Aim

1. 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. 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. 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. 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. 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. 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 Make a note of the protein you have chosen and Go to 7 Protein 3: Factor IX

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. 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. 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. 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. 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. 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: • No primer • Oligo(dC) • Oligo(dG) • Oligo(dT) • An oligo synthesised from known amino acid sequence Make a note of your choice and Go to 17 Primers

13. 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. 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. 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. 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. 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. You may choose to use: Klenow fragment of DNA polymerase I or Reverse transcriptase or DNA polymerase I Make a note of your choice and then check first strand synthesis here. DNA polymerases

18. 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. 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. 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 DNA polymerase I

21. 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. 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. Click here Click here Click here Click here Click here Click here First strand synthesis

22. 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. 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. 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. 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. 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. 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-U- DNA 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. 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 Make a note of your choice and then check 2nd strand synthesis here. Alkali denature

29. 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. 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. 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. 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. 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. 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. 35 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 to ‘Alkali denature’ Return to ‘RNase H treatment’ DNA polymerase I

36. 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. 37 You may check 2nd strand synthesis the same way as 1st strand synthesis. Click here Click here Click here Click here Click here Click here Second strand synthesis

38. 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. 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. 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. 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. 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. 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. 44 Advantages of yeasta 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 yeasta 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 yeastas your host Consider other options Yeast host

45. 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 systemas your host Consider other options Mammalian cell host

46. 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. 47 pBR322 DNA multicopy plasmid Usually used as a cloning vector Selective markers : - ampicillin resistance (Amp R) - tetracycline resistance (Tet R) Return pBR322

48. 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. Return λgt11

49. 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. 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 Return pJP31