DNA Replication

AS · 13 min

Before a cell divides, it must make an exact copy of every DNA molecule in its nucleus, so that each daughter cell receives identical genetic information. This happens in S phase of the cell cycle by semi-conservative replication: each new DNA molecule contains one strand from the original molecule and one newly made strand. This note explains how replication works, including the roles of DNA polymerase and DNA ligase, why one new strand is made continuously and the other in fragments, and the classic experiment by Meselson and Stahl that showed replication is semi-conservative. Expect "describe" questions worth 5–6 marks and data questions on the Meselson–Stahl experiment.

What semi-conservative means

Definition

Semi-conservative replication is the copying of DNA in which each strand of the original molecule acts as a template for a new strand, so that each of the two new DNA molecules consists of one original (parental) strand and one newly synthesised strand.

"Semi-conservative" means "half kept". Half of each new molecule is conserved from the parent molecule. Because each new strand is built by complementary base pairing with a template, both new molecules have exactly the same base sequence as the original.

The process

Replication takes place in the nucleus during S phase of interphase.

Semi-conservative replication of DNA

Unwinding and separation. The enzyme DNA helicase unwinds the double helix and breaks the hydrogen bonds between complementary bases, separating the two strands. The point where the strands separate is a replication fork.
Exposed bases act as templates. Each separated strand acts as a template. Its bases are now exposed.
Free nucleotides pair up. Free DNA nucleotides in the nucleus (in activated form, as nucleotides carrying extra phosphate groups that provide energy) are attracted to their complementary bases on each template strand: A with T, C with G. Hydrogen bonds form between them.
DNA polymerase joins the nucleotides. DNA polymerase catalyses condensation reactions that form phosphodiester bonds between adjacent nucleotides, building the sugar–phosphate backbone of the new strand. It can add nucleotides only to the 3′ end of a growing strand, so new strands are built in the 5′ to 3′ direction.
Leading and lagging strands. On one template, the new strand can be made continuously towards the replication fork (the leading strand). On the other, it must be made discontinuously, in short fragments, away from the fork (the lagging strand).
DNA ligase joins the fragments. DNA ligase forms phosphodiester bonds between the fragments of the lagging strand, joining them into a continuous strand.
Two identical molecules. Each new DNA molecule winds into a double helix. Each contains one parental strand and one new strand, and both are identical to the original.

Why there is a leading strand and a lagging strand

This is a consequence of two facts you already know:

the two strands of DNA are antiparallel;
DNA polymerase can only add nucleotides to the 3′ end of a strand, so a new strand can only grow in the 5′ → 3′ direction (which means DNA polymerase moves along the template strand in its 3′ → 5′ direction).

At a replication fork, the double helix is opening up in one direction. Consider the two templates:

Template running 3′ → 5′ towards the fork. The new strand grows 5′ → 3′ in the same direction as the fork moves. DNA polymerase can simply follow the fork, adding nucleotides continuously as more template is exposed. This new strand is the leading strand.
Template running 5′ → 3′ towards the fork. The new strand must still grow 5′ → 3′, which is away from the fork. DNA polymerase can only make a short length of new strand, working back away from the fork, before it runs into the strand made previously. As the fork opens further, it must start again near the fork. So this strand is made in short sections called Okazaki fragments, which are then joined by DNA ligase. This new strand is the lagging strand.

Key result

	Leading strand	Lagging strand
Direction of synthesis	5′ → 3′, towards the replication fork	5′ → 3′, away from the replication fork
Synthesis	continuous	discontinuous, in short Okazaki fragments
Enzymes	DNA polymerase	DNA polymerase, then DNA ligase joins fragments
Reason	template is oriented so the new strand can grow in the direction the fork opens	template is oriented so the new strand must grow in the opposite direction to fork movement

Detail beyond the syllabus: primers

DNA polymerase cannot start a new strand from nothing; it can only extend an existing one. An enzyme called primase first makes a short RNA primer complementary to the template, and DNA polymerase extends it. The lagging strand needs a new primer for every Okazaki fragment. The primers are later removed and replaced with DNA. At the very end of a linear chromosome, the gap left by removing the last primer cannot be filled, which is why chromosomes shorten at each replication and why telomeres are needed (see Chromosomes and the cell cycle).

Speed and accuracy

Eukaryotic chromosomes are very long, so replication starts at many points (origins of replication) along each DNA molecule at once, forming many "replication bubbles", each with two forks moving in opposite directions. These eventually meet.
DNA polymerase proofreads as it goes, removing wrongly paired nucleotides. Errors that escape are about 1 in $10^{9}$ to $10^{10}$ bases. Uncorrected errors are mutations (see Gene mutations).

Roles of the enzymes (routine, 4 marks)

State the roles of DNA helicase, DNA polymerase and DNA ligase in DNA replication.

Solution

DNA helicase: unwinds the double helix and breaks hydrogen bonds between complementary bases, separating the strands.
DNA polymerase: joins free nucleotides that have paired with the template strand, catalysing the formation of phosphodiester bonds (condensation) to build the new sugar–phosphate backbone;
adds nucleotides only to the 3′ end, building the new strand in the 5′ → 3′ direction.
DNA ligase: joins Okazaki fragments on the lagging strand by forming phosphodiester bonds between them.

Evidence: the Meselson–Stahl experiment

Before 1958, there were three hypotheses for how DNA might be copied:

Hypothesis	Prediction
Conservative	the original double helix stays intact; the copy is made entirely of new DNA
Semi-conservative	each new molecule has one old strand and one new strand
Dispersive	each strand of each new molecule is a mixture of old and new sections

Meselson and Stahl distinguished these using two isotopes of nitrogen, which is found in DNA bases: the normal light isotope $^{14}\text{N}$ and the heavier isotope $^{15}\text{N}$ . DNA containing $^{15}\text{N}$ is denser. When DNA is centrifuged in a solution of caesium chloride, which forms a density gradient in the tube, each kind of DNA settles at the position matching its density: heavy DNA low in the tube, light DNA higher, and hybrid (intermediate) DNA in between.

Bacteria (E. coli) were grown for many generations in a medium containing only $^{15}\text{N}$ , so all their DNA was heavy. Centrifuged DNA formed one heavy band.
The bacteria were transferred to a medium containing only $^{14}\text{N}$ and allowed to divide once (one round of replication). DNA formed one band at an intermediate (hybrid) position.
After a second generation in $^{14}\text{N}$ , DNA formed two bands: half intermediate and half light.
In later generations, the light band grew and the intermediate band stayed the same size: after three generations, one-quarter intermediate and three-quarters light.

Key result

Interpreting the results

Generation 1, one intermediate band: rules out conservative replication (which would give one heavy band and one light band). Consistent with semi-conservative (every molecule has one $^{15}\text{N}$ strand and one $^{14}\text{N}$ strand) and with dispersive.
Generation 2, half intermediate and half light: rules out dispersive replication (which would give a single band, between intermediate and light). Exactly as predicted by semi-conservative replication: the two original heavy strands are each in one hybrid molecule; all other molecules are entirely light.

Predicting bands (moderate)

Copy and complete the table to show the proportions of DNA at each position after centrifuging, for each hypothesis, for bacteria grown in $^{15}\text{N}$ and then transferred to $^{14}\text{N}$ .

Generation in $^{14}\text{N}$	Conservative	Semi-conservative	Dispersive
0
1
2

Solution

Generation in $^{14}\text{N}$	Conservative	Semi-conservative	Dispersive
0	all heavy	all heavy	all heavy
1	½ heavy, ½ light	all intermediate	all intermediate
2	¼ heavy, ¾ light	½ intermediate, ½ light	all in one band, between intermediate and light

Explanation for semi-conservative generation 2: each of the 2 hybrid molecules from generation 1 separates into one heavy and one light template strand. The heavy strand pairs with a new light strand → hybrid; the light strand pairs with a new light strand → light. So from 2 molecules we get 2 hybrid + 2 light: ½ intermediate, ½ light.

Proportions after several generations (moderate)

Bacteria with only $^{15}\text{N}$ DNA were transferred to a $^{14}\text{N}$ medium. Assuming semi-conservative replication, calculate the percentage of DNA molecules that would be hybrid after 1, 2, 3 and 4 generations.

Solution

Starting from one heavy molecule, after $n$ generations there are $2^{n}$ molecules. The two original heavy strands are never broken down, and each ends up in a different hybrid molecule, so there are always 2 hybrid molecules (from generation 1 onwards).

\text{fraction hybrid} = \frac{2}{2^{n}}

Generation	Total molecules	Hybrid	Light	% hybrid
1	2	2	0	100
2	4	2	2	50
3	8	2	6	25
4	16	2	14	12.5

No heavy DNA is found after generation 0.

Why many origins? (moderate)

(a) In E. coli, DNA replication starts at a single origin and two replication forks move in opposite directions round the circular chromosome of $4.6 \times 10^{6}$ base pairs, each at about $1000$ base pairs per second. Calculate how long replication takes, in minutes. (b) A human chromosome contains a DNA molecule of about $2.5 \times 10^{8}$ base pairs, and replication forks in human cells move at about 50 base pairs per second. Calculate how long replication would take from a single origin, in days. (c) S phase in this cell lasts 8 hours. Calculate the minimum number of origins needed, assuming each origin has two forks.

Solution

(a) Each fork copies half the chromosome: $\dfrac{4.6 \times 10^{6}}{2 \times 1000} = 2300\ \text{s} = 38\ \text{min}$ .

(b) $\dfrac{2.5 \times 10^{8}}{2 \times 50} = 2.5 \times 10^{6}\ \text{s}$ . In days: $\dfrac{2.5 \times 10^{6}}{86\,400} = 29\ \text{days}$ .

(c) 8 hours $= 28\,800\ \text{s}$ . Each origin copies $2 \times 50 \times 28\,800 = 2.88 \times 10^{6}$ base pairs in that time.

\text{origins} = \frac{2.5 \times 10^{8}}{2.88 \times 10^{6}} = 86.8

So at least 87 origins are needed. (Real human chromosomes use many more than this, typically thousands per chromosome.)

Exam-hard: describing replication (6 marks)

Describe how DNA is replicated, including the differences between replication of the leading and lagging strands.

Solution

Any six of:

DNA unwinds / helicase breaks hydrogen bonds between bases; the strands separate;
both strands act as templates;
free (activated) DNA nucleotides are attracted to / pair with exposed bases by complementary base pairing (A–T, C–G);
DNA polymerase joins nucleotides by forming phosphodiester bonds (condensation);
DNA polymerase adds nucleotides only to the 3′ end / builds new strands 5′ → 3′;
because the strands are antiparallel, the leading strand is made continuously towards the replication fork;
the lagging strand is made discontinuously / in short Okazaki fragments, away from the fork;
DNA ligase joins the fragments by forming phosphodiester bonds;
each new molecule has one old and one new strand: semi-conservative;
happens in S phase of interphase.

Watch out

In semi-conservative replication, both strands are templates. Students sometimes say "one strand is copied".
DNA polymerase does not break hydrogen bonds or unwind DNA; that is helicase. DNA polymerase forms phosphodiester bonds between nucleotides.
Hydrogen bonds between the new and template strands form spontaneously by base pairing; no enzyme is needed to form them.
DNA ligase joins fragments of the lagging strand; it is not the enzyme that adds individual nucleotides.
The new strand is built 5′ → 3′. Do not say DNA polymerase "moves 5′ to 3′ along the template": it moves 3′ → 5′ along the template.
In Meselson–Stahl, generation 1 rules out conservative; generation 2 is needed to rule out dispersive.

Exam tip

Use the precise vocabulary: template, complementary base pairing, free nucleotides, DNA polymerase, phosphodiester bonds, 5′ → 3′, leading strand (continuous), lagging strand (discontinuous, Okazaki fragments), DNA ligase, semi-conservative.
If the question says "explain why the lagging strand is made in fragments", the key ideas are antiparallel strands and DNA polymerase only adds to the 3′ end / works 5′ → 3′.
In Meselson–Stahl questions, draw or describe the position of bands (heavy at the bottom of the tube) and use proportions. Explain what each result rules out.
Link replication to the cell cycle: it occurs in S phase, so that each chromosome has two identical sister chromatids before mitosis.

Summary

DNA replication is semi-conservative: each new molecule has one parental strand and one new strand.
Occurs in S phase of interphase.
Helicase unwinds DNA and breaks hydrogen bonds; both strands act as templates.
Free nucleotides pair with complementary bases (A–T, C–G).
DNA polymerase forms phosphodiester bonds, adding nucleotides only to the 3′ end, so new strands grow 5′ → 3′.
Leading strand: continuous, towards the fork. Lagging strand: discontinuous Okazaki fragments, away from the fork, joined by DNA ligase.
Meselson and Stahl ( $^{15}\text{N}$ then $^{14}\text{N}$ , density centrifugation): generation 1 all intermediate (not conservative); generation 2 half intermediate, half light (not dispersive): semi-conservative.

Practice questions

Question

Explain what is meant by semi-conservative replication.
State the stage of the cell cycle in which DNA replication occurs.
Name the type of bond formed by DNA polymerase.
Explain why the two new DNA molecules produced by replication are identical to the original. (2 marks)
Explain why DNA ligase is needed during replication of the lagging strand but not the leading strand. (3 marks)
In the Meselson–Stahl experiment, state the result after one generation in $^{14}\text{N}$ and explain which hypothesis it rules out.
Bacteria grown in $^{14}\text{N}$ for many generations were transferred to $^{15}\text{N}$ for two generations. Assuming semi-conservative replication, state the proportions of light, intermediate and heavy DNA after two generations.
A section of template strand reads 3′ TACGGATC 5′. Write the sequence of the new strand with its ends labelled, and state whether DNA polymerase would move left to right or right to left along this template.
Suggest why it is an advantage for a eukaryotic cell to have many origins of replication on each chromosome.
Describe the semi-conservative replication of DNA and explain how the results of the Meselson–Stahl experiment after two generations support this mechanism rather than the conservative or dispersive mechanisms. (8 marks)

Answers

Each strand of the original DNA molecule acts as a template for a new strand, so each new DNA molecule consists of one original (parental) strand and one newly synthesised strand.
S phase of interphase.
Phosphodiester bond (formed by condensation).
Each new strand is built by complementary base pairing to a template strand (A–T, C–G), so its sequence is complementary to the template and identical to the template's original partner; each new molecule therefore has the same base sequence as the original.
DNA polymerase adds nucleotides only to the 3′ end (works 5′ → 3′), and the strands are antiparallel; the leading strand is made continuously, towards the fork, so no gaps are left; the lagging strand is made away from the fork in short fragments (Okazaki fragments), which must be joined by DNA ligase forming phosphodiester bonds.
One band, at an intermediate position (all DNA hybrid). It rules out conservative replication, which would give two bands: one heavy and one light.
Light 0; intermediate ½; heavy ½. (The two original light strands each end up in a hybrid molecule; all other molecules are fully heavy.)
New strand: 5′ ATGCCTAG 3′. DNA polymerase moves along the template from its 3′ end to its 5′ end, i.e. left to right as written.
Eukaryotic DNA molecules are very long and DNA polymerase works relatively slowly; with many origins, many sections are replicated at the same time, so replication can be completed within S phase (a few hours rather than weeks).
Description (up to 6): helicase unwinds DNA and breaks hydrogen bonds between bases; both strands act as templates; free nucleotides pair by complementary base pairing (A–T, C–G); DNA polymerase forms phosphodiester bonds, adding to the 3′ end (5′ → 3′); leading strand continuous, lagging strand in Okazaki fragments joined by DNA ligase; each new molecule has one old and one new strand. Evidence (up to 4): after two generations in $^{14}\text{N}$ , there are two bands, half intermediate and half light; semi-conservative predicts this because the two original heavy strands each end up in a hybrid molecule and all other molecules are light; conservative would give ¼ heavy and ¾ light (a heavy band persists), which was not seen; dispersive would give a single band between intermediate and light, not two bands.

What semi-conservative means

The process

Why there is a leading strand and a lagging strand

Speed and accuracy

Evidence: the Meselson–Stahl experiment

Practice questions

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