Build DNA, work the enzymes at the fork, redo Meselson–Stahl and race real genomes.
Visualize the double helix unwinding, the leading strand racing ahead, and Okazaki fragments piecing the lagging strand together — semiconservative replication, one nucleotide at a time. Seven missions take you from base pairing to the replication race, with a quiz, a certificate, a printable worksheet and answer key, and an AP / IB section.
Before a cell divides, each strand of its DNA becomes a template for a new partner strand. Read the template (written 3′ → 5′) and build the new strand 5′ → 3′ by pairing A with T and G with C. Build 2 strands, then count the bases.
Tap a base button to fill the highlighted spot (keyboard: type A, T, G or C). Tap any spot to change it.
Count every base in your finished double strand (both strands, 24 bases).
Copying DNA takes a team of enzymes at the replication fork. For each job, choose the enzyme that does it. Your first choice is your prediction; a wrong choice tells you what that enzyme really does. Then match 8 job cards (6 right on the first try completes the mission).
The original animation: press ▶ Play to watch helicase, primase, DNA polymerase and ligase at work, and switch views with the tabs (3D helix, base pairing, Meselson–Stahl, synthesis rate). Scrolling over the picture scrolls the page; use Ctrl + scroll or a two-finger pinch to zoom. Template length and Okazaki size are scaled down for the animation (real fragments: 100–200 nucleotides in eukaryotes, 1,000–2,000 in bacteria), and its Replication forks slider counts origins.
Semiconservative: Each daughter DNA has ONE original parental strand + ONE newly synthesized strand. Proved by Meselson & Stahl (1958).
At each fork one new strand, the leading strand, is built in one piece toward the fork. The other, the lagging strand, is built away from the fork in short Okazaki fragments, and every fragment needs its own RNA primer. Run at least 3 copies, with both a bacterium and a eukaryote.
Real range in bacteria: 1,000–2,000 nucleotides.
| Run | Organism | Template (nt) | Fragment (nt) | Okazaki fragments | Primers: leading / lagging |
|---|
Matthew Meselson and Franklin Stahl grew E. coli on heavy nitrogen (¹⁵N) until all its DNA was heavy, then moved the cells to normal light nitrogen (¹⁴N). After each generation they spun DNA in a cesium chloride density gradient: heavier DNA settles lower in the tube. Three models were on the table. Predict what each one gives, then run the centrifuge.
| Model | Generation 1 | Generation 2 |
|---|
| Generation | Heavy ¹⁵N/¹⁵N (1.724 g/mL) | Hybrid ¹⁵N/¹⁴N (1.717 g/mL) | Light ¹⁴N/¹⁴N (1.710 g/mL) |
|---|
Each origin of replication opens a bubble with 2 forks moving apart. Time ≈ length ÷ (2 × fork speed × origins). Bacterial forks copy about 1,000 nucleotides per second, human forks about 50. Run at least 3 races.
E. coli has a single origin (oriC).
| Run | DNA | Origins | Forks | Time |
|---|
Real cells: E. coli about 40 min; a human cell has about 30,000–50,000 origins, yet S phase lasts about 8 hours because origins fire at different times (early and late), not all at once.
A human cell copies about 6.4 billion base pairs every time it divides. Choosing the right base gets about 1 in 100,000 wrong. Two checks clean up most of the mistakes. Run all three settings, then find the mismatches yourself. Errors that stay become mutations.
Bar length on a log scale: each step down is 100 times fewer errors.
Run: base selection only · + proofreading · + proofreading and mismatch repair.
You are the mismatch-repair crew. The top strand is the old template, the bottom strand is new. Tap the pair that does not follow A–T, G–C.
10 questions drawn from a bank of 20. Get 7 or more right to complete the mission. Every answer comes with an explanation.
| Enzyme / protein | Job |
|---|---|
| Helicase | unwinds the helix by breaking hydrogen bonds between base pairs |
| Topoisomerase | relieves supercoiling ahead of the fork (cuts and rejoins DNA) |
| SSB proteins | keep the single strands apart |
| Primase | makes a short RNA primer with a free 3′ end |
| DNA polymerase III (bacteria) | adds nucleotides to the 3′ end: builds 5′ → 3′, reads the template 3′ → 5′; proofreads |
| DNA polymerase I (bacteria) | replaces RNA primers with DNA |
| DNA ligase | seals nicks with a phosphodiester bond |
| Quantity | Value |
|---|---|
| E. coli chromosome | 4.6 million bp, 1 origin, ~1,000 nt/s per fork → about 40 min |
| Human chromosome 1 | ≈ 248 million bp; ~50 nt/s; 1 origin would take ≈ 29 days |
| Human genome (diploid cell) | ≈ 6.4 billion bp, ~30,000–50,000 origins, S phase ≈ 8 h |
| Okazaki fragments | 1,000–2,000 nt (bacteria) · 100–200 nt (eukaryotes) |
| Error rate | 1 in 10⁵ (base selection) → 1 in 10⁷ (proofreading) → 1 in 10⁹ (mismatch repair) |
| Meselson–Stahl bands | heavy 1.724 · hybrid 1.717 · light 1.710 g/mL |
| Chargaff’s rule | A = T and G = C; human DNA ≈ 30% A, 30% T, 20% G, 20% C |
The missions use the high-school model. These four ideas take it to AP Biology (topic 6.2) and IB Biology (D1.1) depth.
DNA polymerase III on its own falls off the template after adding only a few nucleotides. A ring-shaped sliding clamp (the β clamp in bacteria, PCNA in eukaryotes) is loaded around the DNA by a clamp loader and holds the polymerase in place, so it can add many thousands of nucleotides without letting go. This “processivity” is why the leading strand can be built in one piece. On the lagging strand a new clamp is loaded for every Okazaki fragment.
DNA polymerase checks the shape of each new base pair. If the base it just added is wrong, the mismatched 3′ end does not fit the active site well, so the strand slips into a second site that cuts nucleotides off the 3′ end (3′ → 5′ exonuclease). Polymerase then tries again. This only works because synthesis runs 5′ → 3′: removing the last nucleotide leaves an ordinary 3′-OH end to extend. Proofreading improves accuracy about 100×, from about 1 in 10⁵ to about 1 in 10⁷. Mismatch repair proteins then find leftover errors in the new strand (in E. coli the new strand is recognized because it is not yet methylated) and improve accuracy about 100× again, to about 1 in 10⁹.
E. coli has one origin (oriC) and two forks that meet on the opposite side of its circular chromosome after about 40 minutes. Eukaryotic chromosomes are linear and far longer, so they use many origins: about 30,000–50,000 in a human cell. Each origin is “licensed” in G1 and can fire only once per cell cycle, so no stretch of DNA is copied twice. Origins fire at different times (early and late in S phase), which is why S phase takes about 8 hours rather than the half hour you would get if every origin fired at once.
At the very end of a linear chromosome, the last RNA primer on the lagging strand is removed, but there is no upstream 3′ end for DNA polymerase to extend into the gap. So each round of copying leaves the new strand slightly shorter. Chromosome ends are capped by telomeres, thousands of repeats of a short sequence (TTAGGG in humans) that protect the genes inside. The enzyme telomerase carries its own RNA template and adds new repeats to the 3′ end, making up for the loss. Telomerase is active in stem cells and in most cancer cells, but low in most body cells, whose telomeres shorten as they divide.
Eukaryotic names: polymerase α–primase starts each strand, polymerase ε mainly builds the leading strand and polymerase δ the lagging strand; primers are removed by RNase H and FEN1 rather than DNA polymerase I. The jobs are the same as in bacteria.
Imagine a family with one treasured recipe card that every child wants to take to their own kitchen. Nobody memorizes it — they photocopy it. A cell about to divide faces exactly this problem: both daughter cells need the complete instruction book, the DNA. So before it splits, the cell makes a full copy of every page. The clever part is how it copies, and it all comes from one matching rule.
DNA is a twisted ladder. Each rung is a pair of letters, and the letters always pair the same way: A with T, and G with C. That single rule is the whole trick. Unzip the ladder down the middle and each lonely half-rung announces exactly which letter belongs opposite it — so each old strand becomes a template for rebuilding its missing partner. This is why we call it semiconservative: every new double helix keeps one whole old strand and one freshly built strand.
Now name two quantities: the template length $L$ (how many base pairs must be copied) and the polymerase speed $v$ (letters added per second). The most natural question — how long does copying take? — is just distance over speed:
Try a number: a small bacterial gene of $L = 1000$ base pairs, copied at the bacterial rate $v = 1000$ bases per second, takes about $t = 1000/1000 = 1$ second. Simple.
Two refinements turn that estimate into the real picture. First, the two strands point in opposite directions (they are antiparallel), and DNA polymerase can only add new letters onto a 3′ end. So only the leading strand is built smoothly toward the opening fork; its partner, the lagging strand, has to be stitched backward in short pieces called Okazaki fragments, which ligase later joins up. Second, replication runs outward in both directions from each starting point, and a real genome fires many starting points (origins) at once. Both effects share the workload, so the time shrinks:
The factor of 2 is there because each origin opens a bubble with two forks moving in opposite directions, and $N_o$ is the number of active origins. In the simulation, the Template length slider sets $L$, the Polymerase rate slider sets $v$, and the Replication forks slider (it counts origins) sets $N_o$ — so the Bio-time readout drops every time you raise the speed or add origins.
Drag Polymerase rate between the slow human value (~50 b/s) and the bacterial ~1000 b/s and watch the Bio-time readout change. In the animation the lagging strand trails a little because each Okazaki fragment can only start after the fork has opened more template; in real cells both new strands are finished at the same overall speed.
Toggle Proofreading OFF and watch the uncorrected-error count climb — similar to cancers with a faulty proofreading polymerase (POLE mutations); Lynch syndrome is the mismatch-repair version of the same problem.
Raise Replication forks from 1 toward 6 and watch the Bio-time readout collapse — a hands-on proof of why your chromosomes need thousands of origins rather than just one.
DNA Replication — Watson & Crick (1953) proposed the double-helix structure; Meselson & Stahl (1958) confirmed semiconservative replication; Kornberg (1956) discovered DNA polymerase. Every time a cell divides, it must first copy its entire genome so each daughter cell receives a complete set of genetic instructions. In humans this means duplicating roughly $6.4 \times 10^9$ base pairs in a diploid cell ($3.2 \times 10^9$ per set of chromosomes) — and getting it right within an error rate of about one mistake per billion bases.
| Symbol / Structure | Biological meaning | Unit / type | A-level relevance |
|---|---|---|---|
| A, T, G, C | Adenine, Thymine, Guanine, Cytosine — four nitrogenous bases | — | Base-pairing rules (A–T, G–C) |
| 5′ → 3′ | Direction of DNA strand (carbon numbering on sugar) | — | Polymerase only adds to 3′ end |
| Helicase | Enzyme that unwinds the double helix at the fork | Protein | Breaks H-bonds between bases |
| DNA polymerase III | Main enzyme that adds nucleotides to growing strand | Protein | Builds new strand 5′→3′ |
| Primase | Lays down short RNA primer to start synthesis | Protein | Required because polymerase needs a 3′-OH |
| Ligase | Joins Okazaki fragments on lagging strand | Protein | Seals sugar–phosphate gaps |
| $L$ | Template length | base pairs (bp) | Determines replication time |
| $v$ | Polymerase synthesis rate | bases s⁻¹ | ≈50 bp/s human, ≈1000 bp/s E. coli |
| $\mu$ | Mutation rate per base | dimensionless | ≈10⁻⁹ after proofreading + repair |
| $N_o$ | Number of origins of replication (each gives 2 forks) | integer | E. coli has 1; a human cell about 30,000–50,000 |
Why does the cell need to replicate its DNA? Before any cell can divide, each daughter cell needs a complete and identical copy of the genetic instructions. Replication produces two identical DNA molecules from one — but with a clever twist: each new molecule is half-old, half-new.
The total replication time for a chromosome of length $L$ (in base pairs), with polymerase rate $v$ (bp/s), and $N_o$ origins, each sending two forks outward in opposite directions, is approximately:
The factor of 2 appears because each origin starts two forks that move apart in opposite directions (bidirectional replication). The error rate of replication, accounting for proofreading (P) and mismatch repair (R), is:
The default Replication Fork tab visualizes the entire process happening at the fork: helicase splitting the helix, primase laying primers (orange marks), DNA polymerase III moving along both strands, the leading strand growing continuously, and Okazaki fragments forming on the lagging strand. The Polymerase rate slider controls how fast bases are added — ~50 bp/s in eukaryotes, ~1000 bp/s in bacteria. In the animation the lagging strand trails a little because each Okazaki fragment can only start after the fork has opened more template; in real cells both new strands are finished at the same overall speed.
The Error rate slider injects mismatches. With Proofreading ON (default), most mismatches are caught and removed (you'll see a red flash when polymerase reverses). Toggle proofreading OFF and watch uncorrected errors accumulate in the readout — this is what happens in cells with defective polymerase: cancer-causing mutation rates skyrocket.
The Meselson–Stahl tab reproduces the classic 1958 experiment: bacteria grown in heavy ¹⁵N nitrogen, switched to light ¹⁴N, then DNA spun in a density gradient. After 1 generation, all DNA appears at intermediate density (one heavy + one light strand) — ruling out conservative replication; the generation-2 result (hybrid + light bands) then rules out dispersive, leaving semiconservative.
The longest human chromosome (chromosome 1) contains approximately $L = 2.5 \times 10^8$ base pairs. DNA polymerase in eukaryotes works at $v \approx 50$ bp/s. If there were only one origin of replication ($N_o = 1$):
$$ t = \dfrac{L}{2vN_o} = \dfrac{2.5 \times 10^8}{2 \times 50 \times 1} = 2.5 \times 10^6\ \text{seconds} \approx 29\ \text{days} $$
That is far too slow — a cell could never divide in 24 hours! In reality, human chromosomes contain many origins firing in parallel: about 30,000–50,000 across the genome, a few thousand on chromosome 1. Even if only $N_o \approx 1{,}000$ origins on chromosome 1 fired at once:
$$ t \approx \dfrac{2.5 \times 10^8}{2 \times 50 \times 1{,}000} \approx 2{,}500\ \text{s} \approx 42\ \text{minutes per chromosome} $$
That is far shorter than the observed S phase (about 8 h): origins do not all fire at once but in a timed program, some early and some late. Takeaway: multiple origins are not optional — they are essential to fit replication into the cell cycle.
Everything runs in the browser on classroom laptops, tablets and phones: no login and no personal data. The page uses Google Analytics for anonymous usage counts only; student answers, mission progress and the name on the certificate stay in the browser on the student’s device and are not sent anywhere. Template strands, quiz questions and the mismatch rounds are drawn at random, so neighbors get different items.
Seven missions in one period is brisk. For a slower pace use two periods: Missions 1–4, then Missions 5–7 and the worksheet.
| Time | Phase | Activity |
|---|---|---|
| 0–5 min | Engage | Hook: “A human cell copies 6.4 billion base pairs in about 8 hours and makes only about 6 mistakes. How?” Students write one guess. |
| 5–12 min | Explore | Mission 1 Strand Builder: build two complementary strands and discover A = T, G = C by counting. |
| 12–20 min | Explain | Mission 2 Enzyme Crew, then watch the 2D fork animation (Play) and name each enzyme as it appears. |
| 20–28 min | Explore | Mission 3 Leading vs Lagging and Mission 4 Meselson–Stahl Lab in pairs: one student predicts, the other runs, then swap. |
| 28–36 min | Elaborate | Mission 5 Replication Race and Mission 6 Proofreading: return to the hook question with real numbers. |
| 36–45 min | Evaluate | Mission 7 Final quiz (or homework) and the exit ticket on the worksheet. AP classes: Going further. |
Two printable pages per student, one section per mission: a strand to complete and count, an enzyme matching table, Okazaki and replication-time calculations, the Meselson–Stahl band table, the error-rate chain and an exit ticket. The key has every answer.
NGSS: HS-LS1-1 (the structure of DNA determines the structure of proteins), HS-LS3-1 (the role of DNA and chromosomes in coding the instructions for traits passed from parents to offspring), HS-LS1-4 (cell division: DNA is copied before mitosis). AP Biology: Unit 6 Gene Expression and Regulation, topic 6.2 Replication. IB Biology: D1.1 DNA replication. Practices: developing and using models, analyzing and interpreting data, using mathematics and computational thinking, engaging in argument from evidence (Meselson–Stahl).
❌ Misconception: "DNA polymerase builds both strands in the same direction."
✅ Correction: DNA polymerase can only synthesize in the 5′→3′ direction. Because the two parental strands are antiparallel, only the leading strand can be built continuously toward the fork. The lagging strand template runs the "wrong way", so its new strand must be built away from the fork in short Okazaki fragments, then stitched together by ligase.
📖 Alberts et al., Molecular Biology of the Cell, 7th ed., Ch. 5, "The DNA Replication Fork is Asymmetric".
❌ Misconception: "Semiconservative replication means each new DNA molecule is half-old and half-new mixed together."
✅ Correction: Each daughter DNA contains one entire intact parental strand paired with one entirely new strand — not bits of old and new mixed within a single strand. Students who draw "checkered" daughter DNA are describing the disproven dispersive model. Meselson & Stahl's density gradient ruled this out: dispersive would give a single band at every generation (hybrid density at generation 1, then drifting lighter), but semiconservative gives a clean split into hybrid and light bands by generation 2.
📖 Campbell & Reece, Biology, 12th ed., Ch. 16, Figure 16.10 "Three alternative models of DNA replication".
❌ Misconception: "RNA primers are part of the final DNA molecule."
✅ Correction: RNA primers are temporary. After polymerase III has extended the strand, DNA polymerase I removes the RNA primer base-by-base (using its 5′→3′ exonuclease activity) and replaces those nucleotides with DNA. Ligase then seals the final gap. The finished daughter molecule contains no RNA. This step is the reason eukaryotic chromosome ends shorten with each division — the end-replication problem solved by telomerase.
📖 Jones, Fosbery et al., Cambridge International AS & A Level Biology, 5th ed., Ch. 6.
❌ Misconception: "DNA is copied during mitosis."
✅ Correction: DNA is copied in S phase of interphase, hours before mitosis begins. During mitosis the already-copied chromosomes (two sister chromatids each) are only separated. In a typical human cell S phase lasts about 8 hours.
🔍 Students link “copying” with “dividing”, because both seem to make two of something.
❌ Misconception: "After replication each new DNA molecule is made of two brand-new strands."
✅ Correction: Replication is semiconservative: every new molecule keeps one whole old strand and has one newly built strand. The Meselson–Stahl experiment shows this: one hybrid band at generation 1, then hybrid + light bands at generation 2 (Mission 4).
🔍 Many diagrams color both daughter molecules in the same new color.
❌ Misconception: "DNA polymerase can start a new strand on its own."
✅ Correction: DNA polymerase can only add nucleotides to an existing 3′ end. Primase first lays a short RNA primer (about 10 nucleotides) that provides the 3′-OH. That is why the lagging strand needs a new primer for every Okazaki fragment (Mission 3).
🔍 The name “polymerase” makes it sound as if the enzyme can do the whole job alone.
❌ Misconception: "Helicase breaks the sugar–phosphate backbone to separate the strands."
✅ Correction: Helicase breaks the hydrogen bonds between paired bases (2 per A–T pair, 3 per G–C pair). The covalent backbone of each strand stays whole, so each strand can serve as an intact template. Cutting the backbone ahead of the fork is the job of topoisomerase, which rejoins it straight away.
🔍 “Unzipping” is easy to picture as cutting; the zipper teeth are the hydrogen-bonded bases, not the backbone.
❌ Error: Drawing both daughter DNAs with the same color for both strands.
✅ Correct approach: Always shade one strand of each daughter DNA in the parental color and the other strand in the new color. Exam mark schemes (Cambridge International, AQA) explicitly require students to distinguish "one original / one new" strand per daughter molecule to score full marks on semiconservative diagrams.
🔍 Students confuse the diagram by coloring the whole DNA as "old" or "new" instead of tracking individual strands.
❌ Error: Writing "DNA polymerase reads 5′→3′" instead of "DNA polymerase synthesizes 5′→3′".
✅ Correct approach: DNA polymerase reads the template in the 3′→5′ direction and synthesizes the new strand in the 5′→3′ direction. Both statements describe the same process from opposite perspectives. Be precise: examiners reject the phrase "reads 5′→3′" because it inverts the template directionality.
🔍 The two directions are easy to confuse when students don't actively track which strand they're describing.
❌ Error: Calculating replication time using only one fork for a human chromosome.
✅ Correct approach: Always include the number of replication origins (each gives two forks). For eukaryotes, divide the total length by $2 \times v \times N_o$ where $N_o$ is the number of active origins (tens of thousands in a human cell). A common exam question asks "why must eukaryotes have many origins?" — the answer is the time calculation: one origin would take weeks, multiple origins fit replication into S phase (about 8 h).
🔍 Students forget that bidirectional replication and multiple origins are both essential speed factors.