This cartoon captures the essence of the Watson-Crick model for DNA
replication but it is wrong in detail as we will see today!
1
Nucleic acids are replicable. Among the millions of
different molecules that have been created and studied in laboratories
throughout the history of science, no molecules have been able to
match the ability of DNA and RNA to replicate in an extremely efficient
and accurate manner. Nucleic acid replication either in the laboratory or
in the cell can take place with remarkable speed. By the end of this
topic, we will understand how it is possible that literally hours after a
single DNA molecule begins to replicate, it is possible to create billions
of identical copies of that molecule. DNA replication relies on DNA
polymerases, which in turn are encoded by DNA. In other words, cells
replicate DNA using proteins that are encoded within the cell's DNA!
Ponder this conundrum about the origin of life: If life
evolved from nucleic acids and if proteins replicate nucleic acids, then
how could life have arisen before there were proteins? And if life sprung
from proteins first, then how could proteins replicate before there were
nucleic acids? We will come back to this puzzle in the last lecture when
we consider the very beginnings of life!
Because every living cell must contain a complete DNA
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instruction set, cell division requires the complete replication of each dividing
cell’s genome. In order to avoid the possibility of cell division with incomplete
DNA replication, the cell has evolved elaborate mechanisms to ensure that
division cannot occur until the entire genome is replicated. As we will learn
later in this course, this requirement is the basis of many anti-cancer drugs
that attempt to block the division of tumor cells by blocking the process of DNA
replication.
In what follows, we will consider the Meselson-Stahl experiment,
which proved the Watson-Crick or “semi-conservative” model for the
replication of DNA. We will then turn to the chemistry of DNA synthesis, the
enzyme (DNA polymerase) that mediates DNA synthesis and how it
proofreads its own work, how mutations arise and are corrected, and finally
one of the most useful tools in molecular biology, the Polymerase Chain
Reaction.
2
Your goal is to understand the chemistry of DNA synthesis and how
DNA is replicated with high accuracy. Accordingly, you should be able
to explain what is meant by semi-conservative replication and how this
was proved, how the phosphodiester bond is formed and the energetics
of the reaction, how the replication fork accommodates the constraint
that DNA synthesis takes place only in a 5’ to 3’ direction, how DNA is
replicated accurately and is protected from damage, and finally the
principle of the Polymerase Chain Reaction.
3
As we have seen, the Watson-Crick structure for the
double helix suggested a mechanism for the replication of DNA in which
each strand of the double helix serves as a template to make a copy of
the other: a new Crick is copied from a parental Watson strand and a
new Watson is copied from a parental Crick. As we shall see, this
mechanism is known as “semi-conservative.” How did we come to
know whether the semi-conservative mechanism is indeed how DNA is
replicated?
In 1958 Matthew Meselson and Franklin Stahl, then at
CalTech (Meselson is now here in MCB), reported what has been called
“the most beautiful experiment in biology.” The Meselson-Stahl
experiment elegantly distinguished the semi-conservative mechanism
from an alternative conservative mechanism.
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The Watson-Crick model for the double helix suggested a
specific mechanism for how DNA is replicated.
Imagine that we start with a DNA molecule that is red and
that newly synthesized DNA is blue. If each strand of the helix serves
as a template to synthesize a copy of the other strand (Watson serving
as a template for a daughter Crick and Crick a template for a daughter
Watson), then after one round of replication each of the resulting double
helices would consist of a blue strand and a red strand. This is known
as semi-conservative replication as one strand is preserved during
replication and the other is newly synthesized.
It was conceivable, however, that the parental double helix
(red-red) was left intact and that during replication a new double helix
was created in which both the Watson and Crick strands were newly
synthesized (blue-blue). Thus following replication we would retain the
red-red, parental helix and have created an entirely new blue-blue,
daughter helix. This mechanism is known as conservative replication
because the original double helix is maintained during replication.
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Certainly semi-conservative replication was a more appealing
model as it was difficult to conceive how mechanistically conservative
replication could take place. But this did not mean that semi-conservative
replication is true. How then do we distinguish between the two possibilities
experimentally?
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To do so, Meselson and Stahl grew E. coli cells in growth medium
labeled with a heavy isotope of nitrogen, 15N (a heavy isotope of
nitrogen with mass 15) for many generations. (E. coli needs nitrogen as
a nutrient and incorporates the isotope from the medium into its proteins
and nucleic acids.) Next, they shifted the bacteria for various periods of
time to medium containing 14N (the common form of nitrogen with mass
14). Finally, they extracted DNA from the cells and subjected the DNAs
to high centrifugal forces in tubes containing cesium chloride (using an
instrument called an ultracentrifuge in which tubes are spun at high
velocity). Cesium forms a density gradient under such conditions and at
equilibrium DNA molecules in the gradient localize to a position that
matches their own buoyant density. Thus, parental DNA in which both
strands contain the heavy isotope would have a high density (red-red)
and localize near the bottom of the gradient. DNA in which both strands
are newly synthesized and hence light (blue-blue) would localize near
the top and finally hybrid molecules (red-blue) in which one strand was
heavy and one light would localize at an intermediate position.
What did Meselson and Stahl observe? When shifted to 14N-containing
medium for enough time (one generation) for the DNA molecules to
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undergo one round of replication, the DNA was found to exhibit an
intermediate density consistent with that expected for hybrid DNA instead of
the high density of the parental, heavy-heavy DNA. This fit with the
expectation from the semi-conservative model that each strand was serving as
a template for the other strand.
Next, Meselson and Stahl went a step further: they extracted DNA from cells
that had been growth for two generations in the light, 14N-medium. What was
the expected outcome if the semi-conservative model is correct?
6
To do so, Meselson and Stahl grew E. coli cells in growth medium
labeled with a heavy isotope of nitrogen, 15N (a heavy isotope of
nitrogen with mass 15) for many generations. Next, they shifted the
bacteria for various periods of time to medium containing 14N (the
common form of nitrogen with mass 14). Finally, they extracted DNA
from the cells and subjected the DNAs to high centrifugal forces in
tubes containing cesium chloride (using an instrument called an
ultracentrifuge in which tubes are spun at high velocity). Cesium forms
a density gradient under such conditions and at equilibrium DNA
molecules in the gradient localize to a position that matches their own
buoyant density. Thus, parental DNA in which both strands contain the
heavy isotope would have a high density (red-red) and localize near the
bottom of the gradient. DNA in which both strands are newly
synthesized and hence light (blue-blue) would localize near the top and
finally hybrid molecules (red-blue) in which one strand was heavy and
one light would localize at an intermediate position.
What did Meselson and Stahl observe? When shifted to 14N-containing
medium for enough time (one generation) for the DNA molecules to
undergo one round of replication, the DNA was found to exhibit an
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intermediate density consistent with that expected for hybrid DNA instead of
the high density of the parental, heavy-heavy DNA. This fit with the
expectation from the semi-conservative model that each strand was serving as
a template for the other strand.
Next, Meselson and Stahl went a step further: they extracted DNA from cells
that had been growth for two generations in the light, 14N-medium. What was
the expected outcome if the semi-conservative model is correct?
7
8
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If replication occurred semi-conservatively, then the hybrid
molecules seen after the first round of replication would be expected to
give rise to another hybrid DNA and a light-light DNA. Thus, the
parental heavy strand of the hybrid would serve as a template for a light
strand, resulting in another hybrid molecule. Meanwhile, the light strand
would serve as a template for another light strand, resulting in a lightlight molecule. This is exactly what was observed! If replication
occurred by a conservative mechanism, the heavy-heavy DNA would
have continued to have been observed through both rounds of
replication and no hybrid density molecules would have been seen. The
Meselson-Stahl experiment provided strong support for the semiconservative model and did so elegantly.
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Notice the appearance of DNA with a fully light (that is, light/light )
density appearing between 1.5 and 1.9 generations.
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The principal features of the chemistry of DNA synthesis
are that the substrates are 2’-dexoynucleoside triphosphates (dXTPs),
the incoming nucleotide is specified by Watson-Crick pairing with the
corresponding nucleotide on the template strand, that synthesis occurs
in a 5’ to 3’ direction, that phosphosphodiester bond formation occurs
with the release of pyrophosphate, and finally that the favorable free
energy of the hydrolysis of pyrophosphate drives DNA synthesis.
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First some nomenclature! Nucleosides are composed of a sugar
(ribose or deoxyribose) and a nitrogenous base bonded via a glycosidic
linkage at the 1’ carbon of the sugar. Nucleotides are nucleosides that have
one, two, or three phosphate groups covalently attached at the 5′ hydroxyl.
Nucleoside monophosphates have a single bonded phosphate, diphosphates
contain two phosphate groups, and triphosphates have a third phosphate. The
phosphates are labeled in sequence – alpha, beta, and gamma, with the alpha
phosphate bonded to the sugar.
The substrates for DNA synthesis are 2’-deoxynucleoside
triphosphates.
The repeating unit of the polynucleotide chain, which is a sugar,
phosphate and base, is a nucleotide with only an alpha phosphate (nucleoside
monophosphate). What happens to the beta and gamma phosphates during
the polymerization of deoxynucleoside triphosphates into polynucleotide
chains is at the heart of the mechanism of DNA synthesis.
Practice drawing a nucleoside triphosphate!
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Now that we have an understanding of base pairing and
how the strands in the double helix are held together from the previous
lecture, we are ready to describe the chemistry of DNA synthesis. DNA
is synthesized through a chemical reaction that is repeated thousands
of times in succession, resulting in the polymerization of nucleotides
into a polynucleotide chain. The reaction consists of the addition of the
nucleotide substrate, a nucleoside triphosphate, to a growing strand of
DNA to yield a longer polynucleotide with the release of the beta and
gamma phosphates bonded to each other. This diphosphate molecule
is known as pyrophosphate.
At the heart of the reaction is the formation of a covalent
bond between the terminal 3' oxygen atom of the growing DNA strand
and the phosphate group (alpha) of the nucleotide triphosphate located
closest to the deoxyribose ring. Once the 3’ end of the growing strand
is bonded with the new nucleotide, the 3’ oxygen atom of the incoming
nucleotide forms a covalent bond with the next nucleoside triphosphate,
and the polymerization reaction continues. Because the newly
synthesized DNA strand grows only at its 3’ end, the directionality of
DNA synthesis is designated 5’ to 3’.
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Synthesis in a 5’ to 3’ direction is an invariant rule in
polynucleotide synthesis.
16
DNA synthesis requires two different molecules as
substrates: the growing polynucleotide strand that is being extended
into the product strand, and a 2’-deoxynucleotide in the activated form
of a triphosphate. The products of this reaction also consist of two
molecules: the growing chain that extended by one nucleotide unit and
pyrophosphate.
In addition to the starting materials and products that
undergo covalent bond formation or bond breakage during the course of
the reaction, a third molecule is also required for the reaction to take
place in a manner that is useful to the cell: a long DNA template.
Although this DNA template is not altered during the course of DNA
polymerization, it serves the crucial purpose of defining which
nucleotide must be added to the growing DNA strand at every position
based on base-pair complementarity. Thus, in the above we see that 2’deoxy CTP (dCTP) is selected for incorporation via base pairing with
the G on the template strand.
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The incoming nucleotide (green) is specified by pairing with the G (blue)
on the template strand. It is joined to the growing strand via nucleophilic
attack of the 3’ oxygen (red) at the extreme end of the growing
polynucleotide strand and the alpha phosphate (green) of the dCTP. As
a result a phosphodiester bond is formed between the incoming C
nucleotide and the A nucleotide at the end of the growing strand.
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As a consequence of phosphodiester bond formation, the
beta and gamma phosphate groups of the substrate are released as
pyrophosphate (abbreviated as PPi where i stands for inorganic). The
reaction is described in the following equation for a growing DNA chain
of length n: (dXMP)n + dXTP → (dXMP)n+1 + PPi
The enzyme that catalyses DNA synthesis is DNA
polymerase. It both mediates phosphodiester bond formation and
ensures that each incoming nucleotide is the complement of the
corresponding nucleotide on the template strand.
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As a consequence of phosphodiester bond formation, the
beta and gamma phosphate groups of the substrate are released as
pyrophosphate (abbreviated as PPi where i stands for inorganic). The
reaction is described in the following equation for a growing DNA chain
of length n: (dXMP)n + dXTP → (dXMP)n+1 + PPi
20
It turns out that phosphodiester bond formation with the release
of pyrophosphate (reaction one) is only modestly favorable. Given the critical
importance of this reaction for living systems, what is the major driving force?
The answer lies in a second reaction (two) that utilizes a product of the first –
namely, the hydrolysis of the pyrophosphate to give two molecules of inorganic
phosphate. This second reaction is catalyzed by an enzyme named
pyrophosphatase, which is responsible for making the overall polymerization
reaction highly favorable.
The combined reactions of nucleotide addition to the growing
DNA chain, and the hydrolysis of pyrophosphate can be expressed overall as:
(dXMP)n + dXTP → (dXMP)n+1 + 2 Pi (reaction three) with an overall ΔG0rxn
= - 7.3 kcal/mole. (Pi stands for inorganic phosphate.)
21
What is the significance of a free energy of -7.3 kcal/mol?
Remember that the free energy is related exponentially to the equilibrium
constant by the above equation. A valuable, ballpark relationship to remember
is that a free energy of -2.7 kcal/ mol is roughly equal to an equilibrium
constant about 100. Therefore, a free energy of -7.3 kcal/mol would be
equivalent to an Keq of about 100,000! This means that the addition of the
nucleotide (phosphodiester bond formation) is virtually irreversible.
22
In summary, DNA is synthesized by polymerization of 2’deoxynucleoside triphosphates, the incoming base is specified by
Watson-Crick base pairing with the corresponding base on the template
strand, the phosphodiester bond is created by nucleophilic attack of the
3’ oxygen at the extreme 3’ end of the growing polynucleotide chain on
the alpha phosphate of the incoming substrate nucleotide with the
release of pyrophosphate, synthesis therefore takes place in a 5’ and 3’
direction, and, finally, the driving force for DNA synthesis is the
hydrolysis of pyrophosphate, which is energetically highly favorable.
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DNA in the cytoplasm is recognized as the presence of an infectious
agent by cGAS cyclic GMP AMP synthease. The cyclic nucleotide in
turn activates a protein called STING, which in turn switches on (type 1)
interferons in the immune system.
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Implicit in this cartoon is the notion that replication takes place at
localized sites called forks, where the parental Watson and Crick
strands separate so that each can serve as a template for a new
daughter strand. The replication fork proceeds progressively along the
chromosome with the old strands unwinding ahead of the moving fork
(why are they unwinding?) and hybrid helices of new and old strands
being left behind. But this creates a conundrum! Indeed, the cartoon is
wrong as we are about to see in the way it suggests that both strands
are copied!
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If DNA only grows in a 5’ to 3’ direction, then how can both strands be
copied, given the antiparallel structure of the DNA double helix?
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In order for both strands to be synthesized continuously at the fork, one
daughter strand would need to polymerize in the 5′-to-3′ direction and
the other in the 3′-to-5′ direction. This is not a problem for the template
strand on the left, the strand whose 5’ to 3’ orientation is opposite to that
of the direction of fork movement. Copying DNA synthesis from this
strand proceeds in a 5’ to 3’ direction. This is known as “leading
strand” synthesis and it takes place in a continuous manner.
But what about the strand on the right?
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The 5’ to 3’ orientation of the strand on the right is aligned
with the direction of fork movement. Therefore, DNA copied from this
strand cannot take place continuously in a 5’ to 3’ direction. Instead,
synthesis occurs discontinuously in short bursts of hundreds to
thousands of nucleotides. This is known as “lagging” strand” synthesis.
Lagging strand synthesis takes place in a 5’ to 3’ direction. So the rule
that polynucleotide synthesis proceeds 5’ to 3’ is not violated. Instead,
short stretches of DNA are synthesized, which are then joined to each
other by an enzyme known as a ligase.
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As the replication fork moves up the page, additional single-strand
template is revealed both for leading strand synthesis and for lagging
strand synthesis. Both take place in a 5’ to 3’ direction. But in the case
of lagging strand synthesis, this synthesis is discontinuous with each
newly synthesized segment of DNA being joined to the previously
synthesized segments by ligase to create an intact daughter strand as
indicated above. Ligase is an enzyme that joins two polynucleotide
chains together.
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Lets now consider the enzyme that is responsible for DNA synthesis,
the DNA polymerase. As we shall see, it is a catalyst, it is fast, it has a
long attention span, and it corrects its own mistakes!
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If you combine a DNA template, a nucleotide primer, and
four DNA nucleotide triphosphates in a test tube under physiological
conditions, the DNA primer will hybridize to the DNA template, but
nothing else will happen on any reasonable time scale. But why is this
the case since we know that the reaction in thermodynamically
favorable? The answer lies in the kinetic barrier or activation energy hill
that must be overcome before reactants can go to products. In other
words, there is a rate limiting step with a high ΔGorxn.
For DNA replication to take place efficiently requires the
action of a protein called DNA polymerase. DNA polymerase is an
enzyme that catalyzes (accelerates) a chemical reaction. There are
tens of thousands of enzymes in the cell that are necessary to catalyze
all of the chemical reactions that must occur for its survival. For now,
simply appreciate that DNA polymerase is responsible for accelerating
DNA synthesis to a rate of about 800 nucleotides per second.
The structure of DNA polymerase is known in atomic
detail. Roughly speaking, it resembles a hand in which the catalytic
center is in the palm. Notice that the 3’ end of the growing strand sits in
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the catalytic center and that the next nucleotide to be added is positioned to
base pair with the corresponding nucleotide on the template strand. Notice too
that the two strand are of opposite polarity as we have discussed.
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DNA polymerase is remarkably fast! It synthesizes DNA
at a rate of 800 nucleotides per second. Consider that the chromomes
of E. coli consists of almost five million base pairs. Because in E. coli
replication takes place simultaneously from two replication forks, the
overall rate of DNA synthesis is 1,600 nucleotides per second. Thus,
E. coli is capable of duplicating its entire chromosome in as little as 40
minutes.
To appreciate this consider the following:
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DNA polymerase is remarkably fast! It synthesizes DNA
at a rate of 800 nucleotides per second. Consider that the chromomes
of E. coli consists of almost five million base pairs. Because in E. coli
replication takes place simultaneously from two replication forks, the
overall rate of DNA synthesis is 1,600 nucleotides per second. Thus,
E. coli is capable of duplicating its entire chromosome in as little as 40
minutes.
To appreciate this consider the following:
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Lets scale up the size of DNA in the mind’s eye to a diameter of one
meter, with each nucleotide the size of a textbook. If so, then
replication would take place at 600 kilometers per hour and the DNA
replication machine would be the size of a FedEx truck. Replicating the
entire genome would require that the FedEx truck would travel 400
kilometers in 40 minutes, equivalent to driving to the Bronx, with the
FedEx truck delivering 1,000 textbook-sized nucleotides every second!
36
DNA polymerase rarely falls off the DNA. It is said to be
processive. It does not dissociate form the template after each round of
phosphodiester bond formation. Rather, it stays on the DNA for many
rounds of nucleotide addition. Think of the Fed Ex truck blitzing to the
Bronx at break neck speed without driving off the road. How does it
accomplish this feat?
The answer is that it is anchored to the DNA by a clamp that fully
encircles the helix. In the cartoon above, DNA polymerase is moving to
the right but is held to the DNA from behind by the circular clamp to
which it is attached. The structure of the clamp is known in atomic
detail. Six protein subunits associate in a doughnut-like structure that
surrounds DNA which projects through the hole.
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An extraordinary feature of DNA synthesis is its accuracy.
DNA polymerase copies template DNA with high fidelity. Imagine the
FedEx truck delivering 1,000 textbooks per second and doing so with
few mistakes.
DNA replication needs to be accurate so that new
daughter cells have the same DNA genomes following cell division. The
3’-most base must be correctly paired (hydrogen-bonded) with the base
of the template strand in order for polymerization to proceed rapidly.
DNA polymerase does not recognize specific bases, rather it recognizes
the geometry of a base pair -- and as you know, the A:T and G:C base
pairs are very similar in structure. When a correctly matched base pair
is present in the enzyme active site, the α-phosphate of the incoming
nucleotide is optimally positioned for the nucleophilic attack of the 3’-OH
at the end of the growing strand. In contrast, a mismatched pair of
nucleotides lack the overall conformation achieved by the optimally
aligned hydrogen bonds of base pairing. Thus, the α-phosphate of the
incoming nucleotide will not be properly oriented to facilitate the
nucleophilic attack of the 3’-OH. Since the mismatched nucleotide is
not efficiently added to the primer, the resulting time lag gives it time to
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dissociate from the polymerase.
In the rare case that an incorrect nucleotide is covalently added
to the growing 3’-end, the DNA polymerase has a proofreading function that
allows it to detect the erroneous nucleotide and have it removed. Proofreading
is based on the fact that the misincorporated nucleotide is not properly base
paired, making it susceptible to removal. DNA polymerase has a 3’ to 5’
exonuclease activity so when it cannot add a nucleotide onto this unpaired
nucleotide, it cleaves off the offending nucleotide and then restarts the
polymerization process.
38
`The DNA polymerase has a second domain (pink) where the
exonuclease active site is located. If the wrong nucleotide is
incorporated and hence not quickly extended by the incorporation of an
additional nucleotide, the growing strand slips into the exonuclease site
where the mismatched nucleotide is removed. DNA synthesis then
resumes with the fresh incorporation of a nucleotide in the active site.
39
As we have seen, DNA polymerase is fast, processive and accurate.
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41
42
43
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2’, 3’ dideoxy cytidine triphosphate is recognized by DNA polymerase as
a substrate for DNA synthesis because it lacks a 2’ hydroxyl.
Therefore, DNA polymerase will incorporate it into the end of a growing
polynucleotide chain. However, once incorporated at the 3’ end of a
chain, it can not be extended by the incorporation of an additional
nucleotide (dATP in the example above) because it lacks a 3’ hydroxyl
and nucleophilic attack cannot occur. Instead, DNA synthesis aborts.
Dideoxy nucleotides make it possible to abort DNA synthesis at specific
bases as illustrated here with dideoxy CTP. This discovery was at the
heart of the original method for sequencing DNA. Spiking reactions
with a little bit of a particular dideoxy XTP results in a ladder (a nested
set) of DNA fragments that terminate at the positions corresponding to
the incorporation of the dideoxy nucleotide.
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DNA must endure the slings and arrows of outrageous fortune, not all of
which occur during replication. Potentially mutagenic damage to DNA
also occurs independently of replication. Cells have elaborate systems
for detecting and repairing such damage, a topic that is considered in
greater depth in more advanced courses such as MCB 52. But lets
consider one example.
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Many mutations arise from chemicals in the environment
and from radiation. For example, ultraviolet radiation from the sun
damages pyrimidines. Consider this when you go to the beach!
Certain individual have a condition called Xeroderma pigmentosum in
which the system for repairing damage from ultraviolet light is defective.
Such individuals are hypersensitive to sunlight and readily contract skin
cancer.
Some mutations arise without external influences.
Remember that DNA is bathed in water at 55 molar (~55 M). Water,
which we usually think of as innocuous, can cause hydrolytic damage.
47
Many mutations arise from chemicals in the environment
and from radiation. For example, ultraviolet radiation from the sun
damages pyrimidines. Consider this when you go to the beach!
Certain individual have a condition called Xeroderma pigmentosum in
which the system for repairing damage from ultraviolet light is defective.
Such individuals are hypersensitive to sunlight and readily contract skin
cancer.
Some mutations arise without external influences.
Remember that DNA is bathed in water at 55 molar (~55 M). Water,
which we usually think of as innocuous, can cause hydrolytic damage.
48
These water molecules are capable of inflicting hydrolytic
damage to bases, such as the deamination of cytosine as depicted
above. Replacement of the exocyclic amino group with a keto group as
a consequence of deamination converts cytosine to uracil.
Why is this mutagenic?
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Uracil has the same base pairing specificity as thymine; it
pairs with adenine. As we shall see, uracil occurs normally in RNA
where it pairs with thymine.
50
If left unrepaired, the U pairs with A during replication and
after a further round of replication the A pairs with T, resulting in
permanent replacement of the G:C base pair with a A:T base pair.
A repair enzyme prevents this from happening. Uracil is
not normally found in DNA. A repair enzyme detects uracil in the
genome. If it encounters one, it removes it and replaces it with cytosine.
51
Keep this statement, which I consider profound, in mind when you take
LS 1b in the spring.
52
The field of molecular biology, more than perhaps any of
the other natural sciences, has been driven by the invention of
techniques that have made it possible to make the discoveries that are
the subjects of these lectures. Many of the inventors of these
techniques have been recognized with Nobel Prizes. Sometimes the
invention of a powerful new method is more significant than a great
discovery because it facilitates the making of many great discoveries.
Examples are DNA sequencing, DNA cloning, and the Polymerase
Chain Reaction, which is the final topic for today. Undoubtedly many of
you have carried out PCR during a science project in high school. We
consider it now because it rests on the concepts that we have been
considering in today’s lecture.
Briefly put, the Polymerase Chain Reaction (PCR) makes
it possible to amplify a specific segment of DNA from a genome.
Suppose you are a forensic scientist and you want to determine
whether a biological sample comes from a particular suspect. PCR
enables you to make millions of exact copies of a particular sequence
that is known to contain a sequence that differs from one individual to
another. Or suppose you want to determine when Neanderthals
diverged from humans and you only have a minute sample of our
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ancient relative. Well, you can use PCR to amplify DNA from the specimen
and compare the sequence of that DNA to that of contemporary hominids.
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PCR involves the following steps:
1. We design and synthesize short DNA primers to sequences that
flank the region we wish to amplify.
2. We denature the genomic DNA by raising the temperature and the
cool the DNA allowing the primers (red) to anneal.
3. We then extend the primers (blue) with DNA polymerase.
4. We repeat the cycle multiple times with further rounds of
denaturation, annealing, and DNA synthesis.
5. As a consequence, the target sequence is amplified exponentially.
This is described more completely in the animation on the
web page (do play the animation!). Notice in the animation that in the
early cycles, the DNA polymerase overshoots the region of interest but
with further cycles the two primers ensure that only the region between
them is amplified. Thus the early products are diluted out by the
exponential increase in the desired product produced over multiple
cycles.
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Watch the PCR animation on the course web site.
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