An enzyme essentially required to restrict the host DNA and cleave the foreign DNA in bacteria , Exploited for manipulating the any desired DNA

2. INTRODUCTION
• A Protein enzyme produced by bacteria that cleaves sugar–
phosphate bond at specific sites /restriction sites along the DNA
Molecules.
These are synthesized and isolated
from bacteria where they carry host
defense function or restriction
modification system by cleaving
foreign DNA, and preventing its own
DNA by methylating the genome

3. Host Defense system of Bacteria
• Normally, a bacterium uses a restriction enzyme to
defend against bacterial viruses called bacteriophage
or phages.
• When a phage infects a bacterium, it inserts its DNA
into the bacterial cell so that it might be replicated.
• So, the name has been given as restriction enzyme i.e
restricting replication of the phage DNA by cutting it
into many pieces.
• However genomic DNA of Host bacteria is not cleaved
as their DNA is modified by restriction modification
system of bacteria.

5. Restriction Modification system /host
defense system
• A system which prevents the DNA from
being cleaved by Restriction endonuclease .
• Prevention is due to methylation of DNA at
A & C position of DNA which change the
confirmation of DNA and escape the binding
of RE at methylated sites.
• Accordingly there may three types of DNA
• Non methylated DNA
• Methylated DNA
• Hemi methylated DNA
• Methylation is due to methyl transferase
activity of restriction enzyme which
methylate the DNA after replication. So, RE
has two properties
• Recognition & cleavage of internal
phosphodiester bond -
• Transfer of methyl at adenine and cytosine of
the host DNA-(Mtase)

6. Action of Mtase
MTases transfer the methyl group from S-adenosyl methionine (SAM) to the
c5/N4 amino group of cytosine or to the N6 amino group of adenine. S-Adenosyl
methionine (SAM) is a common co-substrate involved in methyl group transfers,
Although these anabolic reactions occur throughout the body, most SAM is
produced and consumed in the liver..

8. Source
• Isolated from bacterial cells
and are used in the
laboratory to manipulate
fragments of DNA, such as
those that contain genes.
• Hence, these are utilized as
molecular tools in RDT
/Genetic engineering
• So called
MOLECULAR
SCISSOR

9. unknown till 1968 ,but restriction of foreign DNA replication & expression
was known.
Meselson & Yuan reported one enzyme in the K2 strain of E.coli and coined
the term Restriction Endonuclease. which are able to digest the unmethylated
DNA.
However methylated DNA is protected.
Methylation of DNA involves adding a methyl-group (CH3) to the DNA
base such that the restriction enzyme will not recognize it.
The process of methylation has been shown to be carried out by DNA
sequence-specific methyl-transferase enzymes.
In plants and animals the primary methylated base is 5-methylcytosine
(m5C) while in bacteria the major methylated base is N6-methyladenine
(mA) but N4-methylcytosine (mC) is also found.

10. Discovery
• Stuart Linn and Werner Arber in 1968 showed in vitro restriction of phage
DNA by an E. coli cell extract.
• Soon after this discovery, Hamilton Smith and Kent Wilcox in the year 1970
isolated the first restriction enzyme, endonuclease R (later renamed as
HindII), from Hemophilus influenzae.
• In the subsequent year Kathleen Dana and Daniel Nathans pioneered the
applications of restriction enzymes by showing specific cleavage of SV40
DNA by HindII.
•
Werner Arber, Hamilton Smith and Daniel
Nathans were awarded Nobel Prize in
Physiology or Medicine in 1978 for the
discovery and applications of restriction
enzymes.

11. Specific recognition sequence –
Palindrome
Each restriction enzyme recognizes a short specific
nucleotide in the DNA molecule called as recognition
sequences/sites randomly distributed throughout the DNA. The
frequency of such sequences can be calculated by using
formula
Frequency of recognition sites in a DNA = (1/4)n,
If the proportion of all the nucleotides are uniform or
equal.1.e 25/100=1/4
Suppose a DNA has hexacutter i.e 6bp recognition
sequence and t percentage of each nucleotide is 25 %
Then the recognition sequence will occur in the DNA at a
frequency of (1/4)6 i.e 1/4096 thus such DNA will produce
afragment of average 4 kb
If not then percentage of each base is calculated as per the
given values and n is the number of nucleotides present in the
recognition sites

12. Problem
• Hind III recognize 6bp AAGCCT. If the genomic
DNA is cleaved . What will be the average size
of fragment produced.
• A. when % is equal for all then 1/4096
• B, but if GC =40% then AT =60%
• Therefore A=30%, T=30%, G=20% C=20%
• The occurrence will be
• 30/100 X 30/100 X 20/100 X 20/100 X20/100 X 30/100
• =324/100000 = average 1/3086

13. Requirements of RE action
Pallindrome sequence
Mg2+,
SAM, and
ATP

14. Pallindrome
• Restriction enzymes identify a special sequence of DNA known
as a palindromic sequence.
Palindromic sequences
occur when the 5'-to-3'
sequence is the same on
each DNA strand.
They are identical when
one strand read left to
right , is the same as
another strand read right
to left .
Restriction endonuclease
cuts the DNA at these
palindromic sites.

15. Nomenclature of RE
• Nathans and Smith 1975, proposed a consistent,
standard nomenclature scheme for RE .
• It includes both the genus and the species of the
bacterium from which it was isolated, the strain
number, and the order in series in which the
enzyme was found. For example the restriction
enzyme designated Bam HI
• Genus: Bacillus B
• Species: amyloliquifaciens am
• Order of discovery : Third III

16. Output
• RE produces either blunt end or staggered end DNA
fragment with overhanging ends
• For example
• EcoRV EcoRI
• GAT ATC G AATTC
• CTA TAG CTTAA G
• Under non standard reaction conditions (low ionic
strength,high pH ) some RE will be capable of cleaving
in similar way but no identical to their recgnition
sequence . This altered specificity is called star activity
and is the general prpperty of RE so cut at non
canonical sites under extreme condition.

17. Types of Restriction Enzyme
• RJ Roberts conferred the term
• Isoschizomer (same cutter) on restriction enzymes that recognized
the same DNA sequence
• Ex. Sph I CGTAC/G and BbuI CGTAC/G
• Neoschizomers – two different restriction enzymes with same
recognition but different cleavage at similar reaction condition
• ex. Sma I GGG/CCC and XmaI G/GGGCCC
• Isocaudomers slightly different sequence but same produce the
same ends Ex. Sau 3 A & Bam HI gives 5’GATC3’ sticky ends
• BamHI -NNG GATCCNN
• NNCCTAG GNN
• Sau3A -NNN GATCNNN
• NNNCTAG NNN
•

18. List of RE with source and cutting sites

19. Characteristics Type I RE Type-II RE Type III RE
Nature Bifunctional i.e both
endonculease &
methyltransferase
Unifunctional i.e
separate
endonuclease and
methyltransferase
Bifunctional
Polypepteide Three different sub unit Two similar subunits
Heteodimer
Two different
subunits
Cofactor
requirements
ATP, Mg++
SAM (S/adenosyl
methionine
Mg++ ATP, Mg++
Cleavge Sites Random ,up to 1000 bp
away from RE site
At /near 24-26 bp 3’to
restriction sites
Example Eco B EcoRI EcoPI

21. Mechanisms
• The RE first recognizes the sequence and protein binds it and direct
hydrolysis by nucleophilic attack at the phosphoester bond
•
• Direct hydrolysis by nucleophilic attack at the phosphorous atom
•
3’OH and 5’PO43’ is produced. Mg2+ is required for the catalytic activity of the
enzyme. It holds the water molecule in a position where it can attack the phosphoryl
group and also helps to polarize the water molecule towards deprotination.

22. Extension
Frequency of restriction sites
• The number of restriction sites for any RE can be
calculated by using formula 4n
• Where n is no. of base in the recognition sequence .
• For example A DNA is digested with 6 cutter RE then
the frequency of cut will be after every 4096 base pair.
So one can also calculate the no. of fragments if
genome size is given
• .

23. • Also the no. of cut/fragments depends on the
percentage of each base in the genomic DNA
under digestion, For Instance if GC is 50% the
AT is also 50% so each base has ¼ therefore it
will be =1/4 X ¼ X ¼ X ¼ X ¼ X ¼ ( each base is
25/100=1/4) and six times multiplication is
related to no. of base in the restriction sites .
•

24. EcoRI
• GAATTC- RE Hexacutter
• If GC =40% in a DNA then,
AT=60%
Therefore
G=20%
C=20%
A=30%
T=30%
Probaiblity of occurrence will be
G A A T T C
2/10 X 3/10 X 3/10 X 3/10 X 3/10 X 2/10
324/1000000 =1/3086 (Approx )

25. ENZYMES
FOR
DNA
MANIPULATIONS….