。逻辑斯蒂：写出 $dN/dt = rN(1-N/K)$，代入，计算。即使算术有误，正确公式也可获得部分分。
- Density-dependent vs density-independent: the keyword is "regardless of density."密度制约与非密度制约：关键词是"与密度无关"。 If the effect size changes with population density, it is density-dependent. If the effect size is constant regardless of density, it is density-independent.若效果大小随种群密度变化，则为密度制约。若效果大小与密度无关保持不变，则为非密度制约。
- Memorise the r/K contrast table (§5).记住 §5 中的 r/K 对比表。 Exam questions often give a list of traits and ask "r-selected or K-selected?" Cue: many offspring + no care = r; few offspring + extensive care = K.考题常给出特征列表并询问"r 对策还是 K 对策？"提示词：大量后代 + 无照顾 = r；少量后代 + 充分照顾 = K。
- Conservation biology connection:保护生物学联结： K-selected species are more vulnerable to extinction because low $r$ means slow recovery after population decline. This directly explains why elephants, whales, and condors are endangered while cockroaches are not.K 对策物种更容易灭绝，因为低 $r$ 意味着种群衰退后恢复缓慢。这直接解释了为何大象、鲸鱼和神鹫濒危，而蟑螂却不会。
Flashcards闪卡
Practice Quiz综合测验
Readiness Checklist准备就绪清单
Tick each item when you can do it cold, without notes, on a first attempt.能在无笔记、首次尝试下完成，再勾选每一项。
- Define population density and calculate $D_p = N/A$ for a given example. Distinguish clumped, uniform, and random dispersion with one real example each. 🇺🇸 NGSS HS-LS2-2定义种群密度并计算给定例子的 $D_p = N/A$。用各一个真实例子区分集群、均匀和随机分布。🇺🇸 NGSS HS-LS2-2
- Interpret an age-structure pyramid: identify whether the population is growing, stable, or declining, and explain your reasoning. 🇨🇦 ON SBI4U F3.2解读年龄结构金字塔：判断种群是增长、稳定还是衰退，并解释推理过程。🇨🇦 ON SBI4U F3.2
- State the exponential growth equation $dN/dt = rN$ and calculate $dN/dt$ given $r$ and $N$. Explain why this produces a J-shaped curve.陈述指数增长方程 $dN/dt = rN$，并在给定 $r$ 和 $N$ 时计算 $dN/dt$。解释为何产生 J 型曲线。
- State the logistic growth equation and calculate $dN/dt$ at any given $N$ and $K$. Identify where on the S-curve growth rate is maximum ($N = K/2$) and zero ($N = K$). 🇺🇸 NGSS HS-LS2-1陈述逻辑斯蒂增长方程，并在给定 $N$ 和 $K$ 时计算 $dN/dt$。在 S 型曲线上确定增长率最大（$N = K/2$）和为零（$N = K$）的位置。🇺🇸 NGSS HS-LS2-1
- Distinguish density-dependent from density-independent factors with two examples of each. Explain why density-dependent factors act as negative feedback toward $K$. 🇨🇦 ON SBI4U F3.3各举两例区分密度制约与非密度制约因素。解释密度制约因素为何对 $K$ 起负反馈作用。🇨🇦 ON SBI4U F3.3
- Compare r-selected and K-selected species using at least four traits (offspring number, parental care, lifespan, time to maturity). Give one real-world example of each. 🇨🇦 AB Bio 30 D3.4k用至少四个特征（后代数量、亲本照顾、寿命、成熟时间）比较 r 对策与 K 对策物种。各举一个真实例子。🇨🇦 AB Bio 30 D3.4k
- Explain why K-selected species are more vulnerable to extinction than r-selected species, linking low $r$ to slow population recovery. 🇨🇦 BC Life Sciences 11 Big Idea 2解释为什么 K 对策物种比 r 对策物种更容易灭绝，将低 $r$ 与种群恢复缓慢联系起来。🇨🇦 BC Life Sciences 11 大概念 2
- Describe the four stages of the demographic transition model and explain what drives the change from Stage 1 to Stage 4. 🇨🇦 ON SBI4U F1.1描述人口转变模型的四个阶段，并解释从第 1 阶段到第 4 阶段变化的驱动因素。🇨🇦 ON SBI4U F1.1
- Explain why global human population can still increase in absolute terms even while the per-capita growth rate is declining, using $\Delta N = rN$.使用 $\Delta N = rN$ 解释为何即使人均增长率下降，全球人口的绝对数量仍可增加。
- Name the five main causes of biodiversity loss (HIPCO). Identify habitat destruction as the leading cause and link carrying capacity reduction to species decline. 🇺🇸 NGSS HS-LS2-7, HS-LS4-5说出生物多样性丧失的五大主因（HIPCO）。确认栖息地破坏为主要原因，并将环境容纳量降低与物种衰退联系起来。🇺🇸 NGSS HS-LS2-7、HS-LS4-5
- Honors SBI4U / Bio 30 Calculate $\Delta N$, growth rate $gr = \Delta N/\Delta t$, and per-capita growth rate $cgr = \Delta N/N$ from data. Apply the mark-recapture formula $N \approx n_1 n_2 / m$ and state its assumptions. 🇨🇦 ON SBI4U F2.2; AB Bio 30 D3.1k–D3.2k荣誉 SBI4U / Bio 30 从数据计算 $\Delta N$、增长率 $gr = \Delta N/\Delta t$ 和人均增长率 $cgr = \Delta N/N$。应用标记重捕公式 $N \approx n_1 n_2 / m$ 并陈述其假设条件。🇨🇦 ON SBI4U F2.2；AB Bio 30 D3.1k–D3.2k
Connections to Other Units与其他单元的联结
Population biology is the quantitative culmination of the HS Biology sequence. The growth models you learned here integrate carrying capacity (which depends on energy flow and food web structure from Unit 9 Ecology), life-history trade-offs (which are shaped by natural selection from Unit 7 Evolution), and biodiversity concerns (which connect to Unit 8 Biodiversity and Classification). The demographic transition (§6) connects to Human Anatomy and Physiology (Unit 10) through the public health and medical advances that lowered death rates.种群生物学是高中生物学序列的定量顶点。你在此学到的增长模型整合了环境容纳量（取决于第 9 单元生态学的能量流动和食物网结构）、生活史权衡（由第 7 单元进化中的自然选择塑造）和生物多样性问题（与第 8 单元生物多样性和分类相联系）。人口转变（§6）通过降低死亡率的公共卫生和医疗进步与《人体解剖与生理》（第 10 单元）相联系。
Built on these prior units.建立在以下先修单元之上。
Unit 7 (Evolution): natural selection is the mechanism that produces different life-history strategies (§5); without evolution, r/K theory has no mechanistic underpinning. Unit 9 (Ecology): carrying capacity is set by food web structure, energy transfer efficiency, and abiotic limiting factors — all Unit 9 content. Unit 8 (Biodiversity): the HIPCO drivers of extinction (§7) connect directly to the reasons biodiversity is difficult to maintain. Unit 1 (Cell Structure): at a cellular level, population growth rate ultimately depends on cell division rate — connecting the unit-12 $r$ parameter back to Unit 4 cell cycle content.第 7 单元（进化）：自然选择是产生不同生活史策略的机制（§5）；没有进化，r/K 理论就没有机制基础。第 9 单元（生态学）：环境容纳量由食物网结构、能量传递效率和非生物限制因素决定 — 都是第 9 单元内容。第 8 单元（生物多样性）：灭绝的 HIPCO 驱动因素（§7）直接与难以维持生物多样性的原因相联系。第 1 单元（细胞结构）：在细胞层面，种群增长率最终取决于细胞分裂速率 — 将第 12 单元的 $r$ 参数与第 4 单元细胞周期内容联系起来。
Feeds into AP Biology and IB Biology.衔接 AP Biology 与 IB Biology。
AP Biology Unit 8 (Ecology) requires logistic growth calculations, predator-prey dynamics, and conservation biology analysis at a deeper level. IB Biology HL Topic C (Ecology) covers carrying capacity, population dynamics, and conservation biology in detail with worked data-analysis questions. Both courses extend the quantitative population work here into community ecology and global change contexts. This guide provides the mathematical vocabulary ($r$, $K$, $dN/dt$) you will need from the first day of those courses.AP Biology 第 8 单元（生态学）要求在更深层次进行逻辑斯蒂增长计算、捕食者-猎物动态和保护生物学分析。IB Biology HL Topic C（生态学）详细涵盖环境容纳量、种群动态和保护生物学，并配有数据分析练习题。两门课程都将此处的定量种群研究延伸至群落生态学和全球变化背景。本指南提供了从这两门课程第一天起就需要的数学词汇（$r$、$K$、$dN/dt$）。