DNA Replication: Semiconservative Copying and Its Machinery

One generation after Matthew Meselson and Franklin Stahl switched heavy-nitrogen E. coli to ordinary nitrogen, all of the bacteria’s DNA sat in a single band, halfway between heavy and light.

If each strand of the double helix is a template for its partner, copying could keep the old strands together, split them, or mix old with new. This lesson works out what each idea predicts, then follows the copying machinery, its error checks, and the ends of linear chromosomes. It builds on DNA as the genetic material and on the dNTPs from nucleotide metabolism.

Three models, one experiment: the second generation decides

Picture copying a two-page letter: keep the original pair and make a fresh one, give each old page a new partner, or shred both and paste scraps together. The three models:

• Conservative: the parental double helix stays intact, and the copy is entirely new.
• Semiconservative: each daughter helix has one parental strand and one new strand.
• Dispersive: each daughter helix is a patchwork of old and new segments on both strands.

Meselson and Stahl grew E. coli B for many generations on ammonium chloride made with heavy nitrogen-15, so all their DNA was heavy. Then they added a tenfold excess of ordinary nitrogen-14 ammonium chloride and sampled the culture as it grew. They read the samples by equilibrium density-gradient centrifugation: DNA spun for hours in concentrated caesium chloride (CsCl) settles where the solution matches its own density, so heavy, hybrid and light DNA form separate bands.

Worked problem. E. coli grown for many generations in 15N medium is moved to 14N. Predict the bands after one, two and three generations under each model. Assume all DNA starts fully heavy, rounds are complete and synchronous, new strands are fully light, molecules are equal in length, no heavy nitrogen is recycled, and dispersive mixing is even. H is a heavy strand, L a light one.

Step 1. Semiconservative. Start with one HH duplex. After one round the two H strands separate and each gets an L partner: 2 HL. After two rounds each HL gives one HL and one LL: 2 HL + 2 LL, so 50% hybrid and 50% light. After three rounds: 2 HL + 6 LL, so 25% hybrid and 75% light.

Step 2. Conservative. One round gives the old HH plus a new LL: 50% heavy, 50% light. Two rounds: 1 HH + 3 LL, so 25% heavy and 75% light. Three rounds: 1 HH + 7 LL, so 12.5% heavy and 87.5% light. No hybrid band ever forms.

Step 3. Dispersive. Every molecule is half old after one round, a quarter old after two and an eighth old after three. So there is always one band, sitting 1/2, then 1/4, then 1/8 of the way from light to heavy.

Step 4. Compare with the data. After one generation Meselson and Stahl saw only hybrid DNA. That rules out the conservative model (heavy plus light, never hybrid) but not the dispersive model, which also predicts one band halfway. After two generations they saw hybrid and light DNA in equal amounts, which rules out the dispersive prediction of one band a quarter of the way to heavy.

As a check, the average heavy content is the same in all three models (1/2, 1/4, 1/8), so only the distribution of label among molecules tells them apart. The result was measured in E. coli; 1957 autoradiography of bean (Vicia faba) chromosomes by J. Herbert Taylor and colleagues is a separate line of evidence.

The replication fork: why one strand is made in pieces

A replication fork is the Y-shaped place where the parental helix is opened and both strands copied. Two rules govern it. First, every DNA polymerase adds a nucleotide only to a free 3’-OH: the 3’ hydroxyl attacks the incoming dNTP, releasing pyrophosphate (see enzyme action). So chains grow only 5’ to 3’. Second, no DNA polymerase can start a chain: it needs a primer, a short paired stretch with a 3’ end to extend.

The template strands run in opposite directions. On one, the new strand grows 5’ to 3’ in the direction the fork moves, in one long piece: the leading strand. On the other, growing 5’ to 3’ means growing away from the fork, so each newly opened stretch needs a new primer, extended back toward the previous piece: the lagging strand, made as short Okazaki fragments. It is like copying a line of text uncovered right to left when you can only write left to right: copy a stretch, jump back, copy the next. Each fragment still grows 5’ to 3’.

In 1968 Reiji Okazaki, Tsuneko Okazaki, Kiwako Sakabe, Kazunori Sugimoto and Akio Sugino caught DNA as it was made: they grew E. coli B at 20°C to slow replication and gave pulses of radioactive thymidine as short as five seconds. They sized the labelled chains on alkaline sucrose gradients. After a five-second pulse most label was in a slow component of about 11S; with longer labelling it moved into long DNA. They suggested units of about 1,000 to 2,000 nucleotides, and left the direction of synthesis and the joining of the pieces for further proof. In eukaryotes, Okazaki fragments are shorter, about 100 to 200 nucleotides.

The enzymes: E. coli first, then eukaryotes

E. coli copies its circular chromosome from a single origin, oriC, with two forks moving apart.

1. DnaA binds oriC and opens a short AT-rich stretch of the helix.
2. DnaB, a ring-shaped helicase loaded with help from DnaC, uses ATP to unwind the helix, moving 5’ to 3’ along the lagging-strand template. Single-strand binding protein (SSB) coats the separated strands.
3. Gyrase, a topoisomerase, relieves the overwinding that builds up ahead of the fork.
4. DnaG primase makes short RNA primers.
5. DNA polymerase III holoenzyme extends both strands. Its β sliding clamp, a ring around the DNA, keeps it on the template for thousands of nucleotides.
6. DNA polymerase I removes each RNA primer with its 5’ to 3’ exonuclease while filling the gap with DNA, and DNA ligase seals the remaining nick.

Pol I was purified first, by Arthur Kornberg’s laboratory, but a 1969 mutant with very little Pol I activity still grew; Pol III is the replicative polymerase.

Worked problem. The E. coli K-12 chromosome is about 4.6 million base pairs, copied by two forks from one origin, each moving at about 1,000 nucleotides per second. How long does one round take?

time = (4.6 x 10^6 bp / 2 forks) / 1,000 bp per second per fork = 2,300 s, about 38 minutes

Yet in rich medium E. coli can divide about every 20 minutes. The forks do not speed up; new rounds begin at oriC before earlier rounds finish, so a fast-growing cell carries several sets of forks and each division still receives a finished copy.

Eukaryotes keep the logic but change the parts, and their linear chromosomes use many origins. In G1, MCM2-7 helicase rings are loaded onto origins in an inactive form; in S phase, Cdc45 and the GINS complex join MCM to form the active CMG helicase, which moves 3’ to 5’ along the leading-strand template, the opposite strand from E. coli’s DnaB. Pol α-primase makes a short RNA primer and extends it briefly with DNA. In the usual summary, Pol ε does most leading-strand synthesis and Pol δ most lagging-strand synthesis, both on the PCNA clamp. In a replisome rebuilt from purified budding-yeast proteins, though, Pol δ helped establish leading strands before handing off to Pol ε. Nucleases including FEN1 and RNase H remove lagging-strand primers, and DNA ligase I joins the fragments.

Getting it right, and the ends of linear chromosomes

Errors are caught in layers. First, base selection: the polymerase active site closes well around a correct pair and poorly around a mispair, leaving roughly one error in 10^4 to 10^5 nucleotides. Second, proofreading: a 3’ to 5’ exonuclease removes a mispaired nucleotide from the primer’s 3’ end before synthesis continues. In E. coli Pol III this activity lies in the ε subunit. Proofreading improves accuracy by about 100-fold, and by up to 1,000-fold depending on the polymerase and the mispair. Third, mismatch repair fixes errors after the fork passes (see DNA repair and recombination). Together these bring the overall rate in E. coli to roughly one error in 10^9 to 10^10 base pairs per round. At those rates a 4.6-million-base-pair chromosome gains a new error in roughly one copy in 200 to 2,000. These are orders of magnitude that vary with assay and sequence.

Not every polymerase proofreads: eukaryotic Pol α lacks a functional proofreading exonuclease.

On a circular chromosome every gap left by a removed primer has DNA behind it with a 3’ end to extend. On a linear one, when the last RNA primer at the tip of a lagging strand is removed, no upstream 3’ end is left, so the new strand is shorter than its template: the end-replication problem. Recent work found two sides to it: in yeast replisomes rebuilt on linear DNA, lagging-strand synthesis stopped in a zone more than 26 nucleotides from the template end, and leading-strand synthesis cannot regenerate the single-stranded 3’ overhang that telomeres carry.

Chromosome ends carry telomeres, short repeated sequences: TTGGGG in the ciliate Tetrahymena, TTAGGG in humans. In 1985 Carol Greider and Elizabeth Blackburn found that Tetrahymena extracts add TTGGGG repeats to synthetic telomere primers, an activity destroyed by heat and protease. In 1989 they reported that the enzyme’s RNA contains CAACCCCAA, suggested to be the template. Telomerase is a reverse transcriptase (it makes DNA from an RNA template) that carries its own RNA, and it extends the G-rich 3’ strand. The C strand is filled in separately; in engineered human cells this needed the CST complex with Pol α-primase. Blackburn, Greider and Jack Szostak shared the 2009 Nobel Prize in Physiology or Medicine for work on telomeres and telomerase.

Design problem

E. coli are given a 5-second pulse of tritiated thymidine, and their DNA is denatured and sized. Right after the pulse, most label is in pieces of about 1,000 to 2,000 nucleotides. After a 60-second chase in unlabelled thymidine, the label is in long DNA. In a ligase-defective mutant, the short pieces persist after the chase. What do you conclude, and what control would you add?