PART I · SHORT RESPONSE第一部分 · 短答题AP-style MCQ + ON/BC short answer · 23 marksAP 风格选择题 + 安/卑省考短答 · 共 23 分
Section A · Short ResponseA 部分 · 短答题
Mix of multiple-choice and short-answer items. For MCQs, circle the letter; show enough work that a marker could verify. For short-answer items, use biological terminology precisely. No calculator on Q1-Q3; calculator permitted on Q4-Q5.本节包含选择题与短答题。选择题请圈出字母答案，并写出足以让阅卷人核对的过程。短答题请准确使用生物学术语。Q1-Q3 不可使用计算器；Q4-Q5 可用计算器。
Q1EASY易🇺🇸 US美AP-style MCQAP 风格选择题§1 Population characteristics种群特征 · HS-LS2-1[3 marks][3 分]
A nature reserve contains 240 white-tailed deer distributed across an area of $60\ \text{km}^2$. Which value correctly describes the population density of deer in this reserve?某自然保护区内有 240 只白尾鹿，分布在 $60\ \text{km}^2$ 的面积内。下列哪个值正确描述了该保护区内鹿的种群密度？
A rabbit population of $400$ individuals has an intrinsic rate of increase $r = 0.5$ per year. Assuming exponential growth, what is the instantaneous growth rate $dN/dt$ at this moment?一个有 $400$ 只兔子的种群，内禀增长率 $r = 0.5$/年。假设指数增长，此时种群瞬时增长率 $dN/dt$ 是多少？
In the logistic growth model, a population of size $N$ grows according to $dN/dt = rN(1 - N/K)$, where $K$ is the carrying capacity.在逻辑斯蒂增长模型中，大小为 $N$ 的种群按 $dN/dt = rN(1 - N/K)$ 增长，其中 $K$ 为环境容纳量。
(a)State the value of $dN/dt$ when $N = K$. Explain what this means biologically.写出当 $N = K$ 时 $dN/dt$ 的值，并解释其生物学含义。[2]
(b)At what population size is $dN/dt$ maximized? State your answer in terms of $K$.在哪个种群数量时 $dN/dt$ 最大？用 $K$ 表示你的答案。[1]
(c)Sketch the shape of the logistic growth curve (S-curve) and label $K$ and the inflection point.画出逻辑斯蒂增长曲线（S 型曲线）的形状，并标注 $K$ 和拐点。[1]
Q4MEDIUM中🇨🇦 BC卑BC Provincial-style卑诗省考风格§4 Population regulation种群调节 · Biology 12[7 marks][7 分]
A coastal sea otter population crashed from $3\,000$ to $500$ individuals after an oil spill. Three years later, when the population had recovered to $1\,500$, a disease outbreak killed $30\%$ of individuals. Meanwhile, sea otters in a neighbouring undisturbed bay maintained a stable population near their regional carrying capacity.一个沿海海獭种群在石油泄漏后从 $3\,000$ 只骤降至 $500$ 只。三年后种群恢复至 $1\,500$ 只时，一场疾病暴发杀死了 $30\%$ 的个体。与此同时，邻近未受干扰海湾的海獭种群在接近当地环境容纳量处维持稳定。
(a)Classify the oil spill and the disease outbreak as density-dependent or density-independent factors. Justify each classification.将石油泄漏和疾病暴发分别归类为密度制约或非密度制约因素，并为每个分类提供依据。[4]
(b)Predict what would happen to the disease mortality rate if the recovered population continued to grow and approached its carrying capacity. Explain using the concept of density dependence.预测若种群持续增长并接近环境容纳量，疾病死亡率将如何变化。请用密度制约的概念加以解释。[3]
Two species are described below. Species X: produces $2$ offspring per year, provides extended parental care, lives $25$ years, reproduces late in life. Species Y: produces $10\,000$ eggs per season, provides no parental care, lives $1$ year.以下描述两个物种。物种 X：每年产 $2$ 个后代，提供长期亲本照顾，寿命 $25$ 年，繁殖期晚。物种 Y：每个季节产 $10\,000$ 枚卵，不提供亲本照顾，寿命 $1$ 年。
(a)Classify each species as r-selected or K-selected. Justify both classifications using traits listed above.将每个物种分类为 r 对策种或 K 对策种，并用以上特征为两种分类提供依据。[4]
(b)Explain why Species X is more vulnerable to extinction than Species Y following a sudden 80% reduction in population size.解释为何在种群数量突然减少 80% 后，物种 X 比物种 Y 更容易灭绝。[2]
PART II · EXTENDED RESPONSE第二部分 · 简答题AP-feeder FRQ + honors · 32 marksAP 衔接简答题 + 荣誉级 · 共 32 分
Section B · Extended ResponseB 部分 · 简答题
Show every step of reasoning. Identify the equation or principle used before applying it. Use correct biological terminology throughout. Conclude each question with a sentence in context. Calculator permitted on Q6-Q9.每一步推理都要写出。在运用方程或原理前先注明。全程使用正确的生物学术语。每题以一句结合情境的完整句子作结。Q6-Q9 可用计算器。
An isolated island is colonized by $50$ individuals of a bird species with an intrinsic rate of increase $r = 0.4$ per year. Assume exponential growth and use the equation $dN/dt = rN$.一座孤立岛屿被 $50$ 只某种鸟类个体定居，其内禀增长率 $r = 0.4$/年。假设指数增长，使用方程 $dN/dt = rN$。
(a)Calculate the instantaneous growth rate $dN/dt$ at $t = 0$.计算 $t = 0$ 时的瞬时增长率 $dN/dt$。[2]
(b)Explain why $dN/dt$ increases even if $r$ stays constant as the population grows.解释为何在 $r$ 保持不变的情况下，随着种群增长，$dN/dt$ 仍会增大。[2]
(c)Describe two conditions that must hold for this population to continue growing exponentially.描述该种群持续呈指数增长所必须满足的两个条件。[2]
A lake supports a trout population with a carrying capacity $K = 800$ fish and an intrinsic rate of increase $r = 0.6$ per year. The current population is $N = 300$ fish.一个湖泊能维持一个环境容纳量 $K = 800$ 条、内禀增长率 $r = 0.6$/年的鳟鱼种群。当前种群数量为 $N = 300$ 条。
(a)Calculate $dN/dt$ using the logistic growth equation $dN/dt = rN(1 - N/K)$.用逻辑斯蒂增长方程 $dN/dt = rN(1 - N/K)$ 计算 $dN/dt$。[3]
(b)Determine $dN/dt$ if the population were instead at $N = 600$. Compare this result to part (a) and explain the difference.若种群数量改为 $N = 600$，求 $dN/dt$。将此结果与 (a) 部分对比，并解释差异。[3]
(c)Explain what the factor $(1 - N/K)$ represents biologically and how it limits growth.解释因子 $(1 - N/K)$ 在生物学上代表什么，以及它如何限制增长。[2]
Q8MEDIUM中🇺🇸 US美AP-feeder FRQAP 衔接简答题§6 Human population growth人口增长 · HS-LS2-2[8 marks][8 分]
The human population grew slowly for most of history, then entered a period of rapid exponential-like growth after industrialization. Global population reached $1$ billion around 1800 CE and $8$ billion by 2023.人类种群在历史大部分时间内增长缓慢，工业化后进入了一段快速类指数增长期。全球人口约在公元 1800 年达到 10 亿，并于 2023 年达到 80 亿。
(a)Describe the four stages of the demographic transition model and explain what drives the transition from Stage 2 to Stage 3.描述人口转变模型的四个阶段，并解释从第 2 阶段向第 3 阶段转变的驱动因素。[4]
(b)Explain how a declining growth rate is compatible with continued population increase. Use the concept of population momentum in your answer.解释增长率下降与种群持续增加如何能同时发生。在回答中使用人口惯性的概念。[2]
(c)Identify one ecological impact of rapid human population growth on other species' populations.说明人口快速增长对其他物种种群的一种生态影响。[2]
A critically endangered parrot species survives in a single isolated forest fragment with a population of $N = 18$ individuals. Ecologists propose two interventions: (1) captive breeding followed by release, and (2) habitat corridor creation linking the fragment to a larger forest with an established population of the same species.一种极度濒危的鹦鹉物种仅存活于一处孤立森林斑块中，种群数量为 $N = 18$ 只。生态学家提出两种干预方案：(1) 圈养繁育后放归；(2) 建设生境廊道，将该斑块与具有同物种已建立种群的较大森林相连。
(a)Identify and explain two population-level risks that make this parrot highly vulnerable to extinction, beyond habitat loss alone.识别并解释使该鹦鹉高度濒临灭绝的两个种群层面风险（除栖息地丧失以外）。[4]
(b)Evaluate which intervention (captive breeding or habitat corridor) is more likely to address the risks you identified in part (a). Support your answer with specific biological reasoning.评估哪种干预方案（圈养繁育还是生境廊道）更有可能解决 (a) 部分所识别的风险。用具体的生物学推理支持你的答案。[4]
(c)Explain why the concept of minimum viable population (MVP) is central to conservation planning for this species.解释为何最小可存活种群（MVP）概念对该物种的保护规划至关重要。[2]
PART III · MODELING / APPLIED第三部分 · 建模与应用AB Diploma + Universal · 25 marks阿省毕业考 + 通用题型 · 共 25 分
Section C · Modeling and ApplicationsC 部分 · 建模与应用
Define terms and identify equations at the start of each question. Show all calculations with units where appropriate. Conclude each question with a one-sentence summary in context. Calculator permitted throughout Part III.每题开始时定义术语并写出所用方程。计算时写出完整过程和适当单位。每题以一句结合情境的总结句作结。第三部分全程可用计算器。
Q10MEDIUM中🇨🇦 AB阿AB Diploma-style阿尔伯塔毕业考风格§3 + §4 Carrying capacity and regulation (applied)环境容纳量与调节（应用） · Bio 30 C2[8 marks][8 分]
A prairie ecosystem supports a bison population. Wildlife biologists estimate the carrying capacity is $K = 2\,000$ bison. The current population is $N = 1\,600$ bison with $r = 0.3$ per year.草原生态系统支撑着一个野牛种群。野生生物学家估计环境容纳量为 $K = 2\,000$ 只野牛。当前种群数量为 $N = 1\,600$ 只，$r = 0.3$/年。
(a)Calculate $dN/dt$ using the logistic growth equation. Show all steps.用逻辑斯蒂增长方程计算 $dN/dt$，写出所有步骤。[3]
(b)A severe drought kills $40\%$ of the bison. Identify whether this is a density-dependent or density-independent event. Then recalculate $dN/dt$ for the post-drought population.一场严重旱灾导致 $40\%$ 的野牛死亡。判断这是密度制约还是非密度制约事件。然后重新计算灾后种群的 $dN/dt$。[3]
(c)Compare the growth rates from (a) and (b). Explain what the logistic model predicts will happen to the population after the drought.比较 (a) 和 (b) 的增长率。解释逻辑斯蒂模型对灾后种群的预测。[2]
A conservation organization monitors two isolated wolf populations. Population A: $N = 40$, $K = 200$, $r = 0.25$/yr. Population B: $N = 160$, $K = 200$, $r = 0.25$/yr. Both are subject to the same ecosystem.一个保护组织监测两个孤立的狼种群。种群 A：$N = 40$，$K = 200$，$r = 0.25$/年。种群 B：$N = 160$，$K = 200$，$r = 0.25$/年。两者处于相同生态系统中。
(a)Calculate $dN/dt$ for both Population A and Population B. Show your work.计算种群 A 和种群 B 各自的 $dN/dt$，写出计算过程。[4]
(b)Although Population B is larger, explain why Population A may actually have a higher growth rate per individual (per capita growth rate $dN/dt \div N$). Calculate both per capita growth rates to support your answer.尽管种群 B 数量更大，解释为何种群 A 的个体增长率（人均增长率 $dN/dt \div N$）实际上可能更高。计算两者的人均增长率以支持你的答案。[3]
(c)Wolves are K-selected organisms. Explain why Population A faces greater extinction risk than Population B, and identify one management strategy to reduce that risk.狼是 K 对策生物。解释为何种群 A 比种群 B 面临更大的灭绝风险，并提出一种降低该风险的管理策略。[2]
🇺🇸 US NGSS美国 NGSSHS-LS2-1 · HS-LS2-2
🇨🇦 Ontario安大略SBI3U B2 · B3
🇨🇦 British Columbia不列颠哥伦比亚Biology 12: population dynamics, growth models, carrying capacity生物 12：种群动态、增长模型、环境容纳量
🇨🇦 Alberta阿尔伯塔Bio 30 Unit C · C1.1k · C1.2k · C2.1k
Full Syllabus Map lives in ../Study Guides/Unit_12_Population_Biology.html. Bio 30 (AB) uses diploma-exam framing for population growth and regulation questions.完整大纲对照表见 ../Study Guides/Unit_12_Population_Biology.html。阿省 Bio 30 以毕业考风格考查种群增长与调节内容。