PART I · SHORT RESPONSE第一部分 · 短答题AP-style MCQ + ON/BC/AB short answer · 18 marksAP 风格选择题 + 安/卑/阿省考短答 · 共 18 分
Section A · Short ResponseA 部分 · 短答题
Mix of multiple-choice and short-answer items. For MCQs, circle the letter; write enough reasoning that a marker could verify your choice. For short-answer items, use complete sentences where indicated. No calculator required for Q1-Q3.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推理过程。短答题在有提示时请用完整句子作答。Q1-Q3 无需计算器。
Which of the following is NOT one of the three core statements of cell theory?下列哪项不是细胞学说三条核心内容之一？
(A)All living organisms are composed of one or more cells.所有生物体由一个或多个细胞组成。
(B)The cell is the basic structural and functional unit of all living things.细胞是所有生物体的基本结构和功能单位。
(C)All cells arise from pre-existing cells by cell division.所有细胞均由已有细胞通过细胞分裂产生。
(D)All cells contain a membrane-bound nucleus housing their genetic material.所有细胞都含有包裹遗传物质的膜性细胞核。
Q2EASY易🇨🇦 ON安ON Provincial-style安大略省考风格§1 Prokaryote vs Eukaryote原核 vs 真核 · SBI3U B3.2[3 marks][3 分]
A student examines a single-celled organism under a light microscope. It lacks a nucleus and has no membrane-bound organelles, but it does have a cell wall and ribosomes.学生在光学显微镜下观察一个单细胞生物。该生物无细胞核，也无膜性细胞器，但有细胞壁和核糖体。
(a)Identify whether this organism is a prokaryote or a eukaryote. Justify your answer using two pieces of evidence from the observation.判断该生物是原核生物还是真核生物。用观察中的两条证据说明理由。[2]
(b)State one structural feature that all cells, prokaryotic and eukaryotic, share.写出原核细胞和真核细胞共有的一个结构特征。[1]
A red blood cell (animal cell) is placed in a solution that is more concentrated than the cell's cytoplasm.将一个红细胞（动物细胞）放入浓度高于细胞质的溶液中。
(a)Identify this type of solution as hypotonic, isotonic, or hypertonic relative to the cell.判断此溶液相对于细胞是低渗、等渗还是高渗溶液。[1]
(b)Describe the direction of net water movement across the membrane and predict what will happen to the cell. Name the process responsible.描述水分子跨膜净移动的方向，预测细胞的变化，并说明所涉及的过程名称。[2]
(c)State whether this water movement is active transport or passive transport. Give one reason.说明此水分子移动是主动运输还是被动运输。给出一条理由。[1]
Q4MEDIUM中🇨🇦 BC卑BC Provincial-style卑诗省考风格§3 Organelles细胞器 · Life Sciences 11[4 marks][4 分]
The table below lists four organelles found in eukaryotic cells.下表列出了真核细胞中的四种细胞器。
Organelle细胞器
Primary Function (choose from the list)主要功能（从列表中选择）
Mitochondrion线粒体
Ribosome核糖体
Chloroplast叶绿体
Vacuole (large central, plant)液泡（植物大中央液泡）
Function list: (i) Site of aerobic cellular respiration / ATP production. (ii) Site of protein synthesis (translation). (iii) Site of photosynthesis / conversion of light energy to chemical energy. (iv) Storage of water, nutrients, and waste; maintains turgor pressure in plant cells.功能列表：(i) 有氧细胞呼吸 / ATP 生成的场所。 (ii) 蛋白质合成（翻译）的场所。 (iii) 光合作用 / 将光能转化为化学能的场所。 (iv) 储存水分、营养物质和废物；维持植物细胞的膨压。
(a)Complete the table by matching each organelle to its correct function (i)-(iv). Write the function letter next to each organelle. [4 marks, 1 each]将每种细胞器与对应功能 (i)-(iv) 配对，完成上表。在每种细胞器旁写上功能字母。[4 分，每项 1 分][4]
Q5MEDIUM中🇨🇦 AB阿AB Diploma-style阿尔伯塔毕业考风格§7 Levels of Organization生命组织层次 · Biology 20 Unit D GO1[4 marks][4 分]
Arrange the following levels of biological organization in the correct order from simplest to most complex: organ system, organelle, organism, tissue, organ, cell.将下列生命组织层次按从简单到复杂的正确顺序排列：器官系统、细胞器、生物体、组织、器官、细胞。
(a)Write the complete sequence in order. [2 marks]按顺序写出完整序列。[2 分][2]
(b)A student claims that a single-celled organism like an amoeba skips several of these levels. Evaluate this claim. [2 marks]一名学生认为，像变形虫这样的单细胞生物跳过了其中若干层次。评价这一说法。[2 分][2]
PART II · EXTENDED RESPONSE第二部分 · 简答题AP-feeder FRQ + honors · 35 marksAP 衔接简答题 + 荣誉级 · 共 35 分
Section B · Extended ResponseB 部分 · 简答题
Show every step of reasoning. Write in complete sentences for explanation parts. State the biological principle or structure before applying it to the context. No calculator required for Q6-Q8; calculator permitted on Q9.每一步推理都要写出。说明类小问请用完整句子作答。先陈述生物学原理或结构，再将其应用到具体情境中。Q6-Q8 无需计算器；Q9 可用计算器。
The cell membrane is described as a "fluid mosaic." It controls what enters and leaves the cell through a variety of transport mechanisms.细胞膜被称为"流动镶嵌"模型。它通过多种运输机制控制物质进出细胞。
(a)Describe the two components of the "fluid mosaic" model of membrane structure. In your answer, explain what makes the membrane "fluid" and what forms the "mosaic." [3 marks]描述细胞膜"流动镶嵌"模型的两个组成部分。在回答中，说明是什么使细胞膜具有"流动性"以及什么构成了"镶嵌"。[3 分][3]
(b)Distinguish between simple diffusion and facilitated diffusion. In each case, state whether energy (ATP) is required and give one example of a molecule transported by each mechanism. [3 marks]区分简单扩散和协助扩散。分别说明是否需要能量（ATP），并各举一个通过该机制运输的分子的例子。[3 分][3]
(c)Explain why the sodium-potassium pump is an example of active transport rather than diffusion. [2 marks]解释为什么钠钾泵是主动运输而非扩散的例子。[2 分][2]
Q7MEDIUM中🇨🇦 ON安ON Provincial-style安大略省考风格§3 Organelles: Animal vs Plant细胞器：动物细胞 vs 植物细胞 · SBI3U B3.2[8 marks][8 分]
A student is comparing a typical animal cell and a typical plant cell.一名学生正在比较一个典型动物细胞和一个典型植物细胞。
(a)Identify THREE organelles or structures present in plant cells but absent (or not typically present) in animal cells. For each, state its function. [3 marks]列出三种存在于植物细胞但动物细胞中没有（或通常没有）的细胞器或结构。对每种，说明其功能。[3 分][3]
(b)Both cell types contain mitochondria. Explain why, and describe what process occurs in the mitochondria that is essential for both cell types. [2 marks]两种细胞都含有线粒体。解释原因，并描述线粒体中发生的对两种细胞都至关重要的过程。[2 分][2]
(c)Ribosomes are found in all cells, including prokaryotes. Explain what this tells us about the universal role of ribosomes in living organisms. [3 marks]核糖体存在于所有细胞中，包括原核细胞。解释这告诉了我们核糖体在生物中的普遍作用。[3 分][3]
Q8HARD难🇨🇦 BC卑BC Provincial-style卑诗省考风格§4 Nucleus and Genetic Control细胞核与基因调控 · Life Sciences 11[9 marks][9 分]
The nucleus is often called the "control centre" of the eukaryotic cell.细胞核常被称为真核细胞的"控制中心"。
(a)Describe the structure of the nucleus, identifying at least THREE distinct components and the function of each. [3 marks]描述细胞核的结构，至少列出三个不同组成部分及各自的功能。[3 分][3]
(b)Explain the pathway by which genetic information in the nucleus ultimately leads to the production of a protein in the cytoplasm. Name the two key processes and the molecules involved. [3 marks]解释细胞核中的遗传信息最终如何导致细胞质中蛋白质产生的途径。说明两个关键过程及涉及的分子。[3 分][3]
(c)A mutation in the DNA of a cell's nucleus can affect all proteins the cell produces. Explain why this is the case using your understanding of the flow of genetic information. [3 marks]细胞核 DNA 中的突变会影响该细胞产生的所有蛋白质。利用你对遗传信息流的理解，解释为什么会这样。[3 分][3]
A neuron (nerve cell) and a red blood cell are both derived from the same fertilized egg, yet they look and function very differently. A neuron has long projections and abundant mitochondria; a mature red blood cell has no nucleus and is packed with hemoglobin.神经元（神经细胞）和红细胞都来源于同一个受精卵，但它们的形态和功能截然不同。神经元有长突起且含大量线粒体；成熟红细胞无细胞核，且充满血红蛋白。
(a)Explain the concept of cell differentiation. Why do two cells with identical DNA end up with such different structures and functions? [3 marks]解释细胞分化的概念。为什么两个具有相同 DNA 的细胞最终会有如此不同的结构和功能？[3 分][3]
(b)Link the structural features of the neuron (long projections, abundant mitochondria) to its function. Explain why each feature is adaptive. [3 marks]将神经元的结构特征（长突起、大量线粒体）与其功能联系起来。解释每种特征为何是适应性的。[3 分][3]
(c)Evaluate the statement: "A mature red blood cell is not a complete cell because it lacks a nucleus." In your answer, address whether it fits the definition of a cell and what the adaptive advantage of lacking a nucleus might be. [4 marks]评价以下说法："成熟红细胞因缺少细胞核而不是完整的细胞。"在回答中，说明它是否符合细胞的定义，以及缺少细胞核可能具有的适应优势。[4 分][4]
PART III · MODELING / APPLIED第三部分 · 建模与应用AB Diploma + Universal · 28 marks阿省毕业考 + 通用题型 · 共 28 分
Section C · Modeling and ApplicationsC 部分 · 建模与应用
Define any terms or symbols before using them. Conclude each explanation question with a one-sentence summary linking structure to function. For Q10, show all arithmetic steps and include units. Calculator permitted throughout Part III.使用任何术语或符号前先进行定义。每道说明题以一句将结构与功能相联系的总结句作结。Q10 请写出所有运算步骤并带上单位。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§5 Microscopy and Cell Size显微镜与细胞大小 · Biology 20 Unit D GO1[9 marks][9 分]
A student views an onion cell under a compound light microscope. The objective lens has a magnification of $\times 40$ and the eyepiece has a magnification of $\times 10$. The student measures the cell image on a projected screen and finds it is $3.2\ \text{mm}$ long.学生用复式光学显微镜观察洋葱细胞。物镜放大倍数为 $\times 40$，目镜放大倍数为 $\times 10$。学生测量投影屏幕上细胞的图像，发现其长度为 $3.2\ \text{mm}$。
(a)Calculate the total magnification of the microscope. [1 mark]计算显微镜的总放大倍数。[1 分][1]
(b)Using the formula $\text{magnification} = \dfrac{\text{image size}}{\text{actual size}}$, calculate the actual length of the cell in micrometres ($\mu\text{m}$). Show all steps. [$1\ \text{mm} = 1000\ \mu\text{m}$] [3 marks]使用公式 $\text{放大倍数} = \dfrac{\text{图像大小}}{\text{实际大小}}$，计算细胞的实际长度，以微米（$\mu\text{m}$）表示。写出所有步骤。[$1\ \text{mm} = 1000\ \mu\text{m}$] [3 分][3]
(c)Consider a spherical cell with radius $r$. As a cell grows, its volume increases as $V = \frac{4}{3}\pi r^3$ and its surface area increases as $SA = 4\pi r^2$. Explain, using the surface-area-to-volume ratio concept, why cells cannot grow indefinitely large. [3 marks]考虑半径为 $r$ 的球形细胞。细胞生长时，体积按 $V = \frac{4}{3}\pi r^3$ 增大，表面积按 $SA = 4\pi r^2$ 增大。利用表面积体积比的概念，解释为什么细胞不能无限增大。[3 分][3]
(d)State one way a cell can overcome a low surface-area-to-volume ratio other than dividing. [2 marks]写出一种细胞在不进行分裂的情况下克服低表面积体积比的方法。[2 分][2]
Q11MEDIUM中🇨🇦 AB阿AB Diploma-style阿尔伯塔毕业考风格§6 + §7 Specialization and Levels of Organization (applied)特化与生命组织层次（应用） · Biology 20 Unit D GO1[9 marks][9 分]
The human digestive system provides an excellent model for studying biological organization and cell specialization. Cells lining the small intestine have finger-like projections called microvilli that dramatically increase their surface area.人体消化系统是研究生命组织层次和细胞特化的优秀模型。覆盖小肠内壁的细胞具有称为微绒毛的指状突起，可显著增加其表面积。
(a)Identify the level of biological organization of each of the following: a single intestinal cell; the lining of the small intestine (a sheet of similar cells); the small intestine (including multiple tissue types); the digestive system. [4 marks]确定下列各项在生命组织层次中所处的水平：单个肠细胞；小肠内壁（由同类细胞构成的细胞层）；小肠（包含多种组织类型）；消化系统。[4 分][4]
(b)Explain how the microvilli on intestinal cells are an example of structure fitting function. In your answer, connect this to the concept of cell specialization. [3 marks]解释肠细胞上的微绒毛如何体现结构适应功能的原则。在回答中，将其与细胞特化的概念联系起来。[3 分][3]
(c)A genetic mutation causes intestinal cells to lose their microvilli. Predict the physiological consequence for the organism. Explain your reasoning. [2 marks]某基因突变导致肠细胞失去微绒毛。预测这对生物体产生的生理后果，并解释你的推理。[2 分][2]
A student conducts an osmosis investigation. Potato cylinders of equal size and mass are placed in sucrose solutions of different concentrations for 30 minutes, then reweighed. The results are recorded below.学生进行渗透作用实验。将等大等质的马铃薯圆柱体放入不同浓度的蔗糖溶液中 30 分钟，然后重新称重。结果记录如下。
Sucrose concentration (mol/L)蔗糖浓度（mol/L）
Initial mass (g)初始质量（g）
Final mass (g)最终质量（g）
% change in mass质量变化百分比
0.0
5.0
5.6
0.2
5.0
5.3
0.4
5.0
5.0
0.6
5.0
4.6
0.8
5.0
4.2
(a)Complete the "% change in mass" column for each row. Use the formula: $\%\text{ change} = \dfrac{\text{final} - \text{initial}}{\text{initial}} \times 100\%$. [2 marks]填写每行的"质量变化百分比"列。使用公式：$\%\text{ 变化} = \dfrac{\text{最终} - \text{初始}}{\text{初始}} \times 100\%$。[2 分][2]
(b)Estimate the isotonic concentration for the potato tissue (the concentration at which no net change in mass occurs). Explain your reasoning from the data. [2 marks]估算马铃薯组织的等渗浓度（质量净变化为零时的浓度）。根据数据解释你的推理。[2 分][2]
(c)Explain, in terms of osmosis and water potential, why the potato cylinders lost mass in the 0.6 mol/L and 0.8 mol/L solutions. [3 marks]用渗透作用和水势的概念解释，为什么马铃薯圆柱体在 0.6 mol/L 和 0.8 mol/L 溶液中质量减少。[3 分][3]
(d)Predict and explain what would happen if the same experiment were conducted with plant cells that have rigid cell walls versus animal cells. How would the results differ, and why? [3 marks]预测并解释：如果用具有坚硬细胞壁的植物细胞与动物细胞进行同样的实验，结果会有什么不同，为什么？[3 分][3]
🇨🇦 Alberta阿尔伯塔Biology 20 Unit D GO1 · Biology 30 Unit C GO1
Full Syllabus Map lives in ../Study Guides/Unit_1_Cell_Structure_and_Function.html. Biology is primarily qualitative; labelled-diagram reasoning, compare/contrast, and FRQ explanation are the dominant question types. Magnification calculations (Q10) apply in all four regions.完整大纲对照表见 ../Study Guides/Unit_1_Cell_Structure_and_Function.html。生物学以定性为主；标注图示推理、比较对比和 FRQ 说明是主要题型。放大倍数计算（Q10）适用于四个地区。