D. Molecular Genetics · Molecular GeneticsDNA Replication: How DNA Copies Itself
D3.1 D2.1 A2.2 A1.8 A1.12About 75 minutes
Learning goals- I can explain how the Meselson-Stahl experiment showed that DNA replication is semi-conservative.
- I can describe the roles of helicase, topoisomerase, single-strand binding proteins, primase, DNA polymerases I, II and III, and DNA ligase.
- I can explain why one strand is made continuously (leading) and the other in Okazaki fragments (lagging).
- I can compare replication in prokaryotes and eukaryotes, including telomeres and telomerase.
- I can calculate replication times from genome sizes and fork speeds.
Words to know
semi-conservative replication
Copying in which each new DNA molecule has one original (parent) strand and one new strand.
origin of replication
A specific sequence where replication begins and the DNA opens into a replication bubble.
replication fork
The Y-shaped region where the parent DNA is being unwound and copied.
helicase
The enzyme that unwinds the double helix by breaking hydrogen bonds between base pairs.
topoisomerase
An enzyme that relieves the twisting (supercoiling) ahead of the fork by cutting and rejoining DNA (DNA gyrase in bacteria).
primase
An enzyme that makes a short RNA primer, giving DNA polymerase a 3′ end to build on.
DNA polymerase III
In E. coli, the main enzyme that adds nucleotides to the 3′ end of a growing strand; it also proofreads.
DNA polymerase I
In E. coli, the enzyme that removes RNA primers and replaces them with DNA.
DNA polymerase II
In E. coli, a polymerase mainly involved in repairing damaged DNA.
DNA ligase
The enzyme that seals nicks in the backbone by forming a phosphodiester bond.
leading strand
The new strand made continuously in the same direction the fork moves.
lagging strand
The new strand made in short pieces, away from the fork.
Okazaki fragment
A short piece of new DNA on the lagging strand, about 1000 to 2000 nucleotides in bacteria and 100 to 200 in eukaryotes.
telomere
A repeated non-coding DNA sequence at the end of a eukaryotic chromosome (TTAGGG in humans).
telomerase
An enzyme with its own RNA template that adds telomere repeats to chromosome ends.
Billions of letters, copied in hours
Every time one of your cells divides, it copies about 6.4 billion base pairs of DNA, in a few hours, with roughly one uncorrected error per billion bases. Watson and Crick saw the key idea in 1953: "It has not escaped our notice that the specific pairing we have postulated immediately suggests a possible copying mechanism for the genetic material." Each strand is a template for a new partner. But the details took decades to work out, and they explain some puzzling features, such as why one strand is copied in pieces and why chromosome ends need special protection.
Three possible models
In the 1950s there were three hypotheses:
- Conservative: the original double helix stays together, and an entirely new double helix is made.
- Semi-conservative: the two strands separate; each becomes a template, so each new molecule is half old and half new.
- Dispersive: each strand of the new molecules is a patchwork of old and new pieces.
Meselson and Stahl, 1958
Matthew Meselson and Franklin Stahl grew E. coli for many generations in a medium containing heavy nitrogen, ¹⁵N, so all their DNA was heavy. They then moved the bacteria to normal ¹⁴N medium and took samples after each generation (about 20 minutes). The DNA was spun at high speed in a cesium chloride density gradient, where each kind of DNA settles at the level that matches its density.
- After 1 generation: one band, at hybrid density. This rules out the conservative model, which predicts one heavy band and one light band.
- After 2 generations: two bands, hybrid and light, in equal amounts. This rules out the dispersive model, which predicts a single band getting gradually lighter.
Only semi-conservative replication fits both results. Some biologists call it "the most beautiful experiment in biology".
Replication in E. coli, step by step
Bacterial replication is the best-understood model, so we learn it first. The Ontario curriculum names DNA polymerase I, II and III, which are bacterial enzymes.
1. Initiation. Replication starts at one origin of replication (oriC) on the circular chromosome. Proteins bind and open a small region, forming a replication bubble with two forks that move in opposite directions (bidirectional replication).
2. Unwinding. Helicase moves along the DNA and breaks the hydrogen bonds between base pairs, separating the strands. Single-strand binding proteins (SSBs) coat the exposed strands so they do not re-pair or get damaged. Unwinding over-twists the DNA ahead of the fork (try untwisting the middle of a twisted rope); topoisomerase (called DNA gyrase in bacteria) cuts the DNA, lets it untwist, and rejoins it.
3. Priming. DNA polymerase cannot start a new strand from nothing; it can only add to an existing 3′ OH. Primase makes a short RNA primer (about 10 nucleotides) complementary to the template.
4. Elongation. DNA polymerase III adds nucleotides to the 3′ end of the primer, matching each to the template (A with T, G with C). The building blocks are deoxynucleoside triphosphates (dATP, dTTP, dGTP, dCTP). Each one loses two phosphates as it is added, and that releases the energy to make the new phosphodiester bond. In E. coli, pol III adds about 1000 nucleotides per second.
5. Two strands, two strategies. DNA polymerase can only build 5′→3′, reading the template 3′→5′. The two template strands are antiparallel, so:
- On one template, 5′→3′ synthesis heads toward the fork. Pol III follows the helicase and makes one long, continuous leading strand.
- On the other template, 5′→3′ synthesis heads away from the fork. As the fork opens more template, primase lays down a new primer and pol III makes a short piece backwards until it reaches the previous piece. These pieces are Okazaki fragments, after Reiji and Tsuneko Okazaki, who discovered them in the late 1960s. This is the lagging strand.
6. Finishing. DNA polymerase I removes each RNA primer and replaces it with DNA. That leaves a nick, a missing bond in the backbone between neighbouring fragments. DNA ligase seals it with a phosphodiester bond. (DNA polymerase II is mainly used to repair damaged DNA; you will meet repair in the next lesson.)
7. Termination. In E. coli the two forks meet on the opposite side of the circle, and the two daughter circles are separated.
| Enzyme or protein | Job |
|---|
| Helicase | unwinds the helix by breaking hydrogen bonds |
| Single-strand binding proteins | keep the separated strands apart and protected |
| Topoisomerase (DNA gyrase) | relieves over-twisting ahead of the fork |
| Primase | makes RNA primers |
| DNA polymerase III | main builder: adds nucleotides 5′→3′; proofreads |
| DNA polymerase I | removes RNA primers and fills the gaps with DNA |
| DNA polymerase II | mainly DNA repair |
| DNA ligase | seals nicks with a phosphodiester bond |
Worked exampleExample 1: Leading or lagging?
A replication fork is moving to the right. The upper template strand is written 3′-TACGGA...-5′ (left to right). (a) Will the new strand on this template be leading or lagging? (b) Write the first six bases of the new strand with its ends labelled.
- The new strand must be antiparallel to its template, so it runs 5′ (left) → 3′ (right).
- It grows at its 3′ end, which is on the right, the same direction the fork is moving. So it is made continuously: the leading strand.
- Pair the bases: T→A, A→T, C→G, G→C, G→C, A→T.
- New strand: 5′-ATGCCT...-3′.
(a) Leading strand. (b) 5′-ATGCCT...-3′.
Worked exampleExample 2: How long does replication take?
The E. coli chromosome has 4.6 × 10⁶ base pairs and one origin. Each fork moves at about 1000 nucleotides per second. Estimate the time to copy the chromosome.
- Replication is bidirectional: two forks share the work, so each copies 4.6 × 10⁶ ÷ 2 = 2.3 × 10⁶ base pairs.
- Time = 2.3 × 10⁶ ÷ 1000 per second = 2300 s.
- 2300 s ÷ 60 ≈ 38 minutes.
About 38 minutes (2300 s).
Eukaryotes: bigger genomes, linear chromosomes
Eukaryotic replication uses the same principles (helicase, primase, RNA primers, 5′→3′ polymerases, leading and lagging strands, ligase), with important differences:
- Many origins. Human chromosomes are huge and eukaryotic forks are slower (tens of nucleotides per second), so each chromosome has many origins, tens of thousands in the genome. Bubbles grow and merge.
- Different polymerases. Eukaryotes use several DNA polymerases named with Greek letters (for example, α, δ and ε). Don't mix them up with the bacterial pol I, II and III.
- Shorter Okazaki fragments, about 100 to 200 nucleotides, roughly the length of DNA in one nucleosome.
- Histones must be removed ahead of the fork and reassembled on both daughter molecules.
- Linear chromosomes have an end-replication problem.
The end-replication problem and telomerase
At the very end of a linear chromosome, the last RNA primer on the lagging strand is removed, but there is no upstream DNA for a polymerase to extend into that gap. So each round of replication leaves the new strand slightly shorter.
Chromosome ends are protected by telomeres: thousands of base pairs of a short repeat (TTAGGG in humans) that contain no genes. Losing a little telomere each division does no harm at first. But after many divisions, telomeres become very short and the cell stops dividing, which is one factor in cellular aging.
Telomerase is an enzyme that carries its own RNA template and adds telomere repeats back onto chromosome ends. It is active in stem cells and germ-line cells (which must keep dividing) and is switched back on in most cancer cells, which helps them divide without limit. Elizabeth Blackburn, Carol Greider and Jack Szostak (who grew up in Canada and studied at McGill University) shared the 2009 Nobel Prize in Physiology or Medicine for discovering how telomeres and telomerase protect chromosomes.
Just for funWhy are Okazaki fragments never on time?
Because they're always on the lagging strand.
Hands-on activityAct out a replication fork
You need: a zipper (on a jacket or bag), two different colours of paper, scissors (use them carefully), tape, a few small sticky notes, a marker, and one or two family members.
- Cut 16 paper "nucleotides" of each colour (about 3 cm × 6 cm). Write A, T, G or C on them so that you have 4 of each letter in each colour. Colour 1 = parent strands, colour 2 = new nucleotides.
- Tape 8 parent nucleotides in a line on a table as the top template: 3′ T A C G G A T C 5′ (write 3′ and 5′ at the ends). Below it, tape the complementary bottom template 5′ A T G C C T A G 3′, touching it.
- One person is helicase: they slowly "unzip" by sliding the bottom strand down, starting at the left end and moving right, two bases at a time, like the zipper.
- One person is DNA polymerase III: on the top template, add new nucleotides (colour 2) so the new strand runs 5′→3′. Notice you can follow helicase smoothly: leading strand.
- On the bottom template, you must also build 5′→3′, which is away from the fork. Wait until 2 or 3 bases are exposed, place a small sticky note as an RNA primer, then add new nucleotides backwards. Repeat each time helicase opens more. These are Okazaki fragments.
- Finally, play pol I (swap each primer for a DNA nucleotide) and ligase (put a strip of tape across each join).
What to look for: the leading strand is one smooth piece; the lagging strand needs several primers and joins. At the end, each double strand has one colour-1 and one colour-2 strand: semi-conservative.
Your turn
1Describe the bands Meselson and Stahl would have seen after one generation in ¹⁴N if replication were conservative.
Two bands: one heavy (the original ¹⁵N/¹⁵N molecule) and one light (the completely new ¹⁴N/¹⁴N molecule), with no hybrid band.
2Why is an RNA primer needed?
DNA polymerase can only add nucleotides to an existing 3′ OH group; it cannot start a new chain. Primase can start from nothing and gives the polymerase that 3′ end.
3Explain why the lagging strand is made in fragments.
Polymerase builds only 5′→3′. On the lagging template that direction points away from the fork, so the polymerase can only make short pieces as new template is exposed, starting a new fragment each time.
4Match the enzyme to its job: helicase, ligase, DNA polymerase I.
Helicase: unwinds DNA by breaking hydrogen bonds. Ligase: seals nicks (forms phosphodiester bonds). DNA pol I: removes RNA primers and replaces them with DNA.
5What would happen to replication if DNA gyrase (topoisomerase) were blocked by a drug?
The DNA ahead of the fork would become more and more over-twisted until the helicase could not unwind it, so replication would stall. (Some antibiotics, such as ciprofloxacin, work this way in bacteria.)
6Why do eukaryotic chromosomes need many origins of replication?
They are very long and eukaryotic forks are slow. With a single origin, copying would take far too long (weeks to years), so many origins work at once.
More practice: warm-up
1Which enzyme is the main builder of new DNA in E. coli?
2What is the repeat sequence of human telomeres?
3In which direction is new DNA always synthesized?
4Name the short DNA pieces on the lagging strand.
5What does semi-conservative mean?
Each new DNA molecule has one old strand and one new strand.
More practice: on your own
1Predict the bands Meselson and Stahl saw after three generations in ¹⁴N. Give the ratio.
From 8 molecules, 2 are hybrid and 6 are light: bands at hybrid and light in a ratio of 1 : 3 (hybrid : light).
2A template strand reads 5′-GACTTA-3′. Write the new complementary strand with its ends labelled, in the 5′→3′ direction.
Lined up: 3′-CTGAAT-5′. Written 5′→3′: 5′-TAAGTC-3′.
3Explain the role of the triphosphate on each incoming nucleotide.
As the nucleotide is added, it loses two phosphates (pyrophosphate). Breaking those high-energy bonds releases the energy needed to form the new phosphodiester bond.
4Why does a linear chromosome get slightly shorter each time it is copied?
At the end of the lagging strand, the final RNA primer is removed, but there is no upstream 3′ end for DNA polymerase to extend into the gap. The new strand is left shorter than its template (the end-replication problem).
5Why is telomerase activity in cancer cells a possible target for cancer drugs?
Most normal body cells have little telomerase, while most cancer cells use it to keep their telomeres long and divide without limit. Blocking telomerase could make cancer cells' telomeres shorten until they stop dividing, while affecting most normal cells less.
Challenge
1A human diploid cell has 6.4 × 10⁹ base pairs. Assume a fork moves 50 nucleotides per second. (a) How long would replication take with a single fork? (b) With 30 000 origins, each with two forks working at the same time? (c) The real S phase takes about 8 hours. Suggest why.
(a) 6.4 × 10⁹ ÷ 50 = 1.28 × 10⁸ s ≈ 4.1 years. (b) 30 000 × 2 = 60 000 forks; 6.4 × 10⁹ ÷ (60 000 × 50) ≈ 2133 s ≈ 36 minutes. (c) Origins do not all fire at once: they fire at different times during S phase (tightly packed heterochromatin is often copied later), and forks pause at obstacles, so the whole process takes hours.
2A mutant E. coli strain has a temperature-sensitive DNA ligase that stops working at 42 °C. Predict what DNA molecules would accumulate if the cells are moved to 42 °C during replication, and explain.
Short Okazaki fragments (about 1000 to 2000 nucleotides) would build up, because the lagging-strand pieces could no longer be joined. The leading strand would still be long. (This kind of experiment helped confirm discontinuous replication.)
3Explain how the structure of DNA (antiparallel strands) together with one enzyme property (5′→3′ synthesis only) logically requires both a leading and a lagging strand.
At a fork both templates are exposed together and unwind in one direction. Because they are antiparallel, a 5′→3′ polymerase moves toward the fork on one template but away from it on the other. On the first, synthesis can continue as the fork opens (leading). On the second, synthesis must restart repeatedly as new template appears, creating fragments (lagging). If polymerases could build 3′→5′ too, both strands could be continuous.
Quiz
1. After one generation in ¹⁴N medium, Meselson and Stahl saw a single hybrid band. Which model did this rule out?
2. Which enzyme relieves the over-twisting of DNA ahead of the replication fork?
3. What does DNA polymerase I do in E. coli?
4. Why is the lagging strand synthesized discontinuously?
5. Which enzyme uses its own RNA template to lengthen chromosome ends?
6. A template strand is 3′-AGCTTC-5′. The new strand is:
Big idea (D3.1): DNA is copied semi-conservatively by a team of enzymes. Because polymerase only builds 5′→3′ and the strands are antiparallel, one new strand is continuous and the other is made in Okazaki fragments.
How to help: Ask your teen to "be" each enzyme and say in one sentence what it does. Then ask the big "why" question: why does the lagging strand exist? A strong answer mentions both antiparallel strands and 5′→3′ synthesis.
Common difficulties:
- Mixing up pol I and pol III, or helicase and topoisomerase.
- Thinking polymerase reads the template 5′→3′ (it reads 3′→5′ and builds 5′→3′).
- Forgetting replication is bidirectional when calculating times (two forks per origin).
- Confusing bacterial DNA pol I, II, III with eukaryotic enzymes.
The activity uses scissors; supervise as needed.