Basis of course

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

BNS216• Phases of course

– Isolation of a specific DNA sequence (gene or cDNA)

– 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

BNS216– Production of recombinant protein

• Expression vectors

– Detecting genes and transcripts• Northern hybridization• Southern hybridization• RTPCR

– Manipulation of eukaryotic organisms

BNS216

• 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

BNS216

• Isolation of a cDNA– Purification of mRNA– DNA from mRNA– Construction cDNA library– Screening the library Analysis of DNA

• DNA sequencing• Why you need to know the sequence• Automated method with fluorescent

dideoxynucleotides

BNS216

• 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 cDNAs to be

expressed– How expression vectors are regulated

BNS216• 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 4th February

Assessment

• In course tests– Four tests

• Test one: Isolation of a gene or cDNA• Test two: DNA sequencing and PCR• Test three: Expression vectors and nucleic acid

detection • Test four: Manipulating eukaryotic organisms

BNS216 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 BNS216

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

• nvCJD

• 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 1972-1974 (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 chromosomalDNA

Fragment DNA

Gene library

E. coli

Vector

Mix and ligate

Transform

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 libraryfor appropriate gene(cellulase gene)

Isolate plasmid

Isolating a cellulase gene

Isolate chromosomalDNA

Fragment DNA

Restriction endonucleases• Enzymes that cut DNA at specific

sequences

• Discovered in the early 1950s

• 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


GAATTCCTTAAG

CTGCAGGACGTC

CCCGGGGGGCCC

G AATTCCTTAA G

CTGCA GG ACGTC

CCC GGGGGG CCC

EcoRI

PstI

SmaI

Sticky ends

Blunt ends


• Named after organism• e.g. EcoRI = 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

GAATTCCTTAAG

GAATTCCTTAAG

GAATTCCTTAAG

CH3

CH3

GAATTCCTTAAG

CH3

CH3

EcoRI DNA methylase

EcoRI

G AATTCCTTAA G

Foreign DNA Host DNA

Gene library

E. coli

Vector

Mix and ligate

Transform

GAATTCCTTAAG

GAATTCCTTAAG

G AATTCCTTAA G

G AATTCCTTAA G

GAATTCCTTAAG

GAATTCCTTAAG

GAATTCCTTAAG

Forming hybrid or recombinant DNA moleculesusing restriction enzymes and DNA ligase

GAATTCCTTAAG

GAATTCCTTAAG

GAATTCCTTAAG

Recombinant or hybrid DNA

DNA ligaseDNA ligase

Digest with EcoRI

Mix DNA

Inserting chromosomal DNA into a vector

GAATTCCTTAAG

GAATTCCTTAAG

GAATTCCTTAAG

GAATTCCTTAAG

GAATTCCTTAAG

Vector

Chromosome

Cut with EcoRI and add DNA ligase

Recombinant vector

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– pUC or pBluescript-based plasmid vectors– Contains two selection genes ampicillin

(antibiotic) and LacZ; codes for -galactosidase

pUC18

LacZ’ encodes -galactosidase

Ampr confers ampicillin resistance

Origin of replication

EcoRIHindIII BamHI

Cells containingpBluescript areampicillin resistanceand blue on X-Gal

Bromo-chloro-indoyl--galactopyranosidase or X-Gal (Clear)

Bromo-chloro-indoyl (Deep blue insoluble)

+

galactose

-galactosidase

Inserting chromosomal DNA into a vector

GAATTCCTTAAG

GAATTCCTTAAG

GAATTCCTTAAG

GAATTCCTTAAG

GAATTCCTTAAG

Vector

Chromosome

Cut with EcoRI and add DNA ligase

Recombinant vector

Ampicillin resistant; -galactosidase negative (White on X-Gal)

LacZ gene codes for -galactosidase

Ampicillin resistance gene

GAATTCCTTAAG

Wild type vector

Ampicillin resistant; -galactosidase active (Blue on X-Gal)

LacZ 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 platedon ampicillin and X-Gal

Containswild typeplasmd

Containsrecombinantplasmd

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

Gene library

E. coli

Vector

Mix and ligate

Transform

Size of gene library

N = ln(1-P) ln (1-A/B)

N = Number of clones P = 95 % probability of finding geneA = Average size of DNA fragmentsB = Total size of genome

E. coli has genome of 4,800,000 nucleotidesAverage size of insert is 10,000 nucleotidesNumber 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 libraryfor 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)

DNA replication generatingconcatamers

Lambda infects E. coli

DNA injectedinto E. coli

Lambda DNA is linear in virusCos sequence is 12 nucleotides and single strandedThe two cos sequences are complementary

Replicates by rolling ciricle in E. colito produce concatemerscos lambda DNA cos etc etc

cos sequences hybridise in E.colito form circular genome

Lambda head genes transcribed and translatedto produce head proteins

Endoglucanase A cuts at cossequence

Endoglucanase A expressed. Cuts DNA at cos sequenceand assists packaging lambda DNA into viral capsid(head proteins)

Tail genes then expressed

Tail bind to heads to form virus

Lambda virus produces lysozyme that hydrolysesbacterial cell wall releasing viruses to attack other bacterial cells

DNA replication generatingconcatamers

Lambda infects E. coli

DNA injectedinto E. coli

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 packagingTwo E. coli mutants. One synthesises the tails the other the heads and A protein1. Take DNA and mix with E. coli extracts containing heads and A protein

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

+

Cos sequence

Internal DNA

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 Chromosomal DNA

Cut out 3 regionsand purify arms

Cut chromosome withsame enzyme as lambda

Mix DNA and addDNA ligase

Will not package Recombinant lambda will package. Why?

cos

DNA will formconcatamers

DNA packaging is size dependent

DNA too large to package >25 kb

DNA packages 18-25 kb

DNA too smallto package <18 kb

cos

Endonuclease Acleaves cos sequence

Cosmid vectors• Problem with vectors

is that you can’t transform large pieces of DNA into E. coli

• Cosmid– Similar to plasmid

• Ampicillin resistance gene

• 5 kb in size• Unique BamHI site• Cos sequence

Cos

Ampr

BamHI

Origin

Cosmid vector

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

Cos

Ampr

BamHI

GGATCC GGATCCCCTAGG CCTAGG

GATCC G G CCTAG

GGATCC GGATCC GGATCC GGATCC

GGATCC GGATCC GGTACC GGATCC

Cut with BamHI

GCCTAG

GATCC G

Cut with BamHI

Mix DNA at veryhigh concentrationand add DNA ligase

Origin

Packaged in vitro

Cosmid vector

How do you selectfor E. coli cells containingcosmids?

They are resistantto ampicillin

Lambda particleinjects cosmid in E. coli.E. coli is viable asno lambda genes in cosmid so it acts asa normal plasmid

Origins of

replication

CentromereTelomere Telomere

Accepts up to 1 Mb of chromosomal DNA

BamHI sites

sup4 gene

centromereoriginof replication

Trp1gene

SnaBI site

Yeast Artificial Chromosome (YAC)Accepts up to 1 Mb of DNA

Telomeres

Ura3 gene

0.1-1 Mb

Telomere TelomereTrp1 Ori Cent sup4’ chromosomal DNA sup4’ Ura3

Cut withSnaBI andBamHI

Mix and addDNA ligase

Transform recombinant YACs into mutant yeastthat lack Ura3 and Trp1 genes (I.e. can’t make tryptophanand uracil) so the amino acid and nucleotide mustbe added for yeast to grow

Cut with SnaBI

Telomeres

BamHI sites

sup4 gene

Ura3 gene

centromereoriginof replication

Trp1gene

SnaBI site

Yeast Artificial Chromosome Chromosomal DNA

Plate yeast on medium lacking uracil and tryptophan. Yeast colonies that grow contain YACand those that are pink contain recombinant YACas this indicates inactivation of Sup 4

Colony containswild type YAC

Colony containsrecombinant YAC

Transformed yeast plated on media lacking tryptophan and uracil

What vectors for what libraries?

Vector Insertsize

What libraries

Plasmid <10 kb Bacteria

Lambda phage 18-25 kb Yeast

Cosmid 34-45 kb Intermediateeukaryotes

YAC etc 0.1 – 1 Mb Higher Eukaryotes

Human library requires 14000 YAC clones

Human library requires >1,000,000 plasmid 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 1st 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 1st enzyme non-functional thus enzyme

not produced

Add mutagen

E. coli culture Mutated E.coli culture

Plate out on media containing

leucine

Transfercoloniesto medialacking leucine

Colony 4 requiresleucine for growth(leucine auxotroph)

Complementation(generating a leucine auxotroph)

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 1st 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?

Purify asparaginasefrom Erwinia

Blood of rabbit contains antibodies to Erwinia asparaginase butno other Erwinia protein

Erwinia

Antibody Production

1 2 3

4 5 6

7 8 9

Screening E. coli library of Erwinia DNA

Plate out E. colicells on nylon filter

1 2 3

4 5 6

7 8 9

Lyse bacteria with lysozymeProtein binds to filter and thus a print of the bacterium’sproteins replaces colony

1 2 3

4 5 6

7 8 9

Add inert protein(e.g. bovine serumalbumin) to blockprotein binding siteson nylon filter

1 2 3

4 5 6

7 8 9

Add antibodieslabelled withradioactive iodine

1 2 3

4 5 6

7 8 9

1 2 3

4 5 6

7 8 9

Expose filter to X-ray filmto detect where antibody has bound (radioactive andthus blackens X-ray film)

Screening E. coli library of Erwinia DNA

www.staff.ac.ncl/h.j.gilbert

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

1 2 3

4 5 6

7 8 9

1 2 3

4 5 6

7 8 9

1 2 3

4 5 6

7 8 9

NaOH added to lyse bacteriaand denature DNA intosingle strand. Filter is heated to 70 to immobilisesingle strand DNA

Inert DNA to blockvacant DNA binding

sites on filter

Single stranded print ofbacterium’s DNA insteadof a bacterial colony

Plate out E. colicells on nylon filter

Nucleic acid hybridization

1 2 3

4 5 6

7 8 9

1 2 3

4 5 6

7 8 9

Add radioactivenucleic acid probethat binds to target gene

1 2 3

4 5 6

7 8 9

Expose filter to X-ray filmto detect where nucleic acidhas bound (radioactive andthus blackens X-ray film)

Nucleic acid hybridization


• 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)


• 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 sequenceMet-Asn-Lys-Trp-Glu-Met

ATG AAT AAA TGG GAA ATGATG AAT AAA TGG GAG ATGATG AAT AAG TGG GAA ATGATG AAT AAG TGG GAG ATGATG AAC AAA TGG GAA ATGATG AAC AAA TGG GAG ATGATG AAC AAG TGG GAA ATGATG AAC AAG TGG GAG ATG

Met = ATG; Asn = AAT or AAC; Lys = AAA or AAG; Trp = TGG; Glu = GAA or GAG

How many probes must we make?

1st Test in BNS216

• Test is next Thursday February 26th

• 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 cDNA• cDNA: complementary or copy DNA• It is derived from mRNA• double stranded DNA with identical

sequence to mRNA in one strand• Why clone cDNA

– Important in expressing eukaryotic proteins in bacteria

– Introns not removed in bacteria– Therefore not protein produced

Eukaryotic organism

Transcription

RNA

Splicing

Translation

Functionalprotein

Bacteria

Translation

Inactive protein

mRNA

Bacteria

Transcription

mRNA

Translation

Functionalprotein

cDNA

Isolate specific cDNA(encodes growth hormone)

• How do isolate mRNA encoding growth hormone

• Isolate cells that express growth hormone– Pituitary

Isolate specific cDNA(encodes growth hormone)

• Extract mRNA

• Convert to cDNA

• Construct cDNA library

• Screen library for members containing cDNA encoding growth hormone

Pituitary

AAAAAA

AAAAAA

AAAAAA

AAAAAA

AAAAAA

Isolate total mRNA

Encodes GH

TTTTTT

TTTTTT

Purify mRNA

AAAAAA

AAAAAA

Ribosomal RNA etc

Matrix ispolydTcellulose

Pass total RNA overmatrix. Only mRNAwill bind. Other RNAswill pass directlythrough

AAAAAA

AAAAAA

TTTTTTAAAAAA

TTTTTTAAAAAA

Low salt buffer

TTTTTT

TTTTTT

AAAAAA

AAAAAA

+

cDNA 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

5’ 3’

5’ 3’

5’ 3’

5’ 3’3’ 5’

mRNA

primer

Reversetranscriptase

Primer

• Small single stranded DNA sequence

• Binds to 3’ end of all mRNA

• What is the sequence?– PolydT

AAAAAATTTTTT

5’ 3’

AAAAAATTTTTT

5’ 3’

3’5’

AAAAAATTTTTT

5’ 3’

3’5’

AAAAAATTTTTT

5’ 3’

3’5’

AAAAAATTTTTT

5’ 3’

3’5’

AAAAAA 5’ 3’

TTTTT Add primer

Reverse transcriptase + dNTPs

mRNAcDNA

mRNA

mRNA

RNAase H

DNA polymerase + dNTPs

Completed DNA synthesis

Double stranded cDNA

Collection of cDNA molecules

• Insert cDNAs into a vector– Generate recombinant vectors

• Insert recombinant vectors into a bacterium such as Escherichia coli

• This comprises a cDNA library derived from pituitary mRNA

• Screen cDNA library for those containing growth hormone encoding cDNA

cDNA

GAATTCCTTAAG

GAATTCCTTAAG

GAATTCCTTAAG

Add linkers at high concentrationsand DNA ligase

GCTTAAG

AATTC G

Cut with EcoRI

Insert into a vectorcut with same enzymefollowed by DNA ligase

cDNA is blunt-ended thus ligatesto other DNA unless concentration high

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 (cDNA molecules)

• Once ligated to cDNA cut with restriction enzyme to create sticky ends

cDNA

GAATTCCTTAAG

GAATTCCTTAAG

GAATTCCTTAAG

Add linkers at high concentrationsand DNA ligase

GCTTAAG

AATTC G

Cut with EcoRI

Insert into a vectorcut with same enzymefollowed by DNA ligase

cDNA is blunt-ended thus ligatesto other DNA unless concentration high

Transform intoE. coli

Screen for members oflibrary that containgrowth hormone cDNA

Basis of course

Documents

Transcript of Basis of course