DNA Replication: Study Guide (Biology 6.2)

Why DNA must be copied

Before a cell divides, it must make a complete copy of its DNA so that each daughter cell receives the same genetic information. This copying is DNA replication. It ensures the continuity of hereditary information from one cell generation to the next, and, through gametes, from parents to offspring.

The structure you studied in Topic 6.1 makes copying possible. Because A pairs with T and G pairs with C, each strand carries the information needed to rebuild its partner. Separate the strands, and each one can act as a template.

Semiconservative replication

In replication, the two strands of the parent molecule separate. Each one serves as the template for a new complementary strand. So each of the two new DNA molecules contains:

• one original strand (from the parent), and
• one newly made strand.

This is called semiconservative replication: half of each parent molecule is conserved in each daughter molecule.

Before this was known, two other models were proposed (Figure 1). In the conservative model, the parent molecule stays intact and a completely new molecule is made. In the dispersive model, every strand of every new molecule is a patchwork of old and new pieces.

The evidence. In 1958, Matthew Meselson and Franklin Stahl grew the bacterium E. coli on a medium containing a heavy isotope of nitrogen (¹⁵N), so that all its DNA became “heavy”. They then moved the bacteria to a medium with normal, lighter nitrogen (¹⁴N) and separated DNA by density in a centrifuge. After one round of replication, all the DNA had an intermediate density, which ruled out the conservative model. After two rounds, there were equal amounts of intermediate and light DNA, which ruled out the dispersive model. Only the semiconservative model predicts both results. Worked example 2 uses the same reasoning with new data.

The direction rule: 5′ → 3′ only

The enzyme that builds new DNA is DNA polymerase. It reads the template strand and adds nucleotides that pair with it. Two features of this enzyme explain almost everything else in the topic:

1. It adds nucleotides only to a free 3′ end. As you saw in Topic 1.6, each new nucleotide bonds to the 3′ –OH of the growing strand. So new DNA is always synthesised in the 5′ → 3′ direction. Because the strands are antiparallel, the polymerase moves along the template from its 3′ end towards its 5′ end.
2. It cannot start a new strand from nothing. It can only extend a strand that already exists. A short RNA primer, complementary to the template, is laid down first. DNA polymerase then adds DNA nucleotides to the primer’s 3′ end.

Background (name not required): the primers are made by an enzyme called primase, which is a kind of RNA polymerase. Unlike DNA polymerase, it can start a chain from scratch.

At the replication fork

Replication begins at a specific site called an origin of replication. The strands separate there, opening a “bubble” with a replication fork at each side. Each fork moves away from the origin as more DNA is unwound. Figure 2 shows one fork.

Enzyme	Job at the fork
Helicase	Unwinds the double helix by breaking the hydrogen bonds between paired bases, separating the two template strands
Topoisomerase	Works on the still-paired DNA ahead of the fork. Unwinding at the fork makes the DNA in front twist more tightly (supercoil). Topoisomerase cuts, untwists and rejoins the DNA to relieve this strain
RNA primers	Short RNA pieces give DNA polymerase a 3′ end to build on
DNA polymerase	Adds DNA nucleotides, complementary to the template, to the 3′ end of the primer or growing strand
Ligase	Joins the fragments on the lagging strand by forming the final covalent bond in the sugar–phosphate backbone

Leading and lagging strands

The two template strands at a fork run in opposite directions, but DNA polymerase can only build 5′ → 3′. So the two new strands are made in different ways:

• Leading strand. Its 5′ → 3′ direction points into the fork. As helicase opens more template, DNA polymerase simply keeps going. One primer is enough, and synthesis is continuous.
• Lagging strand. Its 5′ → 3′ direction points away from the fork. The polymerase can only build backwards from the fork, a short stretch at a time. Each time the fork opens a new stretch of template, a new primer is laid down and a new short fragment is made. Synthesis is discontinuous. (Background: these fragments are often called Okazaki fragments.)

To finish the lagging strand, the RNA primers are removed and the gaps are filled with DNA by a DNA polymerase. Ligase then seals each remaining break in the backbone, giving one continuous strand.

Every new DNA molecule ends up with one template strand and one new strand: replication is semiconservative at every fork.

Background: DNA polymerase also checks (“proofreads”) each new base and removes most mismatches. The few errors that remain are one source of the mutations you will study in Topic 6.7.

Worked example 1: writing the new strands at a fork

Question. At a replication fork, the parental DNA reads:

• Top strand: 5′-GCTTAGCATCGA-3′ (left to right)
• Bottom strand: 3′-CGAATCGTAGCT-5′ (left to right)

The fork is moving from left to right.

(a) Which template will give the leading strand, and which the lagging strand? (b) Write the full sequence of each new strand, from 5′ to 3′. (c) On the lagging strand, at which end of each fragment is the RNA primer?

(a)

1. A new strand runs antiparallel to its template and grows 5′ → 3′.
2. On the bottom template (3′ at the left), the new strand runs 5′ on the left to 3′ on the right. It grows to the right, towards the fork: this is the leading strand.
3. On the top template (5′ at the left), the new strand runs 3′ on the left to 5′ on the right. It grows to the left, away from the fork: this is the lagging strand.

(b)

1. Leading strand (pairs with the bottom template): 5′-GCTTAGCATCGA-3′. It has the same sequence as the top strand, because both pair with the bottom strand.
2. Lagging strand (pairs with the top template): pair each base, then read it 5′ → 3′, which is right to left. Result: 5′-TCGATGCTAAGC-3′. It has the same sequence as the bottom strand.

Check. Read the lagging strand from its 3′ end: C-G-A-A-T-C-G-T-A-G-C-T. Pair it with the top strand from its 5′ end: G–C, C–G, T–A, T–A, A–T, G–C, C–G, A–T, T–A, C–G, G–C, A–T. All twelve pairs are correct.

(c) Each fragment is built from its primer onwards, in the 5′ → 3′ direction, so the primer is at the 5′ end of every fragment, the end nearest the fork. For example, if the fragment nearest the fork began with a four-base primer, the primer would read 5′-UCGA-3′ (RNA uses U, not T).

Interpretation. Both daughter molecules end up with the same sequence as the parent: one is top strand + new copy of the bottom strand, the other bottom strand + new copy of the top. That is semiconservative replication in practice.

Worked example 2: testing the models with density data

Question. A fictional bacterium is grown for many generations on light nitrogen (¹⁴N), so all its DNA is light. It is then moved to a medium containing only heavy nitrogen (¹⁵N). Samples of DNA are separated by density.

Rounds of replication after transfer	Bands seen (percentage of DNA)
1	one band, intermediate density (100%)
2	intermediate (49%) and heavy (51%)
3	intermediate (26%) and heavy (74%)

(a) Predict the bands after rounds 1 and 2 for each model. (b) Which model do the data support? (c) After 3 rounds, what fraction of all DNA strands are original light strands?

(a) In this experiment the new strands are heavy and the original strands are light.

Model	Round 1	Round 2
Conservative	50% light, 50% heavy	25% light, 75% heavy
Semiconservative	100% intermediate	50% intermediate, 50% heavy
Dispersive	one band, half heavy nitrogen (intermediate)	one band, 75% heavy nitrogen (between intermediate and heavy)

How the semiconservative row works: after round 1, each of the 2 molecules has one light and one heavy strand. In round 2, each light strand pairs with a new heavy strand (2 intermediate molecules), and each heavy strand pairs with a new heavy strand (2 heavy molecules): 2 out of 4 = 50% each.

(b) Round 1 has one intermediate band and no light band, which rules out the conservative model. Round 2 has two separate bands, not one band at a single new position, which rules out the dispersive model. The percentages (49 : 51 and 26 : 74) match the semiconservative predictions (50 : 50 and 25 : 75) within measurement error. The data support the semiconservative model.

(c) After 3 rounds, one molecule has become 2³ = 8 molecules, with 16 strands. Only the 2 original strands are light: 2 ÷ 16 = 1/8 (12.5%) of all strands. They are found in 2 of the 8 molecules (25%), matching the intermediate band.

Worked example 3: why so many starting points?

Question. Use these model values for two fictional organisms. Each origin produces two forks that move in opposite directions.

• Bacterium: one circular chromosome of 3.6 × 10⁶ base pairs, one origin, each fork copies 900 nucleotides per second. Lagging-strand fragments are about 1 200 nucleotides long.
• Eukaryote: one linear chromosome of 8.0 × 10⁷ base pairs, each fork copies 40 nucleotides per second.

(a) How long does the bacterium take to copy its chromosome? (b) About how many lagging-strand fragments, and so primers and joining steps, are needed? (c) How long would the eukaryotic chromosome take with one origin in the middle, and how long with 1 600 evenly spaced origins?

(a) The two forks meet halfway round the circle, so each fork copies 3.6 × 10⁶ ÷ 2 = 1.8 × 10⁶ bp. Time = 1.8 × 10⁶ ÷ 900 = 2 000 s, which is about 33 minutes.

(b) Each fork has one lagging strand that copies 1.8 × 10⁶ nucleotides, so the total lagging DNA is 3.6 × 10⁶ nucleotides. Number of fragments ≈ 3.6 × 10⁶ ÷ 1 200 = 3 000. Each fragment needs its own primer, and ligase must make roughly 3 000 joins. The leading strands need only one primer per fork.

(c)

1. One origin: each fork copies 8.0 × 10⁷ ÷ 2 = 4.0 × 10⁷ bp. Time = 4.0 × 10⁷ ÷ 40 = 1.0 × 10⁶ s ≈ 11.6 days.
2. 1 600 origins: each origin is responsible for 8.0 × 10⁷ ÷ 1 600 = 50 000 bp, so each of its two forks copies 25 000 bp. Time = 25 000 ÷ 40 = 625 s ≈ 10 minutes.

Interpretation. Eukaryotic forks in this model are much slower and the chromosome is much longer, so many origins are essential to copy the DNA in reasonable time. The model ignores the ends of the chromosome and assumes all origins start together.

Common misconceptions

• “Replication makes one old molecule and one brand-new molecule.” That is the conservative model. Each new molecule has one old and one new strand.
• “DNA polymerase can add nucleotides at either end.” Only at the 3′ end, so new strands always grow 5′ → 3′.
• “DNA polymerase starts each strand.” It needs an RNA primer with a free 3′ end.
• “The lagging strand is made 3′ → 5′.” Both strands are made 5′ → 3′. The lagging strand is made in fragments because its 5′ → 3′ direction points away from the fork.
• “Helicase breaks the covalent backbone.” Helicase breaks the hydrogen bonds between bases. Topoisomerase is the enzyme that cuts and rejoins the backbone to relieve twisting.
• “Topoisomerase works behind the fork.” It acts on the unreplicated DNA ahead of the fork, where overwinding builds up.
• “Ligase joins the base pairs.” Ligase forms covalent bonds in the backbone between fragments; hydrogen bonds between bases form by themselves.
• “The primers stay in the DNA.” RNA primers are removed and replaced with DNA before the fragments are joined.

Where this leads

Next you will see how the same template-copying idea is used to make RNA from DNA: continue with Topic 6.3, Transcription and RNA Processing. Before you move on, try the practice questions, then use the revision notes and the checklist. The previous topic is Topic 6.1, DNA and RNA Structure.