Gene library.ppt is a good start to learn

Genes are manipulate to give powerful genes

1. Construction of Gene Library
Tileye Feyissa

2. The isolation of genes that encode proteins is often the goal of a
biotechnology experiment
In prokaryotes, structural genes have continuous coding domain
In eukaryotes, the exons are separated by introns
Therefore, different cloning strategies have to be used

5. To achieve this, the complete DNA of an organism is cut with a
restriction enzyme
Each fragment is inserted into a vector
Then the specific cell line (clone) that carries the target DNA
sequence must be identified, isolated, and characterized
This process of subdividing genomic DNA into clonable elements and
inserting them into host cells is called creating a library (clone bank,
gene bank)
In a prokaryote, the target DNA is frequently minuscule, 0.02%
The problem is then how to clone and select the targeted DNA
sequence

6. One way to create DNA library is by treating the DNA from a
source organism with a four-cutter restriction endonuclease (e.g.
Sau3AI)
This enzyme theoretically cleaves the DNA approximately once in
every 256 bp
The conditions of the digestion reactions are set to give a partial,
not a complete digestion
In this way, all possible fragment sizes are generated

7. However, restriction endonuclease sites are not randomly located
Some fragments may be too large to be cloned
In these cases, an incomplete library is available for selection
So it may be difficult or even impossible to find a specific target DNA
sequence
This problem can be overcome by forming a library with another
restriction enzyme
After a library is created, the clones with the target sequence
must be identified

8. Screening DNA library
Three popular methods of identification are used
-DNA hybridization with a labeled DNA probe
-Immunological screening for the protein product
-Screening for protein activity

9. Screening by DNA hybridization
The presence of a target DNA sequence can be determined by this
method
It depends on the formation of stable base pairs between the
probe and the target sequence
Double stranded DNA can be converted into single-stranded DNA
by heat or alkali treatment
The target DNA is denatured and the single strands are
irreversibly bound to a matrix (nitrocellulose or nylon)
This binding is carried out at a high temperature

10. Then the single strands of a DNA probe are incubated with the
bound DNA sample
The probes are labeled with either a radioisotope or another
tagging system
If the sequences of nucleotide in the DNA probe is
complementary to a nucleotide sequence in the sample, then
hybridization occurs

11. The hybridization can be detected by autoradiography or other
visualization procedures, depending on the nature of the probe label
If the nucleotide sequence of the base does not base pair with a
DNA sequence in the sample
Then no hybridization occurs and the assay gives a negative
result

12. Probes range in length from 100 to more than 1000 bp
Both larger and smaller probes can be used
Depending on the conditions of the hybridization rxn, stable base
pairing requires a match of > 80% within a segment of 50 bases

13. DNA probes can be labeled in various ways
-One strategy is by random primer method
-This method utilizes a mixture of synthetic random
oligonucleotides
-These oligonucleotides are formed from all possible combinations
sequences of six nucleotides that act as primers for DNA synthesis
-On the bases of the chance occurrence of complementary
sequences, some of the oligomers in the sample will hybridize to
complementary sequences on the unlabeled probe DNA template

14. After the oligomer sample is mixed with the denatured probe
template DNA,
The four dNTPs and the Klenow fragment are added
The Klenow fragment retains both DNA polymerase and 3’
exonuclease activities but lacks the 5’ exonuclease activity that is
normally associated with E. coliDNA polymerase I
With the available 3’-OH groups of the bound random primers and
the strands of the probe as templates, new DNA synthesis occurs
If a radioactive label is used, then one of the dNTPs contains the
isotope
32
P in the -position phosphate
Autoradiography can be used to determine whether the labeled
probe sequences hybridize to sequences of a target DNA sample

16. For nonisotopic detection of hybridization
biotin
digoxygenin
horse radish peroxidase
can be attached to one of the four dNTPs that is
incorporated during the DNA synthesis step

18. When a probe with this kind of label hybridizes to the sample DNA,
Detection is based on the binding of an intermediary compound
(e.g. streptavidin) that carries an appropriate enzyme
Depending on the assay system, the enzyme can be used for the
formation of either a chromogenic molecule that can be visualized
directly or
A chemiluminescent response that can be detected by
autoradiography

19. There are at least two possible sources of probes for screening a
genomic library
1.Cloned DNA from a closely related organism (a heterologous
probe) can be used
-In this case, the conditions of the hybridization rxn can be
adjusted to permit considerable mismatch between the probe and
the target DNA to compensate for the natural differences between
the two sequences
2. A probe can be produced by chemical synthesis
-The nucleotide sequence of a synthetic probe is based on the
probable nucleotide sequence that is deduced from the known
amino acid sequence of the protein encoded by the target gene

20. Genomic DNA libraries are often screened by plating out the
transformed cells on the growth medium of a master plate
Then transferring samples of each colony to a solid matrix such as
a nitrocellulose or nylon membrane
The cells are lysed, deproteinized, the DNA is denatured and
bound to the matrix
At this stage, a labeled probe is added
If hybridization occurs, signals are observed on an autoradiograph
The colonies from the master plate that correspond to samples
containing hybridized DNA are then isolated and cultured

24. Because most libraries are created from partial digestions, a
number of colonies (clones) may give a positive response to the
probe
The next task is to determine which clone, if any, contains the
complete sequence of the target gene
Preliminary analyses that use the results of gel electrophoresis
and restriction endonuclease mapping reveal the length of each
insert and identify those inserts that are the same and those that
share overlapping sequences
By using overlapping sequences, it may be possible to join
sections of the gene in additional cloning exp’ts

25. Alternatively, if an insert in any one of the clones is large enough
to include the full gene,
Then the complete gene can be recognized after DNA sequencing
Because this will have start and stop codons and contiguous set of
nucleotides that code for the target protein
Unfortunately, there is no guarantee that the complete sequence of
a target gene will be present in a particular library
If the search for an intact gene fails, then another library can be
created with a different restriction endonuclease and screened with
either the original probe or probes derived from the first library

26. Alternatively, libraries that contain DNA fragments larger than the
average prokaryotic gene can be created to increase the chance that
some members of the library will carry a complete version of the
target gene

27. Screening by immunological assay
If a DNA probe is not available, this is alternative method to screen
a library
E.g. if a cloned DNA sequence is transcribed and translated,
The presence of the protein, or even parts of it, can be determined
by an immunological assay
All cell lines of the library are grown on master plates
A sample of each colony is transferred to a matrix, where the cells
are lysed and the released proteins attach to the matrix
The matrix with the bound proteins is treated with an antibody
(primary antibody)

28. The primary antibody specifically binds to the protein encoded by
the target gene
Following the interaction of the primary antibody with the target
protein (antigen),
Any unbound antibody is washed away and the matrix is treated
with a second antibody that is specific for the primary antibody
The secondary antibody has an enzyme such as alkaline
phosphatase attached to it
After the matrix is washed, a colorless substance is added

29. If the secondary antibody has bound to the primary antibody,
The colorless substrate is hydrolyzed by the attached enzyme and
produces a colored compound that accumulates at the site of the rxn
The colonies on the master plate that correspond to positive
results (colored spots) on the matrix contain either an intact gene or
a portion of the gene that is large enough to produce a protein
product that is recognized by the primary antibody
After detection by immunoassay of genomic DNA libraries,
The positive clones must be characterized further to determine
which, if any, carry a complete gene

30. Screening by protein activity
If the target gene produces an enzyme that is not normally made by
the host cell,
A plate assay can be devised to identify members of a library that
carry the functional gene encoding that enzyme
E.g. the genes for -amylase, endoglucanase, and -glucosidase
from various organisms have been isolated by plating the genomic
library in E. colionto medium supplemented with a specific substrate

31. Then using a selective stain to identify those colonies that are
capable of utilizing the substrate
The gene that is sought encodes the product that is essential for
the growth of a mutated host cell, then the library can be formed by
transformation into these mutant cells

32. The cells that are able to grow on minimal medium in the absence
of the required substrate must carry a functional form of the target
gene on the cloning vector
Variations of this form of genetic complementation have been used
to isolate a variety of important genes
These genes include those for biosynthesis of antibiotics and the
formation of nitrogen-fixing nodules on the roots of certain plants

33. Cloning DNA sequences that encode eukaryotic proteins
Special techniques are required for cloning eukaryotic structural
genes
Prokaryotic hosts are not able to remove introns from
transcribed RNA
Therefore, this mRNA is not translated correctly in a bacterial
host cell
A eukaryotic DNA sequence needs prokaryotic transcriptional
and translational control sequences to be properly expressed

34. A functional eukaryotic mRNA has a G cap at the 5’ end and usually a
string of up to 200 adenine residues (poly (A) tail)
The poly (A) tail can be used to separate the mRNA fraction of a
tissue from the rRNA and tRNA
Extracted cellular eukaryotic RNA is passed through a column
packed with cellulose beads to which is bound short chains of
thymidine residues
Each thymidine residues are about 15 nucleotides long (Oligo(dT),
dT15)
The poly(A) tails of the mRNA bind by base pairing to the oligo(dT)
chains
The tRNA and rRNA which lack poly(A) tails, pass through the column

35. The mRNA is eluted from the column by treatment with a buffer
that breaks the A:T hydrogen bonds
Thereby releasing the bound mRNA
Before the mRNA can be cloned into a vector,
They must be converted to double-stranded DNA

36. This synthesis is accomplished by using two different kinds of
nucleic acid polymerases
Reverse transcriptase and
Klenow fragment of DNA pol I
After the mRNA fraction is purified,
short unbound sequences of oligo(dT) molecules
enzyme reverse transcriptase and
the four dNTPs are added to the sample
The oligo(dT) base pair with the poly(A) tail regions and provide an
available 3’-hydroxyl group to prime the synthesis of a DNA strand
Reverse transcriptase uses RNA strand as a template

37. The synthesis of DNA strand by reverse transcriptase in vitrois
often incomplete
However, before synthesis ceases, the DNA strand usually turns
back on itself for a few nucleotides to form a hairpin loop
The second DNA strand is synthesized by the addition of the
Klenow fragment of E.coliDNA pol which uses the first DNA strand
as a template
After the rxn is complete, the sample is treated with the enzyme
RNaseH
RNaseH degrades the mRNA
S1 nuclease opens the hairpin loops and degrades single-stranded
DNA extensions

38. At the end of this procedure, the sample contains a mixture of
partial and complete double-stranded complementary (cDNA) copies
of the more prevalent mRNAs in the original sample
This cDNA populations can be cloned by blunt-end ligation or other
joining mechanisms into a plasmid cloning vector to form a cDNA
library
Then screen the DNA library by one of the screening methods

39. DNA manipulative enzymes
The range of DNA manipulative enzymes can be grouped into five
broad classes
This classification is based on the type of rxn that they catalyze
1.Nucleases
2.Ligases
3.Polymerases
4.Modifying enzymes
5.Topoisomerases

40. Although most enzymes can be assigned to a particular class, a
few display multiple activities that span two or more classes
Many polymerases combine their ability to make new DNA
molecules with an associated nuclease activity
Many similar enzymes able to act on RNA are known (e.g.
ribonuclease)

41. Gene cloning requires that DNA be cut in a very precise and
reproducible fashion
Each vector must be cleaved at a single position to open up the
circle so that new DNA can be inserted
It is also necessary to cleave the DNA that is to be cloned
There are two reasons for this
Restriction endonucleases

42. 1.If the aim is to clone a single gene, which may consist of only 2
or 3 kb of DNA, that gene will have to be cut out of the large
DNA
2.Large DNA may have to be broken down to produce fragments
of small enough to be carried by the vector

43. Blunt ends and sticky ends
The exact nature of the cut produced by a restriction endonuclease
is of considerable importance in design of a gene cloning expt
Many restriction endonucleases make a simple double-stranded
cut in the middle of the recognition sequence, resulting in blunt or
flush end
e.g. PVuII and AluI

44. A large number of restriction endonucleases cut DNA in a slightly
different way
The cleavage is staggered by two or four nucleotides resulting in
short single-stranded overhungs at each end
These are called sticky or cohesive ends
The restriction endonucleases with different recognition
sequences may produce the same sticky ends. E.g. BamHI, BgllI,
Sau3A

47. Restriction Sites are not evenly spaced

48. Sticky ends increase the efficiency of ligation
Although ligation of two blunt-ended fragments can be carried
out in the test tube, it is not very efficient
This is because ligase is unable to ‘catch hold’ of the molecule to
be ligated
Has to wait for chance association to bring the ends together
If possible, blunt end ligation should be performed at high DNA
concentrations
In contrast, ligation of complementary sticky ends is much more
efficient

49. Compatible sticky ends can base pair with one another by
hydrogen bonding
Forming a relatively stable structure for the enzyme to work on
If phosphodiester bonds are not synthesized fairly quickly, the
sticky ends will fall apart again
These transient, base-paired structures do, however, increase the
efficiency of ligation by increasing the length of time the ends are in
contact with one another

51. Putting sticky ends onto blunt-ended molecule
Compatible sticky ends are desirable on the DNA molecules to be
ligated together in a gene cloning exp’t
These sticky ends can be provided by digesting both the vector and
the DNA to be cloned with the same restriction endonucleases, or
With different enzymes that produce the same sticky end
However, it is not always possible to do this
A common situation is where the vector molecule has sticky ends,
but the DNA fragments to be cloned are blunt-ended
Under these circumstances, one of three methods can be used to
put the correct sticky ends onto the DNA fragments

52. Linkers
Are short pieces of double-stranded DNA
Of known nucleotide sequence
Are synthesized in the test tube
It is blunt-ended, but contains a restriction site (e.g. BamHI)
DNA ligase can attach linkers to the ends of larger blunt-ended
DNA

53. This particular rxn can be performed very efficiently
Synthetic oligonucleotides such as linkers can be made in a very
large amounts and added into the ligation mix at a high concentration
More than one linker will attach to each end of the DNA molecule,
producing the chain structure

54. However, digestion with BamHI cleaves the chains at the
recognition sequences
Producing a large number of cleaved linkers and the original DNA
fragment, now carrying BamHI sticky ends
This modified fragment is ready for ligation into a cloning vector
restricted with BamHI
There is one potential drawback with the use of linkers

56. Consider what would happen if the blunt-ended molecule
contained one or more BamHI recognition sequences
If this was the case, the restriction step needed to cleave the
linkers and produce sticky ends would also cleave the blunt-ended
molecule
The resulting fragments will have correct sticky ends,
But that is no consolation if the gene contained in the blunt-ended
fragment has now been broken into pieces

58. The second method of attaching sticky ends to blunt-ended
molecule is designed to avoid this problem
Adaptors
Short synthetic oligonucleotides
Has one sticky end
The idea is to ligate the blunt end of the adaptor to the blunt ends of
the DNA fragment, to produce a new molecule with sticky ends
This may appear to be simple method but in practice a new problem
arises
The sticky ends of individual adaptor molecules could base pair
with each other to form dimers
So that the new DNA molecule is still blunt-ended

61. The sticky ends could be recreated by digestion with a restriction
digestion
But that would defeat the purpose of using adaptors in the first place
The answer to the problem lies in the precise chemical structure of
the ends of the adaptor molecule
One end, 5’ terminus, carries a phosphate group (5’-P)
The other, the 3’ terminus, has a hydroxyl group (3’-OH)
In the double helix the two strands are antiparallel
So each end of a double-stranded molecule consists of one 5’-P
terminus and one 3’-OH terminus
Ligation normally takes place between the 5’-P and 3’-OH ends

63. Adaptor molecules are synthesized so that the blunt end is the
same as the ‘natural’ DNA, but the sticky end is different
The 3’-OH terminus of the sticky end is the same as usual, but the
5’-P terminus is modified
It lacks the phosphate group, and a 5’-OH terminus

64. DNA ligase is unable to form phosphodiester bridge between 5’-OH
and 3’-OH ends
The result is that, although base pairing is always occurring between
the sticky ends of adaptor molecules, the association is never
stabilized by ligation
Adaptors can therefore be ligated to a DNA molecule but not to
themselves
After the adaptors have been attached, the abnormal 5’-OH terminus
is converted to the natural 5’-P form by treatment with the enzyme
polynucleotide kinase,
Producing a sticky-ended fragment that can be inserted into an
appropriate vector

66. Producing sticky ends by homopolymer tailing
The technique of homopolymer tailing offers a radically different
approach
It is a polymer in which all the subunits are the same
A DNA strand made up entirely of, say, deoxyguanosine is an
example of a homopolymer, and is referred to as
polydeoxyguanosine or poly (dG)
Tailing involves using the enzyme terminal deoxynucleotidyl
transferase to add a series of nucleotides onto the 3’-OH termini of
a double-stranded DNA
If this rxn is carried out in the presence of just one
deoxyribonucleotide, a homopolymer tail is produced
To be able to ligate two tailed molecules together, the
homopolymers must be complementary

68. Frequently poly(dC) tails are attached to the vector and poly(dG) to
the DNA to be cloned
Base pairing between the two occurs when the DNA molecules are
mixed
In practice, the poly(dG) and poly(dC) tails are not usually exactly the
same length, and the base-paired recombinant molecules that result
have nicks as well as discontinuities

69. Repair is therefore, a two step process using Klenow polymerase
to fill in the nicks followed by DNA ligase to synthesize the final
phosphodiester bonds
This repair rxn does not always have to be performed in the test
tube
If the complementary homopolymer tails are longer than about 20
nucleotides, then quite stable base-paired associations are formed

71. A recombinant DNA molecule, held together by base pairing
although not completely ligated, is often stable enough to be
introduced into the host cell in the next stage of the cloning exp’t
Once inside the host, the cell’s own DNA polymerase and DNA
ligase repair the recombinant DNA molecule, completing the
construction begun in the test tube