In which we learn how to obtain precise control over the coding content of DNA
The principle of site-directed mutagenesis is that a mismatched oligonucleotide
is extended, incorporating the "mutation" into a strand of DNA that can
be cloned. In this lecture, I will present a number of current methods in use.
First, let's talk about the approaches in very general terms, because that will allow
us to organize the specific methods in our minds. When we talk about making a specific
mutation, let's call the molecule that we are starting with, the one without the
mutation, the "parent"
molecule. It might look like this:
The top strand is blue and the bottom strand is magenta. Let's be a bit more specific
and point to a specific nucleotide pair that we intend to change, as follows:
So let us say that in the parent molecule there is a GC base pair
that we want to change into an AT base pair. We want the mutated version to look
The first general approach
is to take the parent molecule and convert it to the mutated version by polymerase
chain reaction. The mutation is made by having a mismatch between the parental template
and one (or more) oligonucleotide primers. You can start with a very small amount
of the parent molecule, and by PCR make a tremendous amount of the mutated version,
so much in fact that the chances of cloning the mutated version from the product
is essentially 99.999%. There is always a bit of room for error, of course, so you
must carefully confirm your work by DNA sequencing.
To master this approach, we will need to learn a bit more about the design of primers.
When you start thinking about making primers that are slightly mismatched with the
template, you have to know what you can get away with and what you cannot.
Let's consider a very simple case, in which you want to make a
change in the end of a DNA molecule. Suppose you have a PCR fragment that looks like
this, where the dots indicate an extended sequence that is not shown:
The two oligonucleotides you used to make the fragment look like
After we had digested it with the two enzymes, the bits on the
ends would be lost and the product would be ready to clone:
Please note that we added a "GC" base pair to each end
to make the enzymes work better - that is a subject for a future lecture, so don't
worry about it just now. The important thing is that we managed to change the ends
of the DNA, just by adding a bit of sequence to the 5' ends of each oligonucleotide.
We can do more than append a sequence - we could also change the parental sequence
at the end without making the product any longer. Remember that our original oligonucleotide
Now let's have some fun, and make a change in the oligo on the
Both strands are affected, because the new version is simply copied
into its complementary nucleotides on the bottom strand. So you see, we can make
changes in the sequence that are internal.
Why did we extend the oligo on the left so that it was 21 nt in length? Well, we
wanted to be sure that it would anneal correctly to the template in the very first
cycle. If we introduce a mismatch, we want to be sure that there are an adequate
number of matching nucleotides at the 3' end of the primer (18 is a safe number,
in most cases). So you see that in this mutagenesis approach, the first annealing
would be imperfect, and the 5' end of the oligo on the left would be single stranded
for three nucleotides.
Of course, the mutation is copied into the product by extension from the oligo on
the right, so once the PCR reaction is underway, the annealing will be perfect over
the entire 21 nt of the primer:
So you see, it is fairly straightforward to change a DNA sequence if it can be covered
by an oligonucleotide during polymerase chain reaction.
Suppose you want to do something a bit more challenging - creating
a point mutation in the middle of a DNA sequence, at the position marked with an
"*" in the figure:
The ways of doing this in the old days were unspeakable, but now
we can simply get on the phone and order four oligonucleotides; two of which are
flanking and two of which cover and introduce the mutation into the amplified material:
We perform two PCR reactions to obtain the two halves of our final
product, and combine them in a third reaction, using the two "outside"
oligonucleotides to generate a chimeric product.
How does this happen? During the PCR process, the right side of
the first molecule can prime the synthesis from the left side of the second.
Now we can simply cut the PCR product with EcoRI and BamHI, and
drop it into the vector, in place of the original version. Or, we can continue to
manipulate the DNA by PCR.
To some extent, this is just using the end-based method we described in the first
place, but doing it twice, and then combining the results into a single product.
Here's a different approach, which would be appropriate if the
DNA template is circular, for example in a plasmid:
We may start with a circular plasmid, and use two oligonucleotides to change a small
region by PCR (see asterisk). The 5' ends of the oligonucleotides are shown not annealed
- they do not base pair because they are mutagenized. The two oligos are situated
in such a way that they re-copy the entire plasmid.
They point towards each other, but only going the "long way
around." That is like deciding to go to Los Angeles for the day, but instead
of heading down the 405 you go up the Pacific Coast Highway to Alaska, snowshoe over
to Denmark, hop a train to Capetown, boat over to Tierra del Fuego, and bicycle up
through South and Central America to Los Angeles! It's the long way to communte!
In this PCR example however, it makes sense because it means you don't need to combine
two pieces for cloning. What you obtain in the end is a linear fragment, suitable
for reclosure and cloning:
One comment however, is that the 5' ends of a PCR fragment are
exactly the 5' ends of the oligonucleotides. You will need to have a 5' phosphate
if you intend to use DNA ligase to reclose the circle, and so if your oligonucleotide
does not have a 5' phosphate (which would be typical) then you need to apply a phosphate
to each end using the enzyme T4 polynucleotide kinase.
One more thing - if you want this to work, you need to use very few template molecules
in the reaction, perhaps 1000. If you start with, let's say just to be really gross,
a nanogram of template, then you will have too much parental DNA lingering by during
your transformation. That is, you will find that very few of your transformed bacteria
actually have the mutation in the plasmid.
You see that the blue strand has been mutated and is now mismatched
with the magenta strand. At this point in the DNA, the mismatch would make a small
bubble of single-stranded DNA. What will the bacteria do with this? When the DNA
is replicated (typically as part of a plasmid), semiconservative replication will
cause two different daughter molecules. One looks just like the parent, and the other
has the mutation fixed in both strands:
That's great! Now half of the products will be mutated! Well, that
would be true if we could efficiently make the heteroduplex in the first place (which
is a bit of a dubious assumption), and if the heteroduplex could be transformed into
cells with the same high efficiency as the parental version (which may not be true,
since it has a bit of single-stranded DNA in it), and many other worries of a similar
nature. We need to have some way of disfavoring the parental version in this contest.
As we are making the mutation in one strand, we need to link that strand to some
persuasive form of selection, like this:
That is, we make the mutated (blue) strand a "happy strand"
in some way or another, and the parental (magenta) strand into an "unhappy strand".
By the strength of this linkage, we select for the mutated version by disfavoring
the parental version.
When using this approach, it is common to employ single stranded DNA as a template
(the magenta strand), because then you can simply apply a mutagenic oligonucleotide
and make a second strand. The second strand you make will not have a parental strand
with which to compete. Think of it as follows:
Start with a double-stranded plasmid with the parental sequence
Then, you make a single-stranded DNA containing just the inner
strand of the plasmid, which would look like this:
Just one of the parental strands
Then, apply an oligonucleotide that anneals to the single strand,
and carries a mutation. Extend the primer with a DNA polymerase such as Klenow fragment:
Synthesize a second strand, incorporating the mutagenic primer
Note that there is an A/C mismatch at the top of the figure. Once
extension is complete, it is double-stranded and might look like this. Of course,
the Klenow enzyme does not make the blue strand covalently closed. There will be
a "nick" where the synthesis ends, but don't worry about that - once this
is transformed into bacteria, the bacterial host will repair the nick.
Heteroduplex intermediate, ready to transform into bacteria
Remember once again, that the mutagenized strand will be copied
into half of the daughter molecules during replication.
So, that's just terrific, and I hear you wondering how we managed to get the single
stranded template with which we started the method. It is a minor digression...
As we learned previously, an origin of DNA replication is a required element for
ensuring plasmid maintenance. Origins of replication do come however, in several
different "colors and styles." Most commercial plasmids are based on the
ColE1 origin, a natural "high copy number origin" which fosters the accumulation
of several hundred copies of a plasmid per bacterium. It is also not uncommon to
find a second conditional origin of replication in some plasmids, derived from a
filamentous bacteriophage such as M13, fd, or f1. These origins of replication have
two important features:
Why would we want to make single-stranded DNA? One reason would
be to make a single stranded template for a sequencing reaction (a matter we will
discuss later in the course), or a single stranded DNA probe. Site directed mutagenesis
is sometimes facilitated by having a single-stranded plasmid to work with.
In any case, a slight digression to discuss the life-cycle of the
filamentous bacteriophage might be in order:
Those poor male bacteria! They have to contend with invading filamentous
phage - something that Rogaine just can't cure! We'll be talking more about male
and female bacteria in a later lecture ("sex" in bacteria isn't quite the
same concept as in eukaryotes).
What is significant here is that the virion of the filamentous phage
(i.e. the viral particle) carries a single-strand of DNA - not a double helix. In
the cell, this single-stranded genome (2.) is used as a template to synthesize a
double-stranded replicative form (RF), which is essentially a plasmid (3.). The replicative
form is used as a template to generate new single-stranded genomes (4.) that are
packaged into virions (5.) to generate new phage. The cell doesn't die - it just
grows more slowly and continues to secrete phage indefinitely.
The practical side of this story - if you use a cloning vector
that is based on a filamentous bacteriophage (such as M13mp18 which is an engineered
version of the phage M13) or merely contains an origin of replication from a filamentous
bacteriophage (such as f1), then you can induce single-stranded DNA replication and
collect the products in the form of secreted phage particles (which may be precipitated
from the growth medium with polyethylene glycol). In the case of a plasmid that only
contains an f1 origin of replication, and not the remaining genes from the phage,
it is necessary to infect the plasmid containing cell with a filamentous "helper
phage" that will activate the f1 origin of replication in the plasmid and foster
Suppose you had a parental DNA that had a unique restriction site
in it. If you mutagenized the restriction site at the same time that you made a mutation
in your gene of interest, then the parental strand would be sensitive to the enzyme
and the other strand containing the mutation would not.
The company Clontech has such a method, called the Transformer
Site-Directed Mutagenesis Kit. Digestion of the heteroduplex with the restriction
enzyme debilitates the parental strand, because it introduces a "nick".
The DNA can then be transformed into a bacterial strain. The efficiency can be increased
by extracting the pooled DNA from these cells and digesting a second time. This will
eliminate the products of replication in the bacteria that are purely parental (homoduplex),
and will spare the ones that are purely mutagenized (homoduplex). These plasmids
can then be reintroduced into bacteria, and most of the surviving plasmids should
be the mutagenized form.
Let me give you another example. There is a restriction enzyme named Dpn I that will
cleave the sequence GMeATC where MeA means that the adenylate
nucleotide is methylated. Dpn I will not cleave the unmethylated sequence GATC. We
can methylate such sequences in a plasmid by growing the plasmid in a "dam+"
strain of bacteria. Suppose then that we prepare a single stranded DNA template in
such a "dam+" strain. The parental strand would be methylated at every
GATC sequence (that is, approximately every 200 to 300 nt). When we apply an oligonucleotide
primer to this template and extend it using Klenow fragment, however, the new DNA
that is synthesized in vitro will be unmethylated. We therefore create a marked difference
between the parental strand (methylated) and the mutagenized strand (unmethylated).
Once we have completed synthesis of the mutagenized strand, what would happen if
we tried to digest the heteroduplex with Dpn I?
The answer is that the parental strand would be nicked (cleaved)
in many places, but the mutagenized strand would not. By putting this extra damage
into the parental strand, it is less favored during replication in the bacteria.
Here's another method, and this one involves taking advantage of
the enzyme that we discussed in the first lecture that removes uridylate nucleotides
from DNA (where they don't belong!)
How do we get a parental DNA that contains numerous uracil bases incorporated in
place of thymidine bases? The answer is that we grow the plasmid in a strain that
makes deoxyuridine triphosphate (a strain that is "dut-", meaning dUTPase
deficient) and does not surveil the DNA for uracil to excise (a strain that is "ung-",
meaning uracil N-glycosylase deficient). The bacteriologists usually don't say "minus"
and "plus" by the way - they would just call such a strain "dut,
ung", meaning that those two loci had mutated alleles.
So we make the single stranded parental DNA in a dut, ung strain of
bacteria, apply the mutagenic oligonucleotide in vitro, and extend it using the usual
DNA substrates and Klenow fragment. The newly synthesized DNA will not contain uracil
bases, because we did not use dUTP as one of the substrates - only dATP, dGTP, dCTP,
What would happen if we treated this DNA with uracil N-glycosylase? Well, the parental
strand would be shredded and the mutagenized strand would be untouched.
We don't actually have to add the enzyme ourselves - we could just
take the heteroduplex and transform a wild type bacterium with it - one that was
not dut, ung that is. The wild type bacteria would shred the parental
strand specifically, because its uracil N-glycosylase would find the U-rich DNA to
"Altered States" - sounds like something that ought to
have been invented by Kary Mullis, but you'll find it at Promega Inc.
Sound's easy? Here's a diagram that help to explain it.
Start with a double stranded plasmid containing your DNA insert, the sequence you
wish to mutagenize:
Note that there is a G/C base pair that we want to mutagenize to
an A/T base pair, in our dark blue sequence (the parental insert). Also, there is
a green sequence representing tetracycline resistance, and a red sequence that is
a defective version of the ampicillin resistance gene. Since the Amp gene is defective,
we will say that it is AmpS meaning "sensitive".
Now we prepare a single stranded version of the plasmid, perhaps by simply denaturing
them in alkali. We anneal THREE oligonucleotides to the circle: One to the DNA parental
insert, that causes the mutation in our gene of interest (from a G to an A in this
example), one to the tetracycline resistance gene that will debilitate it by the
introduction of a mutation, and one to the ampicillin "sensitive" gene
that will repair it by the introduction of a mutation.
These oligonucleotides are extended clockwise around the plasmid
using DNA polymerase Klenow fragment, so it looks like this:
Note that this heteroduplex has three points of mismatch, in three entirely different
places in the plasmid. The "inner strand" that contains the parental sequence
is unchanged, but the outer strand will contain the three alterations. What happens
when the bacteria replicates this?
The answer is that two types of products will appear. First, the replicative products
of the inner strand:
These will carry an intact tetracycline resistance gene and a nonfunctional
ampicillin resistance gene. The cells that inherit these plasmids will die in ampicillin.
On the other hand, the products of the outer strand:
These will carry a functional ampicillin resistance gene, a nonfunctional tetracycline
resistance gene, and more importantly, the G->A mutation in the DNA insert.
So, by transforming the synthetic product into E. coli and growing on ampicillin,
we favor the mutated strand. Here's an example from the Promega site, showing
the repair of the beta galactosidase gene
As you see, when the bacterial streaks are plated in ampicillin
and the colorimetric substrate X-gal (panel on left), you get most showing blue color
indicating repair of the gene. This is an indication of good concordance between
ampicillin resistance and beta galactosidase gene mutation (repair). You also see
that very few are tetracycline resistant (panel on right).
Now why should we want to debilitate the tetracycline gene? So that we can use the
method to make additional changes, and while we're doing that we will repair the
tetracycline gene and debilitate the ampicillin gene. That is, we can make a sequence
of changes in our insertion, toggling between ampicillin resistance and tetracycline
A piece of cake!
Department of Biology
California State University Northridge
Northridge CA 91330-8303
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2000, 2001, 2002