Download Gene cloning tutorial

1
The aim of this exercise is to work together, as a group, to design a strategy for
the production of a medically important protein using recombinant DNA
technology. You are provided with a series of cards. These begin with a general
introduction (cards 2-3) and the outline of the characteristics of the three
particular proteins (cards 4-6). After choosing which protein you want to
produce you should work through the remaining cards and produce a strategy
after discussion in the group. The cards give full details of the procedures
involved in cloning and expressing a gene. At the end of each card you are
given a choice as to what to do next. Some of your decisions will lead to dead
ends as you may have done something incorrectly, whilst others will eventually
lead to the production of your protein. You must also consider the fact that for
the purpose of simplicity it is assumed that all techniques work with 100%
efficiency - this is not the case in real life!!!
Go to 2
Aim
2
Recombinant DNA technology has proven to be extremely useful in the
treatment of several medical disorders. For example, the human insulin gene
has been cloned into a plasmid vector and expressed in E. coli. Large amounts
of insulin can then be produced and used to treat diabetes. Other examples of
proteins produced by recombinant methods are growth factors, blood clotting
agents and vaccines. Producing proteins by recombinant methods can be
cheaper and safer than previously used methods. Protein extracted from
human or animal sources may be contaminated e.g. with viruses. Moreover,
those sources may be in short supply.
Go to 3
Introduction
3
The first step in producing a medically important protein is to clone the wild
type gene. The gene must then be transformed into a host cell where it can be
expressed, and then the gene product purified. The most popular expression
systems are E. coli, yeast and cultured mammalian cells. Each host has its
own pros and cons which must be considered when choosing a system for the
expression of a particular protein. For example, many eukaryotic proteins have
to undergo complex post-translational modifications in order to become
biologically active. Many of these processes are specific to higher eukaryotic
cells, and do not take place in E. coli or yeast.
Go to 4
Introduction
4
Somatostatin is a simple 14 amino acid peptide hormone which inhibits the
secretion of other peptide hormones, such as growth hormones, insulin and
glucagon.
It
is
synthesised in several tissues including
the
brain,
hypothalamus and pancreatic islets. Somatostatin is important in the treatment
of a variety of human growth disorders, including acromegaly, a condition
characterised by uncontrolled bone growth. The amino acid sequence of this
protein is known and antibodies are available. The gene encoding
somatostatin is 1542bp in length. This contains the coding region, a signal
sequence for secretion and a single intron, which is present in the signal
sequence.
Learn more about Somatostasin
Go to 5
Protein 1: Somatostatin
5
Human epidermal growth factor (hEGF) is a single chain polypeptide
consisting of 53 amino acids. It is synthesised in the duodenum and the
salivary glands, and small amounts of the protein can be isolated in urine,
thus the amino acid sequence is known and antibodies can be produced. This
peptide hormone is a promoter of epithelial cell proliferation, and inhibits
gastric acid secretion. Thus, it may be possible to use recombinantly
produced hEGF in the treatment of duodenal ulcers.
Learn more about duodenal ulcers
Go to 6
Protein 2: hEFG
6
Factor IX is a 415 amino acid plasma glycoprotein which has an essential role
in blood clotting. Production of the protein is vital for the treatment of
haemophilia. Factor IX is synthesised in liver hepatocytes where it undergoes
three distinct types of post-translational modification. These modifications are
very complex and very specialised. The active protein can be purified in small
amounts from blood plasma, therefore the amino acid sequence is known and
antibodies can be produced. The gene encoding Factor IX is 34kb in length.
The gene consists of 8 exons and 7 introns, which make up 90% of the
sequence. Thus it is an enormously complicated gene.
Learn More about Factor IX
Protein 3: Factor IX
Make a note of the protein you
have chosen and Go to 7
7
You now know some of the specific characteristics of your chosen protein
which will help you in choosing the best host, vectors and techniques
necessary for producing your recombinant protein. You must now clone your
gene. This is done by inserting a mixture of DNA fragments, one of which will
contain the gene of interest, into separate vector molecules. These molecules
are then introduced into E. coli by transformation. Usually only one
recombinant molecule will go into each cell. Therefore, the gene of interest can
be identified by screening the resultant colonies. Once the gene has been
identified, it can be subcloned into an appropriate vector for transformation into
either yeast or mammalian cells dependent on requirements.
Go to 8
Cloning the Gene
8
To clone your gene you may use either a cDNA library or a genomic library.
To learn about cDNA libraries click here
To learn about genomic libraries click here
To use a cDNA library click here
To use a genomic library click here
Choice of Library
9
Complementary DNA is obtained by copying mRNA. The cDNA gives an exact
copy of the gene's coding sequences, but lacks introns and transcription
signals. One advantage of using mRNA to obtain your DNA sequences, is that
any given cell type expresses only a subset of its chromosomal genes.
Therefore, if you obtain your mRNA from a source which expresses your gene
of interest, there will be a higher chance of identifying that specific gene.
For more details on constructing
a cDNA library click here.
To return click here.
cDNA library
10
A genomic library is a collection of clones sufficient in number to include all the
genes of a particular organism. The larger the organism, the bigger the library.
Therefore, bacterial and yeast genomic libraries are commonplace, and it is
relatively easy to identify a given gene. However animal libraries are so large,
due to the enormous size of the genomes, that it is a mammoth task to identify
any one gene.
Genomic libraries are prepared by purifying total cell DNA and then making a
partial restriction digest, resulting in fragments that can be cloned into a
suitable vector.
In order to obtain a representative human library, a λ-based vector called a
cosmid is used. Cosmids can be used to clone inserts of up to about 40 kb.
To see a simple diagram showing this process click here.
To return click here.
Genomic library
11
mRNA is purified from cells by oligo(dT) cellulose chromatography. The mRNA
molecules bind to the oligo(dT), which is linked to the cellulose column, via their
polyA tails, while the remainder of the RNA species flows through the column.
The bound mRNAs are then eluted from the column. When the mRNA has been
purified, double-stranded DNA must be synthesised.
You must choose one of the following from which to collect mRNA:
Liver hepatocytes which synthesize blood clotting factors
or
Pancreatic Islets which synthesize insulin, glucagon and somatostatin
or
Duodenal cells which synthesize epidermal growth factor.
Make a note of the cell type you have chosen and Go to 12
Source of mRNA
12
Most DNA polymerases can function only if the template possesses a
double-stranded region which acts a primer for the initiation of
polymerization. If your template is single-stranded, a synthetic primer must
be added for DNA synthesis to occur. You may choose to use:
1. No primer
2. Oligo(dC)
3. Oligo(dG)
4. Oligo(dT)
5. An oligo synthesised from known amino acid sequence
Make a note of your choice and Go to 17
Primers
13
An oligonucleotide consisting only of cytosine (dC) residues, will anneal via
complementary base pairing to a run of guanosine residues (polyG). A polyG
tail can be added onto the 3' end of a double- or single-stranded DNA / RNA
molecule by terminal deoxynucleotidyl transferase.
C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C
Return
Oligo(dC)
14
An oligonucleotide consisting only of guanosine (dG) residues, will anneal via
complementary base pairing to a run of cytosine residues (polyC). A polyC tail
can be added onto the 3' end of a double- or single-stranded DNA / RNA
molecule by terminal deoxynucleotidyl transferase.
G-G-G-G-G-G-G-G-G-G-G-G-G-G-G-G-G-G-G-G-G-G-G-G-G-G-G-G
Return
Oligo(dG)
15
An oligonucleotide consisting only of thymidine (dT) residues, will anneal via
complementary base pairing to a run of adenine residues (polyA). A polyA tail
can be added onto the 3' end of a double- or single-stranded DNA / RNA
molecule by terminal deoxynucleotidyl transferase. A polyA tail is found
naturally at the 3’ end of most RNA molecules.
T-T-T-T-T-T-T-T-T-T-T-T-T-T-T-T-T-T-T-T-T-T-T-T-T-T-T-T
Return
Oligo(dT)
16
Oligonucleotides can be synthesised which correspond to a specific DNA
sequence. The DNA sequence may already be known, or can be determined
from a known amino acid sequence, although the degeneracy of the genetic
code must be taken into consideration when designing an oligonucleotide from
an amino acid sequence.
G-C-A-T-A-G-T-C-C-A-G-C-G-T-T-A-C-T-C-T-G-A-A-T-C-A-C-G
Return
Oligo (amino acid sequence)
17
DNA polymerases are enzymes that synthesize a new strand of DNA
complementary to an existing template. Most polymerases can function only if
the template possesses a double-stranded region which acts as a primer for
initiation of polymerisation. There are a range of different polymerases each
with different characteristics.
5' - A-T-G-C-A-A-T-G-C-A- 3' --- TEMPLATE
3'-C-G-T- 5' --- PRIMER
You may choose to use:
Klenow fragment of DNA polymerase I
or
Reverse transcriptase
or
DNA polymerase I
DNA polymerases
Make a note of your choice and then
check first strand synthesis here.
18
The Klenow fragment of DNA polymerase I contains the polymerase function of
the enzyme. It can only use DNA as a template. Since it does not contain the 5'
- 3' exonuclease activity of DNA polymerase I, Klenow can be used to
synthesise a complementary DNA strand on a single-stranded template without
degrading the cDNA.
Learn more about the Klenow fragment of
DNA polymerase I
Return
Klenow fragment
19
Reverse transcriptase is an enzyme involved in the replication of several kinds
of virus. Reverse transcriptase is unique in that it can use RNA as a template
as well as DNA. Like other DNA polymerases, reverse transcriptase requires a
primer.
Learn more about the reverse
transcriptases
Return
Reverse Transcriptase
20
DNA polymerase I has a polymerase function and nuclease activity. The
enzyme attaches to a short single stranded region (nick) in a mainly doublestranded DNA molecule, then synthesises a complementary DNA strand,
degrading the existing strand as it proceeds. It degrades single-stranded DNA.
Learn more about DNA polymerase I
Return
DNA polymerase I
21
To check DNA synthesis parallel reactions are set up.
Reaction 1: experimental - the sample which will go on to the 2nd strand synthesis
Reaction 2: test sample - to check efficiency of 1st strand synthesis.
In Reaction 2 one of the dNTP's will be radioactive. DNA synthesis will result in the
radiolabelled dNTP being incorporated into the DNA. A sample of the radiolabelled
reaction can then be run on a 1.4% alkaline agarose gel. Autoradiography is then used
to detect DNA synthesis.
Conditions
Results
Primer ?
Polymerase ?
DNA synthesis ?
None
Any
None
14
Click here
Any
Klenow
None
19
Click here
Any
DNA polymerase I
None
Click here
19
oligo(dT)
Reverse transcriptase
Correct DNA synthesis
Click here
25
oligonucleotide
Reverse transcriptase
DNA synthesis
Click here
24
oligo(dG) /oligo(dC)
Reverse transcriptase
None
First strand synthesis
(unless added polyC /polyG tail to template)
Go to
14 Click
(25) here
22
First strand synthesis requires the use of an oligo(dT) primer.
To try again click here
To proceed click here
No DNA synthesis
23
An oligonucleotide consisting only of thymidine (dT) residues, will anneal via
complementary base pairing to a run of adenine residues (polyA). A polyA tail
can be added onto the 3' end of a double- or single-stranded DNA / RNA
molecule by terminal deoxynucleotidyl transferase. A polyA tail is found
naturally at the 3’ end of most RNA molecules.
T-T-T-T-T-T-T-T-T-T-T-T-T-T-T-T-T-T-T-T-T-T-T-T-T-T-T-T
Return
Oligo(dT)
24
First strand synthesis requires the use of a reverse
transcriptase, because the template is RNA.
To try again click here
To proceed click here
No DNA synthesis
25
Reverse transcriptase is an enzyme involved in the replication of several kinds
of virus. Reverse transcriptase is unique in that it can use RNA as a template
as well as DNA. Like other DNA polymerases, reverse transcriptase requires a
primer.
Learn more about the reverse
transcriptases
Return
Reverse Transcriptase
26
It must be noted that if you used an oligonucleotide primer, which was
complementary to mRNA sequences at the 5' end of the molecule, then the
cDNA may be missing some of the 5' untranslated region.
To try again with a different primer click here
To proceed click here
cDNA synthesis
27
You now have RNA-DNA hybrid molecules. You must now synthesise the 2nd
DNA strand.
RNA C-G-C-C-A-U-C-U-A-C-G-U-C-U-UDNA G-C-G-G-T-A-G-A-T-G-C-A-G-A-A-
You may:
Proceed immediately with second strand synthesis.
or
Denature the hybrid molecule with alkali.
or
Partially degrade the RNA strand with RNase H
Second strand synthesis
28
You may denature your RNA-DNA hybrid molecules with an alkali which also
hydrolyses the RNA strand. This will then give you single-stranded DNA
molecules. You now need to synthesise the 2nd DNA strand.
If you need a primer select one of:
oligo(dC)
oligo(dG)
oligo(dT)
an oligo synthesized from known amino acid sequence.
Then select a polymerase from:
Klenow fragment of DNA polymerase I
reverse transcriptase
DNA polymerase I
Alkali denature
Make a note of your choice
and then check 2nd strand
synthesis here.
29
An oligonucleotide consisting only of cytosine (dC) residues, will anneal via
complementary base pairing to a run of guanosine residues (polyG). A polyG
tail can be added onto the 3' end of a double- or single-stranded DNA / RNA
molecule by terminal deoxynucleotidyl transferase.
C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C
Return
Oligo(dC)
30
An oligonucleotide consisting only of guanosine (dG) residues, will anneal via
complementary base pairing to a run of cytosine residues (polyC). A polyC tail
can be added onto the 3' end of a double- or single-stranded DNA / RNA
molecule by terminal deoxynucleotidyl transferase.
G-G-G-G-G-G-G-G-G-G-G-G-G-G-G-G-G-G-G-G-G-G-G-G-G-G-G-G
Return
Oligo(dG)
31
An oligonucleotide consisting only of thymidine (dT) residues, will anneal via
complementary base pairing to a run of adenine residues (polyA). A polyA tail
can be added onto the 3' end of a double- or single-stranded DNA / RNA
molecule by terminal deoxynucleotidyl transferase. A polyA tail is found
naturally at the 3’ end of most RNA molecules.
T-T-T-T-T-T-T-T-T-T-T-T-T-T-T-T-T-T-T-T-T-T-T-T-T-T-T-T
Return
Oligo(dT)
32
Oligonucleotides can be synthesised which correspond to a specific DNA
sequence. The DNA sequence may already be known, or can be determined
from a known amino acid sequence, although the degeneracy of the genetic
code must be taken into consideration when designing an oligonucleotide from
an amino acid sequence.
G-C-A-T-A-G-T-C-C-A-G-C-G-T-T-A-C-T-C-T-G-A-A-T-C-A-C-G
Return
Oligo (amino acid sequence)
33
The Klenow fragment of DNA polymerase I contains the polymerase function of
the enzyme. It can only use DNA as a template. Since it does not contain the 5'
- 3' exonuclease activity of DNA polymerase I, Klenow can be used to
synthesise a complementary DNA strand on a single-stranded template without
degrading the cDNA.
Learn more about the Klenow fragment of
DNA polymerase I
Return to ‘Alkali denature’
Return to ‘RNase H treatment’
Klenow fragment
34
Reverse transcriptase is an enzyme involved in the replication of several kinds
of virus. Reverse transcriptase is unique in that it can use RNA as a template
as well as DNA. Like other DNA polymerases, reverse transcriptase requires a
primer.
Learn more about the reverse
transcriptases
Return to ‘Alkali denature’
Return to ‘RNase H treatment’
Reverse Transcriptase
35
DNA polymerase I has a polymerase function and nuclease activity. The
enzyme attaches to a short single stranded region (nick) in a mainly doublestranded DNA molecule, then synthesises a complementary DNA strand,
degrading the existing strand as it proceeds. It degrades single-stranded DNA.
Learn more about DNA polymerase I
Return to ‘Alkali denature’
Return to ‘RNase H treatment’
DNA polymerase I
36
You may partially degrade the RNA strand of your RNA-DNA hybrid with
Ribonuclease (RNase) H. This leaves small fragments of RNA associated with
your DNA strand. These fragments of RNA can act as primers.
Select a polymerase from:
Klenow fragment of DNA polymerase I
reverse transcriptase
DNA polymerase I
Make a note of your choice
and then check 2nd strand
synthesis here.
RNase H treatment
37
You may check 2nd strand synthesis the same way as 1st strand synthesis.
Conditions
Results
Treatment ?
Primer ?
Polymerase ?
DNA synthesis ?
RNase H
None
Reverse transcriptase
(RT)
Incorrect synthesis
Click here
None
Klenow (K)
Incorrect synthesis
Click here
None
DNA polymerase I (D)
Correct DNA synthesis
Click here
None
RT / K / D
DNA synthesis
Click here
Any
RT / K / D
Correct DNA synthesis
Click here
None / Any
RT / K / D
None
Click here
Alkali
None
Second strand synthesis
37
DNA polymerase I is required for successful RNase H treatment. This
is because Klenow lacks exonuclease activity and reverse
transcriptase will not work unless the RNA strand is fully hydrolysized.
To try again click here
To proceed click here
Incorrect DNA synthesis
39
If a primer is not added to the 1st strand before 2nd strand synthesis, the cDNA
can form a transient self-priming structure in which a hairpin loop at the 3' end
is stabilised by enough base pairing to allow initiation of 2nd strand synthesis.
Once initiated, subsequent synthesis of the 2nd strand stabilises the hairpin
loop. Thus, the resultant double-stranded molecule has the hairpin loop intact,
therefore it has to be removed before ligation can occur. The hairpin loop is
digested with S1 nuclease, however the S1 nuclease treatment can also digest
much of the 5' coding sequences, thus producing an incomplete cDNA.
To try again click here
To proceed click here
Hairpin loop
40
Having created a cosmid library and isolated one cosmid which contains your
specific gene, within about 40 kb, you must now sub-clone the insert DNA into
an alternative vector in order to isolate the gene and express the protein.
Proceed
Genomic library
41
The next step in producing large amounts of protein, is to clone the gene of
interest. Usually the cDNAs or fragments of genomic DNA are inserted into
cloning vectors, which are then transformed into E. coli. E. coli is the organism
used for constructing libraries because of its high transformation efficiency and
simple selection procedures, thus making it possible to screen thousands of
colonies. When the vector carrying the gene of interest is identified, the gene
can be subcloned into the expression vectors of other organisms. These
vectors are then introduced into the appropriate host, where the protein will
eventually
be
expressed.
For
simplicity,
you
will
be inserting
your
cDNA/genomic DNA directly into shuttle vectors. These are vectors which can
replicate and be selected for in E. coli and one other organism. This means that
you can identify your gene in E. coli, but will not have to sub-clone into an
appropriate expression vector. Therefore you must now decide which host
organism will be suitable for you to express your gene in eventually. Do not
forget to consider your protein's characteristics when choosing your host.
Choose a host
Cloning the gene
42
You may choose one of the following to act as your host organism:
E. coli
Yeast
A mammalian cell system
Choosing a host
43
Advantages of E. coli a host for the production of heterologous proteins:
(1)
It is easily, rapidly and cheaply grown in large quantities.
(2)
The manipulation of DNA is well defined and relatively easy.
(3)
There is a wide range of both plasmid and phage vectors, that
can be introduced into bacterial cells at a high efficiency.
(4)
Simple eukaryotic proteins can be produced in very high
yields.
Disadvantages of E. coli a host for the production of heterologous proteins :
(1)
Recognise eukaryotic proteins as “foreign” - therefore will
degrade them.
(2)
Does not carry out eukaryotic post-translational modifications possibly inactive protein.
(3)
Does not fold eukaryotic proteins correctly - possibly inactive
protein.
(4)
Cannot express eukaryotic genes that contain introns.
Use E. coli as your host
Consider other options
E. coli host
44
Advantages of yeast a host for the production of heterologous proteins:
(1)
It is easily, rapidly and cheaply grown in large quantities - but
it is eukaryotic.
(2)
Relatively easy to manipulate DNA, a wide range of plasmid
vectors.
(3)
Can do simple post-translational modifications.
(4)
Can fold simple proteins.
(5)
It has a secretory system - proteins can be secreted into the
medium - easier to purify.
Disadvantages of yeast a host for the production of heterologous proteins:
(1)
Yeast is a lower eukaryote. Therefore it cannot do complex
post-translational modifications - possibly inactive
proteins.
(2)
Inefficient at removing introns - poor expression.
Use yeast as your host
Consider other options
Yeast host
45
Advantages of mammalian cell systems for the expression of proteins:
(1)
Eukaryotic proteins should be correctly folded, appropriately
modified - completely functional.
(2)
Efficient at removing introns - can use genomic genes (with
introns).
(3)
Wide range of plasmid and viral based vectors.
Disadvantages of mammalian cell systems for the expression of proteins:
(1)
Relatively difficult to grow cultured cells in large amounts, also
expensive.
(2)
Poor transformation efficiencies.
(3)
Sometimes need specific cell lines to do specialised
modifications.
(4)
Stringent controls required for detection of contaminants e.g.
viruses.
Use a mammalian cell system as your host
Consider other options
Mammalian cell host
46
You must now choose the vector that you will use. You may choose either:
pBR322
or
λgt11
Make a note of your choice and proceed
E. coli vectors
47
pBR322
DNA multicopy plasmid
Usually used as a cloning vector
Selective markers :
- ampicillin resistance (Amp R)
- tetracycline resistance (Tet R)
BamHI
PstI
Amp R
Tet R
pBR322
Return
pBR322
48
λgt11
Phage
- Defective in lysis. Therefore upon induction of
gene expression the products accumulate in the cell.
Marker gene, lacZ, encodes for β-galactosidase.
- β-galactosidase breaks down X-gal to give a deep
blue product - blue plaques.
- Inactivation of β-galactosidase by insertion of DNA
into the 3’end of lacZ gene - white plaques
N.B. For correct expression of the protein, the coding
sequences must be inserted in the correct reading frame.
EcoRI
Return
λgt11
Lac Z
49
You must now choose the vector that you will use. You may choose either:
pJP31
or
YEp213
Make a note of your choice and proceed
Yeast vectors
50
pJP31
Multicopy, DNA plasmid
Shuttle vector - bacterial and yeast origins of replication
Selective markers :
- Yeast : auxotrophic marker - LEU2 gene
- Bacteria : Ampicillin resistance (Amp R)
Secretory, expression vector
PstI
EcoRI
EcoRI
ADH
terminator
Return
HinDIII
pJP31
 - f actor
leader
 - f actor
promoter
EcoRI
pJP31
PstI
51
YEp213
Multicopy, DNA plasmid
Shuttle vector - bacterial and yeast origins of replication (ORI)
2 µm sequences (ORI and STB)
Selective markers :
- Yeast : auxotrophic marker - LEU2 gene (LEU2)
- Bacteria : Ampicillin resistance (Amp R)
- Bacteria : Tetracycline resistance (Tet R)
Cloning vector
EcoRI
PstI
LEU2
ORI
PstI
Return
STB
EcoRI
EcoRI
YEp213
Amp R
Tet R
PstI
YEp213
BamHI
52
You must now choose the vector that you will use. You may choose either:
pMAMneo
or
Amplicon
Make a note of your choice and proceed
Mammalian cell vectors
53
pMAMneo
DNA plasmid, multicopy
Shuttle vector : containing
- E. coli origin of replication
- Ampicillin resistance marker (Amp R)
- SV40 origin of replication
- Selective marker : neo - confers resistance to G418
antibiotics (neo).
- Expression cassette : MMTVpromoter sequences (MMTV P),
SV40 transcription terminator and polyadenylation signal (pA)
Can express wild type genes, although you would have to remove the
gene's own transcriptional signals in order to get a high level of
expression, which is not cell specific.
N.B. lysis of the cell occurs after only a few days due to excessive
amounts of DNA - therefore only get transient expression.
See plasmid map
Return
pMAMneo
54
MMTV P
SalI
pA
SV40 P
pMAMneo
PstI
neo
Amp R
pA
Return
pMAMneo plasmid map
55
Amplicon
Shuttle vector, containing
- E. coli origin of replication
- Ampicillin resistance marker (Amp R)
- Selective marker : dhfr gene - confers resistance to
methotrexate (MTX) in DHFR- cell lines (DHFR).
- SV40 expression cassette : promoter sequences (P), intron
(I) and splice site, transcription terminator and
polyadenylation
signal (pA).
Can express wild type genes, although you would have to remove the
gene's own transcriptional signals in order to get a high level of
expression, which is not cell specific.
Prolonged exposure to increasing concentrations of MTX results in the
amplification of the vector - stable cell line with a high level of
expression.
See plasmid map
Return
Amplicon
56
SV40
BglII
P
I
pA
PstI
Amp R
Amplicon
P
DHFR
I
pA
Return
Amplicon plasmid map
57
You should now have your insert DNA (either as cDNA molecules or genomic
DNA) and you should have chosen your host-vector system. The next step is to
prepare your vector DNA for ligation with the DNA fragments. You must choose
an enzyme to cut the vector where you would like to insert your foreign DNA.
Each restriction endonuclease has a specific recognition site. Some enzymes
make a simple double-stranded cut in the middle of the site - blunt end. While
others cut the two strands at different positions - staggered cut. This results in
the DNA fragments having single-stranded overhangs - sticky ends. Ligation is
more efficient if the vector and insert DNA have complementary 'sticky ends'.
Choose an enzyme
Restriction enzymes
58
Enzyme
Recognition sequence
Complementary ends
EcoRI (E)
G'AATTC
E
BamHI (B)
G'GATCC
Bg / S / B
PstI (P)
CTGCA'G
P
BglII (Bg)
A'GATCT
Bg / S / B
HindIII (H)
A'AGCTT
H
SalI (Sa)
G'TCGAC
Sa
Sau3A (S)
'GATC
Bg / S / B
N.B. ' indicates where the cleavage site is within the recognition sequence.
To review E. coli vectors click here
To review Yeast vectors click here
To review mammalian cell vectors click here
Make a note of which enzymes you have used and proceed.
Restriction enzymes
59
Return
BamHI
PstI
Amp R
Tet R
pBR322
EcoRI
λgt11
E. coli vectors
Lac Z
60
PstI
EcoRI
EcoRI
EcoRI
PstI
LEU2
ORI
PstI
ADH
terminator
HinDIII
STB
EcoRI
pJP31
EcoRI
YEp213
Amp R
 - f actor
leader
 - f actor
promoter
Tet R
PstI
EcoRI
PstI
Return
Yeast vectors
BamHI
61
MMTV P
SV40
SalI
P
pA
I
pA
SV40 P
PstI
pMAMneo
PstI
Amp R
Amplicon
neo
Amp R
P
pA
DHFR
I
Return
Mammalian cell vectors
pA
BglII
62
After your vector DNA has been cut with a restriction enzyme, it can be treated
with alkaline phosphatase. This enzyme removes the 5' phosphate groups of
the vector DNA, thus preventing its recircularization. The foreign DNA to be
inserted still has 5' phosphate groups. Therefore for the vector to efficiently
recircularize it must incorporate an insert. A control for the efficiency of the
phosphatase reaction (we will assume it is 100% in this case) is to use half the
reaction in a ligation mix with no insert DNA. If the efficiency of the phosphatase
is 100% then you would get no religation of the vector in the absence of insert
DNA, therefore you would see no colonies on your plate.
You must now prepare your insert DNA.
If you are working with cDNA click here.
If you are using genomic DNA click here.
Alkaline phosphatase
63
Your cDNA is blunt-ended. To increase the efficiency of your ligation you will
need to add sticky ends to your cDNA. Homopolymer tailing is one way of
adding sticky ends. A homopolymer is a polymer in which all the subunits are
the same, e.g. a DNA strand can be made up entirely of dGTP, thus giving a
poly(dG) tail. The enzyme terminal deoxynucleotidyl transferase adds
nucleotides onto the 3' OH group of double-stranded molecules. A
complementary tail must also be added onto the vector DNA for ligation to
occur.
You may add sticky ends your cDNAs using either adaptors or linkers.
Note your choice, then click here.
cDNA preparation
64
Adaptors are short synthetic oligonucleotides. One end is blunt and the other
end is sticky. The blunt-end ligates to DNA - thus giving it sticky ends. The
sticky end sequences are complementary to the overhangs left by restriction
enzymes e.g. BamHI. A disadvantage is that the adaptors tend to stick together
- thus giving blunt ends. You must remove unincorporated adaptors before
ligation (by running down a column) otherwise ligation will be inhibited due to
excess adaptors.
Return
Adaptors
65
Linkers are short synthetic double stranded DNA oligonucleotides. They are
blunt-ended. Typical linkers contain restriction sites. They are attached to the
ends of the cDNA by blunt-ended ligation (linkers are added in high
concentrations to increase the efficiency of blunt-ended ligation). To make sticky
ends, the linkers are cut with the appropriate restriction enzyme. You must
remove unincorporated linkers before ligation, by running down a column,
otherwise ligation will be inhibited due to excess linkers.
If you choose to use linkers, you must select a restriction enzyme to make
sticky ends. To review the available restriction enzymes click here.
Return
Linkers
66
Enzyme
Recognition sequence
Complementary ends
EcoRI (E)
G'AATTC
E
BamHI (B)
G'GATCC
Bg / S / B
PstI (P)
CTGCA'G
P
BglII (Bg)
A'GATCT
Bg / S / B
HindIII (H)
A'AGCTT
H
SalI (Sa)
G'TCGAC
Sa
Sau3A (S)
'GATC
Bg / S / B
N.B. ' indicates where the cleavage site is within the recognition sequence.
Return
Restriction enzymes
67
You have isolated cosmid DNA carrying your gene, digested with BamHI and
isolated the insert DNA. The insert DNA must now be partially digested with a
restriction enzyme which gives complementary ends to the vector DNA. The
DNA is partially digested in order to give a range of fragment sizes, and also to
overcome the risk of the gene of interest containing an internal restriction site.
Choose a restriction enzyme:
Enzyme
Recognition sequence
Complementary ends
EcoRI (E)
G'AATTC
E
BamHI (B)
G'GATCC
Bg / S / B
PstI (P)
CTGCA'G
P
BglII (Bg)
A'GATCT
Bg / S / B
HindIII (H)
A'AGCTT
H
SalI (Sa)
G'TCGAC
Sa
Sau3A (S)
'GATC
Bg / S / B
N.B. ' indicates where the cleavage site is within the recognition sequence.
Note your choice, then click here.
Genomic DNA preparation
68
Your vector DNA and insert DNA are ligated by the action of DNA ligase. In your
ligation mix you will have :
- unligated vector
- unligated insert DNA
- vectors recircularised without insert (if you have not
dephosphorylated your vector)
- recombinant DNA molecules
You now need to transform your E. coli with your ligation mix and then
select for recombinant molecules.
Proceed
Ligation
69
cDNA and genomic libraries are usually constructed in E. coli due to the high
efficiency of transformation. During the process of transformation each
bacterial cell should only take up one DNA molecule, which is then amplified
within the cell. Consequently, the cell, once plated onto solid media, will form a
colony which will only have one type of plasmid. It is then possible to screen
the colonies to isolate the one which contains the gene of interest.
If you are using a DNA plasmid (pBR322, pJP31, YEp213, Amplicon or
pMAMneo), click here.
If you are using λgt11, click here.
E.coli transformation
70
There are two major methods of transformation :
(1) Treat bacterial cells with CaCl2, which causes the DNA to precipitate on
the surface of the bacterium. Uptake of DNA into the cell is then
stimulated by heat shock. Once you have transformed your cells with
your ligation mix you must select for recombinant plasmids.
(2) Electroporation : Bacterial, yeast and mammalian cells can be subjected
to a short electrical pulse, which allows the DNA to enter the cell. It is
thought that the electric pulse induces the transient formation of pores
in the cell membrane through which the DNA enters the cell.
Proceed
E. coli transformation
71
Antibiotics, such as ampicillin (Amp) and tetracycline (Tet), can be added to the
solid agar media on which you plate your transformants. Cells containing a
recircularised vector (with or without an insert) will be able to grow on the
media, due to the appropriate antibiotic resistance gene carried on the vector.
To review your plasmid vector click here.
To plate onto ampicillin click here if you are using an E.coli or yeast vector or
here if you are using a mammalian vector.
To plate onto tetracycline click here.
Antibiotic selection
72
PstI
PstI
EcoRI
EcoRI
EcoRI
LEU2
ORI
PstI
ADH
terminator
STB
EcoRI
pJP31
EcoRI
YEp213
BamHI
HinDIII
 - f actor
leader
Amp R
 - f actor
promoter
Tet R
PstI
PstI
BamHI
Amp R
Tet R
EcoRI
PstI
MMTV P
pBR322
SV40
SalI
pA
I
pA
SV40 P
PstI
pMAMneo
PstI
BglII
P
Amp R
Amplicon
neo
Amp R
P
pA
DHFR
I
pA
Return
DNA plasmids
73
If you have plated onto ampicillin, this table will tell you the outcome so far. If
you have (5x106) colonies, click here. If you have none, click here.
Plasmid
Vector
sticky ends
Insert DNA
sticky ends
Phosphatase
treated
Number of
colonies
pBR322
PstI
P
yes
None
PstI
E /H /S /B /Bg /Sa
yes
None
PstI
E /H /S /B /Bg /P
no
5 x 106
BamHI
B /Bg / S
yes
5 x 106
BamHI
E /P / H
yes
None
BamHI
B /Bg /S /E /H /P
no
5 x 106
PstI /EcoRI
B /Bg /S /E /H /P
yes / no
None
BamHI
B /Bg /S
yes
5 x 106
BamHI
E /P /H
yes
None
BamHI
B /Bg /S /E /H /P
no
5 x 106
PstI /EcoRI
B /Bg /S /E /H /P
yes / no
None
HindIII
H
yes
5 x 106
HindIII
B /Bg /S /E /P
yes
None
HindIII
B /Bg /S /E /P /H
no
5 x 106
YEp213
pJP31
N.B. Sticky ends for insert DNA can be obtained by cutting linker or genomic DNA with a
restriction enzyme, or adding adaptors to cDNA which have suitable complementary tails.
P= PstI, B= BamHI, E= EcoRI, Bg= BglII, S= Sau3A, H= HindIII, Sa= SalI
Ampicillin selection
74
If you have plated onto ampicillin, this table will tell you the outcome so far. If
you have (5x106) colonies, click here. If you have none, click here.
Plasmid
Vector
sticky ends
Insert DNA
sticky ends
Phosphatase
treated
Number of
colonies
pMAMneo
PstI
P
yes
None
SalI
Sa
yes
5 x 106
SalI
B /Bg /S /E /H /P
yes
none
SalI
B /Bg /S /E /H /P /Sa
no
5 x 106
PstI
P
yes
None
BglII
E /H /P
yes
None
BglII
Bg /B /S
yes
5 x 106
BglII
Bg /B /S /E /H /P
no
5 x 106
Amplicon
N.B. Sticky ends for insert DNA can be obtained by cutting linker or genomic DNA with a restriction enzyme, or
adding adaptors to cDNA which have suitable complementary tails. P= PstI, B= BamHI, E= EcoRI, Bg= BglII,
S= Sau3A, H= HindIII, Sa= SalI
Ampicillin selection
75
If you have plated onto tetracycline, this table will tell you the outcome so far. If
you have (5x106) colonies, click here. If you have none, click here.
Plasmid
Vector
sticky ends
Insert DNA
sticky ends
Phosphatase
treated
Number of
colonies
pBR322
PstI
P
yes
5 x 106
PstI
E /H /S /B /Bg /Sa
yes
None
PstI
E /H /S /B /Bg /Sa/P
no
5 x 106
BamHI
B /Bg /S
yes
None
BamHI
E /P /H
yes
None
BamHI
B /Bg /S /H /E /P /Sa
no
5 x 106
PstI /EcoRI
E /H /S /B /Bg /Sa /P
yes / no
None
BamHI
B /Bg /S
yes
None
BamHI
E /P /H
yes
None
BamHI
E /H /S /B /Bg /Sa /P
no
5 x 106
pJP31
PstI /EcoRI /HindIII
E /H /S /B /Bg /Sa /P
yes / no
None
pMAMneo
PstI /SalI
Bg /B /S /H /E /P /Sa
yes / no
None
Amplicon
PstI /BglII
Bg /B /S /H /E /P /Sa
yes / no
None
YEp213
N.B. Sticky ends for insert DNA can be obtained by cutting linker or genomic DNA with
a restriction enzyme, or adding adaptors to cDNA which have suitable complementary
tails. P= PstI, B= BamHI, E= EcoRI, Bg= BglII, S= Sau3A, H= HindIII, Sa= SalI
Tetracycline selection
76
There are two methods by which phage λ can be introduced into E. coli cells.
Firstly, purified phage DNA can be mixed with competent E. coli cells and DNA
uptake is induced by heat shock. Secondly, you can infect the cells with mature
phage particles, this method is more efficient. The phage particles are
produced in vitro before being added to a culture of E. coli.
Proceed
Transfection
77
If you cut your phage DNA with EcoRI and your insert DNA had
complementary ends, then your DNA should have been inserted into the lacZ
gene, thus making it inactive. Recombinants are distinguished by plating cells
onto media containing X-gal and IPTG (induces β-galactosidase); plaques
containing normal phage are blue, recombinant plaques are white.
Vector
sticky ends
Insert DNA
sticky ends
Phosphatase
treated
White plaques
Blue plaques
EcoRI
E
yes
1 x 105
none
Click here
EcoRI
P /E / H/ B /Bg /S
yes
none
none
Click here
EcoRI
E
no
none
1 x 105
Click here
N.B. E= EcoRI, P= PstI, H= HindIII, B= BamHI, Bg= BglII, S= Sau3A
Insertional inactivation
78
You have colonies / plaques on your plates. Now you would chose a random
selection of colonies and obtain plasmid DNA from each via a "mini-prep"
method. The plasmid DNA is then cut with a selection of enzymes in order to
check for inserts - this is known as restriction mapping.
(1) If you dephosphorylated your vector DNA then all your colonies will contain
plasmids with inserts (assuming 100% efficient removal of 5' phosphate
groups), click here.
(2) If you did not dephosphorylate your vector DNA, click here.
Mini-prep
79
You did not get any colonies on your plates. Therefore you either :
(1) did not have the correct antibiotic resistance marker on your plasmid.
(2) disrupted an antibiotic marker with your insert, therefore the cells are
sensitive to the antibiotic in the plates. If your vector has two antibiotic
markers check your transformants on the other antibiotic plates.
(3) did not have complementary sticky ends and had dephosphorylated
your vector DNA, therefore the DNA could not religate and so was
degraded.
(4) used an enzyme that did not linearize your vector, but cut it so many
times that the chance of all the fragments joining together in the correct
position and with an insert is negligible.
Click here to try again
Failure
80
Since you did not dephosphorylate your vector DNA then a large proportion of
your clones contain plasmid which has recircularized without an insert. Therefore
you do not have a representative library.
Click here to try again
Ligation
Failure
81
You have a representative library. Now you must screen your library for your
specific gene.
You may use either:
Colony hybridization
or
Immunological screening
Screening
82
If you know the amino acid sequence of your protein then it is possible to
synthesis a DNA oligonucleotide which corresponds to that sequence. The
oligonucleotide, which is radiolabelled, can then be used as a probe to screen
for the gene. Firstly the colonies are replica plated onto nitrocellulose filters. The
colonies are then treated so that the cell walls are broken down and the DNA,
which is made single-stranded, is bound to the filter. The filters are then
incubated with the radiolabelled oligonucleotide. The radioactive oligo will bind
to its complementary sequences, whilst all unbound oligo will be washed off.
The filter is then exposed to x-ray film and the bound oligo will indicate which
colony contains the plasmid with the correct insert.
To learn more about colony hybridization click here or here.
To use colony hybridization, click here.
To consider immunological screening click here.
Colony hybridization
83
You must now identify your specific gene. The protein coded by the gene can be
detected by immunological screening. Antibodies specific for your protein are
obtained by injecting the protein into the bloodstream of a rabbit. The immune
system of the rabbit will synthesise antibodies that bind to the protein, which are
then purified from the blood. The colonies obtained from your transformation are
transferred to a membrane, lysed and then incubated with your specific
antibody which can be detected using a second antibody linked to an enzyme,
for example alkaline phosphatase. The colonies which contain your protein will
then be identified by an enzyme-catalysed reaction where a colourless
substrate gives rise to a coloured product.
To learn more about immunological screening click here.
To use immunological screening click here.
To consider colony hybridization click here.
Immunological screening
84
If you used cDNA and obtained your mRNA from cells that synthesise your
protein (for somatostatin, pancreatic islets, for Factor IX, liver hepatocytes and
for hEGF, duodenal cells) you have a very high probability of identifying your
gene, click here. If you obtained mRNA from any other source you have a low
chance of success, click here. If you used genomic DNA previously isolated in a
cosmid vector you have a high chance of success of finding somatostatin, click
here, but the genes encoding EGF and Factor IX are too large to be cloned in
E. coli intact so you must clone via cDNA, click here.
Colony hybridization
85
You have now isolated your gene of interest. You must now check the
expression of the gene. Yeast or mammalian expression vectors, carrying the
gene, will need to be transformed into the appropriate host strain.
N.B. (1) Many eukaryotic genes will have signal sequences at the 5' end of the
gene. which are required to target the protein for secretion. These sequences
need to be proteolytically cleaved to give an active protein. Usually when using
E. coli or yeast hosts, the signal sequence is cleaved in vitro and the host's
own signal sequences are added.
N.B. (2) If you have used genomic DNA to obtain your gene, then you will
need to sequence the DNA to identify the 5' transcriptional start sequences.
These are removed, so that the gene will be transcribed under the control of
the vector promoter sequences found on expression vectors. This ensures a
high level of expression, which is not cell specific.
E. coli vector, check expression.
Yeast vector, transform into yeast.
Mammalian vector, transfect into mammalian cell line.
Gene expression
86
Check table to see if your gene is identified by immunological screening:
Vector
Signal ?
pBR322
no
lgt11, cDNA
yes
lgt11, genomic DNA
no
YEp213
no
pJP31
no
pMAMneo
no
amplicon
no
If you have a signal, click here.
If you do not have a signal, click here.
Immunological screening
87
You have no positive signals on your autoradiograph which indicates that your
protein is not expressed. Reasons for this are :
(1) You have cloned your gene into pBR322 or YEp213. Both of these vectors
are cloning vectors and not expression vectors and therefore your gene will not
be expressed. You must choose an expression vector. Click here to choose an
E.coli vector or here to choose a yeast vector.
(2) You have used a mammalian or yeast expression vector. These expression
signals will not work in E. coli, therefore you must isolate your gene by colony
hybridization. Click here.
(3) You have cloned genomic DNA into the λgt11 E. coli vector. Since E. coli
cannot remove introns, the gene is not expressed correctly. You must an
alternative host. Click here.
Protein not expressed
88
There are a number of different methods of transforming yeast cells. For
example, yeast cells can be made competent for DNA uptake by treatment with
lithium acetate. Then in the presence of a carrier DNA and polyethylene glycol
the DNA is taken up into the cell after heat shock treatment. Yeast cells can
also be transformed by electroporation.
Complementation selection of plasmid containing cells is then performed. Yeast
plasmids usually carry an auxotrophic marker gene, in this case the LEU2
gene. The yeast strain you have transformed requires leucine for growth since
it has a mutant leu2 gene. Therefore only cells which have plasmids will be
able to grow on leucine free media.
Proceed
Yeast transformation
89
There are a number of methods for transfecting mammalian cells. A commonly
used method is calcium phosphate co-precipitation. DNA is mixed with a
carefully buffered solution containing phosphate. Addition of CaCl2 results in the
formation of a fine precipitate of CaPO4 and DNA. The precipitate is pipetted
onto a monolayer of cells growing in a petri-dish and left on the cells for several
hours, during which time 10-1-10-4 cells take up DNA. The precipitate is then
removed from the cells, which are then incubated in fresh media for 30-48 hrs
transient expression. While incubation in selective media for 10-14 days stable
expression. It is also possible to transfect mammalian cell by electroporation.
Before going ahead with the transfection procedure you must choose a cell line.
Click here.
Mammalian cell transfection
90
Some proteins require specialised post-translational modifications which are
specific to certain cell types. Therefore if you tried to express these proteins in
other cell lines the resultant proteins would be inactive.
You must choose the cell line appropriate for the expression of your protein.
(1) hepatic cell line
(2) fibroblast cell line
(please note which cell line you have used)
After transfection you must select for cells which have taken up DNA. Click here.
Mammalian cell lines
91
There are two methods of selecting for plasmid-containing cells, dependent on the
plasmid marker used.
(1) pMAMneo carries the neo marker gene, which confers resistance
against the antibiotic G418. Therefore :
- mammalian cells plus plasmid (neo) will grow in the presence of G418.
- mammalian cells minus plasmid (neo) will not grow in presence of G418.
(2) Amplicon carries a dhfr gene which confers resistance to the drug,
methotrexate (MTX). Therefore if you are using a mammalian cell line
which does not have a DHFR gene:
- mammalian cells plus plasmid (dhfr) will grow in the presence of MTX.
- mammalian cells minus plasmid (dhfr) will not grow in presence of MTX.
Proceed
Cell selection
92
You have now transformed your vector, which carries your gene of interest, into
the appropriate host strain, and identified plasmid-containing colonies.
You must now check that the protein is expressed. Click here.
Summary
93
Western blotting can be used to detect the expression of a protein. A total
protein extract is obtained from a transformed cell. The proteins are then
separated by polyacrylamide gel electrophoresis. The proteins from the gel are
then transferred to a membrane and then probed with a labelled antibody
specific to that protein. Western blotting can also establish whether the protein
expressed is of the correct length.
Check the expression of your protein :
If using E. coli
- Click here
If using yeast
- Click here
If using mammalian cells
- Click here
Western blotting
94
You have cloned your gene into a λ expression vector, which should give you a
fusion protein of β-galactosidase and your protein. You will have to cleave the
fusion protein with cyanogen bromide in order to remove the β-galactosidase
part of the fusion protein.
Check the table to see if you have expressed a protein.
Plasmid
Enzyme site
Insert DNA
Protein
Expression ?
Go to
pBR322
PstI /BamHI
cDNA / genomic
Any
no
100
lgt11
EcoRI
cDNA
Somatostatin
yes
95
hEGF
yes
95
Factor IX
yes
95
Somatostatin
no
100
lgt11
Summary
EcoRI
genomic
95
Recombinant proteins are tested to see if they are biologically active by a
number of means dependent on the type of protein. For example, Factor IX
protein activity is determined by a blood clotting assay. EGF activity is
determined by a competitive receptor binding assay.
Check table to see if your protein is biologically active.
Plasmid
Enzyme site
Insert DNA
Protein
Active ?
Go to:
lgt11
EcoRI
cDNA
Somatostatin
yes
102
hEGF
no
101
Factor IX
no
101
Somatostatin
yes
102
genomic
Active protein?
96
You have cloned your gene into a yeast expression vector.
Check the table to see if you have expressed a protein.
Plasmid
Enzyme site
Insert DNA
Protein
Expression ?
Go to:
YEp213
BamHI
cDNA / genomic
Any
no
100
pJP31
HindIII
cDNA
Somatostatin
yes
97
hEGF
yes
97
Factor IX
yes
97
Somatostatin
yes
97
pJP31
Summary
HindIII
genomic
97
Recombinant proteins are tested to see if they are biologically active by a
number of means dependent on the type of protein. For example, Factor IX
protein activity is determined by a blood clotting assay. Whereas EGF activity is
determined by a competitive receptor binding assay.
Check table to see if your protein is biologically active.
Plasmid
Insert DNA
Protein
Active ?
Go to
pJP31
cDNA
Somatostatin
yes
102
pJP31
cDNA
hEGF
yes
102
pJP31
cDNA
Factor IX
no
101
pJP31
genomic
Somatostatin
yes
102
Active protein?
98
You have expression of all proteins (somatostatin, EGF and Factor IX), in all cell
lines used (hepatic and fibroblast). You must now check biological activity. Click
here.
Summary
99
Plasmid
Protein
Cell line
Activity
Go to
pMAMneo
Somatostatin
hepatic
transient
102
hEGF
transient
102
Factor IX
transient
102
transient
102
hEGF
transient
102
Factor IX
no
101
stable
102
hEGF
stable
102
Factor IX
stable
102
stable
102
hEGF
stable
102
Factor IX
no
101
pMAMneo
Amplicon
Amplicon
Somatostatin
Somatostatin
Somatostatin
fibroblast
hepatic
fibroblast
N.B. (1) Transient expression - due to high level of DNA in cell, the cell lyses after a few days, therefore the recombinant
protein is only expressed at a high level for a few days.
N.B. (2) Stable expression - recombinant DNA integrates into the genome, therefore there is stable expression of the
protein.
Active protein?
100
You have not been able to express your gene in this host - vector system. The
reasons for this are :
(1) You have cloned into pBR322 or YEp213. These are cloning vectors
which have no expression signals.
(2) Also the insert DNA does not have any expression signals therefore
there is no expression of the protein.
You must go back and choose another host or expression vector
- Click here.
N.B. Another possible reason for the absence of expression is that the
gene may have been inserted into the vector in the incorrect orientation,
therefore is unable to use the vectors promoter sequences. If the sequence or
restriction map of the gene is known then restriction mapping can be used to
determine whether this is the case.
Failure
101
You have managed to express your protein but it is not biologically active
therefore can not be used for medical purposes. There are a number of possible
reasons why your protein is inactive.
For example:
(1) You have expressed hEGF or Factor IX in E. coli.
Both these
proteins require post- translational modifications for them to be active - E. coli
cannot do these post-translational modifications hence they are inactive. You
need to choose a different host system
- Click here
(2) You have expressed Factor IX in yeast. Factor IX is a very complex
protein which requires a number of post -translational modifications which are
specific to mammalian hepatic cells. Yeast can only do simple post-translational
modifications, thus Factor IX is inactive when expressed in yeast. You need to
choose a different host system.
- Click here
(3) You have expressed your Factor IX in a non-hepatic cell line.
Therefore, the protein is not processed correctly.
Failure
- Click here
102
Well done !!!!
You have successfully produced your protein and it is
biologically active.
Producing a cell that synthesises large amounts of a desired protein is only the
first stage in achieving a useful process. It is important to be able to recover the
protein by a simple, economical method that results in high yields of a
biologically active protein. Cell cultures can be scaled up and grown in large
fermentors, and then the protein can be purified using a number of methods.
Success