Download CHAPTER 13 DNA manipulation

13
CH AP TER
DNA manipulation
techniques
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Doudna, a Howard Hughes
Medical Institute Investigator
and Professor of Molecular and
Cell Biology and Chemistry
at the University of California,
Berkeley, and co-discoverer of
the revolutionary CRISPR-Cas9
gene-editing technology. In this
chapter, we will explore some of
the tools for manipulating DNA.
(Image courtesy of Dr Jennifer
Doudna.)
This chapter is designed to enable students to:
■ recognise that DNA can be manipulated in various ways, such as by cutting,
joining, and copying
■ identify various techniques used to manipulate DNA
■ distinguish between endonucleases and exonucleases
■ become familiar with the role of the polymerase chain reaction in producing
copies of DNA
■ develop awareness of the use of plasmids as vectors to transform bacterial
cells.
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FIGURE 13.1 Dr Jennifer
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A recipe for gene editing
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Charpentier, Director of the
new Max Planck Institute of
Infection Biology, co-inventor,
with Dr Jennifer Doudna, of
the CRISPR gene-editing
technology. (Image courtesy of
Hallbauer & Fioretti.)
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FIGURE 13.2 Dr Emmanuelle
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CRISPR stands for clustered
regularly interspaced short
palindromic repeats. These are
short repeated segments of DNA,
with each repeated segment being
separated by a length of spacer
DNA.
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The first attempts at genetic manipulation in the 1970s involved adding DNA to
the plant and animal genomes.
Early techniques used to add DNA to target cells included:
• the use of so-called ‘gene guns’ in which gold atoms coated with DNA were
blasted into cells, a technique used mainly to insert genes for pesticide or
herbicide resistance into plant cells
• the use of modified viruses to act as vectors to carry a functional gene into
cells with a defective gene. Viruses used as vectors include retroviruses and
adenoviruses.
These techniques were clearly very limited because there was no control
over where any added DNA would be inserted into the genome of the target
cells. Instead, DNA was simply added randomly into the genome. What would
be expected to happen if the added DNA was by chance inserted into a gene
that was essential for cell survival, causing the gene to become non-functional?
What would be expected to happen if the added DNA was by chance inserted
into a cancer-suppressing gene so that this gene was disabled?
Another major limitation was that, although these early techniques could
carry a functional replacement copy of a gene into cells with a defective gene,
these techniques were not able to mend a defective gene by editing or disabling it. Editing is a process of correction, such as occurs when an editor of a
manuscript corrects a misspelt word, or adds or replaces a word or phrase, or
deletes a sentence.
What was missing was an efficient and reliable technique for making precise
and targeted changes to the genome of living cells. Such a process is called
gene editing or genome editing or DNA editing. Gene editing refers to
changing a genomic DNA sequence in some way. This might involve making a
single base change to a gene, such as a base substitution, addition or deletion.
Gene editing on a larger scale might involve disabling, replacing or adding a
gene, or changing an upstream DNA sequence that regulates a gene. However,
in 2012, a ‘game-changing’ discovery was made.
In 2012, an article in the journal Science
(Jinek, M. et al. Science 337, 816–821) described a
gene-editing technique that is called CRISPR-Cas9
(often just shortened to CRISPR and pronounced
‘crisper’). Interestingly, CRISPR is part of the
adaptive immune system of bacteria that chops
any DNA of invading viruses, and has been operating in the microbial world for many thousands
of millions of years.
What was most significant was the recognition that CRISPR-Cas9 technology could
be readily adapted to provide an inexpensive and easy-to-use means of genome editing
for use in an endless range of genes from any
organism. The co-inventors of this exciting geneediting technique were two women scientists,
Dr Jennifer Doudna (refer back to figure 13.1) and
Dr Emmanuelle Charpentier (see figure 13.2).
This tool, borrowed and adapted from the bacterial adaptive immune system, is now being put
to use in editing faulty genes and in silencing
genes in plants and animals. Many applications
in agriculture are being explored (see chapter 15,
page 706). Research is already underway on the
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use of the CRISPR technology as a safe and reliable means of editing the defective alleles responsible for human diseases, such as cystic fibrosis and sickle-cell
anaemia.
How does CRISPR-Cas9 work?
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How does the CRISPR technology work?
The cell containing the DNA to be edited is transfected with the Cas9 nuclease
enzyme and the guide RNA. The guide RNA hybridises with its complementary DNA sequence and this identifies the position where the Cas9 nuclease
enzyme will cut the DNA. The cut is made by the Cas9 nuclease at a location
just upstream from a specific three-base sequence called PAM.
Genome specific
sgRNA sequence
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PAM = protospacer adjacent motif.
PAM is a three-base sequence
(NGG) in one strand of the DNA
(N stands for any base). The PAM
sequence is just downstream from
where the Cas9 nuclease makes
one of its two cuts in the DNA.
What gives the CRISPR-Cas9 technology its potential remarkable range of use?
Firstly, guide RNA molecules can be designed with a 20-base sequence that
can home in on any segment of genomic DNA from any organism. Secondly,
the Cas9 enzyme can cut any double-stranded DNA. This is a reminder of the
universality of DNA as the material of inheritance.
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Cas9 is an endonuclease
(endo = within), which can
cut double-stranded DNA
from within the DNA. This is
in contrast to exonucleases
(exo = outside), which can
only remove nucleotides from
the ends of DNA molecules.
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ODD FACT
The CRISPR gene-editing technique simply requires molecular ‘scissors’ to
cut the target DNA and a ‘guide’ to direct the ‘scissors’ to the site where the cuts
in the DNA will be made.
These two tools are as follows:
1. The ‘scissors’ are Cas9 nuclease, a bacterial enzyme that:
(i) can unwind double-stranded DNA
(ii) can cut both strands of double-stranded DNA at a precise location.
2. The ‘guide’ is a segment of an artificially synthesised single strand of RNA.
This guide RNA is designed to include a 20-base sequence that is complementary to part of the target DNA.
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The guide RNA is typically denoted
by sgRNA, where sg = synthetic
guide.
sg RNA
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Cas9 nuclease
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Genomic
DNA
PAM
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Site-specific
dsDNA break
Genomic
DNA
FIGURE 13.3 Stylised diagram showing the components of CRISPR-Cas9
technology. Note the single-stranded guide RNA (sgRNA) with its leading segment
(shown in orange) that is complementary to the DNA of interest. The sgRNA enables
the Cas9 nuclease enzyme (shown as the large blue shape) to be positioned to
cut both strands of the DNA close to the PAM site (shown in red). The two pairs of
scissors indicate the DNA sites cut by the Cas9 nuclease.
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After the Cas9 nuclease enzyme has cut both strands of the DNA of interest, the
breaks can be repaired and, in this process, two kinds of gene editing can occur.
1. In one type of repair process, called a gene ‘knock in’, a specially designed
DNA sequence is inserted into a precise location in the genome (see
figure 13.4). The aim of this type of repair is to edit faulty alleles and restore
their normal function. For example, using this process, the gene mutation
responsible for cystic fibrosis has been corrected in cultured stem cells
taken from individuals with cystic fibrosis.
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NGG
NCC
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Knocked-in sequence of interest
3ʹʹ
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FIGURE 13.4 Diagram showing a ‘knock in’ type of repair after the Cas9
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nuclease has produced a double-stranded break. The faulty gene is edited and
its normal function restored by inserting a specially designed segment of DNA
that corrects the defect. In this case, the defect was a missing segment of DNA.
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2. In the second type of repair, called a gene ‘knock out’, the re-joining that
repairs the break in the DNA involves a process that is subject to error. This
process can result in a random insertion or deletion (indel) of one or two
bases, producing a frameshift mutation. A frameshift mutation can either
disable a gene or can produce a STOP signal. So, the aim of this type of
repair is to disable or silence a gene (see figure 13.5).
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FIGURE 13.5 Diagram
showing a ‘knock out’ type
of repair in which the gene
function is disrupted either as
a result of a frameshift or as a
result of creating a premature
stop triplet.
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NGG
NCC
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Small deletion
What about the future?
CRISPR-Cas9 technology has extraordinary potential for widespread use in
clinical, agricultural, and research settings, for example:
• for a myriad of research purposes:
– to ‘knock out’ genes, one at a time, in order to identify their function
– to introduce specific mutations in a DNA sequence.
• to edit a faulty allele of a gene in a person with a severe inherited disease
• to snip out the faulty segment of a gene and replace it with a working copy
• to activate or to repress a gene
• to add a new gene to the genome.
This potential is highlighted by the fact that major pharmaceutical companies
have made significant investments in CRISPR-Cas9 technologies. For example, in
October 2015, Vertex Pharmaceuticals paid in excess of US$100 million to use the
gene-editing technology of CRISPR Therapeutics. Their intention is to develop
treatments for cystic fibrosis and sickle-cell diseases. More recently, Bayer
Pharmaceuticals and CRISPR Therapeutics set up a joint venture aimed at developing therapies to treat blood disorders, blindness and congenital heart disease.
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In a special review in the international journal Nature (25 May 2015), CRISPR
technology was introduced as follows:
Weblink
CRISPR
CRISPR: THE GOOD, THE BAD AND THE UNKNOWN
‘A DNA-editing technology called CRISPR has rapidly become one of the
most popular ways to alter genomes. Concerns about its risks temper
excitement about its usefulness. It has already been used to modify
human embryos, and the technology could alter wild animal populations; it works in everything from wheat to mice . . .’
CRISPR-Cas9 technology itself is neutral — it is a cut and paste or cut
and replace technique that can operate on the genomic DNA of a human
cell or a mouse cell or a mosquito cell or a wheat cell or a yeast cell, and
the list goes on . . .
What about the good, the bad and the unknown? Like any technology,
CRISPR will be identified as good or bad depending on the uses to which
techit is put and their immediate and longer term consequences. As the tech
nology is applied more widely, more unknowns will emerge.
In April 2015, a group of Chinese scientists reported the results of their
use of CRISPR technology on human embryos in an attempt to edit a
faulty gene. (The embryos came from an in-vitro clinic and, because each
was the product of fertilisation by two sperm, these embryos were not
capable of completing development.) The results of their experiments
appear to have raised many questions and few answers.
Issues being debated include: Should gene editing be allowed on human
genercells, such as eggs, sperm and embryos, whose DNA can pass to future gener
ations? Or, should gene editing be restricted to somatic cells where that DNA
is not passed to the next generation? These and related ethical issues will
continue to engage scientific communities, governments and the public.
In late 2015, the first International Summit on Human Gene Editing
was held in the United States in Washington, DC. The following headline
from a Canadian newspaper that reported on this meeting encapsulates
the situation: CRISPR gene-editing tool has scientists thrilled — but
nervous. (Source of headline: Crowe, Kelly 2015 ‘CRISPR gene-editing
tool has scientists thrilled — but nervous’, CBC News, 30 November.)
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Weblink
CRISPR in the media
CRISPR-Cas9 is one technique that involves the manipulation of DNA at the
level of genes or even the genome. In the following sections, we will explore
some of the other tools that can be used to manipulate DNA.
KEY IDEAS
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CRISPR-Cas9 technology is an efficient, easy-to-use, and inexpensive tool
for precise gene editing.
CRISPR-Cas9 technology can be applied to every eukaryotic species,
including humans.
The key components of CRISPR-Cas9 technology are a guide RNA and an
endonuclease enzyme, Cas9.
Early gene therapy techniques added DNA segments to genomes but at
unpredictable locations.
CRISPR-Cas9 technology enables precise genomic editing.
CRISPR technology can not only add DNA but can also edit, modify or
disable or delete DNA from the genome.
Like any new technology with so many potential applications, aspects of
the use of CRISPR technology raise ethical considerations.
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QUICK CHECK
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Tools to manipulate DNA
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1 Identify one capability provided by CRISPR-Cas9 technology that was not
possible with earlier gene therapy techniques.
2 Identify the following statements as true or false:
a CRISPR technology can only be applied to certain species.
b When CRISPR is used to add DNA to a genome, it is added in a random
location.
c CRISPR can be used to correct faulty genes.
d CRISPR can be used to silence a gene.
e The scissors in CRISPR technology consist of a bacterial endonuclease
enzyme.
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Carpenters work with wood and sculptors work with clay. Scientists who
use gene manipulation technology are sometimes called ‘genetic engineers’.
Genetic engineers work with the genetic material DNA.
Each operator — carpenter, sculptor or genetic engineer — requires tools
to manipulate the raw material. To construct a bookcase, a carpenter needs
tools such as a ruler, saw and hammer so that the wood may be measured, cut
and joined. Similarly, the genetic engineer requires tools to cut, join, copy and
separate DNA. The following section describes the tools and techniques that,
when combined, enable genes to be manipulated in various ways.
Table 13.1 shows some of the ‘tools’ used in gene manipulation. The same
tools can be used regardless of the source of the genetic material.
Action
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TABLE 13.1 Tools
ools for gene manipulation.
• Separate fragments by size
electrophoresis (see Sorting DNA fragments)
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restriction enzymes (see Cutting DNA into
fragments)
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‘Tool’
• Cut DNA into fragments at precise
locations
• Find particular DNA fragments
probes (see Finding DNA fragments of
interest)
• Join DNA fragments
ligase enzyme (see Joining DNA fragments)
• Make specific DNA fragments
DNA synthesiser & reverse transcriptase
(see Making DNA)
• Amplify DNA
Polymerase chain reaction (PCR)
(see Amplifying DNA)
Cutting DNA into fragments
One class of enzymes can cut double-stranded DNA into a reproducible
number of fragments. These enzymes are called restriction enzymes or
cutting enzymes. These cutting enzymes occur naturally in microbes, mainly in
bacteria, but also in some archaea.
The main feature of cutting enzymes is that they do not snip the two strands
of a DNA molecule at random. Each cutting enzyme snips DNA at certain positions only, and is restricted to cutting at specific locations only. (This is why
enzymes in this class are called restriction enzymes.)
Each cutting enzyme has its own name. These are strange-looking names
such as Bgl I, Hap II, and Taq I, which are pronounced as ‘Buggle one’, Hap
two’ and ‘Tack one’. The box below explains how the names of cutting enzymes
are organised.
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ODD FACT
The first restriction enzyme was
discovered at Johns Hopkins
Medical School in 1970.
What is significant about restriction enzymes is that their actions on DNA
are predictable as they cut DNA only at their specific restriction sites, and they
act in the same way on every occasion. Before the first restriction enzyme was
discovered in 1970, it was not possible to cut DNA into fragments in predictable and reproducible ways.
ECO RI — WHAT’S IN A NAME?
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The names of cutting (restriction) enzymes look very strange indeed . . .
Hap II, Eco RI, Bgl II, Not I, Pst I Hin dIII
The name of a restriction enzyme has two parts. The first part is a set
of three letters, such as Eco or Bgl or Pst. The three letters are derived
from the scientific name of the microbe in which the restriction enzyme
occurs.
The first letter is the first letter of the genus and the next two letters are
the leading letters of the trivial (specific) name:
• Eco is a restriction enzyme found in Escherichia coli
coli.
• Bgl is a restriction enzyme found in Bacillus globigii.
globigii
• Hin is a restriction enzyme found in the Haemophilus influenza.
The second part of the name is a Roman numeral such as I or II or
III. What does this mean? A particular species of microbe may contain
several restriction enzymes. Each different restriction enzyme from the
same microbe is given a Roman numeral as follows: I for the first enzyme
isolated from that species, II for the second, and so on.
In some cases an additional letter is added, such as in Hin dII, and the
letter ‘d’ denotes a particular strain or type (serotype) of the bacterium.
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The position where a cutting enzyme can snip double-stranded DNA is the
recognition sequence of that restriction enzyme. A restriction site is a particular order of nucleotides (see figure 13.6). Some restriction enzymes cut
the two strands of a DNA molecule at points directly opposite each other to
produce cut ends that are ‘blunt’. Other cutting enzymes cut one strand at one
point, but cut the second strand at a point that is not directly opposite. The
overhanging cut ends made by these cutting enzymes are called ‘sticky’ or
‘protruding’. These sticky ends are complementary.
Cutting
enzyme
Aha I
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The expected distance between
cut sites of a restriction enzyme
is given by the formula 4n, where
n is the number of bps in the
restriction site.
TTT AAA
AAA TTT
Result of
action
TTT
AAA
Aha I
AAA
TTT
Hae III
FIGURE 13.6 Cutting by
Aha I, Hae III and Hin dIII.
Which restriction enzymes
produce ‘sticky ends’? Are the
sticky ends produced by Hin
dIII complementary?
Recognition
site
Hae II
Hin dIII
GGCC
CCGG
AAGCTT
TTCGAA
GG
CC
CC
GG
Hin dIII
A
AGCTT
TTCGA
A
The number of nucleotide base pairs (bps) in a restriction site varies from
four to about eight base pairs (bps). This length determines how frequently
a restriction enzyme is likely to cut a random sequence of DNA, or in other
words, the average distance between cuts. For example, restriction enzymes
with a 4-bp recognition sequence, such as Hae III, are expected on average to
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cut about every 256 bps along a sequence. In contrast, a restriction enzyme
with a 6-bp recognition sequence, such as Aha I, will on average cut about
every 4096 bps.
Several thousand restriction enzymes have now been isolated, mainly from
bacteria, but some from archaea. About 300 restriction enzymes are available from commercial suppliers. Genetic engineers simply buy a quantity
of the required enzyme in highly purified form from a commercial supplier
(see figure 13.7). Table 13.2 below shows some commonly used restriction
enzymes.
Alu I
Arthrobacter luteus
Bam HI
Bacillus amyloliquefaciens H
GˇGATC C
C CTAGˆG
Bcl I
Bacillus caldolyticus
TˇGATC A
A CTAGˆT
Bgl II
Bacillus globigii
Eco RI
Escherichia coli R factor
GˇAATT C
C TTAAˆG
Hae III
Hemophilus aegyptus
GGˇCC
CCˆGG
Hemophilus influenzae Rd
AˇAGCT T
T TCGAˆA
Kpn I
Klebsiella pneumoniae
G GTACˇC
CˆCATG G
Not I
Norcadia otitidis-caviarum
GCˇGGCC GC
CG CCGGˆCG
Pst I
Providencia stuartii
C TGCAˇG
GˆACGT C
Sac I
Streptomyces achromogenes
GAGˇCTC
CTCˆGAG
Taq I
Thermus aquaticus
TˇCG A
A GCˆT
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Restriction
enzymes
Summary
screen and
practice questions
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AOS 2
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Concept 1
Unit 4
AOS 2
See more
Understanding
restriction enzymes
Topic 1
Concept 1
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Hin dIII
Unit 4
Recognition site
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restriction enzyme, EcoRI.
(Reprinted from www.neb.com
(2016) with permission from
New England Biolabs, Inc.)
Enzyme
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FIGURE 13.7 Vial of
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TABLE 13.2 Restriction enzymes in common use and their recognition sites.
The arrowheads indicate the site of the cut in each strand of the double-stranded
DNA molecule.
AˇGATC T
T CTAGˆA
For cutting, a sample of DNA is dissolved and the particular restriction enzyme is added. Provided recognition sites are present in the DNA
sample, the DNA molecules will be cut into two or more fragments. The
lengths of the fragments depend on the relative positions of the recognition
sites (see figure 13.8). In some cases, two different restriction enzymes may
be added to a DNA sample. How, if at all, would the resulting fragments be
expected to change when two restriction enzymes are used, as compared to
just one?
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FIGURE 13.8 Each cutting
enzyme snips DNA at a
particular site. Would a
different cutting enzyme,
say Aha I, also be expected
to produce three fragments
of the same sizes as those
produced by Bgl II? (Note that
kbp = kilobase pairs.)
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740 kbp
OK, so now we have a solution containing DNA that has been cut into a number
of predictable fragments using a restriction enzyme. How can we sort out this
mixture of DNA fragments? Sorting — no problem: use electrophoresis.
DNA fragments can be sorted using the technique of electrophoresis.
Electrophoresis sorts DNA fragments according to their lengths, that is, their
sizes.
To separate a mixture of dissolved DNA fragments, the mixture is placed in
a slot at one end of a slab of jelly-like supporting material, known as a gel, that
is immersed in a buffer solution. The gel, which is made of agarose, is then
exposed to an electric field, with the positive (+)) pole at the far end and the
negative (−) pole at the starting end or origin.
Look at figure 13.9, which shows a researcher using a micropipette to load
a sample of DNA into one slot at the cathode end of a gel that is covered by a
buffer solution. Typically eight to ten different samples of DNA can be loaded
onto one gel, with each sample being loaded into a different slot. If you look
carefully, you will see blue colouring, which indicates that DNA samples have
been loaded into almost all the slots. The colour is added simply to assist the
researcher to load the DNA samples into the gel slots.
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DNA
Sorting DNA fragments
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Bgl II
Bgl II
FIGURE 13.9 Apparatus for separating DNA fragments by electrophoresis. Notice
the red (+) anode at the far end and the black (−) cathode at the origin where a
DNA sample is being loaded into the last slot in the gel.
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–
Fragments
loaded
onto gel
Electric
field
applied
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DNA
fragments
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+
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FIGURE 13.10 Diagram
showing the result of
electrophoresis of a mixture
of DNA fragments loaded
into one slot on a gel. The
fragments are separated
according to their sizes. Which
fragments move furthest from
the origin — the longest or the
shortest?
All DNA fragments have a net negative charge, because of the phosphate
groups in their sugar–phosphate backbone. As a result, DNA fragments will
move towards the positive pole when placed in an electric field. The shortest
DNA fragments move most quickly and the longest fragments move most
slowly. Fragments of the same size move at the same rate. The end result of
electrophoresis is a series of parallel bands of DNA fragments at differing distances down the gel. (see figure 13.10). Each band can contain thousands or
millions of DNA molecules of the same size.
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FIGURE 13.11 Gel stained
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DNA fragments are invisible so, after the gel run is complete, the separated
DNA bands must be made visible either through the use of a dye or a labelled
probe. One technique makes use of the dye ethidium bromide (EtBr), which
binds to the major groove of DNA molecules. When illuminated by ultraviolet
light, the DNA bound to EtBr fluoresces pale pink (see figure 13.11).
A standard ladder of DNA fragments of known sizes is usually run through
the gel at the same time as the unknown DNA samples. The sizes of unknown
DNA fragments can be approximated by comparing the positions of their
bands with those of the known standards. Figure 13.11 shows a ladder of
DNA markers of known sizes in lane 1, at the left-hand side of the gel.
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with ethidium bromide after
completion of electrophoresis.
This image shows DNA bound
to EtBr in bands revealed
by pink fluorescence under
ultraviolet illumination. Lane 1
(at left) contains DNA markers
of known sizes and lanes 2
through 8 contain DNA
fragments of unknown sizes.
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Interpreting gel runs
Let’s look at examples of electrophoresis in action:
Example 1: One of the main reasons for using electrophoresis to separate DNA
fragments is to obtain estimates of the size of the DNA fragments in the mixture.
Remember that electrophoresis separates DNA fragments according to their
sizes. DNA fragments of unknown sizes can be separated by electrophoresis and
then their sizes can be approximated by comparing them with DNA molecules of
known size separated on the same gel and under the same conditions.
Examine figure 13.12, which shows a gel run of a mixture of DNA fragments of
unknown sizes from the RH slot, and samples of DNA markers of known size from
the LH slot. Note that the unknown sample of DNA has resolved into four bands.
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The ladder of marker DNAs of known sizes shows values ranging from 23 130
base pairs to 564 base pairs. Estimate the size of the DNA fragments that have
separated into the four bands. Remember that this is just an approximation.
Known
markers
Unknown
Known
fragment
sizes
Estimated
fragment
sizes
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9416
6682
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2322
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FIGURE 13.12 Diagram showing the separation of DNA fragments of unknown
sizes on a gel (at RH side) alongside a separation of DNA markers of known sizes
(at LH side). This technique is used to obtain estimates of the sizes of unknown
DNA fragments in a band by comparing their positions with those of bands of
DNA markers of known sizes.
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Example 2: Samples of the DNA of the F8C gene, which controls blood clotting, were collected from several different people, two males (J and T) and one
female (K).
The F8C gene is located on the X chromosome and has two alleles: the C
allele that determines normal blood clotting, and the c allele that determines
defective blood clotting (haemophilia).
The DNA of each sample was treated with the restriction enzyme Bcl I.
The result of Bcl I treatment is shown in figure 13.13a.
The DNA samples were then subjected to electrophoresis and the positions
of the bands were identified using a fluorescent probe. The result is shown in
figure 13.13b.
Let’s consider the results. Remember that the only bands visible are those
labelled by the fluorescent probe. Remember that females have two X chromosomes, but males have only a single X chromosome.
You can estimate the sizes of the fragments by comparing their positions
with those of the known standards.
How many fragments would result from Bcl I digestion of the C allele?
(Answer: 1)
How many fragments would be produced by Bcl I treatment of the c allele?
(Answer: 2)
What are their expected sizes? (Answer: 879 and 286 bp)
What has happened to the smaller 286 bp fragment? (Answer: The fragment
is there, but it is not visible because the probe does not pair with it.)
Gel
electrophoresis
Summary
screen and
practice questions
C
Unit 4
AOS 2
N
Topic 1
U
Concept 3
Unit 4
AOS 2
See more
Gel electrophoresis
Topic 1
Concept 3
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Note that person K, a female, has both the C allele and c allele, as shown by
the presence of two fragments.
Would all females be expected to show this pattern? (No. Most females would
be expected to be genotype CC and show just only one band, the 879 band.)
Person J, a male, has the C allele only. However, person T, a male, shows
the presence of the relatively shorter c allele and so has the X-linked disorder,
haemophilia.
(a)
1165 bp
FS
DNA from C allele
O
Probe
879 bp
O
DNA from c allele
PR
Probe
(b)
Lane 2
Lane 3
PA
G
E
Lane 1
(a)
3.0
2.0
1.0
0.5 kb
Person K
+
FIGURE 13.13 (a) Diagram showing the DNA of the two alleles of the
F8C gene and their cut sites for the restriction enzyme Bcl I. Note that the
C allele has two restriction sites, while the c allele has three restriction sites.
(b) Result of electrophoretic run of the three different DNA samples.
R
EC
(b)
DNA
fragment
DNA
fragment
U
N
C
O
R
Ligase
FIGURE 13.14 In both (a) and
(b), the DNA fragments have
sticky ends created with the
same cutting enzyme. Ligase
enzyme forms strong covalent
bonds so that the fragments
are joined permanently.
Why is the result a circular
DNA molecule in (b)?
622
D
DNA
fragment
Person J
–
TE
DNA
fragment Ligase
Person T
Lane 4
Joining DNA fragments
In gene manipulation, the joining of pieces of DNA is sometimes required. An
enzyme known as ligase catalyses the joining of pieces of double-stranded
DNA at their sugar–phosphate backbones. The bonds that form in this case
are strong (covalent) bonds. The joining can produce one longer piece of DNA
(see figure 13.14a) or it can produce a circular molecule of DNA if the fragments have complementary ‘sticky ends’ at both ends (see figure 13.14b).
KEY IDEAS
■
■
■
■
■
DNA can be manipulated in various ways, including by being cut into
fragments, and DNA fragments can be sorted and can be joined.
Cutting or restriction enzymes cut DNA into fragments at specific
recognition sites.
Cuts of double-stranded DNA by some restriction enzymes produce ‘sticky’
ends, and some produce blunt ends.
Electrophoresis sorts DNA fragments according to their lengths, measured
in base pairs.
DNA fragments can be joined through the action of the enzyme ligase.
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QUICK CHECK
O
O
FS
3 Identify the following statements as true or false:
a Restriction enzymes cut DNA molecules at random sites along their length.
b The restriction enzyme Aha I produces DNA fragments with sticky ends.
c Electrophoresis separates DNA fragments according to their different
charges.
d In electrophoresis, DNA fragments travel from slots at the origin of a gel
to the cathode.
e A restriction enzyme that can act on human DNA would not be able to
act on mouse DNA.
f In a long DNA sequence, it is reasonable to predict that there would be
more cut sites for Hae III than for Hin dIII.
g In electrophoresis, shorter fragments of DNA travel farther than longer
fragments.
PR
More tools for manipulating DNA
E
Let’s now identify other tools and procedures that are used in DNA
manipulation.
G
Finding DNA fragments of interest
U
N
C
O
R
R
EC
TE
D
PA
Electrophoresis can separate a mixture of DNA fragments according to differences in their sizes, but how can one particular set of DNA fragments be
picked out from a large number of DNA fragments? This is a bit like finding a
needle in a haystack. However, this can be done using a probe.
You might use a person’s dog to locate that person in a crowd. In this case,
the person is the target object and the dog is the ‘probe’ that locates the target.
You could use a Geiger counter to find the source of a radioactive emission. In
this case the probe is the Geiger counter and the target that it will locate is the
source of a radioactive emission.
In both cases, the probe will
home in on the target. Follow
the probe and you have found
your target! Likewise, as people
pass through security checks
at airports, both their bodies
and their luggage are ‘probed’
by scanners that detect targets,
including prohibited metal
objects, such as knives or guns.
In this case, the ‘probe’ does not
physically attach to the target
but identifies it on a screen
display (see figure 13.15).
Likewise, a molecular probe
can locate particular target DNA
fragments from among a mixture
of many DNA fragments. This
kind of probe is commonly a
piece of single-stranded nucleic
acid, either DNA or RNA, with
a base sequence that is comFIGURE 13.15 The luggage scanner is like a probe that can home in on particular
plementary to part of the base
prohibited targets without making physical contact with the targets.
sequence of one of the strands
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of the target DNA. The key to the use of probes is their ability to hybridise with
the complementary sequences on their single-stranded target DNA.
To locate the DNA fragments of interest, the target DNA must first be denatured to separate its two strands, otherwise the probe cannot home in and pair
with it. Figure 13.16 shows a probe and part of the target it can locate. Because
the probe is complementary to part of the target, it is able to pair with it.
Probe:
ATTAGGCGGCTGG
FS
The DNA target is dissociated into single strands:
……ACATTTATTAGGCGGCTGGTTACC……
PR
ATTAGGCGGCTGG
……TGTAAATAATCCGCCGACCAATGG……
O
A probe is added and pairs with the complementary region of the target:
FIGURE 13.16 The base sequence of the probe is complementary to part of its
target. A probe carries a radioactive or fluorescent label, represented by the flag,
so that it (and its target) can be located.
G
E
When in situ hybridisation
is carried out using a probe
with a fluorescent label, the
technique is referred to as
FISH (fluorescence in situ
hybridisation).
O
……TGTAAATAATCCGCCGACCAATGG……
ODD FACT
U
N
C
O
R
R
EC
TE
D
PA
A probe must be labelled in some way so that it
can be easily located. This label is typically a fluorescent marker or it may be a radioactive marker.
The position of the probe and the target that it has
latched onto are signalled by the fluorescent label
on the probe.
Probes can be used to locate specific DNA segments, either within a cell or outside cells, such as
on a gel.
Figure 13.17 shows the chromosomes of a
human cell that have been exposed to a probe that
binds specifically to a unique sequence of DNA in
part of the human Y chromosome. This probe has
a label that fluoresces bright yellow when exposed
to ultraviolet light. Notice the two bright yellow
spots that indicate that the probe has hybridised
at two sites, signalling the presence of two Y chromosomes in the cell. This process is known as
in situ hybridisation. The term in situ means ‘in
place’ and identifies the fact that the DNA is in its
normal place in the chromosomes. In fact, these
cells came from a male who has 47 chromosomes,
including one X and two Y chromosomes, in each
of his somatic cells.
FIGURE 13.17 A probe that fluoresces bright
yellow has been mixed with chromosomes. The
probe is specific for the Y chromosome. At how
many sites has the Y-specific probe become
bound? Notice that the second cell (upper right)
is at the interphase stage of the cell cycle. (Image
courtesy of Dr Arabella Smith.)
624
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PR
O
O
FS
Probes can also be used to locate a particular DNA fragment from a mixture
of fragments that has been sorted by electrophoresis. Figure 13.18a outlines the
procedure called ‘Southern blotting’ in which the DNA fragments are transferred from the gel to a nylon membrane.
The steps in this procedure are as follows:
• The gel is treated with a denaturing agent to make the DNA single stranded.
• The gel is transferred to a salt solution, and a nylon membrane is placed over
the gel.
• Layers of absorbent paper, weighted on top, are placed over the membrane,
and the salt solution is drawn through the gel, carrying the DNA that is
absorbed by the nylon membrane (see figure 13.18b for details). This procedure typically lasts for about 18 hours and, by the end of this time, all the
DNA from the gel has been transferred to the nylon membrane.
• The membrane is then soaked in a solution containing a labelled probe that
consists of a single-stranded nucleic acid probe that binds specifically to the
intended DNA target.
• The probe, and hence the DNA of interest, is located, either by use of X-ray
film in the case of a radioactively labelled probe, or by fluorescence in the
case of a probe with a fluorescent label.
Membrane
X-ray film
N
C
O
R
(b)
U
DNA made single
stranded and
fixed in place on
membrane.
TE
DNA fragments
transferred by
blotting to
membrane.
R
EC
DNA
fragments
on gel after
electrophoresis.
D
PA
G
E
(a)
Labelled probe
added in solution.
Probe binds to
specific target DNA.
DNA–probe
hybridisation
has
occurred.
Probe seen by
darkening of film
(or by fluorescence
detection).
Weight
Paper towel stack
Stack of filter paper
Membrane
Gel
Filter paper
Plastic wrap
Platform
Transfer buffer
FIGURE 13.18 (a) Steps in the use
of a probe to locate a particular
DNA target after electrophoresis.
(b) Details of arrangement for
the transfer of denatured (singlestranded) DNA from the gel to a
nylon or nitrocellulose membrane.
Making DNA fragments
It is easy to isolate the total DNA from human somatic cells. This, however,
consists of the DNA of two sets of 23 chromosomes, each set containing about
3000 million base pairs encoding about 21 000 genes, as well as very long
sequences of noncoding DNA. Some scientists may want short lengths of
DNA to act as probes. Other scientists might want copies of just one specific
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DNA base sequence as part of their research. How can particular segments
of DNA be obtained for these various purposes?
Two methods for making specific DNA fragments are outlined below:
1. Synthesise DNA from nucleotide building blocks. This method makes use of
an instrument called a DNA synthesiser (see figure 13.19 for one example).
To use this method, the base sequence of the required DNA must be known.
Instruments called DNA synthesisers can join nucleotide sub-units in a
predefined order to produce DNA segments with lengths greater than 100
bases. The chemical synthesis of DNA does not require a template strand
nor does it require the enzyme DNA polymerase. The products can be used
as primers or as probes.
PA
G
E
PR
O
O
FS
eLesson
Somatic cell nuclear transfer
eles-2465
TE
FIGURE 13.19 A scientist working in a DNA laboratory. DNA synthesisers can
generate primers and probes as well as lengths of DNA up to about 100 bps.
DNA synthesisers work in conjunction with a computer and DNA synthesis
software.
U
N
C
O
R
R
EC
In 1972, Khorana and his
team reported the first gene
to be artificially synthesised;
this was the gene that
encodes the information for a
transfer RNA (tRNA) in yeast.
In 1977, the first proteincoding gene was artificially
synthesised.
D
ODD FACT
The name of the enzyme reverse
transcriptase describes its action.
‘Reverse’ and ‘transcriptase’ refer
to the fact that the process is the
exact opposite of transcription —
normally in cells, DNA directs the
manufacture of RNA.
626
2. Make a copy of DNA using an mRNA template. To do this, mRNA is isolated from the specific cells in which the gene concerned is active. The
enzyme reverse transcriptase uses the mRNA as a template to build a
single-stranded DNA with a complementary base sequence. This DNA is
known as complementary DNA (cDNA). The reverse transcriptase procedure has been used successfully to make copies of the human growth
hormone gene and also the gene for tissue plasminogen activator (t-PA), a
blood clot‒dissolving enzyme used in the treatment of some forms of heart
attack.
Figure 13.20a outlines a simplified version of the process to produce
single-stranded cDNA using reverse transcriptase enzyme and an mRNA template. Commercial kits are available for the use of reverse transcriptase to
produce cDNA (see figure 13.20b).
To make this DNA double stranded, the enzyme DNA polymerase is then
used. Note that this method will work only for coding DNA: that is, genes that
produce mRNA. This method cannot be used for the bulk of the genome that
consists of noncoding DNA.
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(a)
(b)
mRNA isolated from
cystol of specific cells.
mRNA
Poly-A tail added to this mRNA. (This
segment provides an anchor to which a
‘primer’ can attach.)
AAAAAA
Oligo-dT primer added and
binds to poly-A mRNA tail. Reverse
transcriptase enzyme added. DNA
lengthens by addition of nucleotides. The
order is controlled by the sequence in the
mRNA.
FIGURE 13.20 (a) Simplified
FS
Direction of growth
O
TTTT
ss DNA
TTTT
AA AA
Final double-stranded DNA product
(cDNA).
E
TT TT
PR
Direction of growth
Polymerase enzyme added.
This catalyses the building of
complementary DNA strand.
G
ds DNA
version of the steps involved in
making complementary DNA (cDNA)
using reverse transcriptase enzyme
and an mRNA template (ss = single
stranded). Note that the end product
is single-stranded DNA. How might
it be made double stranded?
(b) One of the commercial kits
(b) One
that provides everything needed to
reverse transcribe RNA to singlestranded cDNA. (Image (b) courtesy
of Thermo Fisher Scientific.)
O
When DNA chain is complete,
mRNA is removed by alkali treatment.
AAAAAA
TTTT
PA
Amplifying DNA fragments
U
N
C
O
R
R
EC
TE
D
In forensic investigations, critical evidence can come from trace amounts of
DNA obtained from a single root hair, a licked stamp, a dried blood stain, a
discarded cigarette butt, a piece of chewing gum, drinking glasses, a soiled
tissue or a trace of semen. For medical purposes, a few fetal cells obtained
from amniotic fluid may be the source of DNA that can reveal if a fetus has an
inherited disease. These DNA traces become useful when they are amplified to
millions of copies using the polymerase chain reaction (PCR).
Figure 13.21 shows an instrument that automates the PCR procedure.
Minute quantities of DNA provide the starting point for PCR. From a few
starting copies of DNA, millions of copies can be quickly produced.
FIGURE 13.21 A Veriti®
96-Well Fast Thermal Cycler,
which can automatically take
up to 96 samples through a
programmed polymerase chain
reaction. (Image courtesy of
Thermo Fisher Scientific.)
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ODD FACT
Tiny bone and tissue
fragments on ancient stone
tools are a source of DNA.
This DNA has been amplified
using PCR to identify the
species from which it came.
Denature
2
FS
Bind
primers
The polymerase chain reaction (see
figure 13.22) depends on the enzyme
DNA polymerase to amplify or make
multiple copies of a sample of DNA. The
polymerase enzyme comes from a bacterial species (Thermus aquaticus) that
lives in hot springs. It is known as the Taq
polymerase.
The procedure is carried out in a test
tube (in vitro) as follows:
1. Denature: The DNA sample is denatured by heating so that it dissociates to
single strands (94 °C
C for two minutes).
2. Bind primers: Short segments of
single-stranded DNA, known as primers,
are added; these primers pair with
regions at either end of the DNA region of
interest (55 °°C
C for two minutes).
3. Extend primers: The polymerase
enzyme uses the primers as a starting
point and extends them so that two
complete double strands are formed.
For this step, a supply of nucleotides must be available (72 °C for
one minute).
About five minutes is required for each
cycle of ‘denature — bind primers —
extend primers’. Over time, an exponential
growth of the DNA from the two original
strands occurs: 4, 8, 16, 32 . . . , so that
after 20 cycles about one million copies of
the DNA exist.
The amplified DNA produced in a
few hours by PCR is enough to provide
material for a DNA profile in a forensic
investigation, or to identify whether an
embryo or fetus has inherited a genotype that causes a serious disorder, such
as sickle-cell anaemia. In contrast, other
prenatal procedures require several days
or longer to obtain enough DNA by cell
reproduction.
4
Kerry Mullis shared the 1993
Nobel Prize in Chemistry for
his development of the use of
the heat-tolerant enzyme Taq
polymerase to make multiple
copies of genes via PCR.
PR
ODD FACT
E
CYCLE 1
G
Denature
PA
4
D
Bind
primers
O
R
R
EC
TE
FIGURE 13.22 Polymerase
chain reaction (PCR) was
developed in 1984. Primers
are short single-stranded
sequences of DNA (20–25 bp)
complementary to regions at
the boundaries of the DNA to
be amplified. Notice that
four DNA strands are present
after one cycle, and eight
strands after two cycles. How
many copies would be present
after eight cycles?
C
N
U
Extend
primers
8
CYCLE 2
KEY IDEAS
■
■
■
■
■
628
O
O
Extend
primers
Probes typically consist of short single-stranded nucleic acids, either DNA
or RNA, that can locate target DNA through complementary base pairing.
Probes carry a label that allows them (and their targets) to be located
either inside or outside cells.
DNA fragments can be artificially made using a DNA synthesiser that
requires a known DNA base sequence.
Reverse transcriptase enzyme can use an mRNA template to build a single
strand of complementary DNA (cDNA).
The polymerase chain reaction (PCR) can amplify minute amounts of DNA.
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QUICK CHECK
DNA amplification
Summary
screen and
practice questions
Unit 4
AOS 2
Topic 1
Concept 2
4 Identify the following statements as true or false:
a Target DNA must be denatured before it can be located with a probe.
b The base sequence of a probe is complementary to part of the base
sequence of its DNA target.
c The reverse transcriptase process can be used to produce cDNA copies
of noncoding DNA.
d A DNA synthesiser does not require enzymes or a template strand.
5 What is meant by the phrase ‘denature — bind primers — extend primers’?
Concept 2
O
See more
DNA amplification
Topic 1
In this section we will examine how DNA can be transported into bacterial
cells resulting in their transformation, and see how recombinant plasmids are
involved in this story.
PR
Transporting DNA into cells
O
AOS 2
FS
Transforming bacteria
Unit 4
PA
G
E
Suppose that a team of genetic engineers wants to insert copies of a DNA fragment isolated from plant cells into bacterial cells, or they wish to insert copies
of a human gene into bacterial cells. Special carriers or vectors can be used to
transport DNA from the external source into the bacterial cells. We can give the
label of ‘foreign DNA’ to any DNA that is transported from the external environment into cells using a vector. Very often, in the laboratory setting, plasmids
are the vectors used to transport foreign DNA into bacterial cells.
Main
chromosome
R
EC
Plasmid
TE
D
Plasmids: vectors of DNA
What are plasmids? In addition to their main chromosome, most bacterial
cells contain one or more plasmids. Plasmids are small circular pieces of
double-stranded DNA ranging in length from about 2000 to 10 000 base pairs
(see figure 13.23) found mainly in bacterial cells. All plasmids are self-replicating
because they contain an origin of replication (ORI) that is a specific DNA
base sequence where DNA replication of a plasmid begins. As a result, plasmids can replicate independently of the main bacterial chromosome.
In addition to naturally occurring plasmids, plasmids can also be constructed
so that they include certain features in addition to an origin of replication.
For example, plasmids that are used to transfer DNA include several features,
including:
1. genes for one or more traits, including a selectable marker gene
The marker gene is often a gene governing antibiotic resistance, with the particular form of the gene being the allele for resistance to an antibiotic, such as
ampicillin (AmpR). Bacteria containing the plasmid with the marker can be
selected from among other bacteria by exposure to the antibiotic. Bacteria that
have taken up the plasmid with the marker are unaffected by the antibiotic, but
bacteria lacking that particular plasmid will be killed by the antibiotic.
2. a multiple cloning site (MCS), also known as a polylinker
This is a region of the plasmid DNA that contains the recognition sites for
several restriction enzymes that are ones producing ‘sticky’ ends. These are
sites where it is possible to insert foreign DNA. Treatment of plasmids with
the specific restriction enzyme cuts the plasmid DNA open, allowing the
incorporation of foreign DNA with complementary sticky ends.
Figure 13.24 shows two of many plasmids that can be used as vectors to
transport foreign DNA into bacterial cells to transform them. Both plasmids
have the required origin of replication (ORI) and also have a marker (AmpR) for
the antibiotic ampicillin. In addition, the pUC19 plasmid has a second marker
(TetR) for a second antibiotic, tetracycline.
U
N
C
O
R
FIGURE 13.23 Stylised
diagram of one bacterial cell,
showing its main chromosome
and one plasmid. Plasmids
are small circular DNA
molecules, additional to
and separate from the main
genetic information held in the
main bacterial chromosome.
Multiple copies of plasmids
may be present in one
bacterial cell.
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25 October 2016 8:28 AM
Hin dIII
(a)
Eco RI Eco RV
Bam I
4000
375
Sal I
651
3607
amp
tet
1000
FS
pBR322
4361 bp
3000
O
O
ori
PR
2000
2295
(b)
E
Nde I
G
0
2500
lacZα
Polylinker
amp
396–454
500
D
pUC19
2000
1000
R
EC
ori
TE
2686 bp
1500
U
N
C
O
R
FIGURE 13.24 Diagrams of plasmids.
(a) Plasmid
Plasmid pBR322 with many cloning sites
present throughout the plasmid. (b) Plasmid
pUC19 has a cluster of many restriction
enzyme recognition sites in the blue sector of
the plasmid labelled as polylinker (or multiple
cloning site). Note the presence in both
plasmids of an origin of replication ((ori) and a
marker gene for ampicillin resistance ((AmpR).
ODD FACT
The allele that encodes
antibiotic resistance (AmpR)
produces the protein
beta-lactamase, which
inactivates ampicillin.
630
First knowledge of bacterial transformation
Transformation of bacteria by natural processes has probably been taking place for at least a thousand million years or
so. However, it was not until 1928 that it was recognised — this
by an English scientist, Frederick Griffith (1877–1941). Griffith
experimented with pneumococcus bacteria; the smooth variety
of these bacteria causes pneumonia, but the rough variety
is harmless. Griffith showed that when the contents of dead
smooth bacteria cells were added to a culture of living rough
bacteria cells, some of the rough cells were transformed into
pneumonia-causing smooth cells that, when injected into mice,
killed them. At that time, Griffith called the agent responsible
for this change the ‘transforming factor’ but its identity was not
known. It was not until 1944 that Oswald Avery (1877–1955)
demonstrated that the transforming factor was DNA.
PA
Pst I
4359 0 29 185
Bacterial transformation
In the laboratory, scientists commonly use a plasmid as the
carrier or vector to introduce new genes into bacterial cells.
When foreign DNA is directly taken up from an external source
by bacterial cells, those bacteria are said to be transformed.
Evidence that bacterial transformation has occurred is seen
in the expression of the foreign marker gene in the phenotype
of the bacteria. For example, some bacteria are susceptible to a
particular antibiotic, such as ampicillin. If these bacteria acquire
a plasmid carrying a gene for ampicillin resistance, these bacteria are transformed and become ampicillin resistant.
Bacterial transformation may occur naturally, as when bacteria swap plasmids among themselves. Naturally occurring
transformation of bacterial species by plasmid transfer has produced increasing numbers of strains of bacteria that are resistant
to multiple antibiotics. Bacterial transformation may also occur
artificially, as a deliberate process carried out by microbiologists
in laboratories.
Bacterial transformation can involve the direct uptake by bacteria of naked fragments of DNA from the environment, such as
DNA fragments from dead bacteria. Very commonly, however,
transformation is achieved when vectors, such as plasmids,
introduce foreign DNA into bacterial cells.
Making recombinant plasmids
Recombinant plasmids are plasmids that carry foreign DNA.
Making recombinant plasmids involves the following steps:
1. The DNA of the plasmid is cut at one point using a specific
cutting enzyme that creates sticky ends. This changes the
form of the plasmid from circular to linear.
2. The foreign DNA fragments are prepared using the same cutting enzyme so
that the foreign DNA has sticky ends that match those of the cut plasmid.
The foreign DNA fragments and the plasmids are mixed, and, in some
cases, their ‘sticky ends’ pair by using weak hydrogen bonds. These are
recombinant plasmids.
Other pairings will also occur, such as cut plasmids resealing themselves
so that they are not recombinant plasmids.
3. The joining enzyme, ligase, is added and this makes the joins permanent
through covalent bonding.
Figure 13.25 shows the process by which recombinant plasmids are formed.
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Plasmid
vector
ODD FACT
C
Smooth pneumococcus
bacteria have an external
capsule that prevents their
destruction by phagocytic
cells of the immune system.
Infection by a single smooth
cell can lead to the death of
a mouse by pneumonia, while
an infection by hundreds of
rough cells that lack capsules
is harmless.
Recognition site
GA
CT ATT
TA
A
G
Recognition sites
GAATTC
CTTAAG
C
PR
G
Gene of interest
O
AA
TT
O
Restriction
enzymatic
digestion
GAATTC
CTTAAG
FS
Foreign DNA
Antibiotic
resistance
gene
G
CTTAA
C
G
TT
AA
Ligation
G A AT
CTTAAT C
G
Recombinant
vector carrying
gene of interest
U
N
C
O
R
Mix vector and DNA
fragment
C
AT T G
G ATTAA
C
R
EC
TE
D
PA
G
E
TTC
AA G
DNA fragment
containing gene
of interest
FIGURE 13.25 Diagram showing the steps in the formation of a recombinant
plasmid. The formation of complementary sticky ends in both the foreign DNA
fragment and the cut plasmid means that pairing between the plasmid and the
foreign DNA can occur.
FIGURE 13.26 The
PlasmidFactory company
logo. (Image courtesy of
PlasmidFactory.)
In reality, commercial companies exist that will produce DNA plasmids
at the highest quality to meet the design requirements of research groups,
as, for example, PlasmidFactory®, a German company founded in 2000 (see
figure 13.26).
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O
R
R
EC
TE
D
PA
G
E
PR
O
O
FS
Getting plasmids into bacterial cells
Once recombinant plasmids have been created, the challenge is to transfer
them into bacterial cells. In general, the success rate for the transfer of recombinant plasmids into bacterial cells is low, but the success rate can be increased
through various techniques:
• One technique is termed electroporation. In this method, cells are briefly
placed in an electric field that shocks them and appears to create holes in
their plasma membranes so that plasmid entry is facilitated.
• Another method is to heat shock the bacterial cells by suspending them in
an ice-cold salt solution, and then transferring them to 42 °C
C for less than
one minute. This treatment appears to increase the fluidity of the plasma
membranes of the bacterial cells and increases the chance of uptake of
plasmids.
An example of the heat-shock treatment is shown in figure 13.27. The
starting culture of bacterial cells is sensitive to the antibiotic tetracycline, and
is denoted TetS.
1. The bacterial culture is placed in an ice bath and chilled.
2. Recombinant plasmids with the tetracycline resistance allele TetR are added
to the bacterial culture and chilled.
3. The bacteria and plasmid mix is placed in hot water at 42 °C for 50 seconds,
producing a heat shock.This is the stage when the plasma membranes of the
bacterial cells are altered, increasing the chance of uptake of plasmids by
the cells.
4. The mix is returned to an ice bath for two minutes.
5. The bacteria are plated on an agar plate containing the antibiotic tetracycline and incubated at 38 °C
°C overnight. Bacteria that have not taken up the
plasmids are killed by the tetracycline. Bacterial cells that have taken up the
plasmids will be selected; these bacteria have been transformed and will
replicate.
C
0 °C
TetR plasmids added
U
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TetS bacteria in cold
salt solution
42 °C
Bacteria and plasmid
mix in hot water (50 sec)
Back to ice bath (2 min)
Incubate overnight
at 38 °C
Only colonies of transformed
TetR bacteria grow
Bacterial cells spread on
agar plate with tetracycline
FIGURE 13.27 Stages in the uptake of plasmids by bacterial calls, resulting in their transformation.
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Recombinant
DNA
Summary
screen and
practice questions
Unit 4
AOS 2
Topic 1
Concept 4
The importance of recombinant plasmids is that the foreign DNA
they incorporate can come from any source: it may be a human gene, a
plant gene, a jellyfish gene, a yeast gene, and so on. The importance of
transformed bacteria is that they express the foreign gene in their phenotype. We will see in chapter 14 how these transformed bacteria can be
factories to produce the protein products of these foreign genes, such as
human insulin.
Now that you have met plasmids, let’s look at how they can be used to make
multiple copies of DNA.
FS
Making multiple copies of DNA
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It is possible to obtain multiple copies of DNA using the polymerase chain
reaction (PCR) as described earlier in the chapter. However, the PCR technique
is used to amplify DNA only when very tiny amounts of DNA are available,
such as in forensic investigations.
In contrast, if DNA is available in relatively large amounts, recombinant plasmids can be used to make multiple copies of large fragments
of DNA or even entire genes. This technique is called gene cloning and
we will explore this technique in detail in chapter 14 (see page 652). Note
that gene cloning using recombinant plasmids uses the fact that plasmids
are self-replicating and can multiply independently of the main bacterial
chromosome.
A simplified diagram showing how plasmids can make multiple copies of
foreign DNA that they carry is shown in figure 13.28. This multiplication is
achieved through (i) replication of the plasmids and (ii) binary fission (division) of the bacterial cells in which the plasmids are located.
In summary, the DNA to be copied (cloned) is joined to a bacterial plasmid
and the plasmid is transferred into a bacterial cell. Once inside the bacterial
cell, the plasmid and its foreign DNA multiply to produce up to 20 copies.
Within about 20 minutes, the bacterial cell itself divides by binary fission to
produce two daughter cells. Within these cells, the plasmids again multiply and
then the bacterial cells divide again. Over a period of several hours, millions of
bacterial cells and even larger numbers of copies of the foreign DNA will be
produced.
O
R
Bacterial cell
Plasmid multiplication
in bacterial cell
FIGURE 13.28 Gene cloning
C
involves the formation of many
copies of a particular fragment
of DNA. In this case, the red
region of the recombinant
plasmid is the foreign DNA
fragment. Multiplication of this
foreign DNA involves both
replication of the plasmid’s
DNA within cells and binary
fission of the bacterial cells. In
this diagram, after one binary
division of the bacterial cell
and plasmid replication, how
many copies of the original
plasmid are present?
U
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Bacterial cell undergoes
binary fission
Plasmid
multiplication
Binary fission
etc.
etc.
etc.
etc.
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BIOLOGIST AT WORK
and debilitating disease, their ability to develop
resistance to antibiotics and the many widespread
plagues and epidemics that arise due to pathogenic
(disease-causing) microorganisms.
As a microbiology PhD student, my thesis and
lab research focuses on one particular pathogenic
bacterium, Clostridium difficile. Clostridium difficile
is the most common cause of antibiotic-associated
hospital-acquired diarrhoea and can result in a
spectrum of disease ranging from mild diarrhoea
to severe and life-threatening diarrhoea, which
may lead to more serious and fatal complications.
Disease is mediated by two large toxins produced
by C. difficile,, which lead to cell death and inflammation. My research specifically focuses on how
C. difficile interacts with the host during infection.
From my research we hope to better understand
how the host responds to infection with C. difficile at
the immunological level, as well as understanding
other systemic disease complications that may arise
during severe infection. To do so, we use models
of infection to mimic human disease and analyse
the pathologies, and immunological and host responses, induced by infection with C. difficile. We
then compare this to infection with genetically
engineered C. difficile, which lacks the ability to
produce the disease-inducing toxins, in the hope
of identifying key differences in host responses
between both groups. Medical research is, however,
not a one-man job and relies on our team of students and staff working and collaborating together
to achieve our research outcomes. By working
together on various aspects of C. difficile disease, we
aim to pinpoint key pathways involved in pathology
and disease progression during C. difficile infection, and hope to identify more suitable treatment
options and markers for disease severity. Hopefully
this will translate to better disease outcome
and fewer fatalities from C. difficile infection.’
PR
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Steven Mileto — PhD student and
microbiologist
PA
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FIGURE 13.29 Steven Mileto is a microbiologist and
a PhD student at Monash University. (Image courtesy
of Steven Mileto.)
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‘When I tell someone without a science background
that I am a microbiologist they often ask, ‘How do
you know what you’re working on? I thought you
always needed a microscope to see and work on
microorganisms’.
It’s always fascinated me that something so small,
something that as a single organism cannot been
seen by the naked eye, can be so beneficial to our
existence, yet can also result in so much harm to
people, animals and plants. Microorganisms aid
in our breakdown of nutrients, assist in the production of renewable energy sources, life-saving
antibiotics, chemicals, compounds, and enzymes
and many food products that we adore. But we often
focus on the negative aspects of microorganisms,
particularly their ability to cause life-threatening
KEY IDEAS
■
■
■
■
■
634
Plasmids are small circular fragments of double-stranded DNA occurring
naturally in bacteria.
Plasmids are vectors that can carry foreign DNA from any source.
Plasmids can be constructed to carry selectable marker genes, such as
antibiotic resistance.
Transformation of bacteria occurs when these cells take up DNA from an
external source and express it.
Recombinant plasmids can be used to make multiple copies of foreign
DNA fragments or even genes.
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QUICK CHECK
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6 Identify the following statements as true or false:
a Plasmids are the main chromosome of a bacterial cell.
b Plasmids are self-replicating entities.
c When bacterial cells and plasmids are mixed, all bacterial cells will be
transformed by taking up the plasmids.
d Bacteria that are ampicillin sensitive will be killed by exposure to this
antibiotic.
e If ampicillin-sensitive bacteria take up plasmids that contain AmpR, these
bacteria will be transformed and will become ampicillin resistant.
f The foreign
eign DNA in a recombinant plasmid could come from mice or
men.
7 Identify two procedures
ocedures that can increase the chance of uptake of plasmids
by bacteria.
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BIOCHALLENGE
The plasmid known as pAMP contains 4539 base pairs
of DNA. These include an origin of replication (ORI), the
AmpR allele of the gene conferring ampicillin resistance.
Indicate the origin of your gel that marks the position
where the samples are loaded.
In addition, this plasmid includes a single occurrence of
each of the following sequences:
At the side of your gel, include an arrow with a positive (+)
and a negative (–)) sign to show the direction of migration of
the various DNA fragments.
and
AAGCTT
TTCGAA
FS
GGATCC
CCTAGG
Number each lane.
First, draw the bands for the sample of known markers in
lane 1 and show their sizes at the LH side of the gel.
The pAMP plasmid was treated with a mixture of the
restriction enzymes Bam HI and Hin dIII.
O
Then draw the bands for the other samples and show
their sizes at the RH side of the gel.
a fragment of 784 base pairs
■
a fragment of 3755 base pairs that included the ORI and
the AmpR gene sequence.
G
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Various
arious samples were subjected to electrophoresis in an
agarose gel (at 1%) as follows:
PA
3 What is the function of the ORI?
E
1 Use the above information to draw a rough sketch of
the pAMP plasmid and show the positions of the
two recognition sites.
2 What kinds of ends are produced by treatment of these
plasmids:
a with Bam HI
b with Hin dIII?
TE
Lane 1: A set of DNA markers of known sizes, 5000 bp,
4000 bp, 3000 bp, 2000 bp, 1000 bp, and 500 bp
R
EC
Lane 2: pAMP plasmids after treatment
eatment with Bam HI and
Hin dIII
eatment with Hin dIII only
Lane 3: pAMP plasmids after treatment
eatment with Bam HI only
Lane 4: pAMP plasmids after treatment
eign DNA with a size of 500 base pairs
5 Fragments of foreign
have ends resulting from Bam HI treatment. Copies
of this foreign DNA were mixed with pAMP plasmids
that had also been treated with Bam HI only. Some
recombinant plasmids were formed.
ough sketch of the recombinant pAMP
a Make a rough
plasmid.
b Only some, but not all, plasmids became rrecombinant.
Suggest a possible reason for this.
c The enzyme ligase is involved in the formation of
rrecombinant
ecombinant plasmids. Identify its role.
PR
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The result of this treatment was two fragments as follows:
6 Recombinant plasmids wer
were mixed with recipient
bacterial cells that were ampicillin sensitive ((AmpS).
Some bacterial cells took up recombinant plasmids, but
others failed to do so.
a Which bacteria were transformed?
b What evidence would indicate that these bacteria were
transformed?
c Briefly explain how you could separate the
transformed bacteria from those that were not
transformed.
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our task is to draw a gel with four lanes showing the
4 Your
relative positions of the bands after the electrophoresis is
completed.
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Unit 4
AOS 2
Topic 1
Chapter review
DNA manipulation
Practice questions
Key words
ligase
multiple cloning
site (MCS)
origin of replication
(ORI)
plasmids
polylinker
prenatal
FS
probe
recombinant plasmids
replication
restriction enzymes
reverse transcriptase
selectable marker
gene
vector
O
electrophoresis
electroporation
F8C
gene cloning
gene editing
genome editing
guide RNA
in situ hybridisation
Questions
PR
O
bacterial transformation
Bcl I
Cas9 nuclease
complementary
DNA (cDNA)
covalent
CRISPR-Cas9
DNA editing
a What
hat kinds of ends does this enzyme produce?
E
Copies
opies of one long fragment of DNA were treated with
Pst I and three smaller fragments were produced:
fragment 1
2.3 kbp
fragment 2
8.1 kbp
fragment 3
4.4 kbp
(One
(O
ne kbp denotes one kilo (thousand) base pairs.)
b H
How
ow long was the original DNA fragment?
c H
How
ow many recognition sites for this enzyme were
present?
d If the three fragments were subjected to
electrophoresis, draw the expected outcome,
showing the + and − electrodes.
5 Demonstrating knowledge and understanding ➜
A young woman from a family with a history of
haemophilia had a test that revealed that she was
a carrier of the haemophilia c allele. This woman
later became pregnant and found out that her fetus
was male. She wished to know if her fetus would be
affected by haemophilia.
a Carefully explain if and how an answer might be
given to her.
One DNA test for haemophilia involves the use
of the restriction enzyme Bcl I.
b For this test, could Bcl I be replaced with another
restriction enzyme, such as Sac I? Explain your
decision.
6 Analysing data and drawing conclusions ➜ A piece
of double-stranded DNA is shown below.
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following:
Example:
Petrol : (is to) engine :: (as) food : (is to) animal
Saw : carpenter :: ................. : genetic engineer
Needle and thread : fashion designer :: ................. :
genetic engineer
2 Recognising similarities ➜ An analogy is defined
as a comparison between two different things on
the basis of some similarity. Statements such as ‘The
circulatory system is like an irrigation system’, and
‘The cell membrane is like the outer wall of a castle’
are examples of analogies.
a Identify
dentify a tool of the genetic engineer that might
be seen as analogous to one of the following
objects. Give a reason for each of your choices.
................ is likee a photocopier because ...............
................ is likee adhesive tape because ................
.................. is likee a scalpel because .......................
b Things
gs compared in an analogy have some
similarities, but they are not similar in every way.
For each analogy, describe one important way in
which the items compared differ.
3 Demonstrating
emonstrating knowledge ➜ A solution contains
many copies of a piece of DNA with three cutting
sites for a particular restriction enzyme. This solution
of DNA is treated with the restriction enzyme under
controlled conditions.
a What is the expected result of this treatment?
b Is this change in DNA an example of a chemical
or a physical change?
4 Applying knowledge and understanding ➜ The
action of the cutting enzyme Pst I is as follows:
G
1 Recognising similarities ➜ Copy and complete the
CTGCAG
GACGTC
CTGCA
G
G
ACGTC
8.4 kbp
A sample of this DNA was treated with the cutting
enzyme Alu I. The cutting site for this enzyme is AGCT.
a After this treatment, the sample was subjected to
electrophoresis. The result is shown in figure 13.30.
The numbers denote the sizes of the fragments in
kbp. What conclusion may be drawn?
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25 October 2016 8:28 AM
–
iii DNA fragments with an approximate size of
1500 bps
iv DNA fragments with a size in excess of
3000 bps.
b If this same DNA that was treated with Bgl II
had instead been treated with Hin dIII, would
the pattern of bands obtained in electrophoresis
have been the same? Explain.
c A different sample of DNA was treated with a
particular restriction enzyme that produced
three fragments of sizes 400, 750 and 2500 bps.
Consider that this DNA mixture was loaded
into lane 5 of the gel shown in figure 13.31.
Draw the expected pattern on completion of
electrophoresis of this DNA in lane 5.
3.9 kbp
O
2
3
4
5
PR
1
3000 bp
1500 bp
500 bp
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a different cutting enzyme, Hin dIII. The cutting
site for this enzyme is AAGCTT. The result after
electrophoresis revealed two DNA fragments of
sizes 5.4 kbp and 3.0 kbp. What conclusion may
be drawn?
c A third sample of this DNA was treated with yet
another cutting enzyme, Not I. The result after
electrophoresis revealed one fragment of
8.4 kbp. What conclusion may be drawn?
7 Demonstrating understanding ➜ One sample (S1)
of DNA was cut into pieces using a cutting enzyme
that produces blunt ends. A second sample (S2) of
DNA was cut using a cutting enzyme that produces
sticky ends. In both cases, the sizes of the fragments
produced were similar.
Scientist AA wanted to join the DNA pieces in
sample S1. Scientist BB wanted to join the DNA
pieces in sample S2. Both scientists carried out
their experiments with the DNA in solution under
identical conditions.
a Explain the terms sticky end and blunt end
end.
b What enzyme would the scientists use to catalyse
the joining?
c The scientists observed that the pieces of DNA
from one sample joined more quickly than those
in the other sample. For which sample would the
faster rate of joining be observed? Explain your
choice.
8 Demonstrating knowledge ➜
a Describe one means by which a scientist can
obtain a copy of a gene.
b Having obtained a copy, describe one means by
which the scientist can obtain millions of copies
of the gene.
9 Analysing data and applying principles ➜ Examine
figure 13.31 and answer the following questions:
a Identify the lane that contains:
i the shortest fragments of DNA
ii DNA fragments that include one restriction
site for Bgl II and were treated with that
restriction enzyme
E
b A second sample of this DNA was treated with
G
FIGURE 13.30
+
O
FS
2.4 kbp
2.1 kbp
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FIGURE 13.31
10 Demonstrating knowledge and understanding ➜
Complete the following sentences:
a In bacterial transformation, bacteria take up
foreign ________.
b A vector commonly used in transformation
experiments is a ________.
c For transformation to occur, the DNA taken
up by bacteria must be ________ in their
phenotypes.
d Sticky ends join temporarily through the
formation of ________.
e Temporary joins can be made permanent
through the use of ________.
f The replication of plasmids is ________ of
the replication of the main bacterial
chromosome.
11 Analysing data and applying principles ➜
Examine figure 13.32, which shows a pAES30
plasmid and the location of recognition sites for
various restriction enzymes. This
plasmid comprises 4715 bps. Note that KmR is
the DNA of the allele for resistance to the
antibiotic kanamycin, and MCS is a multiple
cloning site for the restriction enzymes shown
alongside it.
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Alw NI 3255
Cla I 4370
K
R
m
Nru I 4406
Xho I 4461
pAES 30
Xho I 01
Sig. Seq.
MCS
Bcl I 1935
Apa I 1746
Eco RV 1503
Nar I 1311
Nco I 833
PR
Hpa I 1447
FS
Iq
O
lac
Sac I
Kpn I
Sal I
Pst I
O
ori
Nde I 2562
i treatment with Nde I
ii treatment with Nde I plus Sac I
iii treatment with Nde I plus Sac I plus Cla I.
These plasmids are to be made recombinant
by incorporating a fragment of foreign DNA. The
decision needs to be made about which restriction
enzyme to use for this purpose.
Student P says: ‘Use any one of them. It makes no
difference.’ Student Q says: ‘No way. There are some
we should definitely not use!’
b With which student do you agree? Explain your
choice.
12 Discussion question ➜
CRISPR-Cas9 technology allows for gene editing.
An issue that is the subject of debate is:
Should gene editing be allowed on human cells,
such as eggs, sperm and embryos, whose DNA can
pass to future generations? Or, should gene editing
be restricted to somatic cells where that DNA is not
passed to the next generation?
Discuss this issue with your classmates.
FIGURE 13.32
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terms of the number of fragments and sizes, if it
was subjected to the following treatments:
E
a What would be the effect of such a plasmid in
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