The mechanism of DNA replication and its enzymes

25 min

Copying six billion letters

Before one of your bone marrow cells divides, it copies all 6.4 billion base pairs of its DNA, in about eight hours, with roughly one uncorrected mistake in every billion bases. Then it does it again, a day or two later. No factory works like this. The copying is done by a team of enzymes working on the template rules you met in Lesson 2: each strand of the double helix fixes the sequence of its new partner. This lesson follows that team at work.

Definition

DNA replication: the copying of a DNA molecule to make two identical molecules, before a cell divides.

Semi-conservative replication: replication in which each new DNA molecule consists of one strand from the original molecule (the template) and one newly made strand.

Origin of replication: a specific sequence where replication begins. Replication fork: the Y-shaped region where the double helix is being opened and copied.

The enzymes and proteins

Enzyme or protein	Job
Helicase	Unwinds the double helix at the fork and breaks the hydrogen bonds between base pairs, separating the strands.
Single-strand DNA-binding proteins	Coat the separated strands, stopping them re-pairing and protecting them from breakage.
Primase	Makes a short RNA primer, about ten nucleotides long, complementary to the template; DNA polymerase needs it as a starting point.
DNA polymerase	Adds free nucleotides to the 3′ end of the growing strand, pairing each with the template base, and forms phosphodiester bonds. It can only build in the 5′ to 3′ direction. It also proofreads, removing wrongly paired nucleotides. Another DNA polymerase removes the RNA primers and fills the gaps with DNA.
DNA ligase	Joins the Okazaki fragments of the lagging strand by forming the last phosphodiester bond in each gap.

The raw materials are free nucleotides carrying three phosphate groups (deoxyribonucleoside triphosphates: dATP, dTTP, dGTP and dCTP). As each is added, two of its phosphates are split off, and this releases the energy that forms the new bond.

Step by step at a replication fork

Unwinding. Helicase binds at an origin of replication and moves along the DNA, separating the strands. Single-strand binding proteins hold them apart.
Priming. Primase lays down a short RNA primer on each template.
Elongation. DNA polymerase adds nucleotides complementary to the template (A opposite T, G opposite C) to the 3′ end of the primer. The new strand is antiparallel to its template.
Leading strand. On one template, the new strand grows 5′ to 3′ in the same direction as the fork moves, so it is made continuously after a single primer.
Lagging strand. On the other template, 5′ to 3′ means away from the fork. As the fork opens, primase makes a new primer near the fork and DNA polymerase builds a short fragment backwards until it meets the previous one. These Okazaki fragments are about 1 000 to 2 000 nucleotides long in bacteria and 100 to 200 in eukaryotes.
Finishing. The RNA primers are removed and replaced with DNA, and DNA ligase seals each gap.
Result. Two DNA molecules, each with one old and one new strand.

Worked example 1: writing the new strand

A template strand reads 3′-TACGGATC-5′. Write the new strand made against it, showing its ends.

Pair each base: T→A, A→T, C→G, G→C.
The new strand is antiparallel, so its 5′ end lies opposite the template's 3′ end: 5′-ATGCCTAG-3′.
DNA polymerase built it from the 5′ end towards the 3′ end, reading the template from 3′ to 5′.

Worked example 2: how fast is a fork?

(a) The circular chromosome of the bacterium E. coli has $4.6 \times 10^{6}$ base pairs and is copied from one origin by two forks moving in opposite directions, in 40 minutes. Calculate the speed of one fork in nucleotides per second. (b) A human fork moves at about 50 nucleotides per second. What is the least number of origins needed to copy $6.4 \times 10^{9}$ base pairs in 8 hours?

(a) Each fork copies half the chromosome: $2.3 \times 10^{6}$ base pairs in $40 \times 60 = 2\,400\,\mathrm{s}$ . Speed $= \frac{2.3 \times 10^{6}}{2\,400} \approx 960$ nucleotides per second.
(b) Each origin has 2 forks. Work done by one origin in 8 hours: $2 \times 50 \times 8 \times 3\,600 = 2.88 \times 10^{6}$ base pairs. Origins needed $= \frac{6.4 \times 10^{9}}{2.88 \times 10^{6}} \approx 2\,200$ .

Interpretation: with one origin per chromosome, copying a human genome would take weeks. Eukaryotic chromosomes have many origins, in fact far more than the minimum, because they do not all fire at once.

yourgenome (Wellcome Connecting Science): "DNA replication - 3D". Watch for helicase unzipping the helix, the leading strand made continuously, and the lagging strand looping out as Okazaki fragments are made and joined.

Warning

Common mistakes

"Helicase joins the fragments." Helicase separates the strands; ligase joins the fragments.
"DNA polymerase can build in both directions." It adds only to the 3′ end, so new strands grow 5′ to 3′. That is the whole reason for the lagging strand.
Writing the new strand with the same 5′/3′ orientation as its template. New strands are antiparallel to their templates.

Tip

Enzyme order: "Helicase Holds open, SSBs Stabilise, Primase Primes, Polymerase Polymerises, Ligase Links."

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Practice: the mechanism of replication

Give the role of (a) helicase, (b) single-strand binding proteins, (c) DNA ligase.
Explain why DNA polymerase needs a primer.
Explain why one strand is made continuously and the other in fragments.
Write the new strand made against the template 5′-GGCATTAC-3′, showing its 5′ and 3′ ends.
Where does the energy for forming phosphodiester bonds during replication come from?
Suggest what would happen to a cell's DNA if its DNA ligase did not work.
A bacterial chromosome of $3.0 \times 10^{6}$ base pairs is copied by two forks, each moving at 1 000 nucleotides per second. How long, in minutes, does replication take?
Explain how the proofreading activity of DNA polymerase keeps replication accurate.

Show answersHide answers

(a) Unwinds the helix and breaks the hydrogen bonds between bases. (b) Keep the separated strands apart and protect them. (c) Joins Okazaki fragments by forming phosphodiester bonds.
DNA polymerase can only add nucleotides to an existing 3′ end; it cannot start a new strand from nothing, so primase makes a short RNA primer for it to extend.
Polymerase builds only 5′ to 3′. The strands are antiparallel, so on one template this direction runs towards the fork (continuous, leading strand), and on the other it runs away from the fork, so synthesis must restart in short pieces as the fork opens (lagging strand).
3′-CCGTAATG-5′, which is the same as 5′-GTAATGCC-3′.
From the free nucleotides themselves: each is a triphosphate, and two phosphates are released as it is added, giving the energy to form the bond.
The lagging strand would remain in unjoined fragments, so the new DNA would be broken into pieces; the cell could not complete replication and would probably die.
Each fork copies $1.5 \times 10^{6}$ base pairs: $\frac{1.5 \times 10^{6}}{1\,000} = 1\,500\,\mathrm{s} = 25\,\mathrm{minutes}$ .
After adding each nucleotide, DNA polymerase checks the pairing; if the base is wrongly paired, it removes it and inserts the correct one, reducing errors greatly.

Key points

Helicase unwinds; single-strand binding proteins hold strands apart; primase makes RNA primers; DNA polymerase extends 5′ to 3′ and proofreads; ligase joins fragments.
Leading strand: continuous; lagging strand: Okazaki fragments, each with a primer.
Each new molecule has one old and one new strand: semi-conservative replication.
Eukaryotic chromosomes have many origins of replication.

Sources and credits

Written by Ganzaa.
Video: DNA replication - 3D, yourgenome, embedded from YouTube.

Figure 1, the table, the worked examples and the practice questions are written by Ganzaa.

1Copying six billion letters

2The enzymes and proteins

3Step by step at a replication fork

Copying six billion letters

The enzymes and proteins

Step by step at a replication fork