DNA Replication: Making an Exact Copy

1. Where and when replication happens: S phase31

Yesterday you met the cell cycle: a long interphase made of G1, S and G2 — the busiest part of the cycle, not a rest — followed by a short M phase of mitosis and cytokinesis. Today we zoom right into the S sitting in the middle of interphase. S stands for synthesis, and what is being synthesised is DNA. This is the single moment in the whole cycle when the cell makes a complete second copy of every DNA molecule it owns. Before S phase, a human somatic cell has 46 chromosomes, each one a single DNA molecule. After S phase it still has 46 chromosomes, but each one now contains two identical DNA molecules held together. Hold onto that sentence — the number does not change, the content doubles. We will come back to it at the end of the lesson.

昨天你已经认识了细胞周期 (cell cycle)：漫长的（分裂）间期 (interphase) 由 G1 期、S 期、G2 期组成——它是细胞周期中最繁忙的阶段，绝不是「休息期」——之后才是短短的 M 期（有丝分裂 + 胞质分裂 cytokinesis）。今天我们把镜头推近到间期正中间的那个 S。S 代表 synthesis（合成），合成的东西就是 DNA。整个细胞周期里，只有这一刻细胞会把自己全部的 DNA 分子完整地再复制一份。S 期之前，一个人类体细胞 (somatic cell) 有 46 条染色体，每条染色体是一个 DNA 分子；S 期之后，它仍然是 46 条染色体，但每条染色体里现在装着两个一模一样的 DNA 分子。请记住这句话：**数目不变，内容加倍**。这节课结束时还会用到它。

Replication takes place inside the nucleus, because that is where the chromosomes are. It needs three things present at once: the DNA itself, a supply of free nucleotides floating in the nucleus, and the enzymes that do the work. Those free nucleotides are already there — the cell built up a stock of them during G1, before S phase began, as part of its ordinary metabolism, and keeps topping the supply up while replication runs. This matters more than it sounds, because a very common student picture is of an enzyme manufacturing each new nucleotide to order, one at a time, like a chef cooking a dish only once the customer asks for it. Nothing so clever is going on. The ingredients are already in the pot; the process simply selects the correct one at each position by complementary base pairing.

复制发生在细胞核 (nucleus) 内部，因为染色体在那里。它需要三样东西同时到位：DNA 本身、细胞核内游离的核苷酸 (nucleotide) 原料、以及负责干活的酶。这些游离核苷酸是**早就准备好的**——细胞在 S 期之前的 G1 期就通过正常代谢把它们造好并储备在核内，复制进行时还会继续补充。这一点听着平常，其实很关键，因为很多同学脑子里的画面是「有一种酶按需现造出每一个核苷酸」，像厨师听到点单才现炒一道菜。实际上没有这么「聪明」：原料早已在锅里，过程只是按碱基互补配对 (complementary base pairing) 的规则，在每个位置上挑出正确的那一个。

One more timing point that markers reward every year: replication is finished before mitosis starts. By the end of S phase every chromosome has been duplicated. G2 is then spent growing further, checking the copy, and building the proteins that will later form spindle fibres. Only after all of that does prophase begin. If you write in an exam that DNA replicates during prophase, or at the start of mitosis, you lose the mark immediately — and worse, the rest of your answer becomes impossible to follow, because mitosis contains no copying step anywhere in it.

还有一个时间点，评分老师年年都在看：**复制在有丝分裂开始之前就已经完成了**。S 期结束时，每条染色体都已完成复制；接下来的 G2 期用来继续生长、检查复制质量、并合成将来构成纺锤丝 (spindle fibre) 的蛋白质。这些都做完，前期 (prophase) 才开始。如果你在考试里写「DNA 在前期复制」或者「在有丝分裂一开始复制」，这一分立刻就没了——更糟的是，你后面的答案会整个说不通，因为有丝分裂里从头到尾都没有「复制」这个步骤。

2. Semi-conservative: what 'half old, half new' really means42

The word semi-conservative describes the outcome of replication, and it is worth pinning down exactly, because there are two wrong versions of it that students write every single year. Semi-conservative means this: each of the two new DNA molecules consists of one complete original strand and one complete newly built strand. The original molecule is neither destroyed nor kept whole — it is separated down the middle, and each of its two strands ends up in a different daughter molecule, partnered with a brand new strand. 'Conserve' means keep; 'semi' means half; so half of each new molecule has been kept from the parent.

「半保留」这个词描述的是复制的**结果**，一定要说准，因为每年都有两种典型的说错方式。半保留复制 (semi-conservative replication) 的意思是：复制出的两个 DNA 分子，每一个都由**一条完整的原有旧链**和**一条完整的新合成链**组成。原来的分子既没有被销毁，也没有被整个保留下来，而是从中间被分成两条链，这两条旧链分别进入两个新分子，各自配上一条全新的链。conserve = 保留，semi = 一半，所以「每个新分子保留了亲代的一半」。

Wrong version one: 'each new strand is half old and half new'. No. Each STRAND is entirely old or entirely new; it is each MOLECULE that is half and half. Wrong version two: 'one molecule is completely old and the other is completely new'. That is called the conservative model, and it is not what happens. A good habit is to draw it. Draw the parent molecule as two lines, colour one blue and one black. Then draw the two products. Each must show one blue line and one black line. If your drawing ends up with a blue-blue molecule sitting next to a black-black molecule, you have drawn the conservative model by mistake and you can catch yourself before an exam does.

错误说法一：「每条新链一半是旧的、一半是新的」。不对——**每条链要么整条是旧的，要么整条是新的**；「一半一半」说的是**分子**，不是链。错误说法二：「一个分子全是旧的，另一个全是新的」——那叫全保留复制 (conservative)，事实并非如此。一个好办法是动手画：亲代分子画两条线，一条涂蓝、一条涂黑；再画出两个子代分子，每个都必须是「一蓝一黑」。如果你画出来是「蓝蓝」一个、「黑黑」一个，那你画的就是错误的全保留模型——自己就能在考试之前抓住这个错。

Why does this matter biologically, rather than just as vocabulary? Because keeping one original strand in each product is exactly what guarantees accuracy. The old strand is the template — it dictates, base by base, what the new strand must be. If both original strands were thrown away and DNA were built from scratch, there would be no instruction to follow and no way to know whether the copy was right. Semi-conservative replication is therefore not merely a description of a pattern; it is the reason the copy is faithful. That link — old strand acts as template, therefore the copy is accurate, therefore the daughter cells are identical — is the causal chain markers are hunting for in extended-response answers.

这在生物学上为什么重要，而不只是一个名词？因为「每个产物里都保留一条旧链」正是**保证准确性**的关键。旧链就是模板链 (template strand)，它一个碱基一个碱基地规定新链必须是什么。如果两条旧链都被丢掉、从零开始造 DNA，那就没有任何指令可循，也无从判断复制对不对。所以「半保留」不只是描述一种模式，它本身就是「复制忠实可靠」的原因。这条因果链——旧链当模板 → 复制准确 → 子细胞遗传物质完全相同——正是长题答案里评分老师要找的东西。

3. The steps of replication, in order597

Step 1 — the helix unzips. An enzyme called helicase moves along the DNA and breaks the hydrogen bonds holding each base pair together. Because those bonds are weak, they can be broken without damaging anything else, and the two strands peel apart. Now read that again and notice what is NOT broken: the sugar-phosphate backbone. The strong covalent bonds running along each backbone stay completely intact. This is precisely the design feature you met on Day 1 — strong bonds along the strands, weak bonds across the middle — and this is the moment it pays off. If you write that the backbone breaks, you have described a shattered chromosome, not a replicating one. The Y-shaped region where the strands have come apart and copying is under way is called the replication fork — a term you need if you sit NSW Module 5, and one you can simply recognise if you sit VCE Unit 1.

**第一步——双螺旋「拉开拉链」**。一种叫解旋酶 (helicase) 的酶沿着 DNA 前进，把每一对碱基之间的氢键 (hydrogen bond) 打断。因为氢键很弱，可以在不破坏其他结构的情况下被打开，两条链就此分离。现在重读一遍，注意**没有被打断的是什么**：磷酸—脱氧核糖骨架 (sugar-phosphate backbone)。沿骨架排列的那些很强的共价键 (covalent bond) 完全保持完整。这正是第 1 天讲过的结构精妙之处——链的内部是强键、中间横档是弱键——现在派上用场了。如果你写「骨架断开」，那描述的是一条断碎的染色体，而不是一条正在复制的染色体。两条链已经分开、正在合成新链的那个 Y 形区域叫做复制叉 (replication fork)——NSW Module 5 要求掌握这个名词，VCE Unit 1 认得即可。

Step 2 — each old strand becomes a template. With the bases now exposed, each single strand carries a readable instruction: a strand reading A-T-G-C can only ever be partnered by T-A-C-G. Step 3 — free nucleotides move in and pair up. The nucleotides already floating in the nucleus line up against the exposed bases, and only the complementary one fits and bonds: adenine with thymine (two hydrogen bonds), cytosine with guanine (three hydrogen bonds). This is where the accuracy comes from. It is not that an enzyme 'remembers' the sequence; it is that only the correct base can form the right hydrogen bonds in the right place, so the correct base is the one that stays.

**第二步——每条旧链充当模板链**。碱基现在暴露出来了，每条单链就成了一份可读的说明书：一条读作 A-T-G-C 的链，其配对链只可能是 T-A-C-G。**第三步——游离核苷酸进来配对**。核内本来就漂浮着的核苷酸排到暴露的碱基旁边，只有互补的那一个能配得上并结合：腺嘌呤 (adenine) 配胸腺嘧啶 (thymine)（2 个氢键），胞嘧啶 (cytosine) 配鸟嘌呤 (guanine)（3 个氢键）。**准确性就来自这里**——不是某种酶「记得」序列是什么，而是只有正确的碱基才能在正确的位置形成正确的氢键，所以留下来的自然是正确的那一个。

Step 4 — the new nucleotides are joined up. DNA polymerase bonds each newly arrived nucleotide to the one before it, building a continuous new sugar-phosphate backbone along each template. Step 5 — two complete molecules re-wind. Each template strand now has a full new partner, and each pair twists back into a double helix. Where there was one DNA molecule there are now two, and by the rule from section 2, each contains one old strand and one new one. The photocopier analogy is useful here — one page in, two identical pages out — but notice exactly where it breaks down: a photocopier keeps the original intact and produces a separate copy, which is the conservative model. DNA does not work that way. A zipper being opened, with each half then growing a new set of teeth, is closer to the truth, though no real zipper does that either.

**第四步——把新核苷酸连接起来**。DNA 聚合酶 (DNA polymerase) 把每个新到位的核苷酸与前一个连接起来，沿着每条模板链逐渐建成一条连续的新磷酸—脱氧核糖骨架。**第五步——两个完整分子重新盘绕**。每条模板链现在都配上了一条完整的新链，各自重新扭成双螺旋。原来一个 DNA 分子，现在变成两个；按第 2 节的规则，每一个都是「一条旧链 + 一条新链」。这里可以用**复印机**打比方——放进一张纸，出来两张一样的——但要看清它在哪里失效：复印机保留原件完整、另外产出一份副本，那其实是**全保留模型**，DNA 并不是这样。更接近事实的比喻是「拉链拉开，每半条各自长出一排新的齿」，尽管现实中也没有哪条拉链会这么干。

4. Worked example: building the complementary strand946

If you sit NSW Module 5 or QCAA Unit 4, this is the most reliably examinable skill in the whole topic, and it is free marks once the habit is automatic. VCE students meet it formally in Unit 3 — practise it anyway, because everything later in genetics rests on it. The task: given the base sequence of one strand, write the sequence of the strand that will be built against it. The only rule you need is complementary base pairing — A pairs with T, C pairs with G. Work along the sequence one base at a time and write each partner directly underneath. Do not try to do it in your head, and do not reverse the order of the letters.

如果你考的是 NSW Module 5 或 QCAA Unit 4，这是本章**最稳定会考、也最容易拿分**的一项技能，只要习惯变成自动的，就是白送的分。维州 VCE 学生要到 Unit 3 才正式学到它——但现在就练，因为后面整个遗传学都建立在这一步上。题目形式：给出一条链的碱基序列，要求写出以它为模板新合成的那条链的序列。你只需要一条规则——碱基互补配对：A 配 T，C 配 G。**一个碱基一个碱基地对着写，写在正下方**。不要在脑子里默算，也不要把字母顺序倒过来写。

Worked example. Template strand: 5'-T A C G G A T C-3'. Take them one at a time. T pairs with A. A pairs with T. C pairs with G. G pairs with C. G pairs with C. A pairs with T. T pairs with A. C pairs with G. So the new complementary strand is 3'-A T G C C T A G-5'. Now check your work two ways. First, count: eight bases in, eight bases out — if the lengths differ you have skipped one. Second, scan for illegal pairs: any A sitting opposite a G, or any C opposite a T, is an error. Those two specific wrong pairings are the ones students actually produce under time pressure, so they are worth checking for deliberately.

**例题**。模板链：5'-T A C G G A T C-3'。一个一个来：T 配 A；A 配 T；C 配 G；G 配 C；G 配 C；A 配 T；T 配 A；C 配 G。所以新合成的互补链是 3'-A T G C C T A G-5'。写完用两种方法检查。第一，**数长度**：进去 8 个碱基，出来也必须是 8 个，长度不一致就说明漏了一个。第二，**扫一遍有没有违规配对**：出现 A 对着 G，或者 C 对着 T，就是错的。这两种错配恰恰是同学在考场紧张时真正会写出来的，所以值得专门检查一遍。

About those 5' and 3' labels. You will see them on exam diagrams and on the strands in your textbook, and at this level they simply record that the two strands run in opposite directions — that is what antiparallel means. At Year 11 level you are expected to state that the strands are antiparallel and to notice that the complementary strand's labels are flipped. You are NOT expected to explain the chemistry of the 3' and 5' carbons of deoxyribose, or to reason about the direction in which synthesis proceeds. If a question gives you 5'-...-3' and asks for the complementary strand, write your answer as 3'-...-5' and move on to the next question.

关于 5' 和 3' 这两个标记。你会在考试图和课本的链上看到它们，在这个阶段它们只是记录了**两条链方向相反**这一事实——这就是反向平行 (antiparallel) 的含义。Year 11 阶段，你需要能说出「两条链是反向平行的」，并注意到互补链的标记是反过来写的。你**不需要**解释脱氧核糖 3'、5' 碳原子的化学结构，也不需要推理合成沿哪个方向进行。题目给你 5'-...-3'，问互补链，你就写成 3'-...-5'，然后做下一题。

One extra habit worth building now: if a question gives you a sequence and asks how many hydrogen bonds hold the resulting double-stranded segment together, count 2 for every A-T pair and 3 for every C-G pair. In the worked example above there are four A-T pairs and four C-G pairs, giving (4 x 2) + (4 x 3) = 20 hydrogen bonds. NSW markers like this style of question because it checks whether you genuinely know which pair carries which bond count, rather than just reciting the letters of the pairing rule.

现在再养成一个额外习惯：如果题目给你一段序列，问「形成的这段双链之间共由多少个氢键维系」，就按 A-T 每对 2 个、C-G 每对 3 个来数。上面例题中有 4 对 A-T 和 4 对 C-G，所以 (4×2)+(4×3)=20 个氢键。NSW 的出题老师喜欢这种问法，因为它能检验你是不是**真的记得哪种配对有几个氢键**，而不是只会背 A 配 T、C 配 G 这几个字母。

5. After replication: sister chromatids and the X shape

Now connect the molecule back to the structure you can actually see down a microscope. Before S phase, a chromosome is a single long DNA molecule wrapped up with protein — one thread carrying one copy of the genes on that chromosome. After replication in S phase, that same chromosome contains two identical DNA molecules. When the chromosome later condenses during prophase, those two copies become visible as two rods lying side by side, joined at a narrow waist. Each rod is a chromatid, and because the two are exact copies of one another we call them sister chromatids. The point at which they are held together is the centromere. Joined this way, they produce the familiar X shape you see in every textbook diagram.

现在把分子层面和你在显微镜下真正看得到的结构连起来。S 期之前，一条染色体是**一个**很长的 DNA 分子与蛋白质缠绕在一起——一根丝，携带这条染色体上全部基因的一份拷贝。S 期复制之后，同一条染色体里含有**两个完全相同的 DNA 分子**。等到前期染色体高度螺旋化变粗时，这两份拷贝就显现为并排的两根棒，中间由一个「细腰」连着。每一根棒叫一条染色单体 (chromatid)；由于两者互为精确拷贝，所以称为姐妹染色单体 (sister chromatid)。把它们连在一起的那个位置就是着丝粒 (centromere)。这样连着，就构成了课本插图里那个熟悉的 X 形。

Here is the sentence that earns and loses the most marks in this entire topic: an X-shaped structure is still ONE chromosome. It is one chromosome made of two sister chromatids. It is not two chromosomes. A human cell that has just finished S phase still has 46 chromosomes, not 92 — it has 46 chromosomes and 92 chromatids. Later, when the centromeres divide at anaphase and the chromatids are pulled apart, each separated chromatid is from that moment called a chromosome in its own right. The count per cell doubles only momentarily at anaphase, when the centromeres divide, and cytokinesis then gives each daughter cell 46 — the same number the parent cell started with, so the ploidy of the cell never changes. Replication itself never doubled the chromosome number. Students who believe the number doubles at prophase and halves at anaphase have imported meiosis logic into mitosis, and the whole answer collapses around that one error.

下面这句话是整章**得分和丢分的分水岭**：一个 X 形结构仍然是**一条**染色体。它是由两条姐妹染色单体组成的一条染色体，**不是两条染色体**。刚完成 S 期的人类细胞仍然有 46 条染色体，不是 92 条——它有 46 条染色体、92 条染色单体。之后到了后期，着丝粒分裂、姐妹染色单体被拉开，每条分离出来的染色单体从那一刻起才各自算作一条独立的染色体。细胞内的染色体数目只在后期着丝粒分裂的那一瞬间短暂变成 92 条，胞质分裂后每个子细胞各得 46 条，与亲代细胞相同，因此细胞的倍性 (ploidy) 自始至终没有改变。**复制本身从未使染色体数目加倍。** 那些认为「前期加倍、后期减半」的同学，是把减数分裂 (meiosis) 的逻辑硬套到有丝分裂上，整道题都会因为这一个错误而塌掉。

Two more traps to close off. First, a chromosome does not always look like an X. For most of a cell's life it is unreplicated and uncondensed chromatin — a single thin thread you cannot even distinguish as a separate structure under a light microscope. The X shape is only visible once the replicated chromosome condenses in prophase, and it lasts only until the centromeres divide at anaphase. Between the end of S phase and prophase the chromosome is already replicated but still an uncondensed chromatin thread — replicated does not mean visibly X-shaped. Second, do not confuse sister chromatids with homologous chromosomes. Sister chromatids are two identical copies of the SAME chromosome, produced by replication and joined at a centromere. Homologous chromosomes are a matching pair of separate chromosomes, one inherited from each parent, carrying the same genes in the same order but not necessarily identical versions of those genes — and they are never joined at a centromere.

还有两个坑要堵上。**第一**，染色体并不总是 X 形。细胞一生的大部分时间里，它是未复制、未螺旋化的染色质 (chromatin)——一根细丝，在光学显微镜下根本分辨不出是独立结构。X 形只有在复制后的染色体于前期高度螺旋化之后才看得见，并且到后期着丝粒分裂时就消失。S 期结束到前期之间，染色体虽然已经复制，但仍是未螺旋化的染色质细丝——**已复制并不等于看得见 X 形**。**第二**，不要把姐妹染色单体和同源染色体 (homologous chromosomes) 混为一谈。姐妹染色单体是**同一条**染色体复制出来的两份完全相同的拷贝，由着丝粒连在一起；同源染色体是**两条独立**、彼此配对的染色体，一条来自父方、一条来自母方，以相同的顺序携带相同的基因，但基因的具体版本不一定相同，而且它们**从不**由着丝粒相连。

Finally, the causal chain. This is the assessed skill, so learn it as one connected sequence rather than four separate facts. DNA is replicated semi-conservatively in S phase, so each chromosome now consists of two identical sister chromatids joined at a centromere. At anaphase the centromeres divide and one chromatid of each pair is pulled to each pole. Therefore each daughter nucleus receives one complete, identical copy of every chromosome, and the two daughter cells are genetically identical to each other and to the parent cell — they are clones. Both are still diploid: they carry two copies of every chromosome, 46 in total, arranged as 23 homologous pairs. Only gametes carry a single set, and they are not made by mitosis. Notice what happens if you delete the first step: without prior copying, mitosis would simply share out the existing chromosomes and the number would halve at every division. That is your justification for why replication must come first.

最后是那条**因果链**。这才是真正被考的能力，请把它当作一个连贯的序列来记，而不是四条互不相干的事实。DNA 在 S 期进行半保留复制，**所以**每条染色体现在由两条完全相同、在着丝粒处相连的姐妹染色单体组成；到了后期，着丝粒分裂，每对中的一条染色单体被拉向一极；**因此**每个子细胞核都获得每一条染色体的一份完整而相同的拷贝，两个子细胞彼此之间、以及与亲代细胞之间都遗传物质完全相同——它们是克隆 (clone)。两个子细胞仍然都是二倍体 (diploid)：每种染色体各有两条，共 46 条，构成 23 对同源染色体。只有配子 (gamete) 只带一套染色体，而配子并不是由有丝分裂产生的。注意如果把第一步删掉会怎样：没有事先复制，有丝分裂就只是把现有染色体分一分，每分裂一次数目就减半。这就是「为什么复制必须先发生」的论证 (Justify) 依据。