IB生物HL人体生理 消化吸收 血液循环 精讲
IB Biology HL 的 Human Physiology 是整个课程里分值最重、也最容易丢分的单元之一。它横跨消化吸收、血液循环与气体运输、通气与气体交换、神经传导与突触、稳态与激素调控、免疫与器官移植六大主题。Paper 2 Section B 的 9 分长答题常把其中两三个主题串在一起考，例如"解释运动时血红蛋白解离曲线右移如何提高组织供氧"，既考知识点，也考你用数据推理的表达能力。本文按 Syllabus 顺序梳理六大主题的高频考点、必背数据与计算题型，每个主题配一道典型例题与易错提醒，帮你把这一单元从"背过"变成"会做"。
The Human Physiology unit in IB Biology HL carries the heaviest mark weight in the whole syllabus, and also the most avoidable lost marks. It spans six themes: digestion and absorption, circulation and oxygen transport, ventilation and gas exchange, nerve impulse and synaptic transmission, homeostasis and hormonal control, plus immunity and organ transplantation. Nine-mark extended responses in Paper 2 Section B frequently chain two or three of these themes, for instance asking you to explain how the rightward shift of the haemoglobin dissociation curve increases oxygen delivery to tissues during exercise. This article walks through the six themes in syllabus order, covering high-frequency content, data you must know by heart, calculation types, and a worked example with common pitfalls for each one.
1. 消化与吸收: 小肠绒毛结构与吸收机制
小肠是吸收的主要场所，原因在于其巨大的表面积与丰富的运输结构。环形皱褶、绒毛、微绒毛三级放大把表面积提高约 600 倍。绒毛上皮为单层细胞，缩短扩散距离；内部有密集的毛细血管网与中央乳糜管。葡萄糖和氨基酸的吸收依靠钠离子共转运，属于继发性主动运输: 钠钾泵在基底侧维持细胞内低钠浓度，顶端膜上的共转运蛋白借此把葡萄糖逆浓度梯度拉进细胞，再经易化扩散进入血液。甘油与脂肪酸进入上皮后重新合成甘油三酯，包裹成乳糜微粒，先进入乳糜管(淋巴)，再汇入血液循环，因此脂肪吸收路径与糖、氨基酸完全不同。
The small intestine is the main site of absorption because of its enormous surface area and dense transport structures. Circular folds, villi and microvilli amplify the surface roughly 600-fold. The epithelium is a single cell layer, which shortens the diffusion distance, and each villus contains a dense capillary network plus a central lacteal. Glucose and amino acids are absorbed by sodium co-transport, which is secondary active transport: the sodium-potassium pump keeps intracellular sodium low, so apical co-transporter proteins drag glucose in against its concentration gradient, after which it leaves by facilitated diffusion. Glycerol and fatty acids are reassembled into triglycerides inside the epithelium, packaged into chylomicrons, and enter the lacteal first, so fat follows a completely different route from sugars and amino acids.
典型例题: 小肠绒毛高度 0.5 mm，直径 0.1 mm，估算单个绒毛的柱状侧面积。解法: 把绒毛近似为圆柱，侧面积 = pi x 直径 x 高度 = 3.14 x 0.1 mm x 0.5 mm = 0.157 mm2。若一段小肠有 20000 个绒毛，则总侧面积约 3140 mm2 = 31.4 cm2，这还不包括微绒毛带来的进一步放大。答题时必须写单位，并说明该结构如何缩短扩散距离、增加表面积从而加快吸收速率。
Worked example: a villus is 0.5 mm tall and 0.1 mm wide. Estimate the lateral surface area of one villus. Treat it as a cylinder: lateral area = pi x diameter x height = 3.14 x 0.1 mm x 0.5 mm = 0.157 mm2. With 20000 villi in a section of intestine, total area is about 3140 mm2, or 31.4 cm2, and this excludes the extra amplification from microvilli. Always state units and explain how the structure shortens diffusion distance and increases surface area, so the rate of absorption rises.
易错提醒: 常见错误是把绒毛写成"增加体积"。评分点要求写的是表面积与扩散距离。另外，乳糜管属于淋巴系统而非血液系统，答"脂肪直接进入毛细血管"不给分。
Common pitfall: many candidates write that villi increase volume. Marks are awarded for surface area and diffusion distance. Also, the lacteal belongs to the lymphatic system rather than the blood system, so saying fat enters capillaries directly scores nothing.
2. 血液循环与气体运输: 心输出量与血红蛋白解离曲线
心输出量等于心率乘以每搏输出量。安静时心率约 70 次每分钟、每搏输出量约 70 mL，心输出量约 4.9 L/min。运动时可升到心率 180 次每分钟、每搏量 120 mL，即约 21.6 L/min，接近安静时的 4.5 倍。氧气运输依靠血红蛋白: 一个血红蛋白分子可结合四个氧分子，四个亚基之间的协同结合使解离曲线呈 S 形。肺部氧分压约 100 mmHg 时饱和度约 97.5%，组织氧分压约 40 mmHg 时降到约 75%，也就是说每 100 mL 血液每次循环大约释放 5 mL 氧气。
Cardiac output equals heart rate multiplied by stroke volume. At rest, a rate of about 70 beats per minute and a stroke volume of about 70 mL give roughly 4.9 L per minute. During exercise the rate may reach 180 and stroke volume 120 mL, so output rises to about 21.6 L per minute, close to 4.5 times the resting value. Oxygen transport depends on haemoglobin: each molecule binds four oxygen molecules, and cooperative binding between the four subunits produces the S-shaped dissociation curve. At a lung partial pressure near 100 mmHg saturation is about 97.5 per cent, falling to roughly 75 per cent at a tissue partial pressure near 40 mmHg, so each 100 mL of blood releases about 5 mL of oxygen per circulation.
典型例题: 解释运动时解离曲线右移的生理意义。要点有四条: 一是肌肉呼吸作用产生更多二氧化碳，血液 pH 下降；二是温度升高；三是曲线右移使同一氧分压下血红蛋白饱和度降低；四是卸载量增加，从约 5 mL 提升到接近 8 mL 每 100 mL 血液，从而为呼吸作用提供更多底物。答题时把"右移"与"卸载增加"两个因果环节都写出来才拿满分。
Worked example: explain the physiological benefit of the rightward shift during exercise. Four marking points apply. Working muscle respires faster and produces more carbon dioxide, lowering blood pH; temperature also rises. The curve shifts right, so saturation at any given partial pressure falls. Unloading therefore increases, from about 5 mL to nearly 8 mL per 100 mL of blood, supplying more substrate for respiration. A complete answer states both the shift and the resulting increase in unloading.
易错提醒: 胎儿血红蛋白曲线在母体曲线的左侧，亲和力更高，这有利于从母体血液中夺取氧气，方向不要记反。肌红蛋白曲线更靠左且极其陡峭，作用是储存氧气而非卸载氧气。
Common pitfall: fetal haemoglobin lies to the left of the maternal curve and has higher affinity, which lets it take oxygen from maternal blood; do not invert the direction. Myoglobin lies further left with a much steeper curve, and its role is oxygen storage rather than unloading.
3. 通气与气体交换: 肺泡结构与肺通气量计算
肺泡是气体交换的场所，其结构与绒毛类似: 数量巨大(约 3 亿个)、单层扁平上皮、紧贴毛细血管、毛细血管血流缓慢，共同维持足够的扩散表面积与分压梯度。肺泡内壁分泌表面活性物质，降低水的表面张力，防止肺泡在呼气末塌陷，这也解释了早产儿呼吸窘迫综合征的成因。通气量等于潮气量乘以呼吸频率: 潮气量约 500 mL、频率 12 次每分钟时通气量约 6000 mL/min，其中约三分之一停留在气道内不参与交换，称为解剖无效腔。
Alveoli are the site of gas exchange, and their design mirrors that of villi: roughly 300 million of them, a single squamous epithelium, close contact with capillaries, and slow capillary blood flow, all maintaining a large diffusion surface and a steep partial pressure gradient. The alveolar lining secretes surfactant, which lowers surface tension and prevents collapse at the end of expiration; this also explains respiratory distress syndrome in premature infants. Ventilation rate equals tidal volume multiplied by breathing frequency: a tidal volume of about 500 mL at 12 breaths per minute gives roughly 6000 mL per minute, about one third of which remains in the airways and takes no part in exchange, the anatomical dead space.
典型例题: 一名学生潮气量 450 mL，呼吸频率 18 次每分钟，解剖无效腔 150 mL。求肺泡通气量。解法: 有效潮气量 = 450 - 150 = 300 mL，肺泡通气量 = 300 x 18 = 5400 mL/min。若呼吸频率翻倍而潮气量减半，肺泡通气量会明显下降，这正是浅快呼吸效率低的原因。
Worked example: a student has a tidal volume of 450 mL at 18 breaths per minute, with a dead space of 150 mL. Find alveolar ventilation. Effective tidal volume is 450 minus 150, which equals 300 mL, so alveolar ventilation is 300 x 18, or 5400 mL per minute. If frequency doubles while tidal volume halves, alveolar ventilation falls sharply, which is why rapid shallow breathing is inefficient.
易错提醒: 扩散速率与表面积和分压差成正比、与扩散距离成反比。写"肺泡壁厚有利于交换"会直接丢分，肺泡壁薄(约 0.5 微米)才是有利因素。
Common pitfall: diffusion rate is proportional to surface area and partial pressure difference, and inversely proportional to diffusion distance. Writing that a thick alveolar wall aids exchange loses the mark; thin walls, around 0.5 micrometres, are the advantageous feature.
4. 神经传导与突触传递: 动作电位与跳跃式传导
神经元静息电位约负 70 毫伏，由钠钾泵每次泵出三个钠离子、泵入两个钾离子，加上钾离子漏通道造成，膜内相对为负。刺激使膜电位达到阈值约负 55 毫伏时，电压门控钠通道打开，钠离子内流形成去极化，峰值约正 30 毫伏；随后钠通道失活、钾通道打开，钾离子外流复极，甚至出现短暂超极化。髓鞘包裹的轴突只在郎飞结处去极化，跳跃式传导速度可达每秒 100 米，而无髓鞘轴突只有每秒 1 米左右。
A neurone has a resting potential near minus 70 millivolts, generated by the sodium-potassium pump exporting three sodium ions for every two potassium ions imported, together with potassium leak channels, leaving the inside negative. When stimulation brings the membrane to the threshold near minus 55 millivolts, voltage-gated sodium channels open and sodium influx depolarises the membrane to about plus 30 millivolts. Sodium channels then inactivate and potassium channels open, so potassium efflux repolarises the membrane and may briefly overshoot into hyperpolarisation. In myelinated axons depolarisation occurs only at the nodes of Ranvier, and saltatory conduction reaches about 100 metres per second, against roughly 1 metre per second in unmyelinated axons.
突触传递的步骤是: 动作电位到达轴突末梢，钙离子内流，突触小泡与突触前膜融合，神经递质释放到突触间隙，与突触后膜受体结合，引起钠离子内流并产生兴奋性突触后电位。乙酰胆碱会被乙酰胆碱酯酶分解，终止信号，这也是新烟碱类杀虫剂的作用靶点: 它们与乙酰胆碱受体结合，使昆虫神经持续兴奋而死亡。由于体液免疫和神经传导方向单一，突触保证信号单向传递。
Synaptic transmission proceeds as follows: the action potential arrives at the axon terminal, calcium ions enter, vesicles fuse with the presynaptic membrane, neurotransmitter is released into the cleft, binds receptors on the postsynaptic membrane, and sodium influx generates an excitatory postsynaptic potential. Acetylcholinesterase breaks down acetylcholine and terminates the signal, and this is the target of neonicotinoid insecticides, which bind the acetylcholine receptor and keep insect neurones permanently excited. Because vesicles exist only on the presynaptic side, the synapse guarantees one-way transmission.
典型例题: 为什么动作电位沿轴突传导时幅度不衰减？要点: 动作电位是"全或无"事件，每一段膜都重新打开钠通道并再生一个新的动作电位，而不是原有电位被动衰减；因此幅度不随距离下降，只会被髓鞘损伤或无髓鞘化影响速度。另需记住不应期限制了传导方向，防止反向传播。
Worked example: why does an action potential travel along an axon without losing amplitude? It is an all-or-nothing event. Each patch of membrane opens its own sodium channels and regenerates a fresh action potential rather than a passively decaying voltage, so amplitude does not decay with distance; only speed is affected by myelination. The refractory period restricts conduction to one direction and prevents back-propagation.
5. 稳态与激素调控: 负反馈回路与血糖、体温、水盐平衡
稳态是维持内环境相对稳定的过程，依赖负反馈: 偏离设定值的变化被感受器检测，经协调中枢整合后由效应器逆转该变化。血糖设定值约 5 毫摩尔每升。进食后血糖升高，胰岛 beta 细胞分泌胰岛素，促进肝细胞把葡萄糖转化为糖原，并增加细胞对葡萄糖的摄取；血糖偏低时胰岛 alpha 细胞分泌胰高血糖素，促进糖原分解与糖异生。这一循环由肝脏作为主要效应器完成，1 型糖尿病源于 beta 细胞受损，2 型糖尿病则与靶细胞胰岛素受体敏感性下降有关。
Homeostasis maintains a relatively stable internal environment through negative feedback: a deviation from the set point is detected by receptors, integrated by a coordination centre, and reversed by effectors. Blood glucose is held near 5 millimoles per litre. After a meal, rising glucose stimulates beta cells of the pancreatic islets to secrete insulin, which promotes glycogenesis in liver cells and increases glucose uptake by cells; low glucose stimulates alpha cells to release glucagon, which promotes glycogenolysis and gluconeogenesis. The liver acts as the main effector. Type 1 diabetes results from damage to beta cells, while type 2 involves reduced sensitivity of insulin receptors on target cells.
体温调控由下丘脑负责: 体温升高时皮肤小动脉舒张、汗腺分泌增加；体温下降时小动脉收缩、骨骼肌战栗、立毛肌收缩。水盐平衡依赖抗利尿激素: 血浆渗透压升高被下丘脑渗透压感受器检测，垂体后叶释放抗利尿激素，促使集合管增加水通道蛋白，水分重吸收增加，尿量减少、尿液浓缩(渗透压升高)。
Thermoregulation is controlled by the hypothalamus: rising temperature causes vasodilation of skin arterioles and increased sweating, while falling temperature causes vasoconstriction, shivering and erection of hair. Water balance depends on antidiuretic hormone: raised plasma osmolarity is detected by hypothalamic osmoreceptors, the posterior pituitary releases antidiuretic hormone, aquaporins increase in the collecting duct, more water is reabsorbed, and urine becomes small in volume and high in osmolarity.
典型例题: 一名运动员在高温下运动两小时后尿量显著减少。解释这一现象并说明负反馈的各个环节。要点: 出汗导致失水、血浆渗透压升高(感受器)，下丘脑检测(协调)，垂体释放抗利尿激素增加(信号)，集合管水通透性上升、重吸收增加(效应器)，尿液浓缩(结果)，渗透压回落即完成负反馈。
Worked example: an athlete exercises for two hours in the heat and urine volume falls sharply. Explain this and identify each element of the feedback loop. Sweating causes water loss and a rise in plasma osmolarity as the stimulus, osmoreceptors in the hypothalamus detect it as the receptor, the pituitary releases more antidiuretic hormone as the signal, the collecting duct becomes more permeable and reabsorbs more water as the effect, urine becomes concentrated as the result, and osmolarity returns toward the set point, completing the loop.
易错提醒: 负反馈的关键是"逆转偏离"，不要写成"放大变化"。描述激素通路时不要漏掉感受器与效应器，这两项各占一个评分点。
Common pitfall: negative feedback reverses a deviation; do not describe it as amplifying change. Also, do not omit receptors and effectors when describing hormonal pathways, since each carries its own mark.
6. 免疫与器官移植: 特异性免疫与组织相容性
特异性免疫分为体液免疫与细胞免疫。B 淋巴细胞经克隆选择被抗原激活后分化为浆细胞，分泌抗体，同时留下记忆细胞；辅助 T 细胞通过分泌细胞因子激活 B 细胞与细胞毒性 T 细胞；细胞毒性 T 细胞识别被感染细胞表面的抗原并使其凋亡。抗体是四级结构的蛋白质，由两条重链与两条轻链组成，可变区决定了抗原结合特异性。记忆细胞解释了为什么第二次接触同一抗原时抗体产生更快、浓度更高，这是疫苗接种的免疫学基础。HIV 攻击辅助 T 细胞，因此同时破坏体液免疫与细胞免疫。
Specific immunity divides into humoral and cell-mediated responses. B lymphocytes activated by antigen through clonal selection differentiate into plasma cells that secrete antibodies, leaving memory cells behind. Helper T cells secrete cytokines that activate both B cells and cytotoxic T cells, and cytotoxic T cells recognise antigen on the surface of infected cells and trigger their destruction. An antibody is a quaternary structure protein with two heavy and two light chains, and the variable regions determine binding specificity. Memory cells explain why a second exposure to the same antigen produces antibodies faster and at higher concentration, which is the immunological basis of vaccination. HIV attacks helper T cells and therefore disables both humoral and cell-mediated immunity.
器官移植的成败取决于主要组织相容性复合体的匹配程度。供体与受体的 MHC 分子差异越大，受体免疫系统识别为"非己"并发动细胞毒性 T 细胞攻击的可能性越高，排斥反应越强。免疫抑制剂通过抑制淋巴细胞增殖来延长移植物存活时间，但会增加感染风险。输血时则关注红细胞表面抗原(ABO 与 Rh)，血型不合会引起凝集反应。
Transplant success depends on matching major histocompatibility complex molecules. The greater the difference between donor and recipient MHC, the more likely the recipient immune system is to identify the tissue as non-self and mount a cytotoxic T cell attack, so rejection is stronger. Immunosuppressant drugs suppress lymphocyte proliferation and prolong graft survival but raise infection risk. In transfusion the relevant antigens are those on red blood cells, namely the ABO and Rh systems, and incompatibility causes agglutination.
典型例题: 比较初次与再次免疫应答并解释疫苗加强针的原理。要点: 初次应答潜伏期长、抗体峰值低、以 IgM 为主；再次应答由记忆细胞介导，潜伏期短、峰值高、以 IgG 为主且持续时间长。因此加强针的作用是增加记忆细胞数量与抗体亲和力。
Worked example: compare the primary and secondary immune responses and explain booster vaccination. The primary response has a long lag phase, a low antibody peak and mainly IgM, while the secondary response is driven by memory cells and shows a short lag, a much higher peak, mainly IgG and a longer duration. A booster dose therefore increases both the number of memory cells and antibody affinity.
7. 备考策略与高分技巧
第一，读懂指令词。Define 要求给出准确的一句话定义，Outline 要求给出要点不展开，Explain 要求给出因果链，Compare 要求给出相同点与不同点，Discuss 要求呈现正反两面的证据。9 分长答题一般需要 9 个独立评分点，只写三段泛泛的描述很难拿高分。第二，把数据写进答案。血红蛋白饱和度、心输出量、肺泡通气量、血糖设定值这类数字是你区别于其他考生的关键。第三，图表题先看坐标轴与单位，再读趋势，最后解释机制。第四，动手画图: 小肠绒毛、肺泡、神经元突触都要能在 3 分钟内画出并标注，标注线不要交叉，单位不要遗漏。第五，整理错题本时按主题分类而非按试卷分类，同一主题的高频评分点往往反复出现。
First, decode the command terms. Define asks for a precise one-sentence definition, Outline asks for points without development, Explain demands a causal chain, Compare requires both similarities and differences, and Discuss expects evidence on both sides. A nine-mark extended response usually needs nine independent marking points, so three vague paragraphs will not earn full marks. Second, put numbers into your answers: haemoglobin saturation, cardiac output, alveolar ventilation and the glucose set point are the details that separate strong scripts from average ones. Third, read axes and units before trends, then explain the mechanism. Fourth, practise diagrams: villus, alveolus and synapse should each be drawn and labelled within three minutes, with uncrossed label lines and units included. Fifth, organise your error log by theme rather than by paper, because the same marking points recur across exam sessions.
复习节奏建议: 用三天一轮滚动复习六大主题，每天先闭卷默写该主题的评分点清单，再做一道 9 分长答题并对照评分标准自评。把丢分的句子单独抄写一遍并改写，这比重复阅读课本的收益更高。若能在考前把这六大主题的评分点写成一张 A4 纸，Human Physiology 通常能稳定拿到七成以上分数。
For revision pacing, cycle through the six themes every three days. Each day, write the marking point checklist from memory, then answer one nine-mark question and mark it against the official criteria. Copy out every sentence that lost marks and rewrite it, which pays off far more than rereading the textbook. If you can compress the marking points of all six themes onto a single sheet of A4 before the exam, Human Physiology will usually settle above seventy per cent.
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