aim

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!!!

Aim

1

Go to 2

http://www.rpi.edu/dept/chem-eng/Biotech-Environ/Projects00/rdna/rdna.html

Introduction

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

http://www.accessexcellence.org/RC/VL/GG/inserting.html

Introduction

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.

3

Go to 4

http://library.thinkquest.org/C004535/eukaryotic_cells.html

Protein 1: Somatostatin

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.

4

Go to 5

Learn more about Somatostasin

http://www.vivo.colostate.edu/hbooks/pathphys/endocrine/otherendo/somatostatin.html

Protein 2: hEFG

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.

5

Go to 6

Learn more about duodenal ulcers

http://www.emedicine.com/med/topic591.htm

Protein 3: Factor IX

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.

6

Learn More about Factor IX

Make a note of the protein you have chosen and Go to 7



Cloning the Gene

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

Choice of Library

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

cDNA 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.

http://dbtss.hgc.jp/docs/fullcdna.html

Genomic library

10A 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.

http://www.accessexcellence.org/RC/VL/GG/ecb/genomic_library.html

Source of mRNA

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 factorsorPancreatic Islets which synthesize insulin, glucagon and somatostatinorDuodenal cells which synthesize epidermal growth factor.

Make a note of the cell type you have chosen and Go to 12

http://www.cshprotocols.org/cgi/content/full/2006/2/pdb.prot4047

Primers

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 primer2. 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

Oligo(dC)

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.

13

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(dG)

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.

14

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(dT)

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.

15

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 (amino acid sequence)

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.

16

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

DNA polymerases

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.

17

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 IorReverse transcriptaseorDNA polymerase I

Make a note of your choice and then check first strand synthesis here.

Klenow fragment

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

http://www.vivo.colostate.edu/hbooks/genetics/biotech/enzymes/klenow.html


Reverse Transcriptase

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

http://www.vivo.colostate.edu/hbooks/genetics/biotech/enzymes/rt.html


DNA polymerase I

20

DNA polymerase I has a polymerase function and nuclease activity. The enzyme attaches to a short single stranded region (nick) in a mainly double-stranded 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

http://www.vivo.colostate.edu/hbooks/genetics/biotech/enzymes/dnap1.html

First 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.

To check DNA synthesis parallel reactions are set up.

Reaction 1: experimental - the sample which will go on to the 2nd strand synthesisReaction 2: test sample - to check efficiency of 1st strand synthesis.

21

14 (25)None(unless added polyC /polyG tail to template)

Reverse transcriptaseoligo(dG) /oligo(dC)

24DNA synthesisReverse transcriptaseoligonucleotide

25Correct DNA synthesisReverse transcriptaseoligo(dT)

19NoneDNA polymerase IAny

19NoneKlenowAny

14NoneAnyNone

Go toDNA synthesis ?Polymerase ?Primer ?ResultsConditions

14 (25)None(unless added polyC /polyG tail to template)

Reverse transcriptaseoligo(dG) /oligo(dC)

24DNA synthesisReverse transcriptaseoligonucleotide

25Correct DNA synthesisReverse transcriptaseoligo(dT)

19NoneDNA polymerase IAny

19NoneKlenowAny

14NoneAnyNone

Go toDNA synthesis ?Polymerase ?Primer ?ResultsConditions

Click here

Click here

Click here

Click here

Click hereClick here

No DNA synthesis

22

First strand synthesis requires the use of an oligo(dT) primer.

To try again click hereTo proceed click here

Oligo(dT)


23


Return

No DNA synthesis

24

First strand synthesis requires the use of a reverse transcriptase, because the template is

RNA.



25



Return



cDNA synthesis

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 hereTo proceed click here

26

http://www.changbioscience.com/res/res/r5ssUntranslatedsRegi.htm

Second strand synthesis

27

You may:

Proceed immediately with second strand synthesis.

or

Denature the hybrid molecule with alkali.

or

Partially degrade the RNA strand with RNase H

You now have RNA-DNA hybrid molecules. You must now synthesise the 2nd DNA strand.

DNA G-C-G-G-T-A-G-A-T-G-C-A-G-A-A-RNA C-G-C-C-A-U-C-U-A-C-G-U-C-U-U-

Alkali denature

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 IMake a note of your choice and then check 2nd strand synthesis here.

Oligo(dC)

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.

29

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

30

Oligo(dG)

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

31

Oligo(dT)



Return

32

Oligo (amino acid sequence)

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

33

Klenow fragment

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’



34








35

DNA polymerase I

DNA polymerase I has a polymerase function and nuclease activity. The enzyme attaches to a short single stranded region (nick) in a mainly double-stranded 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



http://www.vivo.colostate.edu/hbooks/genetics/biotech/enzymes/dnap1.html

36

RNase H treatment

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 Ireverse transcriptase DNA polymerase I

Make a note of your choice and then check 2nd strand synthesis here.

37

Second strand synthesis

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

None Klenow (K) Incorrect synthesis

None DNA polymerase I (D) Correct DNA synthesis

Alkali None RT / K / D DNA synthesis

Any RT / K / D Correct DNA synthesis

None None / Any RT / K / D None Click here

Click here

Click here

Click here

Click here

Click here

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.


Incorrect DNA synthesis

37

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.

Hairpin loop

39


Genomic library

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

Cloning the gene

41The 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

Choosing a host

42

You may choose one of the following to act as your host organism:

E. coli

Yeast

A mammalian cell system

E. coli 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 hostConsider other options

Yeast 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 hostConsider other options

Mammalian cell 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 hostConsider other options

E. coli vectors

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

pBR322

47pBR322

DNA multicopy plasmidUsually used as a cloning vectorSelective markers :

- ampicillin resistance (Amp R)- tetracycline resistance (Tet R)

BamHI

PstI

Tet RAmp R

pBR322

Return

λgt11

48λgt11

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

Lac ZReturn

Yeast vectors

49


pJP31

or

YEp213


pJP31

50pJP31

Multicopy, DNA plasmidShuttle vector - bacterial and yeast origins of replicationSelective markers :

- Yeast : auxotrophic marker - LEU2 gene- Bacteria : Ampicillin resistance (Amp R)

Secretory, expression vector PstIEcoRI

EcoRI

EcoRIPstI

HinDIII

pJP31ADH terminator

- factor leader

- factor promoter

Return

YEp213

51YEp213

Multicopy, DNA plasmidShuttle 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 PstI

EcoRI

BamHIPstI

EcoRI

PstI

EcoRI

ORI STB

Tet R

Amp R

LEU2

YEp213Return

Mammalian cell vectors

52


pMAMneo

or

Amplicon


pMAMneo

53pMAMneoDNA plasmid, multicopyShuttle 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 mapReturn

pMAMneo plasmid map

54

SalI

PstI

MMTV P

pA SV40 P

neo

pA

Amp R

pMAMneo

Return

Amplicon

55AmpliconShuttle 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.

ReturnSee plasmid map

Amplicon plasmid map

56

BglII

PstI

PIpA

pAIDHFR

P

Amp RAmplicon

SV40

Return

Restriction enzymes

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

http://www.accessexcellence.org/AE/AEC/CC/restriction.html

Restriction enzymes

58

Enzyme Recognition sequence Complementary endsEcoRI (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.

Make a note of which enzymes you have used and proceed.

To review E. coli vectors click hereTo review Yeast vectors click hereTo review mammalian cell vectors click here

E. coli vectors

59

BamHI

PstI

Tet RAmp R

pBR322

EcoRI

Lac Zλgt11

Return

Yeast vectors

60

PstI

EcoRI

BamHIPstI

EcoRI

PstI

EcoRI

ORI STB

Tet R

Amp R

LEU2

YEp213

PstIEcoRI

EcoRI

EcoRIPstI

HinDIII

pJP31ADH terminator

- factor leader

- factor promoter

Return

Mammalian cell vectors

61

SalI

PstI

MMTV P

pA SV40 P

neo

pA

Amp R

pMAMneo

BglII

PstI

PIpA

pAIDHFR

P

Amp RAmplicon

SV40

Return

Alkaline phosphatase

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.

cDNA preparation

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.

Adaptors

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

Linkers

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

Restriction enzymes

66

Enzyme Recognition sequence Complementary endsEcoRI (E) G'AATTC E


PstI (P) CTGCA'G P






Return

Genomic DNA preparation

67You 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 endsEcoRI (E) G'AATTC E


PstI (P) CTGCA'G P






Note your choice, then click here.

Ligation

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

E.coli transformation

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.

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E. coli transformation

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

http://www.bio.davidson.edu/Courses/Molbio/MolStudents/spring2003/McCord/electroporation.htm

Antibiotic selection

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.

DNA plasmids

72

SalI

PstI

MMTV P

pA SV40 P

neo

pA

Amp R

pMAMneo

BglII

PstI

PIpA

pAIDHFR

P

Amp RAmplicon

SV40

PstI

EcoRI

BamHIPstI

EcoRI

PstI

EcoRI

ORI STB

Tet R

Amp R

LEU2

YEp213

PstIEcoRI

EcoRI

EcoRIPstI

HinDIII

pJP31ADH terminator

- factor leader

- factor promoter

BamHI

PstI

Tet RAmp R

pBR322

Return

Ampicillin selection

73

Plasmid Vectorsticky ends

Insert DNAsticky ends

Phosphatasetreated

Number ofcolonies

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

YEp213 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

pJP31 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 106N.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

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.

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.



Phosphatasetreated

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

Amplicon 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

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

75If 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.



Phosphatasetreated

Number ofcolonies

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 B /Bg /S /H /E /P /Sa no 5 x 106

YEp213 PstI /EcoRI E /H /S /B /Bg /Sa /P yes / no None

BamHI B /Bg /S 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

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

Transfection

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

Insertional inactivation

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.

Vectorsticky

ends


Phosphatasetreated

White plaques Blue plaques

EcoRI E yes 1 x 105 none

EcoRI P /E / H/ B /Bg /S yes none none

EcoRI E no none 1 x 105

N.B. E= EcoRI, P= PstI, H= HindIII, B= BamHI, Bg= BglII, S= Sau3A

Click here

Click here

Click here

Mini-prep

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.

Failure

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

Ligation

80

Failure

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

Screening

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


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.

http://www.pnas.org/cgi/reprint/72/10/3961

http://filebox.vt.edu/users/chagedor/biol_4684/Methods/colhyb.html


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.

http://www.pubmedcentral.nih.gov/articlerender.fcgi?artid=392640


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.

Gene expression

85You 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.


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.

Protein not expressed

87You 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.

Yeast transformation

88There 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

Mammalian cell transfection

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 lines

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.

Cell selection

91There 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

Summary

92You 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.

Western blotting

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 hereIf using yeast - Click hereIf using mammalian cells - Click here

http://www.molecularstation.com/protein/western-blot/

Summary

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 topBR322 PstI /BamHI cDNA / genomic Any no 100lgt11 EcoRI cDNA Somatostatin yes 95

hEGF yes 95Factor IX yes 95

lgt11 EcoRI genomic Somatostatin no 100

Active protein?

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 101Factor IX no 101

genomic Somatostatin yes 102

Summary

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 100pJP31 HindIII cDNA Somatostatin yes 97

hEGF yes 97Factor IX yes 97

pJP31 HindIII genomic Somatostatin yes 97

Active protein?

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 topJP31 cDNA Somatostatin yes 102pJP31 cDNA hEGF yes 102pJP31 cDNA Factor IX no 101pJP31 genomic Somatostatin yes 102

Summary

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.

Active protein?

99

Plasmid Protein Cell line Activity Go topMAMneo Somatostatin hepatic transient 102

hEGF transient 102Factor IX transient 102

pMAMneo Somatostatin fibroblast transient 102hEGF transient 102

Factor IX no 101Amplicon Somatostatin hepatic stable 102

hEGF stable 102Factor IX stable 102

Amplicon Somatostatin fibroblast stable 102hEGF stable 102

Factor IX no 101

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.

Failure

100You 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

101You 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. - Click here

Success

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.

aim

Documents

protein activity

active protein

type of protein

total protein extract

gene cloning tutorial

dna synthesisalkalinonert

complementary dna strand

strand synthesis