Biology · DNA, Replication & Repair

Replication, directionality and the work of replication proteins

Separating paired DNA strands exposes two templates. The challenge is to build complementary chains on both while the helix opens. The templates point in opposite directions, yet the DNA polymerases that copy them extend a new chain in only one chemical direction.

Use the earlier ideas of nucleotides, complementary molecules, enzymes, inheritance and mutation. A nucleotide is a building block of a nucleic acid. A template is an existing sequence that guides construction of a complementary sequence. The symbols 5′ and 3′ are read five prime and three prime; they identify chemically different strand ends. Every short sequence and numerical example here is constructed for explanation unless identified as historical evidence.

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By the end, you can

Explain semiconservative replication and trace 5′→3′ synthesis on both templates at a specified fork.

Each daughter helix keeps one template strand

DNA replication is semiconservative: each daughter double helix contains one strand from the parental molecule and one newly constructed strand. The whole parental helix is not kept intact in one daughter while an entirely new helix goes to the other. Each separated parental strand guides its own complementary copy.

Matthew Meselson and Franklin Stahl tested this in 1958 using Escherichia coli DNA labelled with the stable heavy isotope nitrogen-15, then allowing growth with nitrogen-14. DNA density after successive generations supported semiconservative copying. Nitrogen-15 is a stable isotope, so this was not a radioactive tracer experiment.

Our strand model shows the expected pattern for complete copying rounds from one heavy-labelled helix: two hybrid helices after one round, then two hybrid and two light helices after the second. Labels track strand ancestry. The constructed counts explain the prediction; they are not the original density measurements or a timetable for bacterial growth.

One parental strand in each daughter helix

Constructed ideal complete copying rounds with no newly supplied heavy label, molecule loss or mixing of partial rounds. Molecule counts explain semiconservative ancestry and do not reproduce the 1958 measurements.

Read the diagram in words

Three rows show one initial heavy-labelled DNA helix, two hybrid helices after one complete copying round and four helices after two rounds. Each helix is drawn as two separate strands. Track the original strands H1 and H2 through distinct descendants; new strands N1 to N6 have no heavy label. Old strands use an explicit heavy-labelled label and a dashed style; new strands use an unlabelled label and a solid style. This is a strand-ancestry model, not a density-band chart.

Heavy nitrogen-15 label is stable, not radioactive

Before copying
Molecule identity	First strand identity and label	Second strand identity and label
m0	H1: Original heavy-labelled strand	H2: Original heavy-labelled strand

After one complete round
Molecule identity	First strand identity and label	Second strand identity and label
m1	H1: Original heavy-labelled strand	N1: New unlabelled strand
m2	H2: Original heavy-labelled strand	N2: New unlabelled strand

After two complete rounds
Molecule identity	First strand identity and label	Second strand identity and label
m3	H1: Original heavy-labelled strand	N3: New unlabelled strand
m4	N1: Previously made unlabelled strand	N4: New unlabelled strand
m5	H2: Original heavy-labelled strand	N5: New unlabelled strand
m6	N2: Previously made unlabelled strand	N6: New unlabelled strand

One parental strand in each daughter helix
Completed copying round	Helices with one original heavy-labelled strand	Helices with two unlabelled strands
1	2	0
2	2	2

One parental strand in each daughter helix
From	To	Relationship
m0	m1	Copy on H1
m0	m2	Copy on H2
m1	m3	Copy on H1
m1	m4	Copy on N1
m2	m5	Copy on H2
m2	m6	Copy on N2

One heavy-labelled and one unlabelled strand

Two unlabelled strands

The growing end is always 3′

A replicating DNA polymerase adds a nucleotide to an existing chain's free 3′ hydroxyl group, written 3′–OH. The new strand therefore extends 5′→3′ while the template is read 3′→5′. An incoming deoxyribonucleoside triphosphate supplies a DNA building block; its phosphate chemistry helps drive incorporation. DNA synthesis is chemical construction, not merely matching loose letters.

The polymerase needs a starting chain end. Primase makes a short RNA primer that provides it. RNA and DNA share sufficiently compatible backbones for the primer's 3′ end to be extended with DNA nucleotides. Later, primer removal, DNA replacement and sealing produce a continuous DNA backbone.

At the fork drawn here, movement is to the right. The upper template runs 3′→5′ from left to right, allowing leading-strand synthesis toward the fork. The lower template runs the other way. New segments called Okazaki fragments are initiated near the opening fork and extended away from it. Each segment still grows 5′→3′; lagging-strand synthesis never requires a DNA polymerase to build 3′→5′.

Both new strands grow 5′→3′

One local fork, drawn without the looping used in an actual replisome. The two fragment cartoons are separate snapshots of synthesis, not a claim about simultaneous polymerase occupancy. Arrow labels must preserve chemical direction even where the spatial arrow points left. No molecular size or copying speed is implied.

Read the diagram in words

A right-moving Y fork has paired parental DNA ahead on the right and exposed templates on the left. The upper template runs 3′ at left to 5′ at right. Its new leading strand has 5′ at left and a growing 3′ end near the fork, with a synthesis arrow to the right. The lower template runs 5′ at left to 3′ at right. Two new lagging fragments beneath it each have a 5′ RNA-primer region at their right and a growing 3′ DNA end at their left, with synthesis arrows to the left. Clearly separate fork movement from new-strand synthesis arrows.

Fork movement →

Both new strands grow 5′→3′
Template	Left end	Right end
Leading-strand template	3′	5′
Lagging-strand template	5′	3′

Each template is read 3′→5′

Both new strands grow 5′→3′
Strand	Template	Left end	Right end	Illustrated synthesis direction	Illustrated primer side	Description
Leading strand	Leading-strand template	5′	Growing 3′–OH	5′→3′ synthesis toward fork →	Left	Continuous extension toward this fork.
Earlier Okazaki fragment	Lagging-strand template	Growing 3′–OH	5′ RNA primer	← 5′→3′ synthesis away from fork	Right	Each fragment is extended 5′→3′ away from this fork.
Newer Okazaki fragment	Lagging-strand template	Growing 3′–OH	5′ RNA primer	← 5′→3′ synthesis away from fork	Right	A new primer is made as additional template is exposed.

Both new strands grow 5′→3′
Item	Illustrated position
Paired parental DNA ahead	Right of the fork
Helicase opens the fork	At the fork
DNA polymerase	At the leading strand's growing end
DNA polymerase	At the newer fragment's growing end

RNA primer supplies a starting 3′ end

Different proteins solve different problems

Helicase separates paired strands, and single-stranded-DNA-binding proteins help stabilise exposed templates. Topoisomerases manage the twisting stress associated with opening DNA. Sliding clamps keep copying polymerases associated with DNA. These jobs complement polymerase activity; none should be represented as an enzyme that alone performs the entire replication process.

After RNA primers are removed and replaced with DNA, a nick can remain: adjacent segments have the needed nucleotides but their backbones are not yet joined. DNA ligase seals that backbone connection. A missing stretch of nucleotides is a gap, which requires synthesis before the final join. Distinguishing gaps from nicks explains why both polymerases and ligases are needed.

Replication begins at origins and proceeds through moving forks; chromosomes differ in the number and arrangement of origins and in the proteins involved. Our fork is a local, simplified view. It omits looping and many cooperating factors, and does not mean an entire chromosome is copied by one fork or that every organism uses the same named polymerases.

Worked example

Follow strand ancestry through three copying rounds

Start with one double helix whose two strands carry an old label. Assume complete semiconservative copying rounds, all new nucleotides unlabelled, no additional old label and no loss of molecules. Helix counts are 1, 2, 4 and 8 after zero, one, two and three rounds.

The two originally labelled strands remain in two different helices after the first round. Thus hybrid helices number 2 after each subsequent round: after the third round there are 2 hybrid and 6 wholly unlabelled helices. The hybrid fraction is 2 ÷ 8 = 1/4. The label's dilution does not mean a labelled strand was cut into quarters; more helices now lack it.

Check your understanding

Which end of a growing DNA strand receives the next nucleotide?
Show reasoning
The free 3′–OH end. Addition there extends the new strand 5′→3′ while the template is read 3′→5′.
At a right-moving fork, a template runs 5′→3′ from left to right. Can its new strand grow continuously rightward by ordinary replication polymerase activity?
Show reasoning
No. Its complementary segments must grow leftward, 5′→3′, away from that fork. Repeated priming produces Okazaki fragments as more template becomes exposed.
After primer replacement, all nucleotides are present but adjacent backbone ends remain unjoined. What kind of problem remains?
Show reasoning
A nick remains, and ligase can seal the backbone connection. A true gap would first need missing nucleotides to be synthesised.

What to carry forward

Both templates are copied by 5′→3′ synthesis; antiparallel orientation explains leading and lagging arrangements at a particular fork.

References

Resources supporting this lesson’s explanations and examples.