Basis of course Understand the technology Understand the
Basis of course • • Understand the technology Understand the terminology Gain some practical experience The applications in biotechnology and basic biology next year • Why? – Fluidity next year (don’t need to explain terms and technology while discussing the applications) – Shows what molecular biology projects are like • Disadvantage – Technology dominated. Can make it a bit boring
Lectures • Two per week, however – The Tuesday lecture may be used to discuss data from practicals – In some lecture slots the in course test will be held – Approximately formal 15 lectures
Practicals • I will supervise the first two, Dianne Ford the second two • Objectives of practicals – Get used to pipetting small quantities – Reinforce lectures – Help focus on project choice for nexy year – Schedules soon
BNS 216 • Phases of course – Isolation of a specific DNA sequence (gene or c. DNA) – Analysis of isolated DNA sequence • DNA sequencing – Manipulation of DNA sequence • PCR to introduce restriction enzyme site • PCR to change codon • PCR to detect specific DNA sequence
BNS 216 – Production of recombinant protein • Expression vectors – Detecting genes and transcripts • Northern hybridization • Southern hybridization • RTPCR – Manipulation of eukaryotic organisms
BNS 216 • Isolation of a gene – Construction of a gene library • • • Choice or organism Restriction enzymes DNA ligase Vectors and which ones to use Screening startegies – Prokaryotic and you look for the protein product – Eukaryotic and you look for the DNA
BNS 216 • Isolation of a c. DNA – Purification of m. RNA – DNA from m. RNA – Construction c. DNA library – Screening the library Analysis of DNA • DNA sequencing • Why you need to know the sequence • Automated method with fluorescent dideoxynucleotides
BNS 216 • Manipulation of DNA – PCR and how it relies on knowing the sequence • Introduction of restriction enzyme sites • Changing the amino acid sequence of a protein • Expression of proteins in bacteria • Expression vectors – How they enable foreign genes or c. DNAs to be expressed – How expression vectors are regulated
BNS 216 • Detecting transcription and genes – Northern hybridization – RTPCR – Southern hybridization • Manipulation of eukaryotic multicellular organisms – Transgenic animals • Insertion of foreign genes by microinjection • Inactivating genes by homologous recombination in stem cels – Transgenic plants
Assessment • Exam: – 60 % of module – Answer 3 questions from 5 – Must get >35 % to pass module • Practicals – 20 % of module – Four practicals starting on 4 th February
Assessment • In course tests – Four tests • Test one: Isolation of a gene or c. DNA • Test two: DNA sequencing and PCR • Test three: Expression vectors and nucleic acid detection • Test four: Manipulating eukaryotic organisms
BNS 216 references • Difficult! – Gene Cloning: An introduction T. A. Brown OK quite simple but not in same way I teach it – Principles of Gene Manipulation: An introduction to genetic engineering R. W. Old and S. B. Primrose. Quite detailed, some of which is unnecessary – Use any standard molecular biology or genetics text book, there will a section on BNS 216
What is genetic engineering or recombinant DNA technology? • A suite of technologies that enable you to • Isolate and characterise genes • Produce and characterise proteins • Alter the genetic make up of an organism – New genes – Loss of existing genes
Applications of genetic engineering • Basic understanding of biology – Protein structure and function – Regulation of gene expression – Importance of proteins in whole organisms (gene knockouts or null mutations)
Applications of genetic engineering • Practical applications – Production of industrially important proteins – Change the properties of proteins – Modification of the phenotype of whole organisms – Diagnosis – Primary applications in medicine and agriculture – Others include chemical, paper and detergent industries
Applications • More details on applications • Protein production • Pharmaceutical proteins. – Constant supply and safe – Growth hormone, insulin, Factor VIII and IX, antitrypsin
Proteins • Microbial proteins • Microbes grow poorly but produce valuable enzymes – Hyperthermophiles – Anaerobes – Archaebacteria • Genetic engineering makes these proteins available to industry
Enzymes used in the Food Industry • Glucose isomerase (Food industry) • Xylanases (Paper industry) • Cellulases (Energy and detergent industries) • Phytases (animal feed) • Protein engineering – Rational design – Forced protein evolution
Modifying organisms • Genetically engineered foods – Herbicide resistance – Pesticide resistance • Good for consumer, farmer or biotech. Companies? – Golden rice with increased vitamin A and oil seed rape with better polyunsaturated fats • Good for consumer?
Genetically engineered foods • Risks – The environment, spread of resistant weeds, alter ecological balance? – Human health. Will we get increased antibiotic resistance – Will the transgene be deleterious to human health?
Change phenotype of farm animals • Convert them into bioreactors to produce pharmaceutical proteins. Why? • Change their biochemistry so • More efficient use of nutrients • Better quality end-products e. g. milk and meat – Humanising milk – Increase polyunsaturates
Change phenotype of small animals • Generate animals for human disease influenced by diet • Colon cancer • nv. CJD • Heart disease
Gene therapy • Correct genetic defect – Not in germ line – Not transmissible
Diagnosis • Diagnosis – Human genome sequenced – Identify all genes soon – Immediate diagnosis test • Hungtintons • Muscular dystrophy • Cystic fibrosis • Sickle cell anaemia • Alzeihmer • Breast cancer • Colon cancer • Heart disease – Good or bad?
Diagnosis • Reduce incidence of disease – Pregnancy termination – Pre-implantation selection • Start treatment to prevent disease – Prophylactic mastectomy – Colon removal – Physiotherapy • Stress if diagnosed. Do you want to know? • Insurance and job prospects?
Genetic engineering history • Pioneered by Cohen and Boyer 19721974 (bacterial systems) • Southern hybridization 1975 • DNA sequencing 1977 -1980 • Transgenic animals 1980 • Polymerase chain reaction 1985 • Site-directed mutagenesis 1985
Where do we start? • If we want to do genetic engineering how do we start? • Isolate the gene of interest – Select organism containing gene – Construct a gene library – Select members of the gene library that contain the gene of interest
How do you start doing recombinant DNA technology? • Isolate the gene of interest • Lets isolate (clone) a cellulase gene – Identify organism that contains the gene • • • Rumen compost Soil Leaf litter Decaying wood
Isolating a cellulase gene Isolate chromosomal DNA Fragment DNA
Mix and ligate Vector E. coli Transform Gene library
Properties of vector DNA • • Replicates in bacterium Foreign DNA inserted will be stable Normally extra-chromosomal Easy to select bacterium containing vector (confers antibiotic resistance) • Vectors – Plasmid (extra chromosomal circular DNA) – Bacteriophage – Cosmids – Artificial chromosomes
Gene library Screen library for appropriate gene (cellulase gene) Isolate plasmid
Isolating a cellulase gene Isolate chromosomal DNA Fragment DNA
Restriction endonucleases • Enzymes that cut DNA at specific sequences • Discovered in the early 1950 s • Agent that enables bacteria to be immune to bacteriophage • Host-controlled restriction • Mainly found in bacteria • Over 1200 characterised
Restriction endonucleases • Three types only Type II important in genetic engineering as they cut the sequence they recognise • Target sequences generally palindromic • Recognise 4, 6 or 8 nucleotides
Restriction endonucleases GAATTC CTTAAG Eco. RI CTGCAG GACGTC Pst. I CCCGGG GGGCCC Sma. I G AATTC CTTAA G Sticky ends CTGCA G G ACGTC CCC GGG CCC Blunt ends
Restriction endonucleases • Named after organism • e. g. Eco. RI = Escherichia (E) coli (co) strain R (R). I refers to Ist enzyme isolated from organism • Why doesn’t a bacterial restriction endonuclease digest its own DNA? • The bacterium produces a DNA methylase that recognises same sequence as restriction endonuclease • Methylates target DNA sequence which makes it resistant to endonuclease cleavage
Restriction enzymes and DNA methylase Foreign DNA GAATTC CTTAAG Eco. RI G AATTC CTTAA G Host DNA GAATTC CTTAAG DNA methylase CH 3 GAATTC CTTAAG CH 3 Eco. RI CH 3 GAATTC CTTAAG CH 3
Mix and ligate Vector E. coli Transform Gene library
Forming hybrid or recombinant DNA molecules using restriction enzymes and DNA ligase GAATTC CTTAAG Digest with Eco. RI G AATTC CTTAA G Mix DNA GAATTC CTTAAG DNA ligase GAATTC CTTAAG G AATTC CTTAA G GAATTC CTTAAG Recombinant or hybrid DNA GAATTC CTTAAG
Inserting chromosomal DNA into a vector Chromosome Vector GAATTC CTTAAG Cut with Eco. RI and add DNA ligase Recombinant vector GAATTC CTTAAG
More details on each stage • Chromosomal DNA is only partially cut because? • Don’t know if the restriction enzyme cuts in the gene • Plasmid vector is designed to enable selection for recombinant plasmid – p. UC or p. Bluescript-based plasmid vectors – Contains two selection genes ampicillin (antibiotic) and Lac. Z; codes for galactosidase
Hind. III Eco. RI p. UC 18 Bam. HI Cells containing p. Bluescript are ampicillin resistance and blue on X-Gal Origin of replication Lac. Z’ encodes -galactosidase Ampr confers ampicillin resistance
Bromo-chloro-indoyl- -galactopyranosidase or X-Gal (Clear) -galactosidase Bromo-chloro-indoyl (Deep blue insoluble) + galactose
Inserting chromosomal DNA into a vector Chromosome GAATTC CTTAAG Vector GAATTC CTTAAG Cut with Eco. RI and add DNA ligase Recombinant vector GAATTC CTTAAG Ampicillin resistant; -galactosidase negative (White on X-Gal) Lac. Z gene codes for -galactosidase Ampicillin resistance gene
Wild type vector GAATTC CTTAAG Ampicillin resistant; -galactosidase active (Blue on X-Gal) Lac. Z gene codes for -galactosidase Ampicillin resistance gene
E. coli sensitive to ampicillin Ampicillin resistant; -galactosidase active (Blue on X-Gal) Ampicillin resistant; -galactosidase negative (White on X-Gal)
Bacteria from ligation plated on ampicillin and X-Gal Contains wild type plasmd Contains recombinant plasmd
Gene library • Collection of microbes (e. g. Escherichia coli) each one containing a recombinant vector • Each recombinant vector contains a random region of the target chromosome • The number of microbes in the library is large • Thus any gene in the target organism’s genome is present in at least one member of the gene library
Mix and ligate Vector E. coli Transform Gene library
Size of gene library N = ln(1 -P) ln (1 -A/B) N = Number of clones P = 95 % probability of finding gene A = Average size of DNA fragments B = Total size of genome E. coli has genome of 4, 800, 000 nucleotides Average size of insert is 10, 000 nucleotides Number of clones for 95 % probability is 1700
Size of gene library • If genome is large e. g. human genome (3 x 109) then number of clones to make library becomes unrealistic (1058000) if using a plasmid vector (accepts only 10 kb as larger DNA can’t be transformed) • Therefore need to use vectors that can accept larger pieces of DNA – I. e. if each vector contains a large piece of DNA you don’t need so many clones to make a gene library
Vectors that accept larger DNA • • Plasmid: 10 kb Lambda bacteriophage: 18 -25 kb Cosmid: ~40 kb Yeast or bacterial artificial chromosome: 100 -1000 kb
Gene library Screen library for appropriate gene (cellulase gene) Isolate plasmid
Screening gene library for cellulase gene • Assume bacterial genes will express in Escherichia coli • Escherichia coli does not degrade polysaccharides • Screen library by looking for members that degrade cellulose • Similar approach for other polysaccharidases (amylases, pectinases, xylanases etc)
Vectors • Lambda vector • Infects E. coli replicates and then viruses released • End of genome are 12 bp sequences known as cos sequences. • Cos sequences play an important role in packaging viral DNA into capsids (head of the virus)
Lambda infects E. coli DNA injected into E. coli DNA replication generating concatamers
Lambda DNA is linear in virus Cos sequence is 12 nucleotides and single stranded The two cos sequences are complementary cos sequences hybridise in E. coli to form circular genome cos lambda DNA cos etc Replicates by rolling ciricle in E. coli to produce concatemers
Lambda head genes transcribed and translated to produce head proteins
Endoglucanase A expressed. Cuts DNA at cos sequence and assists packaging lambda DNA into viral capsid (head proteins) Endoglucanase A cuts at cos sequence
Tail genes then expressed
Tail bind to heads to form virus
Lambda virus produces lysozyme that hydrolyses bacterial cell wall releasing viruses to attack other bacterial cells
Lambda infects E. coli DNA injected into E. coli DNA replication generating concatamers Replication by a mechanism called rolling circle
Vectors • More information on vectors • Lambda vector • Libraries contain larger inserts than plasmids (20 -25 kb) • Naked Lambda DNA can’t be transformed into E. coli • Lambda DNA can be packaged into a virus • Virus then infected into E. coli
In vitro packaging Two E. coli mutants. One synthesises the tails the other the heads and A protein 1. Take DNA and mix with E. coli extracts containing heads and A protein Cos sequence Internal DNA + 2. Packaged DNA is then mixed with E. coli extracts containing tails 3. The virulent phage can then be used to infect E. coli to form plaques in a lawn of bacteria
Lambda vector for genomic cloning 1. Lambda genome is 40 kb 2. Lambda vectors contain right arm, left arm and central region. At ends are single stranded cos sequences 3. Genes in central region not essential 4. Restriction sites at boundary of central region and the two arms. 5. Used to clone DNA about 20 kb
Cloning DNA into vectors Left arm Central Right arm cos Cut out 3 regions and purify arms Chromosomal DNA Cut chromosome with same enzyme as lambda Mix DNA and add DNA ligase DNA will form concatamers Will not package Recombinant lambda will package. Why?
DNA packaging is size dependent Endonuclease A cleaves cos sequence cos DNA too large to package >25 kb DNA too small to package <18 kb DNA packages 18 -25 kb
Cosmid vectors • Problem with vectors is that you can’t transform large pieces of DNA into E. coli • Cosmid vector Cos Ampr – Similar to plasmid • Ampicillin resistance gene • 5 kb in size • Unique Bam. HI site • Cos sequence Origin Bam. HI
Cosmid vectors • Accepts as much as 40 kb of chromosomal DNA • Why? – Smuggle cosmid into E. coli by packaging it into lambda virus as it has a cos sequence
Cosmid vector Cos GGATCC CCTAGG Ampr Cut with Bam. HI Origin Bam. HI GATCC G GGATCC CCTAGG GATCC G G CCTAG Mix DNA at very high concentration and add DNA ligase GGATCC GGATCC GGTACC GGATCC Packaged in vitro
Lambda particle injects cosmid in E. coli is viable as no lambda genes in cosmid so it acts as a normal plasmid How do you select for E. coli cells containing cosmids? They are resistant to ampicillin
Telomere Centromere Telomere Origins of replication Accepts up to 1 Mb of chromosomal DNA Yeast Artificial Chromosome (YAC) Accepts up to 1 Mb of DNA origin of replication centromere sup 4 gene Sna. BI site Trp 1 gene Bam. HI sites Ura 3 gene Telomeres
Yeast Artificial Chromosome origin of replication Chromosomal DNA centromere sup 4 gene Ura 3 gene Sna. BI site Telomeres Trp 1 gene Bam. HI sites Cut with Sna. BI Mix and add DNA ligase Cut with Sna. BI and Bam. HI Telomere Trp 1 Ori Cent sup 4’ 0. 1 -1 Mb chromosomal DNA sup 4’ Ura 3 Telomere Transform recombinant YACs into mutant yeast that lack Ura 3 and Trp 1 genes (I. e. can’t make tryptophan and uracil) so the amino acid and nucleotide must be added for yeast to grow
Transformed yeast plated on media lacking tryptophan and uracil Colony contains recombinant YAC Colony contains wild type YAC Plate yeast on medium lacking uracil and tryptophan. Yeast colonies that grow contain YAC and those that are pink contain recombinant YAC as this indicates inactivation of Sup 4
What vectors for what libraries? Human library requires >1, 000 plasmid clones Human library requires 14000 YAC clones
Screening bacterial gene library for a specific gene • Assume bacterial genes will express in Escherichia coli • Screen for gene by looking for protein that – Changes phenotype of E. coli • e. g. confers ability to degrade a polysaccharide (cellulase, xylanase etc) confers green fluorescence (green fluorescent protein) – Changes phenotype of an E. coli mutant. Known as complementation – Screen for protein directly using an antibody
Complementation • Isolate gene from a bacterium that E. coli contains – e. g. Isolate B. subtilis gene coding 1 st enzyme in leucine biosynthsis pathway • Assume bacteria have same pathway for leucine synthesis • Step 1: Isolate a mutant of E. coli unable to synthesise leucine – I. e. gene for 1 st enzyme non-functional thus enzyme not produced
Complementation (generating a leucine auxotroph) Add mutagen E. coli culture Plate out on media containing leucine Mutated E. coli culture Colony 4 requires leucine for growth (leucine auxotroph) Transfer colonies to media lacking leucine
Complementation • Leucine auxotrophs isolated require leucine for growth – Defect in gene coding for leucine synthesis • Which gene in pathway mutated? • Assay leucine auxotrophs for mutant lacking enzyme one
Complementation • Construct a gene library of Bacillus subtilis chromosomal DNA – Digest chromosome – Insert into E. coli vector to make a library of recombinant vectors – Transform these recombinant vectors into the E. coli leucine auxotroph lacking enzyme 1 – Plate out library on media lacking leucine
Complementation • E. coli colony that grows contains recombinant plasmid with Bacillus leucine biosynthetic gene that codes for 1 st enzyme in leucine synthesis • The recombinant plasmid has overcome or complemented the E. coli mutation
Antibody screening • Some bacterial genes code for proteins that don’t change phenotype of E. coli wild type or mutant e. g. asparaginase from Erwinia • Erwinia asparaginase converts asparagine to aspartic acid • Used as a major ingredient of chemotherapy of childhood acute leukaemia
Antibody screening • Porton needed to produce more to supply world demand – Overexpress enzyme in E. coli • How do you isolate gene? • Make a gene library of Erwinia DNA in E. coli • Screen library using antibody specific to Erwinia asparaginase • How do you produce antibody?
Antibody Production Erwinia Purify asparaginase from Erwinia Blood of rabbit contains antibodies to Erwinia asparaginase but no other Erwinia protein
Screening E. coli library of Erwinia DNA 1 2 3 4 5 6 7 8 9 Lyse bacteria with lysozyme Protein binds to filter and thus a print of the bacterium’s proteins replaces colony 1 2 3 4 5 6 7 8 Plate out E. coli cells on nylon filter 9 Add inert protein (e. g. bovine serum albumin) to block protein binding sites on nylon filter 1 2 3 4 5 6 7 8 9
Screening E. coli library of Erwinia DNA 1 2 3 4 5 6 7 8 Add antibodies labelled with radioactive iodine 9 www. staff. ac. ncl/h. j. gilbert 1 2 3 4 5 6 7 8 9 Expose filter to X-ray film to detect where antibody has bound (radioactive and thus blackens X-ray film)
Practical information • Experiment 1: Each phage has an insert of approximately 20 kb • The fact that the cellulase gene is 1 kb not really relevant
Screening a eukaryotic gene library • Will the gene express? – No, lack of promoter, ribosome binding sequence and introns • Use nucleic acid hybridisation
Nucleic acid hybridization 1 2 3 4 5 6 7 8 Na. OH added to lyse bacteria 1 and denature DNA into single strand. Filter is heated to 70 to immobilise 4 single strand DNA 7 9 Plate out E. coli cells on nylon filter 2 3 5 6 8 9 Single stranded print of bacterium’s DNA instead of a bacterial colony 1 2 3 4 5 6 7 8 9 Inert DNA to block vacant DNA binding sites on filter
Nucleic acid hybridization 1 2 3 4 5 6 7 8 Add radioactive nucleic acid probe that binds to target gene 9 1 2 3 4 5 6 7 8 9 Expose filter to X-ray film to detect where nucleic acid has bound (radioactive and thus blackens X-ray film)
Screening a eukaryotic gene library • Where does the probe come from? – Homologous gene from other organism – Oligonucleotide based on protein sequence or known sequence of homologous gene – Differential screen (deal with this later; microarray analysis)
Screening a eukaryotic gene library • Homologous gene from other organism – Mammalian genes are very similar – Thus if trying to get human gene screen with the equivalent gene from another organism – Oligonucleotide based on protein sequence or known sequence of homologous gene – Purify protein and determine sequence – Build a nucleotide sequence which codes for protein sequence
Amino acid sequence Met-Asn-Lys-Trp-Glu-Met = ATG; Asn = AAT or AAC; Lys = AAA or AAG; Trp = TGG; Glu = GAA or GAG How many probes must we make? ATG AAT AAA TGG GAA ATG AAT AAA TGG GAG ATG AAT AAG TGG GAA ATG AAT AAG TGG GAG ATG AAC AAA TGG GAA ATG AAC AAA TGG GAG ATG AAC AAG TGG GAA ATG AAC AAG TGG GAG ATG
1 st Test in BNS 216 • Test is next Thursday February 26 th • One hour replaces lecture • Multiple choice and single word or simple diagram answers • Consists of – Construction and screening of gene libraries
Construction of gene libraries • Gene library – Restriction endonucleases – DNA ligase – Vectors • • Plasmid Bacteriophage (lambda) Cosmid Yeast artificial chromosome
Screening • Bacterial gene library – Depends on protein expression • Phenotype change e. g. cellulase gene • Complementation of E. coli mutant • Detection of gene product using antibody • Eukaryotic gene library – No protein expression expected – Nucleic acid hybridisation – Probe • Homologous gene from other organism • Oligonucleotide probe based on protein sequence
Cloning c. DNA • c. DNA: complementary or copy DNA • It is derived from m. RNA • double stranded DNA with identical sequence to m. RNA in one strand • Why clone c. DNA – Important in expressing eukaryotic proteins in bacteria – Introns not removed in bacteria – Therefore not protein produced
Eukaryotic organism Bacteria Transcription RNA Splicing m. RNA Translation Functional protein Inactive protein
Bacteria c. DNA Transcription m. RNA Translation Functional protein
Isolate specific c. DNA (encodes growth hormone) • How do isolate m. RNA encoding growth hormone • Isolate cells that express growth hormone – Pituitary
Isolate specific c. DNA (encodes growth hormone) • • Extract m. RNA Convert to c. DNA Construct c. DNA library Screen library for members containing c. DNA encoding growth hormone
Pituitary Isolate total m. RNA AAAAAA AAAAAA Encodes GH
Purify m. RNA AAAAAA Matrix is polyd. T cellulose TTTTTT AAAAAA Pass total RNA over matrix. Only m. RNA will bind. Other RNAs will pass directly through Ribosomal RNA etc
TTTTTT AAAAAA Low salt buffer AAAAAA TTTTTT + TTTTTT AAAAAA
c. DNA synthesis • Use three enzymes – reverse transcriptase – RNAase H – DNA polymerase • Reverse transcriptase – Converts RNA into a complementary DNA sequence – Add onto existing double stranded nucleic acid – Can’t initiate DNA synthesis – Use primer to initiate DNA synthesis
primer 5’ 3’m. RNA 5’ 3’ Reverse transcriptase 5’ 3’ 3’ 5’
Primer • Small single stranded DNA sequence • Binds to 3’ end of all m. RNA • What is the sequence? – Polyd. T
m. RNA 5’ m. RNA c. DNA 5’ 3’ 3’ AAAAAA Add primer TTTTT 3’ AAAAAA TTTTTT Reverse transcriptase + d. NTPs 3’ AAAAAA 5’ TTTTTT RNAase H 3’ AAAAAA 5’ TTTTTT DNA polymerase + d. NTPs 3’ AAAAAA 5’ TTTTTT Completed DNA synthesis 3’ AAAAAA 5’ TTTTTT Double stranded c. DNA
Collection of c. DNA molecules • Insert c. DNAs into a vector – Generate recombinant vectors • Insert recombinant vectors into a bacterium such as Escherichia coli • This comprises a c. DNA library derived from pituitary m. RNA • Screen c. DNA library for those containing growth hormone encoding c. DNA
c. DNA is blunt-ended thus ligates to other DNA unless concentration high c. DNA GAATTC CTTAAG Add linkers at high concentrations and DNA ligase GAATTC CTTAAG Cut with Eco. RI AATTC G G CTTAAG Insert into a vector cut with same enzyme followed by DNA ligase
Linkers • Small complementary oligosaccharides – Synthesised in large amounts • Contain sequence cleaved with a restriction enzyme • Blunt ended double strand molecules • Because synthesised at high concentrations can be ligated efficiently to other double stranded molecules (c. DNA molecules) • Once ligated to c. DNA cut with restriction enzyme to create sticky ends
c. DNA is blunt-ended thus ligates to other DNA unless concentration high c. DNA GAATTC CTTAAG Add linkers at high concentrations and DNA ligase GAATTC CTTAAG Cut with Eco. RI AATTC G G CTTAAG Insert into a vector cut with same enzyme followed by DNA ligase
Transform into E. coli
Screen for members of library that contain growth hormone c. DNA
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