Core Biology 备考攻略 规划

## IB Biology 细胞呼吸 光合 遗传 DNA 分子 考点 IB Biology is one of the most content-rich subjects in the IB Diploma Programme, covering everything from the molecular machinery inside cells to the grand sweep of evolution and ecology. 无论你选的是 SL (Standard Level) 还是 HL (Higher Level), Core Topics 1-6 构成了你试卷的主体，而其中 Topics 1 (Cell Biology), 2 (Molecular Biology), 3 (Genetics), 以及 Topic 8 (Metabolism HL only) 是最让考生头疼的板块。本篇文章将带你逐一击破细胞呼吸、光合作用、遗传学、DNA 复制与转录翻译、以及分子生物学核心考点，中英双语讲解，助你 Paper 1 和 Paper 2 双双冲 7。 SL Core Topics
Topic 1: Cell Biology
Topic 2: Molecular Biology
Topic 3: Genetics
Topic 4: Ecology
Topic 5: Evolution & Biodiversity
Topic 6: Human Physiology HL Additional Topics
Topic 7: Nucleic Acids
Topic 8: Metabolism, Respiration, Photosynthesis
Topic 9: Plant Biology
Topic 10: Genetics & Evolution
Topic 11: Animal Physiology ### 1. Cellular Respiration (细胞呼吸) — 能量的货币是 ATP Every living cell needs energy, and ATP (adenosine triphosphate) is the universal energy currency. 细胞通过细胞呼吸 (cellular respiration) 将葡萄糖等有机物中的化学能转化为 ATP。你需要清楚地区分 aerobic respiration (有氧呼吸) 和 anaerobic respiration (无氧呼吸) 的两条路径。 **Aerobic Respiration 关键步骤:** Glycolysis (糖酵解)
Location: Cytoplasm
Input: Glucose (6C)
Output: 2 pyruvate (3C) + 2 ATP (net) + 2 NADH
No O₂ required Link Reaction (连接反应)
Location: Mitochondrial matrix
Input: Pyruvate → Acetyl-CoA
Output: CO₂ + NADH
2 pyruvate processed per glucose Krebs Cycle (克雷布斯循环)
Location: Mitochondrial matrix
Key products: 2 ATP (GTP), 6 NADH, 2 FADH₂, 4 CO₂
Acetyl-CoA enters, oxaloacetate regenerated ETS / Oxidative Phosphorylation
Location: Inner mitochondrial membrane
NADH and FADH₂ drop off electrons
Proton gradient drives ATP synthase
~28-34 ATP produced
O₂ is the final electron acceptor → H₂O **Exam Tip:** HL students must be able to draw and label the structure of a mitochondrion, including cristae (inner membrane folds), matrix, inner and outer membranes, and intermembrane space. 考试中经常出现 "Explain the role of oxygen in aerobic respiration" — oxygen acts as the terminal electron acceptor in the electron transport chain; without it, the ETC backs up, and NADH cannot be reoxidized. **Anaerobic Respiration in Humans:** When oxygen is scarce (e.g., during intense sprinting), pyruvate is converted to lactate (lactic acid) in the cytoplasm, regenerating NAD⁺ so glycolysis can continue. This produces only the 2 ATP from glycolysis — no Krebs, no ETC. In yeast and plants, pyruvate is instead converted to ethanol and CO₂ (alcohol fermentation). Remember: fermentation does NOT produce ATP directly; it only regenerates NAD⁺. ### 2. Photosynthesis (光合作用) — 把光变成糖 Photosynthesis is essentially the reverse of respiration: it uses light energy to convert CO₂ and H₂O into glucose and O₂. The overall equation: 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂. However, this simple summary hides an intricate two-stage process. **Light-Dependent Reactions (光反应):** Occur in the thylakoid membranes of chloroplasts. Photosystem II (PSII) absorbs light at 680 nm, splitting water molecules (photolysis): 2H₂O → 4H⁺ + 4e⁻ + O₂. Electrons pass through an electron transport chain, pumping protons into the thylakoid lumen. Photosystem I (PSI) re-energizes electrons at 700 nm, which then reduce NADP⁺ to NADPH. The proton gradient drives ATP synthase — this is called photophosphorylation (光合磷酸化). Key products: ATP, NADPH, O₂. SL vs HL: SL students need to know the basic flow: light energy → excited electrons → ATP + NADPH produced. HL must distinguish cyclic photophosphorylation (only PSI, produces ATP only) from non-cyclic (both PSI and PSII, produces ATP + NADPH + O₂). **Light-Independent Reactions / Calvin Cycle (暗反应 / 卡尔文循环):** Taking place in the stroma, the Calvin cycle uses ATP and NADPH from the light reactions to fix CO₂ into glucose. Three phases: 1. Carbon Fixation
CO₂ + RuBP (5C) → 2 × GP (3C)
Catalyzed by RuBisCO 2. Reduction
GP → TP (G3P)
Uses ATP + NADPH 3. Regeneration
5/6 TP → RuBP (5C)
Uses ATP For every 6 CO₂ fixed, 1 TP (triose phosphate) molecule is produced that can be converted into glucose, starch, or cellulose. RuBisCO is famously the most abundant enzyme on Earth — and also famously inefficient, sometimes fixing O₂ instead of CO₂ (photorespiration, HL only). **Absorption vs Action Spectra:** The absorption spectrum shows which wavelengths of light chlorophyll a and b absorb (peaks in red ~680 nm and blue ~450 nm). The action spectrum shows the rate of photosynthesis at each wavelength — they correlate strongly, demonstrating that absorbed light drives photosynthesis. Green light (~550 nm) is reflected, which is why leaves appear green. **Limit Factors:** Temperature, light intensity, and CO₂ concentration all limit the rate of photosynthesis. Know how to interpret graphs with plateaus and the concept of the limiting factor (限制因子). Past paper questions love asking: "Which factor is limiting at point X? Justify your answer." ### 3. Genetics (遗传学) — 从孟德尔到现代 IB Genetics covers topics ranging from Mendelian inheritance patterns to modern gene technology. 遗传学是 IB Biology 中最具逻辑推理性的板块，如果你擅长解谜题，这部分会很对你的胃口。 **核心术语必须滚瓜烂熟:** Gene (基因): A heritable factor consisting of a length of DNA that influences a specific characteristic. A gene occupies a specific position (locus) on a chromosome.

Allele (等位基因): One specific form of a gene, differing from other alleles by one or a few bases. Example: the gene for eye color has alleles for blue, brown, green, etc.

Genome (基因组): The whole of the genetic information of an organism. The Human Genome Project completed sequencing all ~3 billion base pairs in 2003. Genotype vs Phenotype: Genotype is the allele combination (e.g., Tt); phenotype is the observable trait (e.g., tall). Phenotype = genotype + environment.

Dominant / Recessive / Codominant: Dominant alleles mask recessive ones in heterozygotes. Codominant alleles are both expressed equally (e.g., AB blood type).

Homozygous vs Heterozygous: Having two identical alleles (TT or tt) vs two different alleles (Tt). **Monohybrid Crosses (单因子杂交):** Use Punnett grids (庞纳特方格) to predict offspring genotypes and phenotypes. For a cross between two heterozygotes (Tt × Tt), the expected ratio is 3:1 in phenotype (dominant:recessive) and 1:2:1 in genotype (TT:Tt:tt). Remember: these are expected ratios — actual offspring may deviate due to chance, especially with small sample sizes. **Dihybrid Crosses (HL):** Two genes on different chromosomes assort independently (Mendel's Law of Independent Assortment). For AaBb × AaBb, the expected phenotypic ratio is 9:3:3:1. 考试中最常见的陷阱是给你 non-Mendelian ratios (like 9:3:4 or 12:3:1), 这时候你要想到 epistasis (上位效应) — HL 必考。 **Linked Genes (连锁基因):** Genes on the same chromosome do NOT assort independently — they are inherited together unless crossing over separates them during meiosis I. The closer two genes are on a chromosome, the lower the recombination frequency. Recombination frequency = (number of recombinant offspring / total offspring) × 100%. This is used to construct gene maps. **Sex Linkage (性连锁):** Genes on the X chromosome (but not on the Y) show distinctive inheritance patterns. Because males are hemizygous (XY, only one X), they express X-linked recessive traits more frequently — think red-green color blindness and hemophilia. Always write genotypes showing the chromosome: X^R X^r (carrier female), X^r Y (affected male). **Pedigree Charts (系谱图) — 必考题型:** When analyzing a pedigree, ask these questions in order: 1. Is the trait dominant or recessive? (Recessive traits often skip generations; dominant traits appear in every generation.) 2. Is it autosomal or sex-linked? (If more males are affected, suspect X-linked recessive. If an affected father passes the trait to all daughters but no sons, suspect X-linked dominant.) 3. Determine genotypes of all individuals using the evidence. ### 4. DNA Structure, Replication, Transcription, and Translation This is the central dogma (中心法则) of molecular biology: DNA → RNA → protein. 这四者的协同运作是 IB HL Paper 2 的绝对核心考点，Section B 长篇论述题 (8-mark essays) 经常从这里出。 **DNA Structure (DNA 结构):** DNA is a double helix of two antiparallel strands. Each strand consists of a sugar-phosphate backbone (deoxyribose + phosphate, linked by covalent phosphodiester bonds) with nitrogenous bases projecting inward. The bases pair via hydrogen bonds: Adenine with Thymine (2 H-bonds), Cytosine with Guanine (3 H-bonds). This is complementary base pairing. The two strands run antiparallel: one 5' → 3', the other 3' → 5'. HL students should be able to draw a simplified nucleotide and label the 3' and 5' carbons. **DNA Replication (DNA 复制):** DNA replication is semiconservative — each new DNA molecule contains one original strand and one newly synthesized strand. This was proven by the Meselson-Stahl experiment using heavy (¹⁵N) and light (¹⁴N) nitrogen isotopes, with centrifugation producing distinct density bands. Key enzymes: - **Helicase (解旋酶):** Unwinds the double helix at the replication fork by breaking H-bonds. - **DNA Gyrase (HL only, DNA 旋转酶):** Relieves supercoiling tension ahead of the replication fork. - **SSB Proteins (单链结合蛋白):** Prevent separated strands from reannealing. - **DNA Primase (引物酶):** Synthesizes short RNA primers to provide a free 3'-OH for DNA polymerase. - **DNA Polymerase III:** Adds DNA nucleotides to the 3' end of the growing strand. Always synthesizes 5' → 3'. The leading strand is synthesized continuously; the lagging strand is synthesized in short Okazaki fragments (冈崎片段). - **DNA Polymerase I:** Removes RNA primers a nd replaces with DNA. - **DNA Ligase (连接酶):** Seals nicks between Okazaki fragments by forming phosphodiester bonds. PCR (Polymerase Chain Reaction): An artificial DNA replication technique. Three steps per cycle: denaturation (~95°C, separate strands), annealing (~55°C, primers bind), extension (~72°C, Taq polymerase synthesizes). After n cycles, 2ⁿ copies are produced. **Transcription (转录):** Transcription is the synthesis of mRNA from a DNA template. Steps: 1. RNA polymerase binds to the promoter region (rich in TATA box sequences). 2. DNA unwinds; RNA polymerase reads the template (antisense) strand 3' → 5'. 3. RNA polymerase synthesizes a complementary mRNA strand 5' → 3', using uracil (U) instead of thymine (T). 4. Terminator sequence signals the end of the gene; mRNA is released. In eukaryotes, the primary transcript (pre-mRNA) undergoes post-transcriptional modification: a 5' cap (methylated guanine) and a 3' poly-A tail are added for stability; introns (non-coding) are spliced out; exons (coding) are joined together. 这一部分 HL 才需要掌握，但 SL 同学了解 splicing 的概念对 IA 写作也很有帮助。 **Translation (翻译):** Translation is the synthesis of a polypeptide chain using mRNA as a template, carried out by ribosomes. Key components: - **mRNA:** Carries the codon sequence. Each codon (3 bases) specifies one amino acid. - **tRNA:** Has an anticodon (complementary to the mRNA codon) and carries the corresponding amino acid. - **Ribosome (核糖体):** Composed of rRNA and proteins; has A (aminoacyl), P (peptidyl), and E (exit) sites. Steps: 1. **Initiation:** Small ribosomal subunit binds to mRNA at the start codon (AUG, codes for methionine). The tRNA with anticodon UAC brings methionine. The large subunit joins. 2. **Elongation:** A new tRNA carrying the next amino acid enters the A site. A peptide bond forms between the amino acids in the P and A sites (catalyzed by peptidyl transferase, a ribozyme made of rRNA). The ribosome translocates one codon forward; the empty tRNA exits through the E site. 3. **Termination:** When a stop codon (UAA, UAG, or UGA) reaches the A site, a release factor binds, causing the polypeptide to detach. No tRNA corresponds to stop codons. The genetic code is universal (across nearly all organisms), degenerate (multiple codons can code for the same amino acid — this is the "wobble" hypothesis), and non-overlapping. Using a codon table to translate mRNA into an amino acid sequence is a guaranteed Paper 1 and Paper 2 skill. ### 5. Molecular Biology (分子生物学) — 酶与代谢 **Enzymes (酶):** Enzymes are globular proteins that act as biological catalysts, lowering the activation energy of reactions without being consumed. The active site has a specific shape complementary to the substrate (lock-and-key model). The induced-fit model refines this: the active site changes shape slightly upon substrate binding, straining bonds and facilitating the reaction. Factors affecting enzyme activity: - **Temperature:** Rate increases with temperature (more kinetic energy → more collisions) until the optimum. Beyond this, the enzyme denatures (tertiary structure disrupted, active site loses shape). Human enzymes typically have optima around 37°C; thermophilic bacteria enzymes can work at 80°C+. - **pH:** Each enzyme has an optimal pH. Pepsin (stomach) works best at pH ~2; trypsin (small intestine) at pH ~8. Extreme pH denatures enzymes by disrupting ionic bonds and hydrogen bonds in the tertiary structure. - **Substrate concentration:** Rate increases until saturation — all active sites are occupied (Vmax reached). **Competitive vs Non-competitive Inhibition:** Competitive (竞争性抑制): Inhibitor resembles substrate and binds to active site. Can be overcome by increasing substrate concentration. Vmax unchanged; Km (apparent) increases. Example: statins inhibit HMG-CoA reductase by mimicking the substrate. Non-competitive (非竞争性抑制): Inhibitor binds to allosteric site, changing active site shape. Cannot be overcome by adding substrate. Vmax decreases; Km unchanged. Example: cyanide binds to cytochrome c oxidase in the ETC. **End-product inhibition (终产物抑制):** A form of negative feedback where the final product of a metabolic pathway inhibits the first enzyme in the pathway. Example: isoleucine inhibits threonine deaminase, the enzyme catalyzing the first step of isoleucine synthesis. This prevents wasteful overproduction. ### 6. Meiosis and Crossing Over (减数分裂与交叉互换) Meiosis is the foundation of sexual reproduction and genetic variation. Unlike mitosis, meiosis involves two consecutive divisions (Meiosis I and Meiosis II) producing four genetically distinct haploid daughter cells from one diploid parent cell. **Meiosis I — Reduction Division:** Prophase I is the critical stage. Homologous chromosomes pair up (synapsis) forming bivalents. Crossing over occurs at chiasmata — non-sister chromatids exchange segments of DNA, creating new allele combinations on the same chromosome. This is the primary source of genetic variation. Metaphase I: bivalents align randomly on the equator (independent assortment). Anaphase I: homologous chromosomes separate to opposite poles — this reduces the chromosome number from diploid (2n) to haploid (n). Telophase I: two haploid cells form, each with chromosomes still composed of two sister chromatids. **Meiosis II — Separation of Sister Chromatids:** Meiosis II resembles mitosis. Sister chromatids separate in Anaphase II, producing four genetically unique haploid gametes. 记住 SL 的核心考点: Meiosis produces genetic variation through (1) crossing over in Prophase I (2) random orientation of bivalents in Metaphase I, and (3) random fusion of gametes at fertilization. HL students should also understand non-disjunction — the failure of chromosomes to separate properly, leading to aneuploidy (e.g., Down syndrome, trisomy 21). ### 7. Cell Membrane Structure and Transport (细胞膜与物质运输) The cell membrane is described by the Singer-Nicolson fluid mosaic model (流动镶嵌模型). It consists of a phospholipid bilayer with hydrophilic phosphate heads facing outward and hydrophobic fatty acid tails facing inward. Membrane proteins are scattered throughout: integral proteins span the membrane; peripheral proteins attach to the surface. Cholesterol molecules (in animal cells) regulate membrane fluidity. **Transport Mechanisms:** Simple Diffusion (简单扩散): Passive movement down a concentration gradient. No protein channel needed. Small non-polar molecules (O₂, CO₂) and lipid-soluble substances cross freely. Facilitated Diffusion (协助扩散): Passive movement through channel or carrier proteins. Specific, saturable. Example: glucose uptake via GLUT4 transporters, potassium ion channels. Active Transport (主动运输): Movement against the concentration gradient using ATP. Carrier proteins act as pumps. Example: Na⁺/K⁺ pump moves 3 Na⁺ out and 2 K⁺ in per ATP. **Osmosis (渗透):** The passive movement of water across a selectively permeable membrane from low solute concentration (high water potential) to high solute concentration (low water potential). In hypertonic solutions, animal cells shrink (crenation); in hypotonic solutions, they swell and burst (lysis). Plant cells become turgid in hypotonic solutions (the cell wall prevents bursting) and undergo plasmolysis in hypertonic solutions. Osmolarity calculations using solute concentration units are a common Paper 1 calculation question. **Vesicular Transport (HL):** Endocytosis brings materials into the cell via membrane invagination (phagocytosis for solids, pinocytosis for liquids). Exocytosis expels materials when vesicles fuse with the plasma membrane. Both require ATP. ### 8. Exam Strategy: Paper 1 & Paper 2 高分技巧 **Paper 1 (Multiple Choice):** - Read every option carefully — distractors often contain half-truths. - For Data-Based Questions (DBQs), extract numbers from graphs and tables directly; do not rely on memory. - Watch for "EXCEPT" and "NOT" in questions. - Time management: ~1 minute per question for SL (30 questions/30 min), ~50 seconds per question for HL (40 questions/60 min). **Paper 2 (Short Answer + Extended Response):** - Section A: Data-based and short answer. Always quote data: "As shown in the graph, the rate increased from 2.5 to 7.8 arbitrary units..." - Section B: HL has extended response questions (通常 8 分). Use structured paragraphs. Include definitions, processes, examples, and where possible, diagrams or annotated sketches. - Command terms matter: "Outline" = brief summary; "Describe" = detailed account; "Explain" = cause and effect with reasons; "Discuss" = multiple perspectives; "Evaluate" = strengths and limitations. **Internal Assessment (IA) Tips:** - Your IA is worth 20% of your final grade. Choose a topic with measurable variables and a clear hypothesis. - Personal engagement (2 marks): Show genuine interest — explain why you chose the topic. - Exploration (6 marks): Include background theory, detailed methodology, and safety/ethical considerations. - Analysis (6 marks): Process raw data appropriately (means, standard deviation, statistical tests like t-test). Use error bars. Discuss uncertainties. - Evaluation (6 marks): Identify weaknesses and suggest realistic improvements. Do not just list errors — explain their impact on results. - Communication (4 marks): Clear structure, correct terminology, proper citations. Recommended Resources:
- IB Biology Study Guide (Oxford / Cambridge)
- BioNinja (bioninja.com.au)
- Khan Academy Biology
- Past Papers (IB Documents) Study Tips:
- Draw annotated diagrams daily
- Use flashcards for vocabulary
- Teach concepts to a friend
- Do at least 3 past papers before the exam 🎓 无论你是 IB 生物的初学者还是冲刺复习的考生，掌握好细胞呼吸、光合作用、遗传学、DNA 复制与表达、以及分子生物学这五大核心板块，你的 Paper 1 和 Paper 2 就有了稳固的基础。需要 IB Biology 一对一辅导或真题讲解，欢迎添加微信：16621398022同微信。让我们一起向 7 分冲刺！

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