2. The Watson–Crick model for DNA
replication assumed that as new strands
of DNA are made, they follow the usual
base-pairing rules of A with T and G with
C. The model also proposed that the two
parental strands separate and that each
then serves as a template for a new
progeny strand. This is called
semiconservative replication because
each daughter duplex has one parental
strand and one new strand. In other
words, one of the parental strands is
“conserved” in each daughter duplex.
However, this is not the only possibility.
Another potential mechanism is
conservative replication, in which the two
parental strands stay together and
somehow produce another daughter helix
with two completely new strands.
Yet another possibility is dispersive
replication, in which the DNA becomes
fragmented so that new and old DNAs
coexist in the same strand after
replication. This mechanism was
envisioned to avoid the formidable
problem of unwinding the two DNA
strands
Three hypotheses for DNA replication.
(a) Semiconservative replication gives two daughter duplex
DNAs, each of which contains one old strand (blue) and
one new strand (red).
(b) Conservative replication yields two daughter duplexes,
one of which has two old strands (blue) and one of which
has two new strands (red).
(c) Dispersive replication gives two daughter duplexes, each of
which contains strands that are a mixture of old and new.
3. In 1958, Matthew Meselson and Franklin Stahl performed a classic experiment to distinguish
among these three possibilities.
They labeled E. coli DNA with heavy nitrogen (15N) by growing cells in a medium enriched in
this nitrogen isotope. This made the DNA denser than normal. Then they switched the cells to
an ordinary medium containing primarily 14N, for various lengths of time. Finally, they
subjected the DNA to CsCl gradient ultracentrifugation to determine the density of the DNA.
Figure depicts the results of a control experiment that shows that 15N- and 14N-DNAs are
clearly separated by this method.
Separation of DNAs by cesium chloride density gradient
centrifugation.
DNA containing the normal isotope of nitrogen (14N) was
mixed with DNA labeled with a heavy isotope of nitrogen
(15N) and subjected to cesium chloride density gradient
centrifugation. The two bands had different densities, so they
separated cleanly. (a) A photograph of the spinning rotor
under ultraviolet illumination. Note that this is a photograph
through a window in the rotor as it spins. The ultracentrifuge
rotor was designed to allow the experimenter to check its
contents without stopping the centrifuge. The two dark bands
correspond to the two different DNAs that absorb ultraviolet
light. (b) A graph of the darkness of each band, which gives
an idea of the relative amounts of the two kinds of DNA.
4. What outcomes would we expect after one
round of replication according to the three
different mechanisms?
If replication is conservative, the two heavy
parental strands will stay together, and
another, newly made DNA duplex will appear.
Because this second duplex will be made in
the presence of light nitrogen, both its strands
will be light. The heavy/heavy (H/H) parental
duplex and light/light (L/L) progeny duplex will
separate readily in the CsCl gradient (Figure
a).
On the other hand, if replication is
semiconservative, the two heavy parental
strands will separate and each will be supplied
with a new, light partner. These H/L hybrid
duplexes will have a density halfway between
the H/H parental duplexes and L/L ordinary
DNA (Figure b).
That this is exactly what happened; after the
first DNA doubling, a single band appeared
midway between the labeled H/H DNA and a
normal L/L DNA.
This ruled out conservative
replication, but was still consistent
with either semiconservative or
dispersive replication.
5. The results of one more round of DNA replication
ruled out the dispersive hypothesis.
Dispersive replication would give a product with one-
fourth 15N and three-fourths 14N after two rounds of
replication in a 14N medium.
Semiconservative replication would yield half of the
products as H/L and half as L/L. In other words, the
hybrid H/L products of the first round of replication would
each split and be supplied with new, light partners,
giving the 1:1 ratio of H/L to L/L DNAs.
Three replication hypotheses. The conservative
model (a) predicts that after one generation equal
amounts of two different DNAs (heavy/heavy [H/H] and
light/light [L/L]) will occur. Both the semiconservative (b)
and dispersive (c) models predict a single band of DNA
with a density halfway between the H/H and L/L
densities. Meselson and Stahl’s results confi rmed the
latter prediction, so the conservative mechanism was
ruled out. The dispersive model predicts that the DNA
after the second generation will have a single density,
corresponding to molecules that are 25% H and 75% L.
This should give one band of DNA halfway between the
L/L and the H/L band. The semiconservative model
predicts that equal amounts of two different DNAs (L/L
and H/L) will be present after the second generation.
Again, the latter prediction matched the experimental
results, supporting the semiconservative model.
6. Again, this is precisely what occurred. To
make sure that the intermediate density peak
was really a 1:1 mixture of the heavy and light
DNA, Meselson and Stahl mixed pure 15N-
labeled DNA with the DNA after 1.9
generations in 14N medium, then measured
the distances among the peaks. The middle
peak was centered almost perfectly between
the other two (50%±2% of the distance
between them). Therefore, the data strongly
supported the semiconservative mechanism.
Results of CsCl gradient ultracentrifugation experiment
that demonstrates semiconservative DNA replication.
Meselson and Stahl shifted 15N-labeled E. coli cells to a 14N medium for
the number of generations given at right, then subjected the bacterial DNA
to CsCl gradient ultracentrifugation. (a) Photographs of the spinning
centrifuge tubes under ultraviolet illumination. The dark bands correspond
to heavy DNA (right) and light DNA (left). A band of intermediate density
was also observed between these two and is virtually the only band
observed at 1.0 and 1.1 generations. This band corresponds to duplex
DNAs in which one strand is labeled with 15N, and the other with 14N, as
predicted by the semiconservative replication model. After 1.9
generations, Meselson and Stahl observed approximately equal quantities
of the intermediate band (H/L) and the L/L band. Again, this is what the
semiconservative model predicts. After three and four generations, they
saw a progressive depletion of the H/L band, and a corresponding
increase in the L/L band, again as we expect if replication is
semiconservative. (b) Densitometer tracings of the bands in panel (a),
which can be used to quantify the amount of DNA in each band.
7. THE CHEMISTRY OF DNA SYNTHESIS
DNA Synthesis Requires Deoxynucleoside Triphosphates and a Primer:Template Junction
For the synthesis of DNA to proceed two key substrates must be present.
First, new synthesis requires the four deoxynucleoside triphosphates—dGTP, dCTP, dATP,
and dTTP (Fig. a). Nucleoside triphosphates have three phosphoryl groups that are attached
via the 5'-hydroxyl of the 2'-deoxyribose. The phosphoryl group proximal to the deoxyribose
is called the α-phosphate, whereas the middle and distal groups are called the β-phosphate
and the γ-phosphate, respectively.
8. The second essential substrate for DNA synthesis is a particular arrangement of single-
stranded DNA (ssDNA) and double-stranded DNA (dsDNA) called a primer:template
junction (Fig. b).As suggested by its name, the primer: template junction has two key
components.
The template provides the ssDNA that directs the addition of each complementary
deoxynucleotide. The template provides only the information necessary to select which
nucleotides are added.
The primer is complementary to, but shorter than, the template. The primer must have an
exposed 3’-OH adjacent to the single-strand region of the template. It is this 3’-OH that will
be extended by nucleotide addition.
10. Hydrolysis of Pyrophosphate Is the Driving Force for DNA Synthesis
The addition of a nucleotide to a growing polynucleotide chain of length n is indicated by the
following reaction:
But the free energy for this reaction is rather small (∆G = –3.5 kcal/mol).
What, then, is the driving force for the polymerization of nucleotides into DNA?
Additional free energy is provided by the rapid hydrolysis of the pyrophosphate into two
phosphate groups by an enzyme known as pyrophosphatase:
The net result of nucleotide addition and pyrophosphate hydrolysisis the breaking of two
high-energy phosphate bonds. Therefore, DNA synthesis is a coupled process, with an
overall reaction of
This is a highly favorable reaction with a ∆G of –7 kcal/mol, which corresponds to an
equilibrium constant (Keq) of105. Such a high Keq means that the DNA synthesis
reaction is effectively irreversible.
11. THE MECHANISM OF DNA POLYMERASE
Correctly paired bases are
required for DNA-
polymerase-catalyzed
nucleotide addition.
(a) Schematic diagram of the
attack of a primer 3’-OH end
on a correctly base-paired
dNTP.
(b) Schematic diagram of the
consequence of incorrect
base pairing on catalysis by
DNA polymerase. In the
example shown, the
incorrect A:A base pair
displaces the α-phosphate
of the incoming nucleotide.
This incorrect alignment
reduces the rate of catalysis
dramatically, resulting in the
DNA polymerase
preferentially adding
correctly base paired dNTPs.
12. DNA polymerases show an impressive ability to distinguish between
ribonucleoside and deoxyribonucleoside triphosphates (rNTPs and dNTPs).
Although rNTPs are present at approximately 10-fold higher concentration in the cell, theyare
incorporated at a rate that is more than 1000-fold lower than dNTPs. This discrimination is
mediated by the steric exclusion of rNTPs from the DNA polymerase active site.
Schematic illustration of the steric
constraints preventing DNA
polymerase from using rNTP
precursors.
(a) Binding of a correctly base-paired
dNTP to the DNA polymerase.
Under these conditions, the 3'-OH
of the primer and the a phosphate
of the dNTP are in close proximity.
(b) Additionof a 2’-OH results in a
steric clash with amino acids (the
discriminator amino acids) in the
nucleotide-binding pocket. This
results in the α-phosphate of the
dNTP being displaced. In this state,
the a-phosphate is incorrectly
aligned with the 3'-OH of the
primer, dramatically reducing the
rate of catalysis.
14. DNA Polymerases Are Processive Enzymes
Processivity is a characteristic of enzymes that operate on polymeric substrates.
In the case of DNA polymerases, the degree of processivity is defined as the average number of
nucleotides added each time the enzyme binds a primer:template junction.
Each DNA polymerase has a characteristic processivity that can range from only a few nucleotides to
more than 50,000 bases added per binding event.
DNA polymerases synthesize DNA in a
processive manner. This illustration shows the
difference between a processive and a
nonprocessive DNA polymerase. Both DNA
polymerases bind the primer:template junction.
Upon binding, the nonprocessive enzyme adds
a single dNTP to the 30 end of the primer and
then is released from the new primer:template
junction.Incontrast,aprocessiveDNApolymeras
eaddsmanydNTPseachtimeitbinds to the
template.
15. Exonucleases activity of DNA polymerase Proofread Newly Synthesized DNA
• A major limit to DNA polymerase accuracy is the occasional (about one in 105 times) flickering of the
bases into the “wrong” tautomeric form (imino or enol).
• These alternate forms of the bases permit incorrect base pairs to be correctly positioned for catalysis.
• When the nucleotide returns to its “correct” state, the incorporated nucleotide is mismatched with the
template and must be eliminated.
• Removal of these incorrectly base-paired nucleotides is mediated by a type of nuclease that was
originally identified in the same polypeptide as the DNA polymerase. Referred to as proofreading
exonuclease, these enzymes degrade DNA starting from a 3' DNA end (i.e., from the growing end of
the new DNA strand).
• This “proofreading” of the newly added DNA gives the DNA polymerase a second chance to add the
correct nucleotide.
Mispaired DNA alters the geometry between the 3'-OH and the
incoming nucleotide because of poor interactions with the
palm region. This altered geometry reduces the rate of
nucleotide addition in much the same way that addition of an
incorrectly paired dNTP reduces catalysis. Thus, when a
mismatched nucleotide is added, it both decreases the rate of
new nucleotide addition and increases the rate of proofreading
exonuclease activity.
16. As for processive DNA synthesis, proofreading occurs without releasing the DNA from the polymerase. When
a mismatched base pair is present in the polymerase active site, the primer:template junction is destabilized,
creating several base pairs of unpaired DNA. The DNA polymerase active site binds such a mismatched
template poorly, but the exonuclease active site has a 10-fold higher affinity for single-stranded 30 ends.
Thus, the newly unpaired 30 end moves from the polymerase active site to the exonuclease active site. The
incorrect nucleotide is removed by the exonuclease (an additional nucleotide may also be removed). The
removal of the mismatched base allows the primer:template junction to re-form and rebind the polymerase
active site, enabling DNA synthesis to continue. ay that addition of an incorrectly paired dNTP reduces
catalysis. Thus, when a mismatched nucleotide is added, it both decreases the rate of new nucleotide
addition and increases the rate of proofreading exonuclease activity.
18. THE REPLICATION FORK
Replication fork.
(Red) Newly synthesized DNA;
(green) RNA primers.
The Okazaki fragments shown are artificially short for illustrative purposes.
In the cell, Okazaki fragments can vary between 100 and 2000 bases
depending on the organism.
19. • The antiparallel nature of DNA creates acomplication for the simultaneous
replication of the two exposed templates at the replication fork. Because DNA
is synthesized only by elongating a 3' end, only one of the two exposed
templates can be replicated continuously as the replication fork moves. On this
template strand, the polymerase simply “chases” the moving replication fork.
The newly synthesized DNA strand directed by this template is known as the
leading strand.
• Synthesis of the new DNA strand directed by the other ssDNA template is more
complicated. This template directs the DNApolymerase to move in the opposite
direction of the replication fork. The new DNA strand directed by this template
is known as the lagging strand.
• The resulting short fragments of new DNA formed on the lagging strand are
called Okazaki fragments and vary in length from 1000 to 2000 nucleotides in
bacteria and from 100 to 400 nucleotides in eukaryotes. Shortly after being
synthesized, Okazaki fragments are covalently joined together to generate a
continuous, intact strand of new DNA.Okazaki fragments are therefore
transient intermediates in DNA replication.
20. THE INITIATION OF A NEW STRAND OF DNA REQUIRES AN RNA PRIMER
• All DNA polymerases require a primer with a free 3'-OH. They can not initiate a new DNA
strand de novo.
• To accomplish this, the cell takes advantage of the ability of RNA polymerases to do What
DNA polymerases cannot: start new RNA chains de novo.
• Primase is a specialized RNA polymerase dedicated to making short RNA primers (5–10
nucleotides long) on an ssDNA template. These primers are subsequently extended by
DNA polymerase. Although DNA polymerases incorporate only deoxyribonucleotides into
DNA, they can initiate synthesis.
• Although both the leading and lagging strands require primase to initiate DNA
synthesis,the frequency of primase function on the two strands is dramatically different.
Each leading strand requires only a single RNA primer. In contrast, the discontinuous
synthesis of the lagging strand means that new primers are needed for each Okazaki
fragment. Because a single replication fork can add hundreds of thousands of nucleotides
to a primer,synthesis of the lagging strand can require hundreds of Okazaki fragments and
their associated RNA primers.
• Primase activity is dramatically increased when it associates with another protein that acts
at the replication fork called DNA helicase. This protein unwinds the DNA at the replication
fork, creating an ssDNA template that can be acted on by primase. The requirement for an
ssDNA template and DNA helicase association ensures that primase is only active at the
replication fork.
21. RNA Primers Must Be Removed to Complete DNA Replication
• To replace the RNA primers with DNA, an enzyme called RNaseH recognizes and
removes most of each RNA primer. This enzyme specifically degrades RNA that is base-
paired with DNA (the H in its name stands for “hybrid” in RNA:DNA hybrid). RNase H
removes all of the RNA primer except the ribonucleotide directly linked to the DNA end.
This is because RNaseH can only cleave bonds between two ribonucleotides.The final
ribonucleotide is removed by a 5' exonuclease that degrades RNA or DNA from their 5'
ends.
Removal of the RNA primer leaves a gap in the dsDNA
that is an ideal substrate for DNApolymerase—a
primer:template junction.DNA polymerase fills this gap
until every nucleotide is base-paired, leaving a DNA
molecule that is complete except for a break in the
phosphodiester backbone between the 3'-OH and 5'-
phosphate of the repaired strand. This “nick” in the
DNA can be repaired by an enzyme called DNA ligase.
DNA ligases use high-energy co-factors (such as ATP)
to create a phosphodiester bond between an adjacent
5'-phosphate and 3'-OH. Only after all RNA primers
are replaced by DNA and the associated nicks are
sealed is DNA synthesis complete.
22. DNA Helicases Unwind the Double Helix in Advance of the Replication Fork
DNA helicases separate the two
strands of the double helix. When
ATP is added to a DNA helicase
bound to ssDNA, the helicase moves
with a defined polarity on the ssDNA.
In the instance illustrated, the DNA
helicase has a 5’ 3' polarity. This
polarity means that the DNA helicase
would be bound to the lagging-strand
template at the replication fork.
23. Single-Stranded DNA-Binding Proteins Stabilize ssDNA before Replication
After the DNA helicase has passed, the newly generated ssDNA must remain free of base pairing
until it can be used as a template for DNA synthesis. To stabilize the separated strands, ssDNA-
binding proteins (SSBs) rapidly bind to the separated strands. Binding of one SSB promotes the
binding of another SSB to the immediately adjacent ssDNA.
This is called cooperative binding and occurs because SSB molecules bound to immediately
adjacent regions of ssDNA also bind to each other. This SSB–SSB interaction strongly stabilizes
SSB binding to ssDNA and makes sites already occupied by one or more SSB molecules preferred
SSB-binding sites.
SSBs interact with ssDNA in a sequence -independent manner. SSBs primarily contact ssDNA
through electrostatic interactions with the phosphate backbone and stacking interactions with
the DNA bases. In contrast to sequence-specific DNA-binding proteins, SSBs make few, if any,
hydrogen bonds to the ssDNA bases.
