High School Biology

Molecular Genetics分子遗传学

DNA is the molecule of heredity. This guide builds from the double helix and base-pairing rules through DNA replication, transcription of DNA into mRNA, translation of mRNA into protein (the central dogma), mutations and their consequences, gene regulation via the operon model, and an introduction to biotechnology tools including PCR and gel electrophoresis. Worked examples and quiz questions connect molecular mechanisms to real biology throughout.DNA（脱氧核糖核酸）是遗传的分子基础。本指南从双螺旋（双螺旋）与碱基配对（碱基配对）规则出发，依次讲解 DNA 复制（复制）、DNA 转录为 mRNA（转录）、mRNA 翻译为蛋白质（翻译，即中心法则），再到突变（突变）及其后果、操纵子模型下的基因调控（基因调控），最后介绍 PCR 与凝胶电泳等生物技术（生物技术）工具。全部例题与测验均将分子机制与真实生物学情境紧密结合。

7 sections7 节内容 US NGSS · ON · BC · ABUS NGSS · ON · BC · AB SBI4U operon & replication mechanism marked HonorsSBI4U 操纵子与复制机制标为荣誉级

Where this lives in your syllabus本单元在各大纲中的位置

HS-LS1-1 "Construct an explanation based on evidence for how the structure of DNA determines the structure of proteins, which carry out the essential functions of life through systems of specialized cells." [Assessment Boundary: Assessment does not include identification of specific cell or tissue types, whole body systems, specific protein structures and functions, or the biochemistry of protein synthesis.] — maps to §1–§4 (DNA structure, replication, transcription, translation at the conceptual level)."基于证据构建解释，说明 DNA 的结构如何决定蛋白质的结构，蛋白质通过特化细胞系统执行生命的基本功能。"[评估边界：不含特定细胞或组织类型、完整体系统、特定蛋白质结构与功能或蛋白质合成的生物化学。]——对应 §1–§4（概念层面的 DNA 结构、复制、转录、翻译）。
HS-LS3-1 "Ask questions to clarify relationships about the role of DNA and chromosomes in coding the instructions for characteristic traits passed from parents to offspring." [Assessment Boundary: Assessment does not include the phases of meiosis or the biochemical mechanism of specific steps in the process.] — maps to §1 (DNA as the hereditary molecule) and §5 (mutations as sources of variation)."提出问题，阐明 DNA 和染色体在将特征性状编码为遗传指令并传递给后代方面的作用。"[评估边界：不含减数分裂的各期或过程中特定步骤的生物化学机制。]——对应 §1（DNA 作为遗传分子）和 §5（突变作为变异来源）。
NGSS keeps molecular genetics conceptual — it explicitly excludes the biochemistry of protein synthesis. The conceptual DNA→protein→specialized-cell explanation is the expectation, not the named-enzyme mechanism.NGSS 将分子遗传学保持在概念层面——明确排除蛋白质合成的生物化学。其期望是概念性的 DNA→蛋白质→特化细胞解释，而非命名酶的机制。

SBI4U Strand D: Molecular Genetics — Overall Expectation D3: "demonstrate an understanding of concepts related to molecular genetics, and how genetic modification is applied in industry and agriculture." Honors SBI4UStrand D：分子遗传学 — 总体期望 D3："展示对分子遗传学相关概念的理解，以及遗传修改在工业和农业中的应用。" 荣誉 SBI4U
SBI4U D3.1 "explain the current model of DNA replication, and describe the different repair mechanisms that can correct mistakes in DNA sequencing" — §2. Named enzymes (helicase, polymerase I/II/III, ligase, Okazaki fragments) are SBI4U expectations. Honors SBI4UD3.1："解释 DNA 复制的现行模型，并描述可纠正 DNA 测序错误的各种修复机制"——即 §2。命名酶（解旋酶、聚合酶 I/II/III、连接酶、冈崎片段）是 SBI4U 的期望。荣誉 SBI4U
SBI4U D3.3 "explain the steps involved in the process of protein synthesis and how genetic expression is controlled in prokaryotes and eukaryotes by regulatory proteins (e.g., the role of operons in prokaryotic cells)" — §3, §4, §6. Honors SBI4UD3.3："解释蛋白质合成过程中的步骤，以及调控蛋白如何在原核和真核细胞中控制基因表达（如操纵子在原核细胞中的作用）"——即 §3、§4、§6。荣誉 SBI4U

Anatomy & Physiology 12 Big Idea 2: "Gene expression, through protein synthesis, is an interaction between genes and the environment." Content: "DNA: the cell's genetic information; replication" and "Gene expression and protein synthesis" — §1–§4.Anatomy & Physiology 12 大概念 2："基因表达（通过蛋白质合成）是基因与环境之间的相互作用。"内容："DNA：细胞的遗传信息；复制"与"基因表达和蛋白质合成"——即 §1–§4。
Anatomy & Physiology 12 Content: "Genomics and biotechnology: Human Genome Project, cloning, recombinant DNA, GMOs, transgenic organisms, genetic modification, gene therapy" — §7.Anatomy & Physiology 12 内容："基因组学与生物技术：人类基因组计划、克隆、重组 DNA、转基因生物、遗传修改、基因治疗"——即 §7。
Life Sciences 11 Content: "microevolution: changes in DNA: mutations" — §5. BC introduces DNA structure conceptually in Science 10 ("DNA structure and function: genes and chromosomes; gene expression") as the Grade-10 foundation.Life Sciences 11 内容："微进化：DNA 的变化：突变"——即 §5。BC 在 Science 10 中以概念方式介绍 DNA 结构（"DNA 结构与功能：基因与染色体；基因表达"）作为 10 年级基础。

Biology 30 Unit C GO3 — "Students will explain classical genetics at the molecular level." Outcome 30–C3.2k: "describe, in general, how genetic information is contained in the sequence of bases in DNA molecules in chromosomes and how the DNA molecules replicate themselves" — §1, §2.Biology 30 Unit C GO3 —"学生将在分子层面解释经典遗传学。"结果 30–C3.2k："一般性描述遗传信息如何包含在染色体 DNA 分子的碱基序列中，以及 DNA 分子如何自我复制"——即 §1、§2。
Biology 30 Outcome 30–C3.3k: "describe, in general, how genetic information is transcribed into sequences of bases in RNA molecules and is finally translated into sequences of amino acids in proteins" — §3, §4. Alberta hedges with "in general" — sits between NGSS (conceptual only) and SBI4U (full mechanism). Honors Biology 30结果 30–C3.3k："一般性描述遗传信息如何转录为 RNA 分子中的碱基序列，并最终翻译为蛋白质中的氨基酸序列"——即 §3、§4。阿尔伯塔以"一般性"作限定——介于 NGSS（仅概念）与 SBI4U（完整机制）之间。荣誉 Biology 30
Biology 30 Outcome 30–C3.6k: "explain how a random change (mutation) in the sequence of bases results in abnormalities or provides a source of genetic variability" — §5. Alberta defers all of molecular genetics to Biology 30 (Grade 12). Honors Biology 30结果 30–C3.6k："解释碱基序列中的随机变化（突变）如何导致异常或提供遗传变异来源"——即 §5。阿尔伯塔将所有分子遗传学延迟至 Biology 30（12 年级）。荣誉 Biology 30

Read me first先读这里

How to use this guide如何使用本指南

Molecular genetics sits at the heart of every senior biology curriculum. All four curricula agree on the conceptual core: DNA carries hereditary information, it is copied by replication, read out by transcription into RNA, and decoded by translation into protein. They diverge sharply on depth. US NGSS HS-LS1-1 explicitly excludes the biochemistry of protein synthesis — the conceptual DNA→protein→specialized-cell explanation is the expectation. Alberta Biology 30 Unit C GO3 uses "in general" throughout, sitting between NGSS and Ontario. Ontario SBI4U Strand D goes deepest: it names polymerases I/II/III, helicase, ligase, Okazaki fragments, codon/anticodon, and the operon model — all carried here with Honors SBI4U flags. BC Anatomy & Physiology 12 covers DNA replication and gene expression at a level between NGSS and SBI4U. The table below locates each section in your curriculum.分子遗传学（DNA / RNA）是所有高中生物大纲的核心主题。四套大纲在概念核心上高度一致：DNA 携带遗传信息，通过复制（复制）拷贝，通过转录（转录）读出为 RNA，再通过翻译（翻译）解码为蛋白质。但在深度上差异显著。美国 NGSS HS-LS1-1 明确排除蛋白质合成的生物化学——概念性的 DNA→蛋白质→特化细胞解释是期望所在。阿尔伯塔 Biology 30 Unit C GO3 全程使用"一般性"限定，介于 NGSS 与安大略之间。安大略 SBI4U Strand D 最为深入：要求命名聚合酶 I/II/III、解旋酶、连接酶、冈崎片段、密码子/反密码子及操纵子模型——本指南均以荣誉 SBI4U 标注。BC Anatomy & Physiology 12 涵盖 DNA 复制与基因表达，深度介于 NGSS 与 SBI4U 之间。下表定位各节在你大纲中的位置。

If you are in…如果你在…	Focus on these sections重点学习	Defer / lighter可推迟 / 减负	Source依据
🇺🇸 US NGSS HS-LS1-1 / HS-LS3-1美国 NGSS HS-LS1-1 / HS-LS3-1	§1 (DNA structure), §3 (transcription concept), §4 (translation concept), §5 (mutations as variation) — the conceptual DNA→protein→cell model§1（DNA 结构）、§3（转录概念）、§4（翻译概念）、§5（突变作为变异）——概念性 DNA→蛋白质→细胞模型	Enzyme-level replication detail (§2 Honors box); operon mechanism (§6 Honors) — NGSS Assessment Boundary excludes the biochemistry of protein synthesis酶级复制细节（§2 荣誉框）；操纵子机制（§6 荣誉）——NGSS 评估边界排除蛋白质合成的生物化学	`NGSS HS Life Science` — HS-LS1-1 PE and Assessment Boundary— HS-LS1-1 表现期望及评估边界
🇨🇦 ON Grade 12 — SBI4U Honors安大略 12 年级 — SBI4U 荣誉	All 7 sections in full: Strand D D3.1 (replication, polymerases, helicase, Okazaki fragments), D3.2 (RNA vs DNA), D3.3 (protein synthesis steps, operons), D3.4 (mutagens), D3.5 (genetic modification), D3.6 (restriction enzymes, plasmids)全部 7 节完整学习：Strand D D3.1（复制、聚合酶、解旋酶、冈崎片段）、D3.2（RNA 与 DNA）、D3.3（蛋白质合成步骤、操纵子）、D3.4（诱变剂）、D3.5（遗传修改）、D3.6（限制酶、质粒）	Nothing — SBI4U Molecular Genetics is the deepest treatment; all sections are core无 — SBI4U 分子遗传学是最深层次；所有章节均为核心	`Ontario SBI3U/4U Biology` — SBI4U Strand D D3.1–D3.7— SBI4U Strand D D3.1–D3.7
🇨🇦 BC Anatomy & Physiology 12BC Anatomy & Physiology 12	§1–§5 core (DNA structure, replication, gene expression, protein synthesis, mutations); §7 (genomics & biotechnology: recombinant DNA, GMOs, gene therapy)§1–§5 核心（DNA 结构、复制、基因表达、蛋白质合成、突变）；§7（基因组学与生物技术：重组 DNA、转基因生物、基因治疗）	Named-enzyme detail in replication (§2 Honors box); operon detail (§6 Honors) — BC A&P 12 covers gene expression at a higher level than NGSS but not at SBI4U named-enzyme depth复制中的命名酶细节（§2 荣誉框）；操纵子细节（§6 荣誉）——BC A&P 12 基因表达深度高于 NGSS 但未达 SBI4U 命名酶深度	`BC Life Sciences 11 / Anatomy 12` — A&P 12 Big Idea 2 + Content— A&P 12 大概念 2 + 内容
🇨🇦 AB Biology 30 Honors阿尔伯塔 Biology 30 荣誉	§1–§5 at the "in general" level (Unit C GO3: C3.2k replication, C3.3k transcription/translation, C3.6k mutations); §7 (C3.4k restriction enzymes and DNA transformation)"一般性"层面的 §1–§5（Unit C GO3：C3.2k 复制、C3.3k 转录/翻译、C3.6k 突变）；§7（C3.4k 限制酶与 DNA 转化）	Named-enzyme mechanism in replication (§2 Honors) and operon mechanism (§6 Honors) — Alberta hedges with "in general"; SBI4U enzyme lists not required at AB level复制中的命名酶机制（§2 荣誉）与操纵子机制（§6 荣誉）——阿尔伯塔使用"一般性"限定；AB 层级不要求 SBI4U 酶的命名列表	`Alberta Biology 20/30` — Biology 30 Unit C GO3 (C3.1k–C3.7k)— Biology 30 Unit C GO3（C3.1k–C3.7k）

Once you have located your row, use the two cards below for the approach that fits your timeline.找到所在行后，用下面两张卡片选择适合你时间安排的方式。

If you are cramming the night before如果你在临阵磨枪

Know the central dogma: DNA → RNA → protein. Know that DNA is a double helix with base pairing (A-T, G-C), that replication is semi-conservative, that transcription happens in the nucleus, and that translation happens at ribosomes. Know the three types of RNA (mRNA, tRNA, rRNA). Know that a point mutation changes one base and can change one amino acid. Read every cram-cheat box; skip the Honors enzyme-level going-deeper sections.掌握中心法则：DNA → RNA → 蛋白质。了解 DNA 是双螺旋，碱基配对规则为 A-T、G-C；复制是半保留复制；转录在细胞核中进行；翻译在核糖体上进行。掌握三种 RNA 类型（mRNA、tRNA、rRNA）。了解点突变改变一个碱基，可能改变一个氨基酸。读每个速记框；跳过荣誉级酶机制的深入内容。

If you are going for the top mark如果你目标顶分

Be precise about codon/anticodon complementarity; know why a silent mutation does not change the amino acid (degeneracy of the genetic code). For SBI4U: name the key replication enzymes (helicase unwinds, primase lays the RNA primer, DNA polymerase III extends 5'→3', DNA polymerase I replaces primers, ligase seals nicks). Explain the lac operon with a diagram: repressor binds operator in the absence of lactose; when lactose is present the inducer molecule releases the repressor. Know PCR's three steps (denaturation, annealing, extension) and what gel electrophoresis separates (DNA fragments by size).精准掌握密码子/反密码子互补性；了解为何沉默突变不改变氨基酸（遗传密码的简并性）。SBI4U 轨道：命名关键复制酶（解旋酶解链、引物酶铺设 RNA 引物、DNA 聚合酶 III 按 5'→3' 延伸、DNA 聚合酶 I 替换引物、连接酶封接缺口）。用图示解释 lac 操纵子：无乳糖时阻遏物结合操作子；有乳糖时诱导分子释放阻遏物。掌握 PCR 三步（变性、退火、延伸）及凝胶电泳所分离的对象（按大小分离 DNA 片段）。

Honors flag.荣誉级标记。 The enzyme-level replication detail in §2 (helicase, polymerases, Okazaki fragments, ligase) and the operon gene-regulation mechanism in §6 carry the Honors SBI4U chip, because Ontario SBI4U D3.1/D3.3 expects this depth while NGSS explicitly excludes it. Alberta Biology 30 sits in between ("in general"). If your curriculum is NGSS, read these sections for conceptual grounding but do not memorize the named-enzyme lists.§2 中酶级别的复制细节（解旋酶、聚合酶、冈崎片段、连接酶）与 §6 中操纵子基因调控机制标注荣誉 SBI4U，因为安大略 SBI4U D3.1/D3.3 要求这一深度，而 NGSS 明确排除。阿尔伯塔 Biology 30 介于两者之间（"一般性"）。若你的大纲是 NGSS，阅读这些部分以获得概念基础，但无需背诵命名酶列表。

Section 1 · DNA structure第 1 节 · DNA 结构

DNA Structure: The Double Helix and Base PairingDNA 结构：双螺旋与碱基配对

The double helix in four facts.双螺旋四要点。

Structure: DNA is a double-stranded helix. Each strand is a polymer of nucleotides; each nucleotide has a sugar (deoxyribose), a phosphate group, and one of four nitrogenous bases.结构：DNA 是双链螺旋。每条链是核苷酸聚合体；每个核苷酸含一个糖（脱氧核糖）、一个磷酸基团和四种含氮碱基之一。
Base pairing (Chargaff's rules): Adenine (A) pairs with Thymine (T) via two hydrogen bonds; Guanine (G) pairs with Cytosine (C) via three hydrogen bonds. The two strands are antiparallel (one runs 5'→3', the other 3'→5').碱基配对（查哥夫定律）：腺嘌呤（A）与胸腺嘧啶（T）通过两个氢键配对；鸟嘌呤（G）与胞嘧啶（C）通过三个氢键配对。两条链反向平行（一条 5'→3'，另一条 3'→5'）。
Information: The sequence of bases encodes genetic information. A gene is a specific sequence of DNA that codes for a functional product (usually a protein). Human cells contain ~3 billion base pairs organized into 23 chromosome pairs.信息：碱基序列编码遗传信息。基因是编码功能产物（通常是蛋白质）的特定 DNA 序列。人类细胞含约 30 亿碱基对，组织为 23 对染色体。
RNA differs from DNA: RNA is single-stranded; uses ribose sugar instead of deoxyribose; uses Uracil (U) instead of Thymine (T); shorter than a full chromosome.RNA 与 DNA 的区别：RNA 是单链；使用核糖而非脱氧核糖；使用尿嘧啶（U）代替胸腺嘧啶（T）；比完整染色体短。

Worked Example 1 · Base-pairing complement例题 1 · 碱基配对互补

A strand of DNA has the sequence 5'-ATGCCTAG-3'. Write (a) the complementary DNA strand and (b) the mRNA strand that would be transcribed from the template strand.一条 DNA 链的序列为 5'-ATGCCTAG-3'。写出 (a) 互补 DNA 链和 (b) 以该链为模板转录所得 mRNA 链。

(a) Complementary DNA strand:(a) 互补 DNA 链： Apply A-T and G-C pairing, antiparallel: 3'-TACGGATC-5' (or written 5'-CTAGGCAT-3' if reversed).应用 A-T 和 G-C 配对规则，反向平行：3'-TACGGATC-5'（反向写为 5'-CTAGGCAT-3'）。

(b) mRNA transcribed from the template strand (3'-TACGGATC-5'):(b) 以模板链（3'-TACGGATC-5'）转录所得 mRNA： mRNA pairs with the template using A-U, T-A, G-C, C-G (U replaces T in RNA). mRNA sequence: 5'-AUGCCUAG-3'. Note: the mRNA sequence matches the coding (non-template) DNA strand, with U replacing T.mRNA 与模板链配对，规则为 A-U、T-A、G-C、C-G（RNA 中 U 替代 T）。mRNA 序列：5'-AUGCCUAG-3'。注意：mRNA 序列与编码链（非模板链）相同，T 改为 U。

Which base pairs with Guanine (G) in DNA?在 DNA 中，哪个碱基与鸟嘌呤（G）配对？

§1 · Q1

Adenine (A)腺嘌呤（A）

Thymine (T)胸腺嘧啶（T）

Cytosine (C)胞嘧啶（C）

Uracil (U)尿嘧啶（U）

G pairs with C via three hydrogen bonds. A pairs with T via two hydrogen bonds. Uracil (U) is an RNA base that replaces T in RNA; it pairs with A in RNA, not with G in DNA.G 与 C 通过三个氢键配对。A 与 T 通过两个氢键配对。尿嘧啶（U）是 RNA 的碱基，在 RNA 中替代 T；它在 RNA 中与 A 配对，而非在 DNA 中与 G 配对。

G pairs with C (three H-bonds). A pairs with T (two H-bonds). U is an RNA base; it replaces T in RNA, pairing with A.G 与 C 配对（三个氢键）。A 与 T 配对（两个氢键）。U 是 RNA 碱基；在 RNA 中替代 T，与 A 配对。

A DNA sample shows that 30% of its bases are Adenine. What percentage is Cytosine?一个 DNA 样本中 30% 的碱基为腺嘌呤。胞嘧啶占几百分比？

§1 · Q2

20%20%

30%30%

40%40%

25%25%

By Chargaff's rules A = T and G = C. If A = 30%, then T = 30%. Together A+T = 60%, so G+C = 40%, meaning G = C = 20%.查哥夫定律：A = T，G = C。若 A = 30%，则 T = 30%。A+T = 60%，故 G+C = 40%，即 G = C = 20%。

A = T = 30%, so A+T = 60%. The remaining 40% is G+C equally: G = C = 20%.A = T = 30%，故 A+T = 60%。其余 40% 为 G+C 各半：G = C = 20%。

Going deeper — the historical discovery of DNA structure (Watson, Crick, Franklin, Wilkins; AB Biology 30 C3.1k)深入 — DNA 结构的历史发现（沃森、克里克、富兰克林、威尔金斯；AB Biology 30 C3.1k）

Alberta Biology 30 C3.1k states: "summarize the historical events that led to the discovery of the structure of the DNA molecule, including the work of Franklin and Watson and Crick." Rosalind Franklin's X-ray crystallography (Photo 51, 1952) revealed the helical structure and dimensions. Watson and Crick used Franklin's data (controversially, without direct permission) along with Chargaff's base-ratio data and Wilkins' crystallographic work to build their double-helix model (1953, Nature). Erwin Chargaff had established by 1950 that A = T and G = C in any DNA sample (Chargaff's rules). The model correctly predicted that the two strands were complementary and antiparallel, immediately suggesting a copying mechanism.阿尔伯塔 Biology 30 C3.1k 要求："总结导致发现 DNA 分子结构的历史事件，包括富兰克林以及沃森和克里克的工作。"罗莎琳德·富兰克林的 X 射线晶体学（51 号照片，1952 年）揭示了螺旋结构与尺寸。沃森和克里克利用富兰克林的数据（存在争议，未直接获得许可）、查哥夫的碱基比例数据和威尔金斯的晶体学研究，构建了双螺旋模型（1953 年，《自然》）。查哥夫于 1950 年前确立：任何 DNA 样本中 A = T，G = C（查哥夫定律）。该模型正确预测两条链互补且反向平行，立即暗示了一种复制机制。

Section 2 · DNA replication第 2 节 · DNA 复制

DNA ReplicationDNA 复制

Semi-conservative replication — the key rule.半保留复制 — 核心规则。

Semi-conservative: when DNA replicates, each new double helix contains one original (parental) strand and one newly synthesized strand. Both daughter DNA molecules are identical to the parent.半保留：DNA 复制时，每个新双螺旋含一条原有（亲本）链和一条新合成链。两个子代 DNA 分子均与亲本相同。
Where: occurs in the nucleus (eukaryotes) before cell division (S phase of the cell cycle).位置：在细胞分裂前发生于细胞核（真核生物）（细胞周期 S 期）。
Steps (conceptual): (1) The two parental strands separate (unwind). (2) Free nucleotides pair up with each template strand by base-pairing rules (A-T, G-C). (3) New strands are synthesized in the 5'→3' direction. (4) Two complete double helices result.步骤（概念性）：(1) 两条亲本链分离（解旋）。(2) 游离核苷酸按碱基配对规则（A-T、G-C）与每条模板链配对。(3) 新链按 5'→3' 方向合成。(4) 产生两个完整双螺旋。

Worked Example 2 · Semi-conservative replication prediction例题 2 · 半保留复制预测

A researcher labels all DNA in a cell with a heavy nitrogen isotope (¹⁵N). The cell then replicates once in a medium containing only light nitrogen (¹⁴N). Predict the composition of the daughter DNA molecules and explain why this supports the semi-conservative model.研究者用重氮同位素（¹⁵N）标记细胞内所有 DNA。该细胞随后在仅含轻氮（¹⁴N）的培养基中复制一次。预测子代 DNA 分子的组成，并解释这如何支持半保留模型。

Result:结果： All daughter DNA molecules are hybrid — one ¹⁵N strand (original) and one ¹⁴N strand (new). In density-gradient centrifugation, all molecules band at an intermediate density, not at the heavy position. This is exactly what Meselson and Stahl found in 1958, ruling out conservative replication (which would give half heavy + half light) and dispersive replication (which would give all intermediate but in a different pattern). Semi-conservative predicts one band at intermediate density after one replication: confirmed.结果：所有子代 DNA 分子均为杂合体 — 一条 ¹⁵N 链（原有）和一条 ¹⁴N 链（新合成）。在密度梯度离心中，所有分子聚集在中间密度带，而非重密度带。这正是梅塞尔森和斯塔尔在 1958 年的发现，排除了保留复制（将产生一半重 + 一半轻）和弥散复制（模式不同）的可能。半保留复制预测复制一次后出现一条中间密度带：得到证实。

What does "semi-conservative replication" mean?"半保留复制"是什么意思？

§2 · Q1

Both strands of the original DNA are retained in one daughter molecule原始 DNA 的两条链均保留在一个子代分子中

Each daughter DNA has one original strand and one new strand每个子代 DNA 含一条原有链和一条新链

The original strands are destroyed and two new double helices are built from scratch原有链被破坏，从头构建两个新双螺旋

Only half of the DNA is replicated each cell cycle每个细胞周期只复制一半 DNA

Semi-conservative means each daughter molecule retains (conserves) one original strand. "Semi" = half, so half is conserved in each copy. Confirmed by the Meselson-Stahl experiment with nitrogen isotopes.半保留意味着每个子代分子保留（conserves）一条原有链。"半"= 一半，即每份拷贝保留一半。由梅塞尔森-斯塔尔的氮同位素实验证实。

Semi-conservative: each daughter cell gets one old strand + one new strand. Option A describes conservative replication; option C describes dispersive replication; option D misreads the term.半保留：每个子代细胞获得一条旧链 + 一条新链。选项 A 描述保留复制；选项 C 描述弥散复制；选项 D 误读了该术语。

DNA replication in eukaryotic cells occurs during which phase of the cell cycle?真核细胞的 DNA 复制发生在细胞周期的哪个阶段？

§2 · Q2

G1 phaseG1 期

M phase (mitosis)M 期（有丝分裂）

G2 phaseG2 期

S phase (synthesis)S 期（合成期）

DNA replication occurs during S phase (Synthesis phase) of interphase, before mitosis. G1 and G2 are growth phases; M phase is when the duplicated chromosomes separate.DNA 复制发生在间期的 S 期（合成期），在有丝分裂之前。G1 和 G2 是生长期；M 期是复制后的染色体分离的阶段。

S phase = Synthesis phase = when DNA is replicated. G1/G2 = growth; M = mitosis (chromosome segregation, not replication).S 期 = 合成期 = DNA 复制时期。G1/G2 = 生长期；M 期 = 有丝分裂（染色体分离，不是复制）。

Going deeper — enzyme roles in replication: helicase, DNA polymerase, primase, ligase, Okazaki fragments (SBI4U D3.1 / D2.1)深入 — 复制中的酶角色：解旋酶、DNA 聚合酶、引物酶、连接酶、冈崎片段（SBI4U D3.1 / D2.1）

Honors SBI4U SBI4U D2.1 names: "polymerase I, II, and III, DNA ligase, helicase, Okazaki fragment." (1) Helicase unwinds and separates the two parental strands at the replication fork. (2) Primase lays down a short RNA primer to provide a starting 3'-OH for DNA synthesis. (3) DNA Polymerase III extends the new strand by adding nucleotides in the 5'→3' direction only; it cannot start a new strand without a primer. (4) Because the leading strand is synthesized continuously toward the fork, but the lagging strand is synthesized away from the fork in short segments called Okazaki fragments. (5) DNA Polymerase I removes RNA primers and replaces them with DNA. (6) DNA Ligase seals the nicks between Okazaki fragments, producing a continuous strand.荣誉 SBI4U SBI4U D2.1 命名："聚合酶 I、II、III，DNA 连接酶，解旋酶，冈崎片段。"(1) 解旋酶在复制叉处解开并分离两条亲本链。(2) 引物酶铺设短 RNA 引物，为 DNA 合成提供起始 3'-OH。(3) DNA 聚合酶 III 仅能以 5'→3' 方向添加核苷酸延伸新链；无引物时无法启动新链。(4) 前导链向复制叉方向连续合成，而后随链以远离复制叉方向合成，形成短片段即冈崎片段。(5) DNA 聚合酶 I 移除 RNA 引物并以 DNA 替换。(6) DNA 连接酶封接冈崎片段间的缺口，形成连续链。

Section 3 · Transcription第 3 节 · 转录

Transcription: DNA → mRNA转录：DNA → mRNA

Transcription in five facts.转录五要点。

Where: in the nucleus (eukaryotes); in the cytoplasm (prokaryotes, which have no nucleus).位置：细胞核中（真核生物）；细胞质中（无细胞核的原核生物）。
Enzyme: RNA polymerase binds to a promoter sequence on DNA and unwinds a small segment.酶：RNA 聚合酶结合 DNA 上的启动子序列并使一小段解旋。
Process: RNA polymerase reads the template (antisense) strand 3'→5' and synthesizes a complementary mRNA strand 5'→3', using RNA nucleotides (U pairs with A in the template).过程：RNA 聚合酶沿模板（反义）链 3'→5' 方向读取，合成互补 mRNA 链（5'→3'），使用 RNA 核苷酸（U 与模板中的 A 配对）。
Product: a single-stranded pre-mRNA molecule. In eukaryotes, non-coding introns are spliced out and coding exons are joined to produce mature mRNA.产物：单链前体 mRNA 分子。在真核生物中，非编码内含子被剪接掉，编码外显子连接为成熟 mRNA。
Three types of RNA: mRNA (messenger) carries the protein-coding sequence; tRNA (transfer) brings amino acids to the ribosome; rRNA (ribosomal) is a structural/catalytic component of the ribosome.三种 RNA：mRNA（信使 RNA）携带蛋白质编码序列；tRNA（转运 RNA）将氨基酸运至核糖体；rRNA（核糖体 RNA）是核糖体的结构/催化成分。

DNA vs RNA — quick comparisonDNA 与 RNA 快速对比

Feature特征	DNADNA	RNARNA
Strands链数	Double-stranded双链	Single-stranded单链
Sugar糖	Deoxyribose脱氧核糖	Ribose核糖
Bases碱基	A, T, G, CA、T、G、C	A, U, G, CA、U、G、C
Location位置	Nucleus (mainly)主要在细胞核	Nucleus & cytoplasm细胞核与细胞质
Function功能	Long-term genetic storage长期遗传信息储存	Carry, translate, build ribosomes携带、翻译、构建核糖体

Where does transcription occur in a eukaryotic cell?转录在真核细胞的哪里发生？

§3 · Q1

In the nucleus在细胞核中

At the ribosome在核糖体上

In the mitochondria在线粒体中

At the cell membrane在细胞膜处

Transcription occurs in the nucleus, where DNA is stored. mRNA is synthesized there and then exits through nuclear pores to the cytoplasm, where it is translated at ribosomes.转录在细胞核中发生，那里储存着 DNA。mRNA 在此合成后通过核孔输出至细胞质，在核糖体上翻译。

Transcription = nucleus (where DNA is). Translation = ribosome (in cytoplasm). These are two separate locations — a classic exam point.转录 = 细胞核（DNA 所在）。翻译 = 核糖体（细胞质中）。两个不同位置 — 经典考点。

The mRNA sequence corresponding to the DNA template strand 3'-TACGAA-5' is:DNA 模板链 3'-TACGAA-5' 对应的 mRNA 序列是：

§3 · Q2

3'-ATGCTT-5'3'-ATGCTT-5'

5'-TACGAA-3'5'-TACGAA-3'

5'-ATGCTT-3'5'-ATGCTT-3'

5'-AUGCUU-3'5'-AUGCUU-3'

mRNA is complementary and antiparallel to the template strand, using U instead of T. Template 3'-TACGAA-5' → mRNA 5'-AUGCUU-3'. T pairs to A (written as A in mRNA), A pairs to U, C pairs to G, G pairs to C, A to U, A to U.mRNA 与模板链互补且反向平行，用 U 替代 T。模板 3'-TACGAA-5' → mRNA 5'-AUGCUU-3'。T 对应 A（mRNA 中写为 A），A 对应 U，C 对应 G，G 对应 C，A 对应 U，A 对应 U。

Remember: in mRNA, U replaces T. RNA is read 5'→3'. Template 3'-T-A-C-G-A-A-5' gives mRNA 5'-A-U-G-C-U-U-3'.记住：mRNA 中 U 替代 T。RNA 按 5'→3' 方向读取。模板 3'-T-A-C-G-A-A-5' 给出 mRNA 5'-A-U-G-C-U-U-3'。

Section 4 · Translation第 4 节 · 翻译

Translation: mRNA → Protein翻译：mRNA → 蛋白质

Translation in five facts.翻译五要点。

Where: at ribosomes in the cytoplasm (or on rough endoplasmic reticulum for secretory proteins).位置：细胞质中的核糖体上（分泌蛋白在糙面内质网上）。
Codons: the mRNA is read in triplets of bases called codons. Each codon specifies one amino acid (or a start/stop signal). AUG is the universal start codon (codes for methionine). UAA, UAG, UGA are stop codons.密码子：mRNA 以三联体碱基（密码子）方式读取。每个密码子对应一种氨基酸（或起始/终止信号）。AUG 是通用起始密码子（编码甲硫氨酸）。UAA、UAG、UGA 是终止密码子。
tRNA and anticodons: each tRNA carries a specific amino acid and has an anticodon complementary to the mRNA codon. The ribosome aligns codon and anticodon, and the amino acid is added to the growing polypeptide chain.tRNA 与反密码子：每个 tRNA 携带特定氨基酸，具有与 mRNA 密码子互补的反密码子。核糖体使密码子和反密码子对齐，氨基酸被添加到不断延长的多肽链上。
Steps: (1) Initiation: ribosome assembles on mRNA at AUG. (2) Elongation: tRNAs deliver amino acids one by one; peptide bonds form. (3) Termination: ribosome reaches a stop codon; polypeptide is released.步骤：(1) 起始：核糖体在 AUG 处组装于 mRNA 上。(2) 延伸：tRNA 逐一递送氨基酸；形成肽键。(3) 终止：核糖体到达终止密码子；多肽链释放。
Genetic code degeneracy: most amino acids are coded by more than one codon. This means many base changes in the third position of a codon (wobble position) do not change the amino acid — producing a silent mutation.遗传密码的简并性：大多数氨基酸由多个密码子编码。这意味着密码子第三位（摆动位）的许多碱基变化不会改变氨基酸 — 产生沉默突变。

Worked Example 3 · Codon translation例题 3 · 密码子翻译

A segment of mRNA reads 5'-AUG-CCU-AGG-UAA-3'. Identify (a) the start codon, (b) the number of amino acids in the polypeptide, (c) the stop codon, and (d) the anticodon for CCU.一段 mRNA 序列为 5'-AUG-CCU-AGG-UAA-3'。请确定：(a) 起始密码子；(b) 多肽链中氨基酸数目；(c) 终止密码子；(d) CCU 的反密码子。

(a) Start codon: AUG.(a) 起始密码子：AUG。

(b) Amino acids: 2.(b) 氨基酸数目：2 个。 AUG codes for Met; CCU for Pro; AGG for Arg; UAA is a stop codon (no amino acid). So 3 codons are translated (AUG, CCU, AGG), giving Met-Pro-Arg. Wait — 3 amino acids (Met, Pro, Arg). UAA terminates. Final polypeptide: 3 amino acids.AUG 编码 Met（甲硫氨酸）；CCU 编码 Pro（脯氨酸）；AGG 编码 Arg（精氨酸）；UAA 是终止密码子（不编码氨基酸）。共翻译 3 个密码子（AUG、CCU、AGG），产生 Met-Pro-Arg。UAA 终止。最终多肽：3 个氨基酸。

(c) Stop codon: UAA.(c) 终止密码子：UAA。

(d) Anticodon for CCU: GGA(d) CCU 的反密码子：GGA (anticodon is complementary and antiparallel to the codon: codon 5'-CCU-3' → anticodon 3'-GGA-5', conventionally written 5'-AGG-3' as the tRNA anticodon).（反密码子与密码子互补且反向平行：密码子 5'-CCU-3' → 反密码子 3'-GGA-5'，通常写为 tRNA 反密码子 5'-AGG-3'）。

Which molecule delivers amino acids to the ribosome during translation?翻译过程中，哪种分子将氨基酸运送至核糖体？

§4 · Q1

mRNAmRNA

rRNArRNA

tRNAtRNA

DNA polymeraseDNA 聚合酶

tRNA (transfer RNA) delivers amino acids. Each tRNA has an anticodon that base-pairs with the mRNA codon and carries the corresponding amino acid. mRNA is the template; rRNA is part of the ribosome structure.tRNA（转运 RNA）递送氨基酸。每个 tRNA 具有与 mRNA 密码子配对的反密码子，并携带对应的氨基酸。mRNA 是模板；rRNA 是核糖体结构的一部分。

tRNA = Transfer RNA = carries amino acids to the ribosome. mRNA = template. rRNA = ribosome component. DNA polymerase is for DNA replication, not translation.tRNA = 转运 RNA = 将氨基酸运至核糖体。mRNA = 模板。rRNA = 核糖体成分。DNA 聚合酶用于 DNA 复制，与翻译无关。

The central dogma of molecular biology states that information flows in the order:分子生物学的中心法则指出，信息按以下顺序流动：

§4 · Q2

Protein → RNA → DNA蛋白质 → RNA → DNA

DNA → RNA → ProteinDNA → RNA → 蛋白质

RNA → DNA → ProteinRNA → DNA → 蛋白质

DNA → Protein → RNADNA → 蛋白质 → RNA

The central dogma (Crick, 1958): DNA → RNA (transcription) → Protein (translation). Note: retroviruses use reverse transcriptase to go RNA → DNA, but this is an exception to the standard flow.中心法则（克里克，1958 年）：DNA → RNA（转录）→ 蛋白质（翻译）。注：逆转录病毒使用逆转录酶实现 RNA → DNA，但这是标准流程的例外。

Central dogma: DNA → RNA → Protein. DNA is replicated (DNA→DNA), transcribed (DNA→RNA), and then translated (RNA→Protein).中心法则：DNA → RNA → 蛋白质。DNA 被复制（DNA→DNA）、转录（DNA→RNA），再被翻译（RNA→蛋白质）。

Section 5 · Mutations第 5 节 · 突变

Mutations: Changes in the DNA Sequence突变：DNA 序列的改变

Mutations — types and consequences.突变 — 类型与后果。

Definition:定义： a mutation is a permanent, heritable change in the nucleotide sequence of DNA. Mutations can be spontaneous (errors in replication) or induced (caused by a mutagen).突变是 DNA 核苷酸序列的永久、可遗传的变化。突变可以是自发的（复制过程中的错误）或诱发的（由诱变剂引起）。
Point mutations (substitutions):点突变（替换）： one base is changed. Three outcomes: (a) Silent — no amino acid change (genetic code degeneracy). (b) Missense — one amino acid changes; may alter protein function. (c) Nonsense — creates a premature stop codon; truncates the protein.一个碱基改变。三种结果：(a) 沉默突变 — 氨基酸不变（遗传密码简并性）。(b) 错义突变 — 一个氨基酸改变；可能影响蛋白质功能。(c) 无义突变 — 产生提前的终止密码子；蛋白质被截短。
Frameshift mutations:移码突变： insertions or deletions of bases (other than multiples of 3) shift the reading frame of all codons downstream. Usually produce a non-functional protein. More severe than point mutations in most cases.碱基的插入或缺失（非 3 的倍数）使下游所有密码子的读码框移位。通常产生无功能蛋白质。大多数情况下比点突变危害更严重。
Mutagens:诱变剂： agents that increase the mutation rate. Chemical mutagens (e.g. base analogs, alkylating agents); radiation mutagens (UV light causes thymine dimers; ionizing radiation breaks DNA strands). SBI4U D3.4 names "radiation and chemicals."增加突变率的因素。化学诱变剂（如碱基类似物、烷化剂）；辐射诱变剂（紫外线导致胸腺嘧啶二聚体；电离辐射断裂 DNA 链）。SBI4U D3.4 提及"辐射和化学物质"。
Consequences:后果： mutations in somatic cells affect only that individual (e.g. cancer if in tumor-suppressor genes). Mutations in germline cells (gametes) are heritable. Most mutations are neutral or harmful; rarely, a mutation is beneficial (source of evolutionary variation — NGSS HS-LS3-1; BC Life Sciences 11 microevolution; AB 30-C3.6k).体细胞中的突变只影响该个体（如肿瘤抑制基因突变导致癌症）。生殖细胞（配子）中的突变可以遗传。大多数突变为中性或有害；极少数突变是有益的（进化变异的来源 — NGSS HS-LS3-1；BC Life Sciences 11 微进化；AB 30-C3.6k）。

A mutation changes the codon GAA (glutamic acid) to GAG. Knowing that GAG also codes for glutamic acid, what type of mutation is this?一个突变将密码子 GAA（谷氨酸）变为 GAG。已知 GAG 同样编码谷氨酸，这是哪种类型的突变？

§5 · Q1

Missense mutation错义突变

Nonsense mutation无义突变

Silent mutation沉默突变

Frameshift mutation移码突变

A silent mutation changes the DNA base but not the amino acid, because the genetic code is degenerate (multiple codons can code for the same amino acid). GAA and GAG both code for glutamic acid, so the protein sequence is unchanged.沉默突变改变 DNA 碱基但不改变氨基酸，因为遗传密码具有简并性（多个密码子可编码同一氨基酸）。GAA 和 GAG 均编码谷氨酸，故蛋白质序列不变。

If the amino acid does not change → silent. If the amino acid changes → missense. If a stop codon is introduced → nonsense. A frameshift requires an insertion or deletion.氨基酸不变 → 沉默突变。氨基酸改变 → 错义突变。引入终止密码子 → 无义突变。移码突变需要插入或缺失。

Why are frameshift mutations generally more disruptive than point mutations?为何移码突变通常比点突变危害更大？

§5 · Q2

They alter the reading frame of every codon downstream of the mutation它们改变突变点下游所有密码子的读码框

They always create a stop codon immediately它们总是立即产生终止密码子

They only affect RNA, not DNA它们只影响 RNA，不影响 DNA

They are caused by UV radiation, not by replication errors它们由紫外线引起，而非复制错误

An insertion or deletion of a non-multiple of 3 shifts the reading frame for all codons 3' of the mutation, changing every amino acid from that point onward. This usually produces a completely different and non-functional protein.非 3 的倍数的插入或缺失使突变点 3' 侧所有密码子的读码框移位，从该点起每个氨基酸都改变。这通常产生完全不同且无功能的蛋白质。

Frameshifts shift the reading frame for all downstream codons — this changes many amino acids, not just one. They don't always create an immediate stop codon, they affect DNA (mutations are changes to DNA by definition), and they can have many causes.移码突变使下游所有密码子的读码框移位 — 这改变了许多氨基酸，而非仅一个。它们不总是立即创建终止密码子，它们影响 DNA（突变在定义上即 DNA 的变化），且可以有多种原因。

Section 6 · Gene regulation第 6 节 · 基因调控

Gene Regulation and the Operon Model基因调控与操纵子模型

Why regulate gene expression?为何要调控基因表达？

Not all genes are active in every cell at all times. Gene regulation controls when, where, and how much of a protein is made. All somatic cells carry the same DNA; differentiation is due to differential gene expression (different genes switched on in different cell types).并非所有基因在每个细胞中随时都处于活跃状态。基因调控控制蛋白质合成的时间、位置和数量。所有体细胞携带相同的 DNA；分化源于差异基因表达（不同细胞类型中不同基因被开启）。

The lac operon (prokaryotic gene regulation) — Honors SBI4U D3.3lac 操纵子（原核基因调控）— 荣誉 SBI4U D3.3

An operon is a cluster of functionally related genes in prokaryotes controlled by a single promoter. The lac operon of E. coli controls lactose digestion. Key components: (1) Promoter — where RNA polymerase binds. (2) Operator — a DNA switch sequence between the promoter and the genes. (3) Structural genes (lacZ, lacY, lacA) — encode the enzymes for lactose metabolism. (4) Repressor protein — encoded by the regulatory gene (lacI); binds the operator and blocks transcription when lactose is absent. When lactose is present, the inducer molecule (allolactose) binds the repressor, changes its shape so it can no longer bind the operator, and RNA polymerase transcribes the structural genes. When lactose is absent, the repressor is active, the operator is blocked, and the genes are OFF.操纵子是原核生物中由单一启动子控制的一组功能相关基因。大肠杆菌的 lac 操纵子控制乳糖消化。关键成分：(1) 启动子 — RNA 聚合酶结合处。(2) 操作子 — 启动子与基因之间的 DNA 开关序列。(3) 结构基因（lacZ、lacY、lacA）— 编码乳糖代谢酶。(4) 阻遏蛋白 — 由调控基因（lacI）编码；无乳糖时结合操作子并阻止转录。有乳糖时，诱导分子（别乳糖）与阻遏蛋白结合，改变其构型使其无法再结合操作子，RNA 聚合酶转录结构基因。无乳糖时，阻遏蛋白有活性，操作子被阻断，基因关闭。

In the lac operon of E. coli, what happens when lactose is absent from the environment?在大肠杆菌的 lac 操纵子中，当环境中没有乳糖时会发生什么？

§6 · Q1

RNA polymerase binds the operator and transcription beginsRNA 聚合酶结合操作子，转录开始

The inducer molecule binds the repressor protein诱导分子与阻遏蛋白结合

The structural genes are transcribed continuously结构基因持续转录

The repressor protein binds the operator and blocks transcription阻遏蛋白结合操作子并阻断转录

When lactose is absent, there is no inducer to inactivate the repressor. The repressor protein binds the operator, physically blocking RNA polymerase from transcribing the structural genes. The lac enzymes are not made because the cell does not need them.无乳糖时，没有诱导物使阻遏蛋白失活。阻遏蛋白结合操作子，从物理上阻止 RNA 聚合酶转录结构基因。lac 酶不被合成，因为细胞不需要它们。

Without lactose → no inducer → repressor stays active → repressor binds operator → transcription blocked → genes OFF. Lactose (via allolactose) must be present to turn the genes ON.无乳糖 → 无诱导物 → 阻遏蛋白保持活性 → 阻遏蛋白结合操作子 → 转录被阻断 → 基因关闭。乳糖（通过别乳糖）必须存在才能开启基因。

All somatic cells in a human body contain the same DNA. Why do liver cells and muscle cells have different properties?人体所有体细胞含有相同的 DNA。为何肝细胞与肌肉细胞具有不同的特性？

§6 · Q2

Different genes are expressed in each cell type through differential gene regulation不同细胞类型通过差异基因调控表达不同的基因

Liver cells and muscle cells have different chromosomes肝细胞与肌肉细胞具有不同的染色体

Mutations gradually accumulate differently in each tissue over time随时间推移，不同组织中逐渐积累不同的突变

Only muscle cells have the full genome; liver cells have a partial genome只有肌肉细胞有完整的基因组；肝细胞有部分基因组

All somatic cells carry the complete genome. What differs between cell types is which genes are expressed — transcription factors and epigenetic marks switch specific genes on or off, producing different proteins and therefore different cell types. This is differential gene expression.所有体细胞携带完整基因组。不同细胞类型的差异在于哪些基因被表达 — 转录因子和表观遗传标记开启或关闭特定基因，产生不同蛋白质，从而形成不同细胞类型。这就是差异基因表达。

Same DNA in all somatic cells. Different cell types = different genes expressed (differential gene expression). Mutations don't drive normal differentiation. All somatic cells have the complete genome.所有体细胞 DNA 相同。不同细胞类型 = 不同基因被表达（差异基因表达）。突变不驱动正常分化。所有体细胞有完整基因组。

Section 7 · Biotechnology第 7 节 · 生物技术

Biotechnology: PCR and Gel Electrophoresis生物技术：PCR 与凝胶电泳

PCR (Polymerase Chain Reaction) — amplifying DNA.PCR（聚合酶链式反应）— 扩增 DNA。

PCR makes millions of copies of a specific DNA segment from a tiny sample. Used in forensics, disease diagnosis, and research. Three repeated steps (each cycle ≈ 30 s to 1 min at different temperatures):PCR 可从微量样品中扩增出数百万份特定 DNA 片段。用于法医鉴定、疾病诊断和研究。三个重复步骤（每个循环约 30 秒至 1 分钟，温度各不同）：

Denaturation (~95 °C):变性（约 95 °C）： the double-stranded DNA is heated; hydrogen bonds break and the two strands separate.双链 DNA 被加热；氢键断裂，两条链分离。
Annealing (~55–65 °C):退火（约 55–65 °C）： the temperature is lowered; short oligonucleotide primers (complementary to sequences flanking the target region) bind to the template strands.温度降低；短寡核苷酸引物（与目标区域两侧序列互补）与模板链结合。
Extension (~72 °C):延伸（约 72 °C）： Taq polymerase (a heat-stable DNA polymerase from Thermus aquaticus) extends each primer by adding nucleotides 5'→3', duplicating the target region. After $n$ cycles, the target DNA is amplified approximately $2^n$ times.Taq 聚合酶（来自嗜热水生菌的耐热 DNA 聚合酶）从每个引物的 5'→3' 方向添加核苷酸延伸，复制目标区域。经 $n$ 个循环后，目标 DNA 扩增约 $2^n$ 倍。

Gel electrophoresis — separating DNA fragments by size凝胶电泳 — 按大小分离 DNA 片段

DNA fragments are loaded into wells at one end of an agarose gel. An electric current is applied: because DNA is negatively charged (phosphate backbone), fragments migrate toward the positive electrode. Smaller fragments move faster and travel farther; larger fragments move slower and stay near the top. The result is a banding pattern where band position indicates fragment size (compared to a DNA ladder of known sizes). Applications: DNA fingerprinting (forensics), paternity testing, gel-purification of PCR products, restriction mapping.DNA 片段被加入凝胶一端的加样孔。施加电流：因 DNA 带负电（磷酸骨架），片段向正极迁移。较小片段移动较快，迁移更远；较大片段移动较慢，停留在靠近顶端处。结果呈现条带图案，条带位置指示片段大小（与已知大小的 DNA 梯状标记比较）。应用：DNA 指纹鉴定（法医）、亲子鉴定、PCR 产物的凝胶纯化、限制性酶切图谱。

In gel electrophoresis, which DNA fragments travel the farthest from the wells?在凝胶电泳中，哪些 DNA 片段从加样孔迁移最远？

§7 · Q1

Largest fragments, because they have more mass to generate momentum最大片段，因为它们有更多质量产生动量

Fragments with the highest G+C contentG+C 含量最高的片段

Smallest fragments, because they move more easily through the gel pores最小片段，因为它们更容易穿过凝胶孔隙

All fragments travel the same distance; only staining intensity differs所有片段迁移相同距离；只有染色强度不同

In gel electrophoresis, DNA migrates toward the positive electrode (DNA is negatively charged). Smaller fragments experience less resistance from the gel matrix and move farther. This is how fragment size is determined — compare position to a DNA ladder of known sizes.在凝胶电泳中，DNA 向正极迁移（DNA 带负电）。较小片段受凝胶基质阻力更小，迁移更远。这就是确定片段大小的方式 — 与已知大小的 DNA 梯状标记的位置比较。

Small fragments travel farthest. All DNA is negatively charged, so direction is the same; size determines distance.小片段迁移最远。所有 DNA 均带负电，故方向相同；大小决定距离。

What is the purpose of primers in PCR?PCR 中引物的作用是什么？

§7 · Q2

To denature the double-stranded DNA使双链 DNA 变性

To define the specific region to be amplified and provide a 3'-OH for Taq polymerase to extend界定待扩增的特定区域，并为 Taq 聚合酶延伸提供 3'-OH

To join the two new strands together after synthesis合成后将两条新链连接在一起

To label the DNA fragments for gel electrophoresis为凝胶电泳标记 DNA 片段

Primers are short single-stranded DNA oligonucleotides complementary to the sequences flanking the target region. They anneal (bind) to the template during the annealing step and provide the 3'-OH end that Taq polymerase requires to begin extending the new strand. Without primers, Taq polymerase cannot start synthesis.引物是与目标区域两侧序列互补的短单链 DNA 寡核苷酸。它们在退火步骤中与模板结合，提供 Taq 聚合酶开始延伸新链所需的 3'-OH 末端。没有引物，Taq 聚合酶无法启动合成。

Primers define the target region and provide a starting point for Taq polymerase. Denaturation is done by heating; ligation is not part of PCR (that is done by DNA ligase in other contexts); gel labeling uses fluorescent dyes, not primers.引物界定目标区域并为 Taq 聚合酶提供起点。变性通过加热完成；连接不是 PCR 的一部分（在其他情境下由 DNA 连接酶完成）；凝胶标记使用荧光染料，而非引物。

Exam Tactics考试策略

Exam Strategy and Common Pitfalls考试策略与常见陷阱

Base-pairing and transcription questions碱基配对与转录题

In DNA: A pairs with T; G pairs with C.DNA 中：A 与 T 配对；G 与 C 配对。 In RNA (and mRNA): U replaces T. If you are asked for an mRNA sequence from a DNA template, replace every T with U.RNA（及 mRNA）中：U 替代 T。若题目要求从 DNA 模板写出 mRNA 序列，将每个 T 替换为 U。
Always state whether you are reading the template strand or the coding strand.始终说明读取的是模板链还是编码链。 The mRNA sequence matches the coding (non-template) strand with T replaced by U, and is antiparallel to the template strand.mRNA 序列与编码链（非模板链）相同（T 改为 U），与模板链反向平行。

Mutation questions突变题

Classify mutations by their effect on the amino acid sequence.根据对氨基酸序列的影响对突变分类。 Silent = no amino acid change. Missense = one amino acid changes. Nonsense = new stop codon. Frameshift = reading frame shifts; usually many amino acids change.沉默突变 = 氨基酸不变。错义突变 = 一个氨基酸改变。无义突变 = 出现新的终止密码子。移码突变 = 读码框移位；通常许多氨基酸改变。
Germline vs somatic mutations: only germline mutations are heritable.生殖细胞突变与体细胞突变：只有生殖细胞突变可遗传。 Somatic cell mutations affect the individual (e.g. cancer) but are not passed to offspring.体细胞突变影响个体（如癌症），但不传给后代。

Central dogma and location questions中心法则与位置题

Transcription = nucleus; translation = cytoplasm (ribosomes).转录 = 细胞核；翻译 = 细胞质（核糖体）。 This is a classic exam trap: "where does protein synthesis occur?" The answer is the ribosome (translation step), not the nucleus (transcription step).这是经典考试陷阱："蛋白质合成发生在哪里？"答案是核糖体（翻译步骤），而非细胞核（转录步骤）。
Replication also happens in the nucleus, during S phase.复制同样发生在细胞核中，在 S 期。 Three DNA-related events in the nucleus: replication (S phase), transcription (G1/G2/S), chromosome condensation (M phase). Translation is the only one that happens outside the nucleus.细胞核中三种 DNA 相关事件：复制（S 期）、转录（G1/G2/S）、染色体凝缩（M 期）。翻译是唯一发生在细胞核外的事件。

Active Recall主动复习

Flashcards闪卡

DNA base-pairing rules?DNA 碱基配对规则？

A pairs with T (2 H-bonds); G pairs with C (3 H-bonds). In RNA, U replaces T.A 与 T 配对（2 个氢键）；G 与 C 配对（3 个氢键）。RNA 中 U 替代 T。

What is the double helix?什么是双螺旋？

Two antiparallel polynucleotide strands wound around each other. Sugar-phosphate backbone on outside; bases pair in the interior. Proposed by Watson and Crick, 1953.两条反向平行的多核苷酸链相互缠绕。糖-磷酸骨架在外；碱基在内部配对。1953 年由沃森和克里克提出。

What is semi-conservative replication?什么是半保留复制？

Each daughter DNA molecule has one parental (old) strand and one new strand. Confirmed by Meselson-Stahl experiment (1958) with ¹⁵N/¹⁴N labeling.每个子代 DNA 分子含一条亲本（旧）链和一条新链。由梅塞尔森-斯塔尔实验（1958 年，¹⁵N/¹⁴N 标记）证实。

Central dogma of molecular biology?分子生物学的中心法则？

DNA → RNA (transcription) → Protein (translation). DNA is also replicated (DNA → DNA).DNA → RNA（转录）→ 蛋白质（翻译）。DNA 也进行自我复制（DNA → DNA）。

Where does transcription occur vs translation?转录与翻译分别在哪里发生？

Transcription: nucleus (eukaryotes). Translation: ribosomes in cytoplasm. mRNA travels from nucleus to ribosome through nuclear pores.转录：细胞核（真核生物）。翻译：细胞质中的核糖体。mRNA 通过核孔从细胞核运至核糖体。

Three types of RNA and their roles?三种 RNA 及其作用？

mRNA: carries the coding sequence from DNA to ribosome. tRNA: delivers amino acids; has anticodon. rRNA: structural and catalytic component of the ribosome.mRNA：携带编码序列从 DNA 到核糖体。tRNA：递送氨基酸；具有反密码子。rRNA：核糖体的结构与催化成分。

What is a codon?什么是密码子？

A triplet of mRNA bases that specifies one amino acid (or a start/stop signal). AUG = start (Met). UAA, UAG, UGA = stop codons.指定一种氨基酸（或起始/终止信号）的 mRNA 三联体碱基。AUG = 起始（Met）。UAA、UAG、UGA = 终止密码子。

Silent vs missense vs nonsense mutation?沉默突变、错义突变与无义突变的区别？

Silent: base changes but same amino acid (degeneracy). Missense: base change → different amino acid. Nonsense: base change → premature stop codon; protein truncated.沉默：碱基变化但氨基酸相同（简并性）。错义：碱基变化 → 不同氨基酸。无义：碱基变化 → 提前终止密码子；蛋白质截短。

What is a frameshift mutation?什么是移码突变？

Insertion or deletion of bases (not a multiple of 3) shifts the reading frame. All downstream codons are altered → usually a non-functional protein.碱基插入或缺失（非 3 的倍数）使读码框移位。下游所有密码子均改变 → 通常产生无功能蛋白质。

The lac operon: when is it ON vs OFF?lac 操纵子：何时开启与关闭？

OFF (no lactose): repressor binds operator; genes blocked. ON (lactose present): allolactose binds repressor → repressor leaves operator → RNA polymerase transcribes structural genes.关闭（无乳糖）：阻遏蛋白结合操作子；基因被阻断。开启（有乳糖）：别乳糖与阻遏蛋白结合 → 阻遏蛋白离开操作子 → RNA 聚合酶转录结构基因。

PCR three steps?PCR 三个步骤？

1. Denaturation (~95 °C): strands separate. 2. Annealing (~55–65 °C): primers bind template. 3. Extension (~72 °C): Taq polymerase extends 5'→3'. Each cycle doubles the target DNA.1. 变性（约 95 °C）：链分离。2. 退火（约 55–65 °C）：引物与模板结合。3. 延伸（约 72 °C）：Taq 聚合酶按 5'→3' 延伸。每个循环使目标 DNA 加倍。

Gel electrophoresis: which fragments travel farthest?凝胶电泳：哪些片段迁移最远？

Smallest fragments. DNA is negatively charged → migrates toward positive electrode. Smaller = less resistance from gel = farther migration.最小片段。DNA 带负电 → 向正极迁移。越小 = 凝胶阻力越小 = 迁移越远。

DNA vs RNA: three differences?DNA 与 RNA 的三个区别？

1. DNA double-stranded; RNA single-stranded. 2. DNA uses deoxyribose; RNA uses ribose. 3. DNA uses T; RNA uses U.1. DNA 双链；RNA 单链。2. DNA 含脱氧核糖；RNA 含核糖。3. DNA 含 T；RNA 含 U。

What is a mutagen? Give two examples.什么是诱变剂？举两个例子。

An agent that increases the rate of mutation. Examples: UV light (causes thymine dimers); ionizing radiation (breaks DNA strands); chemical mutagens (e.g. base analogs, alkylating agents).增加突变率的因素。例如：紫外线（导致胸腺嘧啶二聚体）；电离辐射（断裂 DNA 链）；化学诱变剂（如碱基类似物、烷化剂）。

Mixed Practice综合练习

Practice Quiz综合测验

A DNA template strand reads 3'-AATCGG-5'. What is the mRNA sequence transcribed from this strand?一条 DNA 模板链为 3'-AATCGG-5'。从该链转录所得 mRNA 序列是什么？

3'-TTAGCC-5'3'-TTAGCC-5'

5'-AATCGG-3'5'-AATCGG-3'

5'-TTAGCC-3'5'-TTAGCC-3'

5'-UUAGCC-3'5'-UUAGCC-3'

mRNA is complementary and antiparallel to the template, using U instead of T. Template 3'-A-A-T-C-G-G-5' → mRNA 5'-U-U-A-G-C-C-3'.mRNA 与模板互补且反向平行，用 U 替代 T。模板 3'-A-A-T-C-G-G-5' → mRNA 5'-U-U-A-G-C-C-3'。

Apply base-pairing (T→A written as U in mRNA, A→U, C→G, G→C) reading the template 3'→5' to produce mRNA 5'→3'. Use U, not T, in the mRNA.应用碱基配对（T→A 在 mRNA 中写为 U，A→U，C→G，G→C），沿模板 3'→5' 方向读取，产生 mRNA 5'→3'。mRNA 中使用 U 而非 T。

A cell's DNA has 22% Adenine. What percentage is Guanine?某细胞 DNA 中腺嘌呤占 22%。鸟嘌呤占多少百分比？

28%28%

22%22%

44%44%

25%25%

By Chargaff's rules, A = T = 22%. Together A+T = 44%. The remaining 56% is G+C equally: G = C = 28%.查哥夫定律：A = T = 22%。A+T = 44%。其余 56% 为 G+C 各半：G = C = 28%。

A = T = 22%, so G+C = 100% − 44% = 56%, giving G = C = 28%.A = T = 22%，故 G+C = 100% − 44% = 56%，G = C = 28%。

An mRNA has the codon sequence AUG-GGC-UAA. How many amino acids are in the resulting polypeptide?一条 mRNA 的密码子序列为 AUG-GGC-UAA。所得多肽链含几个氨基酸？

AUG = start (Met); GGC = Gly; UAA = stop codon (no amino acid added). The polypeptide contains 2 amino acids: Met-Gly.AUG = 起始（Met）；GGC = Gly（甘氨酸）；UAA = 终止密码子（不添加氨基酸）。多肽链含 2 个氨基酸：Met-Gly。

Count only the amino-acid-coding codons: AUG (Met) + GGC (Gly) = 2. UAA is a stop codon; it signals termination, not an amino acid.仅计算编码氨基酸的密码子：AUG（Met）+ GGC（Gly）= 2 个。UAA 是终止密码子；它发出终止信号，不对应氨基酸。

A single base is inserted into the middle of a gene. What is the most likely effect on the protein?在某基因中间插入一个碱基。对蛋白质最可能的影响是什么？

Only one amino acid changes; the rest of the protein is normal只有一个氨基酸改变；其余蛋白质正常

The protein is completely unaffected because the genetic code is degenerate因遗传密码具有简并性，蛋白质完全不受影响

All amino acids downstream of the insertion are altered (frameshift)插入点下游所有氨基酸均改变（移码突变）

Only the start codon is affected只有起始密码子受影响

Inserting a single base (not a multiple of 3) causes a frameshift. Every codon 3' of the insertion is read in a shifted frame, changing every amino acid from that point and usually producing a non-functional protein or a premature stop codon.插入一个碱基（非 3 的倍数）导致移码。插入点 3' 侧的每个密码子均在移位的读码框中读取，从该点起每个氨基酸都改变，通常产生无功能蛋白质或提前终止密码子。

A single-base insertion causes a frameshift — not a missense (that's a substitution) and degeneracy doesn't help (degeneracy only applies to synonymous substitutions, not frameshifts).单碱基插入导致移码突变 — 不是错义突变（那是替换），简并性无济于事（简并性仅适用于同义替换，不适用于移码突变）。

After 3 cycles of PCR, how many copies of the target DNA sequence are present (starting from 1 original double-stranded molecule)?从 1 个原始双链分子出发，经过 3 个 PCR 循环后，目标 DNA 序列有多少份拷贝？

Each cycle doubles the number of DNA molecules. After cycle 1: 2; after cycle 2: 4; after cycle 3: 8 = $2^3$. In general, $n$ cycles give $2^n$ molecules from 1 starting molecule.每个循环使 DNA 分子数量翻倍。循环 1 后：2；循环 2 后：4；循环 3 后：8 = $2^3$。一般地，$n$ 个循环从 1 个起始分子产生 $2^n$ 个分子。

PCR doubles target DNA each cycle: 1 → 2 → 4 → 8. Formula: $2^n$ where $n$ = number of cycles.PCR 每个循环使目标 DNA 加倍：1 → 2 → 4 → 8。公式：$2^n$，其中 $n$ = 循环次数。

Self-check自查

Readiness Checklist准备就绪清单

Tick each item when you can do it cold, without notes, on a first attempt.能在无笔记、首次尝试下完成，再勾选每一项。

0 / 11 mastered已掌握 0 / 11

✓State Chargaff's rules (A=T, G=C) and explain what they imply about the double helix structure. Given any one base percentage, calculate the others. 🇺🇸 NGSS HS-LS1-1陈述查哥夫定律（A=T，G=C），并解释其对双螺旋结构的含义。已知任一碱基百分比，计算其余各碱基比例。🇺🇸 NGSS HS-LS1-1
✓Explain semi-conservative replication and describe the Meselson-Stahl experiment that confirmed it. 🇨🇦 AB Biology 30 C3.2k解释半保留复制，并描述证实它的梅塞尔森-斯塔尔实验。🇨🇦 AB Biology 30 C3.2k
✓State the central dogma: DNA → RNA → Protein. Identify where each step occurs in a eukaryotic cell. 🇨🇦 BC A&P 12陈述中心法则：DNA → RNA → 蛋白质。指出每个步骤在真核细胞中发生的位置。🇨🇦 BC A&P 12
✓Given a DNA template strand, write the complementary DNA strand and the mRNA. Remember: U replaces T in mRNA; strands are antiparallel.给定 DNA 模板链，写出互补 DNA 链和 mRNA。记住：mRNA 中 U 替代 T；链为反向平行。
✓Explain codon-anticodon pairing in translation. Identify start codon (AUG) and the three stop codons (UAA, UAG, UGA). Count the amino acids in a given mRNA sequence. 🇨🇦 AB Biology 30 C3.3k解释翻译中的密码子-反密码子配对。识别起始密码子（AUG）和三种终止密码子（UAA、UAG、UGA）。计算给定 mRNA 序列中的氨基酸数目。🇨🇦 AB Biology 30 C3.3k
✓Classify a mutation as silent, missense, nonsense, or frameshift, and predict the consequence for the protein. 🇨🇦 AB Biology 30 C3.6k将突变分类为沉默突变、错义突变、无义突变或移码突变，并预测对蛋白质的后果。🇨🇦 AB Biology 30 C3.6k
✓Describe the three steps of PCR and calculate the number of copies after $n$ cycles ($2^n$). Explain why Taq polymerase is used instead of human DNA polymerase.描述 PCR 的三个步骤，计算 $n$ 个循环后的拷贝数（$2^n$）。解释为何使用 Taq 聚合酶而非人类 DNA 聚合酶。
✓Explain how gel electrophoresis separates DNA fragments, and determine which band on a gel represents the largest fragment. 🇺🇸 NGSS HS-LS1-1 evidence解释凝胶电泳如何分离 DNA 片段，并判断凝胶上哪条条带代表最大片段。🇺🇸 NGSS HS-LS1-1 证据
✓Distinguish germline mutations from somatic mutations in terms of heritability and effect on the organism.从可遗传性和对生物体的影响区分生殖细胞突变与体细胞突变。
✓Explain differential gene expression: why all somatic cells have the same DNA but different cell types express different genes. 🇨🇦 BC A&P 12 Big Idea 2解释差异基因表达：为何所有体细胞具有相同 DNA 但不同细胞类型表达不同基因。🇨🇦 BC A&P 12 大概念 2
✓Honors SBI4U Name the six key replication enzymes (helicase, primase, DNA Pol III, DNA Pol I, ligase, SSB proteins) and explain the role of each. Explain the lac operon: repressor, operator, inducer, and the ON/OFF switch logic. 🇨🇦 ON SBI4U D3.1, D3.3荣誉 SBI4U 命名六种关键复制酶（解旋酶、引物酶、DNA 聚合酶 III、DNA 聚合酶 I、连接酶、单链结合蛋白）并解释各自作用。解释 lac 操纵子：阻遏蛋白、操作子、诱导物及开/关逻辑。🇨🇦 ON SBI4U D3.1、D3.3

Next Steps后续衔接

What This Feeds Into本单元的去向

Molecular genetics is the mechanistic core that every downstream biology topic draws on. The DNA→RNA→protein pathway you learned here underpins Mendelian Genetics (Unit 5 — alleles are different DNA sequences; dominance is about which protein is produced), Cell Division (Unit 4 — chromosomes must be replicated before division), Evolution (Unit 7 — mutations are the ultimate source of genetic variation), and Biotechnology applications throughout advanced biology.分子遗传学是所有后续生物学主题所依赖的机制核心。你在此处学到的 DNA→RNA→蛋白质通路支撑着孟德尔遗传学（第 5 单元——等位基因是不同的 DNA 序列；显性是关于产生哪种蛋白质的问题）、细胞分裂（第 4 单元——分裂前必须复制染色体）、进化（第 7 单元——突变是遗传变异的最终来源）以及整个高等生物学中的生物技术应用。

Within High School Biology.在 HS Biology 内部。

The Cell Structure guide (Unit 1) introduced the nucleus as the site of transcription and ribosomes as the site of translation. Biochemistry (Unit 2) placed proteins in the context of enzyme structure. Cell Division (Unit 4) connects DNA replication (S phase) to chromosome segregation (mitosis/meiosis). Mendelian Genetics (Unit 5) is now interpretable at the molecular level: a dominant allele typically produces a functional protein; a recessive allele may produce a non-functional one. Evolution (Unit 7) rests on the mutation mechanism explained here.《细胞结构》（第 1 单元）介绍了细胞核作为转录场所、核糖体作为翻译场所。《生物化学》（第 2 单元）将蛋白质置于酶结构的背景下。《细胞分裂》（第 4 单元）将 DNA 复制（S 期）与染色体分离（有丝/减数分裂）联系起来。《孟德尔遗传学》（第 5 单元）现在可在分子层面加以解释：显性等位基因通常产生功能性蛋白质；隐性等位基因可能产生非功能性蛋白质。《进化》（第 7 单元）依赖于本指南所解释的突变机制。

Feeds into AP Biology and IB Biology.衔接 AP Biology 与 IB Biology。

This guide is the direct prerequisite for AP Biology Unit 6 (Gene Expression and Regulation) and IB Biology HL Topic B2 (Molecular Biology). Both courses assume fluency with the central dogma, codon reading, and mutation classification from day one. AP Biology extends into post-translational modification, epigenetics, and gene regulation in eukaryotes (enhancers, silencers). IB Biology HL adds gel electrophoresis analysis, Southern blotting, and a deeper treatment of gene cloning. Neither course exists yet in this repo; treat this guide as the prerequisite and revisit it when those products ship.本指南是 AP Biology 第 6 单元（基因表达与调控）和 IB Biology HL Topic B2（分子生物学）的直接先修。两门课程从第一天起就默认学生熟练掌握中心法则、密码子读取和突变分类。AP Biology 进一步涉及翻译后修饰、表观遗传学和真核生物的基因调控（增强子、沉默子）。IB Biology HL 增加了凝胶电泳分析、Southern 印迹法和更深入的基因克隆处理。目前这两门课程均未收录于本仓库中；将本指南视为先修材料，待相关产品上线后再作衔接。