一套真题看懂CIE Biology阅卷规则,少丢冤枉分!

Question 1 Intestinal villi & gut cells小肠绒毛与肠道细胞（Total 14 总分 14）

1 (a) Explain difference between cilium & microvillus [2 marks] 区分纤毛与微绒毛【2 分】

Official full-mark points 官方采分点

1. Structure: Cilia contain microtubules (9+2 arrangement); microvilli contain actin filaments 结构：纤毛由微管构成（9+2 排列）；微绒毛由肌动蛋白丝构成

2. Function: Cilia can beat to move fluid; microvilli only enlarge surface area for absorption and cannot move 功能：纤毛可摆动推动液体流动；微绒毛仅增大吸收表面积，无运动能力

Common losing marks traps 易错丢分坑

❌ Only describe length difference without structural/functional keywords → 0 mark 仅描述长短差异，无结构 / 功能关键词，不得分

❌ Mix up cytoskeleton components → no credit 

混淆细胞骨架组成，直接不得分

Full-mark answer template 满分答题模板

Cilia are made of microtubules in a 9+2 arrangement and can beat to move liquid along epithelium. Microvilli consist of actin filaments, they only increase surface area for nutrient absorption and have no movement.

1 (b) Suggest one role of mucus from goblet cells [1 mark] 杯状细胞黏液的 1 个功能【1 分】

Acceptable answers 可得分答案（任选其一）

1. Protect intestinal epithelial cells from damage by stomach acid /digestive enzymes 保护肠道上皮细胞，免受胃酸、消化酶侵蚀损伤

2. Trap pathogens /bacteria to stop them contacting gut epithelial cells 黏住致病菌，阻止病原体接触肠道上皮细胞

Reject answer (R) 阅卷否决项

Descriptions about respiratory tract mucus get no marks

描述呼吸道黏液相关内容不得分

1 (c)(i) Name blood vessel X [1 mark]

血管 X 名称【1 分】

Answer: arteriole 小动脉

Knowledge memo 知识点速记:Artery → arteriole → capillary → venule → vein 动脉 → 小动脉 → 毛细血管 → 小静脉 → 静脉

1 (c)(ii) Describe formation of tissue fluid [2 marks]

描述组织液形成过程【2 分】

Two marking points 2 个采分点

1. High hydrostatic blood pressure at arteriole end of capillary creates pressure gradient 毛细血管动脉端血液静水压高，形成压力梯度

2. Ultrafiltration occurs: water, glucose, amino acids, small ions are forced out to form tissue fluid; large plasma proteins remain inside capillary 发生超滤作用：水、葡萄糖、氨基酸、小分子离子被压出血管形成组织液；大分子血浆蛋白留在血管内

Trap 避坑提醒

Do NOT state red blood cells /platelets leak out → mark deducted

不可写红细胞、血小板穿出毛细血管，会扣分

1 (d) Outline cell signalling stages of GLP-1 [3 marks]

GLP-1 细胞信号传导过程【3 分，顺序不可颠倒】

Sequential full logic 分步满分逻辑

1. Enteroendocrine cells secrete GLP-1 peptide, transported to target cells (stomach/pancreas) via blood circulation 肠内分泌细胞分泌 GLP-1 多肽，经血液循环运输至靶细胞（胃、胰腺细胞）

2. GLP-1 binds to specific complementary receptors on target cell surface membrane GLP-1 与靶细胞膜上特异性、互补的受体结合

3. Receptor binding triggers intracellular signal transduction cascade, leading to target cell responses 受体结合触发细胞内信号级联传导，最终产生靶细胞生理应答

Mandatory keywords 必写得分关键词

specific complementary receptor, signal transduction cascade

特异性互补受体、信号级联传导

1 (e)(i) Name mitosis stages in sequence [1 mark]

按顺序写出有丝分裂四阶段【1 分】

Answer: Prophase → Metaphase → Anaphase → Telophase

前期 → 中期 → 后期 → 末期

Deduction rule 扣分规则

If interphase /cytokinesis is written, 0 mark 若写出间期、胞质分裂，整空不得分

1 (e)(ii) Evidence Paneth cell is secretory cell that synthesises lots of protein [3 marks] 电镜证据证明潘氏细胞合成分泌大量蛋白【3 分，任选 3 点】

1. Large amount of rough endoplasmic reticulum (RER): attached ribosomes are site of polypeptide/protein synthesis 大量粗面内质网：附着核糖体是多肽、蛋白质合成场所

2. Numerous secretory vesicles: store synthesised anti-pathogen peptides, released via exocytosis 大量分泌囊泡：储存合成的抗菌多肽，通过胞吐释放

3. Mitochondria present: supply ATP for protein synthesis, vesicle transport and exocytosis 存在线粒体：为蛋白合成、囊泡运输、胞吐提供 ATP 能量

Mark scheme rule 阅卷规则

Only writing ER without word "rough" gets no credit

只写内质网 ER，不标注 rough不得分

1 (e)(iii) Explain difference between Paneth cell & intestinal stem cell [1 mark] 潘氏细胞与干细胞形态差异原因【1 分】

Paneth cell is differentiated /specialised for secretion; intestinal stem cell is undifferentiated and only divides to produce new cells.

潘氏细胞是高度分化、特化的分泌细胞；肠道干细胞未分化，仅负责分裂产生新细胞。

Q1 Summary 第 1 题总结

Core topics: cytoskeleton, tissue fluid formation, cell signalling, ultrastructure of secretory cell, cell differentiation 

核心考点：细胞骨架、组织液生成、细胞通讯、分泌细胞超微结构、细胞分化Common lost marks: wrong spelling of arteriole, mixing cilium & microvillus structure

高频丢分：arteriole 拼写错误、纤毛与微绒毛结构混淆

Question 2 Cholera & Immunity霍乱与免疫（Total 12 总分 12）

2 (a) Name cholera-causing bacterium species [1 mark]

霍乱致病菌学名【1 分】

Answer: Vibrio cholerae Note: Spelling must be completely correct, minor spelling error = 0 mark 

注意：拼写必须完全准确，字母写错直接不得分

2 (b) Explain safe drinking water & personal hygiene control cholera [4 marks] 安全饮水 + 个人卫生防控霍乱【4 分，两点各 2 分】

1. Safe drinking water: Cholera transmits via faecal-oral route. Boiling /chlorination kills Vibrio cholerae in contaminated water to stop pathogen intake. 安全饮用水：霍乱为粪口传播途径；煮沸、加氯消毒可杀灭污染水源中的霍乱弧菌，避免摄入致病菌。

2. Personal hand hygiene: Washing hands with soap after defecation /before food preparation removes faecal bacteria on skin, prevents contaminating food and oral infection. 个人手部卫生：便后、备饭前使用肥皂洗手，清除皮肤上携带粪便的细菌，避免污染食物造成经口感染。

2 (c)(i) Compare Oral Cholera Vaccine (OCV) & passive immunisation [4 marks]

口服霍乱疫苗 VS 被动免疫对比表【4 分】

Core memorisation logic 背诵核心逻辑

Active immunity: inject antigen, body makes own antibody + memory cells; slow onset, long protection 

主动免疫：输入抗原，机体自产抗体 + 记忆细胞；起效慢、保护时间长

Passive immunity: directly inject external antibodies; no memory cells, instant effect, short protection

被动免疫：直接输入外源抗体；无记忆细胞，立刻起效、保护周期短

2 (c)(ii) Two cell types for hybridoma & fusion necessity [3 marks] 杂交瘤融合的两种细胞 + 融合必要性【3 分】

Two cell types: B-lymphocyte (plasma cell) + myeloma cell

两种细胞：B 淋巴细胞（浆细胞）+ 骨髓瘤细胞

Reason for fusion: 

1. B cell: produces specific antibody, but cannot divide and survive long in lab culture 

B 细胞：可合成特异性抗体，但体外无法持续分裂、长期存活

2. Myeloma cell: divides infinitely via mitosis (immortal), cannot produce antibody

骨髓瘤细胞：可无限有丝分裂增殖（永生），但无法合成抗体

3. Hybridoma cell combines both advantages: unlimited proliferation + stable secretion of single monoclonal antibody

杂交瘤融合后兼具二者优势：无限增殖、稳定分泌单一单克隆抗体

Lost mark trap 丢分坑

Only write cell names without explaining their separate functions, maximum 1 mark 只写细胞名称，不分别解释两种细胞特性，最多仅得 1 分

Q2 Reminder 第 2 题避坑提醒

Most students mistakenly tick "Yes" for memory cell in passive immunisation → lose 1 mark 绝大多数学生易错勾选被动免疫可产生记忆细胞，直接丢失 1 分

Question 3 Lactase gene & immobilised enzyme

乳糖酶基因与固定化酶（Total 13 总分 13）

3 (a) Differences between LCT and MCM6 gene [3 marks] LCT 与 MCM6 基因的 3 处区别【3 分，任选 3 点】

1. They have different sequences of DNA nucleotides /bases, coding for different polypeptides

DNA 核苷酸 / 碱基序列不同，编码完全不同的多肽链

2. Different quantity /length of introns and exons

内含子、外显子的数量、长度存在差异

3. LCT codes for lactase enzyme; MCM6 codes for regulatory protein with no role in lactase synthesis

LCT 编码乳糖酶；MCM6 编码调控蛋白，不参与乳糖酶合成

4. Different regulatory sequences: introns of MCM6 control expression of LCT gene

调控序列不同：MCM6 的内含子调控 LCT 基因的表达水平

3 (b) Relationship between transcribed strand, primary transcript and mRNA [2 marks] 转录链、初级转录物、mRNA 的关系【2 分，各 1 分】

1. Transcribed DNA strand acts as template; complementary primary transcript RNA is formed during transcription (uracil replaces thymine)

DNA 转录链作为模板，转录过程合成互补的初级转录 RNA（尿嘧啶替代胸腺嘧啶）

2. Primary transcript undergoes RNA splicing: introns are removed, only exons remain to form mature

mRNA 初级转录物经过 RNA 剪接，切除内含子，仅保留外显子序列，形成成熟 mRNA

3 (c) Type of mutation causing lactase persistence [1 mark] 乳糖持续表达的突变类型【1 分】

Answer: base substitution mutation 碱基替换突变 

Explanation: Total number of nucleotides unchanged, only one single base pair altered 核苷酸总数不变，仅单个碱基发生替换改变

3 (d) Draw ring structure of α-glucose [1 mark] 绘制 α- 葡萄糖环状结构【1 分】

Mark rule: Complete 6-member ring, α configuration (C1 hydroxyl group points down), all C-H bonds labelled fully; missing only C5 hydrogen = 0 mark 

阅卷标准：完整六元环、C1 位 α 构型（羟基朝下），全部碳氢键标注完整；仅遗漏 C5 氢直接不得分

3 (e)(i) Why use colorimeter to track reaction progress [2 marks] 实验选用比色计的原因【2 分】

1. Colorimeter provides quantitative numerical absorbance readings, objective data instead of subjective naked-eye observation

比色计输出定量吸光度数值，数据客观，区别于肉眼主观观察颜色深浅

2. High sensitivity: detects faint colour changes, calibration curve can calculate exact rate of reaction

灵敏度更高：可识别微弱色差，绘制标准曲线精准计算反应速率

3 (e)(ii) Advantages of immobilised lactase (refer to Fig 3.2 & 3.3) [4 marks] 结合图表分析固定化乳糖酶优势【4 分】

1. Higher thermostability (Fig3.2): Immobilised lactase retains higher activity at all high temperatures e.g. 55°C, free lactase nearly loses all activity. More tolerant to temperature fluctuations.

热稳定性更强（图 3.2）：55℃等高温下固定化酶活性远高于游离酶，耐受温度波动。

2. Wider pH tolerance range (Fig3.3): Free lactase has zero activity at pH5.5, immobilised lactase maintains activity across full pH range, stable under acidic/alkaline conditions.

pH 耐受范围更广（图 3.3）：游离酶在 pH5.5 完全失活，固定化酶全 pH 区间均有活性，酸碱环境稳定性更强。

3. Higher overall hydrolysis productivity of lactose under industrial conditions

同等工业条件下乳糖水解总产量更高

4. Magnetic microsphere carrier can be recovered and reused, lower production cost

磁性微球载体可回收重复利用，降低生产成本

Mark requirement 阅卷硬性要求

Answers must quote specific data from graphs; pure theory without graph reference max 2 marks 答案必须引用图表具体数据，仅写理论不结合图最多只得 2 分

Q3 Exam skill 第 3 题答题技巧

All graph analysis questions need theory + graph data combination to get full marks 所有图表分析题必须做到「理论描述 + 图表数据」结合，方可拿满分

Question4 Gram-negative bacterial membrane & extracellular vesicles

革兰氏阴性菌膜与胞外囊泡（Total 7 总分 7）

4 (a) Similarities & differences: bacterial outer membrane vs eukaryotic cell surface membrane [3 marks]

细菌外膜 VS 真核细胞膜异同【最多 3 分，相似点最高 2 分，差异最高 2 分】

Similarities 相同点

1. Both contain phospholipid bilayer 均拥有磷脂双分子层

2. Both have integral /channel transport proteins 均存在跨膜、通道转运蛋白

Differences 不同点

1. Bacterial outer membrane has lipopolysaccharides (LPS), eukaryotic membrane does not 细菌外膜含有脂多糖 LPS，真核细胞膜无

2. Eukaryotic cell membrane contains cholesterol, bacterial outer membrane lacks cholesterol 真核细胞膜含有胆固醇，细菌外膜无胆固醇

3. Eukaryotic membrane has glycoproteins /glycolipids glycocalyx, bacterial outer membrane does not 真核细胞膜有糖蛋白、糖脂构成的糖萼，细菌外膜不存在

4 (b)(i) Fewer O-IMVs produced than OMVs [1 mark] O-IMV 产量少于 OMV 的原因【1 分】

O-IMV formation requires breaking peptidoglycan cell wall, more complex process consuming extra energy and risks cell lysis; OMV only buds from outer membrane without breaking cell wall.

O-IMV 形成需要断裂肽聚糖细胞壁，过程更复杂、消耗更多能量，存在细胞裂解风险；OMV 仅外层膜出芽，无需破坏细胞壁。

4 (b)(ii) ATP exists in O-IMV but not OMV [1 mark]

O-IMV含 ATP、OMV 不含的原因【1 分】

OMV only derives from periplasm space with no cytoplasm or ATP; O-IMV encloses cell surface membrane and cytoplasm which contains ATP molecules.

OMV 仅来源于周质空间，无细胞质、不含 ATP；O-IMV 同时包裹细胞膜与细胞质，细胞质内存在 ATP。

4 (b)(iii) DNA inside O-IMVs links to antibiotic resistance [2 marks]

O-IMV 携带 DNA 对抗生素耐药性的影响【2 分】

1. IMVs can fuse with or be taken up by other Gram-negative bacteria, transferring plasmid /resistance gene DNA into recipient cells O-IMV

可与其他革兰氏阴性菌融合、被细菌摄取，将质粒、耐药基因 DNA 传递给受体菌

2. Recipient bacteria obtain resistance alleles, synthesise proteins (antibiotic-degrading enzyme /efflux pump) to survive antibiotic treatment; resistance genes spread horizontally among bacterial population

受体细菌获得耐药等位基因，合成降解抗生素酶、外排泵等蛋白，抵御抗生素；耐药基因在菌群间水平传播

Q4 Difficult point 第 4 题难点提示

Horizontal gene transfer via extracellular vesicles is a newly tested hot topic in recent CIE papers 胞外囊泡介导的水平基因转移是近年 CIE 生物新高频考点

Question 5 Xerophyte leaf adaptations

旱生植物叶片适应性（Total 5 总分 5）

5 (a) Name structure A and tissue B [2 marks]

标注结构 A、组织 B 名称【2 分】

A = (waxy) cuticle 蜡质角质层

B = spongy mesophyll (layer /tissue) 海绵叶肉组织

Deduction note 扣分说明

Writing "spongy mesophyll cell" gets no credit; must write tissue/layer

写海绵叶肉细胞不得分，必须写组织 / 薄层

5 (b) Xerophyte leaf structural modifications [3 marks]

旱生叶片结构改造特征【每空 1 分，答案不可重复】

Core logic 核心逻辑

All xerophyte features function to reduce water loss via transpiration 所有旱生植物结构的核心作用：减少蒸腾作用水分流失

Question 6 Heart cycle & Haemoglobin

心脏周期与血红蛋白（Total 9 总分 9）

6 (a) Sequence of events in left heart (pulmonary → systemic circulation) [5 marks] 含氧血从肺回左心进入体循环全过程【5 分】

1. Oxygenated blood flows back to left atrium via pulmonary vein 含氧血液经肺静脉流入左心房

2. Atrial systole: left bicuspid (mitral) valve opens, blood moves down into left ventricle 心房收缩：左房室瓣（二尖瓣）打开，血液流入左心室

3. Ventricular systole: pressure inside ventricle rises higher than left atrium → bicuspid valve closes 心室收缩：心室压力高于左心房，二尖瓣关闭

4. Ventricular pressure exceeds aortic pressure → aortic semilunar valve opens 心室压力超过主动脉压力，主动脉半月瓣打开

5. Blood is pumped out of heart into aorta to enter systemic circulation 血液泵入主动脉，进入全身体循环

Mandatory scoring rule 必拿分要求

Must mention pressure difference controls valve opening/closing, no full marks without this point

必须提及压力差控制瓣膜开合，缺少该点无法拿满分

6 (b)(i) Quaternary structure of haemoglobin [2 marks]

血红蛋白四级结构【2 分】

1. Consists of four polypeptide subunits: two identical α-globin chains and two identical β-globin chains

由 4 条多肽亚基构成：2 条相同 α 珠蛋白链、2 条相同 β 珠蛋白链

2. Each globin chain contains one haem prosthetic group with Fe²⁺ (ferrous ion); polypeptide chains are held together by hydrogen, hydrophobic and ionic bonds (disulfide bonds not accepted)

每条珠蛋白链结合 1 个含亚铁离子 Fe²⁺的血红素辅基；多肽链依靠氢键、疏水键、离子键维系四级结构（二硫键不记分）

6 (b)(ii) How haemoglobin becomes fully saturated with oxygen [2 marks]

血红蛋白完全饱和结合氧气过程【2 分】

1. Oxygen molecules bind to Fe²⁺ ion inside each haem group; one haemoglobin molecule can bind maximum four O₂ molecules

氧气分子与每个血红素内部的亚铁离子结合；1 分子血红蛋白最多结合 4 分子氧气

2. At high partial pressure of oxygen, haemoglobin changes conformation from tense (T) state to relaxed (R) state, cooperative oxygen binding occurs until full saturation

高氧分压环境下，血红蛋白构象从紧张 T 态转为松弛 R 态，协同结合氧气直至完全饱和

Q6 Common spelling mistakes 第 6 题高频拼写错误

❌ "arota" instead of aorta → lose marking point

❌ 把 aorta 主动脉拼写成 arota，直接丢失采分点

❌ Confuse bicuspid (left heart) & tricuspid (right heart) valve 混淆左心二尖瓣、右心三尖瓣

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