Skip to content

能量、环境、微生物学和免疫学

Pearson Edexcel · International A-Level · 生物 · 知识点 4

查看幻灯片
4.1

Energy in, energy through, energy lost

A wheat field intercepts sunlight, fixes carbon, feeds a food chain — and loses energy at every step. This unit follows the energy from photon to ecosystem, then turns to the smallest participants: the microbes that recycle, the pathogens that take, and the immune system that defends. WBI14 is also the most calculation-heavy paper after Unit 1: expect NPP, growth constants, diversity indices and time-of-death arithmetic.

WBI14 Energy, Environment, Microbiology and Immunity is the first IA2 unit: 1 hour 30 minutes, 90 marks, all compulsory. It assesses Topic 5 (Energy Flow, Ecosystems and the Environment) and Topic 6 (Microbiology, Immunity and Forensics).

4.1

光合作用与叶绿体色素(知识点5.1–5.8,核心实践10)

教学大纲

主题 5 陈述 5.1-5.8 及核心实验 10 (考纲 pp.29-30)。光合作用的总反应式;光依赖反应(水的光解、ADP的光磷酸化、NADP的还原)发生在类囊体膜上;光非依赖反应(卡尔文循环:RuBP催化固定二氧化碳、GALP生成、核酮糖二磷酸再生)发生在基质中;叶绿体结构与这两个阶段的对应关系;吸收光谱和作用光谱;色素的纸层析法及Rf值(核心实验 10);限制因子(光照强度、二氧化碳浓度、温度)及其农业应用调控。

来源:Cambridge International 教学大纲

The overall reaction: $6\mathrm{CO_2} + 6\mathrm{H_2O} \rightarrow \mathrm{C_6H_{12}O_6} + 6\mathrm{O_2}$, driven by light energy. In the thylakoid 类囊体 membranes, the light-dependent reactions 光反应 photolyse water (releasing O₂ and H⁺), photophosphorylate 光合磷酸化 ADP and reduce NADP. In the stroma 基质, the light-independent reactions 暗反应 (Calvin cycle) fix carbon dioxide onto ribulose bisphosphate with rubisco, reduce the products using the ATP and reduced NADP from stage one, and make GALP — some to glucose, most to regenerate RuBP.

Structure matches function: thylakoids stacked for absorbing surface; stroma packed with enzymes for the cycle.

Pigments and spectra

Chlorophyll a and b absorb red and blue light; carotenoids widen the range and protect. The absorption spectrum 吸收光谱 shows what each pigment takes up; the action spectrum 作用光谱 shows what rate of photosynthesis each wavelength drives — their agreement is evidence that these pigments do the work. Chromatography separates pigments by solubility; $\mathrm{Rf} = \text{distance moved by pigment} / \text{distance moved by solvent}$.

Absorption spectra of the chloroplast pigments, with the shaded action spectrum tracking their combined absorption.

Limiting factors 限制因素: light intensity, CO₂ concentration, temperature. A greenhouse raises the weakest of the three — and exam answers say which and why.

4.2

生产力与能量传递(知识点5.9–5.10)

教学大纲

陈述 5.9-5.10 (考纲 p.30):总初级生产力(GPP)、净初级生产力(NPP = GPP减去呼吸消耗)及所用单位(kJ m-2 yr-1);营养级间的生物量传递效率及其计算;能量传递损失的原因(呼吸作用、排遗、排泄、未食用部分)以及食物链为何较短。

来源:Cambridge International 教学大纲

Gross primary productivity 总初级生产力 (GPP) is the energy fixed by photosynthesis per area per time (kJ m⁻² yr⁻¹). Plants spend much of it on their own respiration; what remains is net primary productivity 净初级生产力:

$$\mathrm{NPP} = \mathrm{GPP} - R$$

Only about 10 % of the energy at each trophic level passes upward — lost to respiration as heat, egestion, excretion and parts never eaten. That is why chains rarely exceed four or five levels and why top predators are rare.

Energy leaving a food chain: GPP split by plant respiration, then roughly a tenth passing at each step upward.

Worked check. A grassland fixes 45 000 kJ m⁻² yr⁻¹ and respirers 19 000. NPP $= 26\,000$ kJ m⁻² yr⁻¹. Cattle eating that grass keep about 10 %: 2 600 kJ m⁻² yr⁻¹. Humans eating the cattle keep 260 — the energy arithmetic behind "eating lower on the chain feeds more people".

4.3

生态系统、种群与演替(知识点5.11–5.15,核心实践11)

教学大纲

陈述 5.11-5.15 及核心实验 11 (考纲 pp.30-31):生境、种群、群落和生态系统的术语;控制分布和数量的生物和非生物因素;生态位概念的应用;捕食者-猎物周期;测量密度的方法(样方、样线、标志重捕法);从先锋物种定殖到顶级群落的演替阶段(核心实验 11 调查生境)。

来源:Cambridge International 教学大纲

Four definitions, precisely: habitat 栖息地 (where), population 种群 (one species' individuals), community 群落 (all populations), ecosystem 生态系统 (community + environment). Abiotic factors (light, water, temperature, pH) and biotic factors (competition, predation, disease) set who lives where and how abundantly — together, the niche 生态位.

Predator and prey cycle with a delay: hares rise, lynx follow, hares fall, lynx starve — a lag of about a quarter cycle.

A classic predator-prey cycle: the lynx population tracks the hare population with a delay.

Succession 演替 begins when pioneer species colonise bare ground; each stage changes the soil and microclimate, enabling the next, until the climax community 顶极群落 stabilises. Measuring abundance uses quadrats and transects (random sampling for estimates, systematic along a gradient) and capture-mark-recapture for animals ($N \approx \text{first catch} \times \text{second catch} / \text{marked recaptures}$).

4.4

气候变化的证据、成因及影响(知识点5.16–5.22,核心实践12)

教学大纲

陈述 5.16-5.22 及核心实验 12 (考纲 pp.31-32):气候变化的证据(温度记录、冰芯、树木年轮、泥炭中的花粉);温室气体及温室效应增强(CO2、甲烷、氧化亚氮——来源、大气寿命、全球变暖潜能值);碳循环及降低大气碳含量的方法;数据外推及其可靠性;气候变化对降雨模式、物种分布、酶依赖性生态系统的影响;核心实验 12 探究温度对酶/相关生境活性的影响。

来源:Cambridge International 教学大纲

Evidence comes from direct temperature records, ice cores (trapped air giving ancient CO₂), tree rings (dendrochronology 树轮年代学) and pollen layers in peat. The enhanced greenhouse effect 温室效应: outgoing long-wave radiation is absorbed by CO₂, methane and nitrous oxide and re-emitted downward. The gases differ — methane is potent but short-lived; CO₂ lasts centuries; nitrous oxide combines both — so policy weighs global warming potential 全球增温潜势 against lifetime.

Extrapolating a trend beyond its data is a judgement, not a fact: state the assumption (the trend continues) and its weakness. Effects already examined: shifting rainfall and species ranges, and — close to this course — temperature's effect on enzyme rates in cold-blooded organisms and whole ecosystems.

Worked check. Methane's warming potential is about 28× CO₂'s per kg, but it lasts ~12 years against centuries. A one-tonne methane leak matters 28× a one-tonne CO₂ leak this decade, but fades while the CO₂ persists — the arithmetic of "warming potential vs cumulative load".

4.5

进化与物种形成(知识点5.23–5.26)

教学大纲

陈述5.23-5.26(大纲页32):进化作为由突变、选择(定向选择和稳定化选择)、基因流和遗传漂变引起的等位基因频率变化;隔离减少基因流(地理隔离=异域物种形成;生殖隔离=同域物种形成)导致物种形成;根据证据标准解读有争议的科学结论(气候、进化);重新造林、可持续资源和生物燃料。

来源:Cambridge International 教学大纲

Evolution is a change in allele frequency 等位基因频率. It needs variation (mutation, meiosis) and a filter: directional selection 定向选择 shifts the mean when the environment moves; stabilising selection 稳定选择 trims the extremes in a stable one. Drift changes frequencies by chance in small populations; gene flow (migration) mixes them back.

Speciation 物种形成 needs isolation: allopatric 异域的 (a physical barrier — allopatric = different homeland) or sympatric 同域的 (reproductive isolation in the same place — behavioural, temporal or genetic). Isolated populations diverge until they can no longer interbreed.

4.6

微生物培养与生长曲线(知识点6.1–6.4,核心实践13)

教学大纲

陈述6.1-6.4及核心实验13(大纲页33):培养基(营养琼脂、肉汤);无菌技术的细节;测量微生物生长的方法(细胞计数、浊度、通过稀释涂布法的活菌计数);细菌生长曲线的四个阶段(延滞期、对数/指数期、稳定期、衰亡期)及对数增长速率常数k;核心实验13利用校准曲线研究液体培养物中的生长速率。

来源:Cambridge International 教学大纲

Aseptic technique protects you and the culture: flame the loop to red heat, flame bottle necks, lift lids briefly, seal plates with tape, incubate at 25 °C (never body temperature, so human pathogens cannot multiply). Measure growth by direct cell count (haemocytometer), turbidity 浊度 in broth, or viable counts by dilution plating 稀释涂布 (colonies counted × dilution factor).

The growth curve: lag 延迟期 (enzymes synthesising), log 对数期 (exponential doubling), stationary 稳定期 (deaths = divisions), death 衰亡期. The exponential constant:

$$k = \frac{\log_{10} N_t - \log_{10} N_0}{0.301 \times t}$$

Worked check. From $2\times10^4$ to $1.6\times10^7$ cells in 5 hours: log difference $= 2.903$; $k = 2.903/(0.301\times5) = 1.93$ generations per hour — a doubling time of about 31 minutes.

4.7

病原体、感染途径及结核病/HIV(知识点6.5–6.7)

教学大纲

陈述6.5-6.7(大纲页33):比较细菌和病毒的结构(核酸、衣壳、包膜);感染途径(消化道、呼吸道、性接触、伤口、媒介);感染屏障(皮肤、黏液、胃酸、血液凝固);结核分枝杆菌和HIV如何感染并损害人体,以及HIV如何加重结核病。

来源:Cambridge International 教学大纲

Bacteria: cell wall, plasmids, ribosomes, circular DNA, some with capsule and flagellum. Viruses: nucleic acid (DNA or RNA) in a capsid 衣壳, some with an envelope — no organelles, no metabolism of their own; they hijack host cells. Mycobacterium tuberculosis 结核分枝杆菌 infects lungs, surviving inside macrophages; HIV 人类免疫缺陷病毒 destroys T helper cells, so the immune system collapses and TB reactivates — the classic AIDS-defining illness.

Barriers come first: skin, mucus and cilia, stomach acid, lysozyme in tears, and the clotting cascade sealing wounds.

4.8

免疫反应、抗原与抗体(知识点6.8–6.12)

教学大纲

陈述6.8-6.12(大纲页34):非特异性反应(炎症、溶菌酶、干扰素);抗原;体液反应——B效应细胞分化为分泌抗体的浆细胞,B记忆细胞介导二次反应;细胞反应——T辅助细胞、T杀伤细胞和T记忆细胞;抗体结构(四条多肽链,可变区结合位点)及抗原-抗体复合物;天然与人工、主动与被动免疫;疫苗接种及群体免疫;病原体与宿主之间的进化军备竞赛。

来源:Cambridge International 教学大纲

Non-specific: inflammation (vasodilation and fluid leak bringing phagocytes), phagocytosis, interferon. Specific — antigens 抗原 trigger it:

  • Humoral 体液: B effector cells → plasma cells → antibodies 抗体; B memory cells B记忆细胞 remain for the faster, stronger secondary response.
  • Cellular 细胞: T helper cells coordinate, T killer cells T杀伤细胞 destroy infected cells, T memory cells persist.

An antibody is four polypeptide chains (two heavy, two light) with a variable region 可变区 forming one specific binding site — the shape that fits one antigen. Immunity splits two ways: active 主动 (your own antibodies — natural infection or vaccination) vs passive 被动 (given antibodies — mother to baby, antiserum); natural vs artificial in each. Vaccination works at the scale of populations too (herd immunity 群体免疫) — but pathogens evolve back: the evolutionary race.

Worked check. A vaccine's second dose produces antibody levels ten times the first and within days rather than weeks: clonal selection finds the memory B cells, which divide rapidly into plasma cells — the secondary response the primary response built.

4.9

抗生素与医院获得性感染(知识点6.13–6.15,核心实践14)

教学大纲

陈述6.13-6.15及核心实验14(大纲页34):抑菌剂与杀菌剂类抗生素;抗生素耐药性的产生与传播(自然选择);核心实验14探究抗生素对细菌的作用;如何通过了解医院获得性感染的成因(卫生条件、侵入性操作、耐药菌株)来降低其发生率。

来源:Cambridge International 教学大纲

Bacteriostatic 抑菌的 antibiotics stop bacterial growth; bactericidal 杀菌的 ones kill. Resistance arises by mutation and spreads by selection whenever exposure kills the susceptible and spares the resistant — finishing a course and avoiding unnecessary use both slow it. Hospital-acquired infections thrive where vulnerable patients, invasive devices and resistant strains meet; hand hygiene, sterile procedure and isolating carriers break the chain.

4.10

分解作用、PCR、DNA指纹鉴定与死亡时间推断(知识点6.16–6.20)

教学大纲

陈述 6.16-6.20 (考纲 pp.34-35):微生物在分解和养分循环中的作用;聚合酶链式反应(变性、引物退火、延伸)扩增DNA;凝胶电泳按长度分离DNA片段;用于识别和遗传关系的DNA指纹鉴定;根据体温、尸僵程度、肌肉收缩程度及昆虫演替顺序推断死亡时间。

来源:Cambridge International 教学大纲

Decomposers recycle carbon and nitrogen — the ecosystem's waste system. The polymerase chain reaction 聚合酶链式反应 cycles denaturation (95 °C), primer annealing (50–60 °C) and extension (72 °C), doubling the target DNA each cycle: $n$ cycles give $2^n$ copies. Gel electrophoresis 凝胶电泳 drags negatively charged DNA fragments through gel — short fragments run furthest. The profile 电泳图谱 of band positions identifies individuals (enough loci differ between unrelated people) and measures relatedness (shared bands).

Time of death narrows by four clocks: body temperature (falls ~1 °C per hour, modified by size, clothing, air), rigor mortis 尸僵 (sets then passes), decomposition degree, and insect succession on the body — each stage's arrivals time-stamp the interval.

Worked check. A body found at 22 °C core temperature in a 15 °C room: $(37-22)/(1\ ^\circ\text{C per hour}) \approx 15$ hours — but state the assumptions (still air, average build) before trusting it.

4.10

自我检测

  1. Name the products of the light-dependent reactions used by the Calvin cycle.
  2. A crop fixes 60 000 kJ m⁻² yr⁻¹ and respirers 20 000. Calculate NPP and the energy reaching a third trophic level at 10 % transfer each step.
  3. Explain why the action spectrum is evidence that chlorophyll carries out photosynthesis.
  4. Distinguish directional and stabilising selection with one example of each.
  5. Give two reasons a food chain rarely exceeds five trophic levels.
  6. A culture grows from $5\times10^3$ to $4\times10^6$ in 6 hours. Calculate k and the doubling time.
  7. Explain why a second vaccine dose raises antibody titre far faster than the first.
  8. State the difference between bacteriostatic and bactericidal, and why the distinction matters for a patient with a weak immune system.
  9. Three PCR cycles from one double-stranded template: how many copies?
  10. A body's core is 30 °C. Estimate the time since death under standard assumptions and name one factor that would extend the true interval.

Answers: 1 ATP and reduced NADP; 2 NPP $= 40\,000$; third level $= 400\,\text{kJ m}^{-2}\text{yr}^{-1}$ (10 % of 4 000); 3 wavelengths chlorophyll absorbs best are the wavelengths that drive photosynthesis best; 4 directional — antibiotic resistance shifting the mean; stabilising — human birth weights; 5 energy lost at each step to respiration/heat/egestion, so little remains; producers also lose to respiration before the first transfer; 6 log difference 2.903, $k = 2.903/(0.301\times6) = 1.61$ h⁻¹, doubling ≈ 37 min; 7 memory B cells persist; the second encounter selects clones that divide rapidly into plasma cells — the secondary response; 8 bacteriostatic stops growth (immune system must clear), bactericidal kills outright — a weak immune system needs bactericidal support; 9 $2^3 = 8$ copies; 10 $(37-30)/1 = 7$ hours; heavy clothing or a warm room slows cooling, so true time is longer.

更多 Pearson Edexcel · International A-Level · 生物 知识点

登录或创建账户

IGCSE、A-Level 与 AP