Construction of genomic library in lambda phage. THIS WAS MY PRESENTATION TOPIC IN CLASS. THOUGHT OF SHARING IT AND HOPE IT HELPS.

1. *

2. *
* The use of genetic information is a powerful tool that
today is becoming more readily available to scientists.
* In order to use this powerful tool it necessary to know
how to navigate throughout the entire genome.
* In humans this project is known as Human Genome
Project.

3. *
*Gene library is a collection of different DNA sequence from
an organism, each of which has been cloned into a vector for
ease of purification, storage and analysis.
*Gene library is of two types:
cDNA libraries
(made from cDNA- copy of
mRNA)
Gene libraries
(made from genomic DNA)

4. *
*Purify genomic DNA –
A) Eukaryotes
B) Prokaryotes
*Fragment this DNA : physical shearing and restriction enzyme
digestion.
*Clone the fragments into vectors.
*To make a representative genomic libraries, genomic DNA must
be purified and then broken randomly into fragments that are
correct in size for cloning into the chosen vector.

5. *
* For Eukaryotes :
 Prepare cell nuclei remove protein, lipids and other
unwanted macromolecules by protease digestion and
phase extraction.
* For Prokaryotes :
 Extract DNA directly from cells.

6. *
* Physical shearing :
 Pipeting, mixing
* Restriction enzyme digestion:
 Partial digestion is preferred to get a greater
lengths of DNA fragments.

7. *
* Ends produced (sticky or blunt) & the cleaved ends of the
vector to be cloned.
 Sau3A: 5’-/GATC-3’, less selectivity
 BamH1: 5’-G/GATCC
* Whether the enzyme is inhibited by DNA modifications.
* Time of digestion and ratio of restriction enzyme to DNA is
dependent on the desired insert size range.

8. *
 λ-phage is treated with
restriction enzymes that
produce λ arms with sticky
end.
 These arms contain all the
lytic genetic information that
is needed for replication and
produces room for insertion
of new genetic information.

9. * DNA sequence is obtain from the cell of interest. It is
cleaved with restriction enzymes that produce 20kb
fragments that have complementary sticky ends.
* Both are mixed in equal amounts and are treated with
a DNA ligase that cleaves them together.
* Afterward the entire combined sequence is packed to
the phage head.

11. *
*The genomic library is generated by using λ-phage for
the following reasons:
*A large number of λ phage can be screened
simultaneously (5 x 10E4 phage plagues).
* λ phage as a higher transformation efficiency about
1000 times higher compared to a plasmid.

12. *The vector as to maintain its lytic growth.
* Lysogenic pathway and other viral genes that are not
important are replaced with the DNA to be cloned.

13. *An infected E.Coli will produce what are
know as concatomers (which is the viral
genome) on either site of the
concatomers there is a site called COS
Site.
* Two proteins recognize this site A
protein and Nu protein, which will lead to
the insertion of the λ DNA into the phage
head. The chromosomal DNA that lacks
the COS sites will not enter the phage
head. Once the genetic information is
inserted the tail will assemble.
* A 50kb can be inserted into the phage.

14. *
* To prepare the phage an E.coli cell is infected with a
mutant λ- phage that as a defective “A-protein” (which is
one of two genes that are responsible for packaging
genetic information).
* Therefore the E.coli accumulates empty heads and also
preassembled tails.

15. * Once enough heads and tails are assembled we
lysate the E.coli cells.
* To the mixture of heads and tail we add isolated A
protein (obtained from E.Coli infected with λ-phage).
* In the next step we add the recombinant DNA that
has the λ- phage genetic information (which also
includes COS sites).

16. * At this point we have a mixture containing mutant λ-
phage heads and tails. There is isolated A protein and
recombinant DNA containing λ-phage genetic
information with COS sites.
* Therefore we have all the components necessary to
package the recombinant DNA into the λ-phage head.
Once the information is inserted the tail assembles and
we have an infectious phage that contains the
recombinant DNA sequence.

17. *
*According to genome’s size, we can select a proper vector
to construct a library.
*Vectors Plasmid phage λ cosmid YAC insert (kb):
5 23 45 1000
*The most commonly chosen genomic cloning vectors are λ
replacement vectors which must be digested with
restriction enzymes to produce the two λ end fragment or
λ arms between which the genomic DNA will be digested.

18. *
cos cos
Long (left)
arm
short (right)
arm
Exogenous DNA
(~20-23 kb)

19. Preparation of genomic inserts:
1) Ligation
2)Packing with a mixture of the
phage coat proteins and
phage DNA processing
enzymes.
3)Infection and formation of
plaques.
4)Library constructed.

20. Constructing a Genomic Library in Lambda

21. *The first thing is to digest the genomic DNA by restriction
fragments.
*Most of the different types of restriction enzymes are used.
*But here only one type of restriction enzymes is used which is
BamH1 is cleaving the fragments apart. The 15 kb size is our
gene of interest.
*The lambda phage genome has cos sites at the end. The genome
is getting cleaved and packaged from the cos sites end.
*

22. *Therefore, it produces a long genome during replication. At the
time of packaging the distance from one cos site to another will
be packages as genome in a phage.
*So, the cos sites are the terminal region for packaging.
*The formation of different cos sites is said to be as
concatamers. So, only after the formation of concatamers the
packing will start.

23. *So, basically we will cleave the genomic DNA and also Lambda
phage with the help of different restriction enzymes and we will
introduce our gene of interest into the phage so it will be
packaged.
*Phage will be having different fragments introduced and we will
get multiple species carrying different segments of the genome.
*Further we can infect bacteria with this phage so the genes gets
into the host bacterial cells.

24. *