How Restriction Enzymes - PowerPoint Presentation

Slide1

How

Restriction

Enzymes

are

working

?

Restriction

Enzymes

Natural occurrence of restriction enzymes is in bacteria for protecting them against a virus attack by cutting the virus DNA/RNA into pieces.



restriction

enzymes

are

protecting

bacteria

against

virus

infections

like

an „

immune

system

“.

Specifity

of Restriction EnzymesLike all enzymes restriction enzymes are also substrate specific. they can recognize a specific base

sequence

on a DNA

strand

e.g.

Eco

RI

=

the

first

restriction

enzyme

isolated

out

of

Escherichia coli

.

Eco

RI

recognition

sequence

is

:

5`-GAATTC-3`

5`-GAATTC-3`

is

a

palindromic

sequence

Palindromic

Sequence Complementary Sequence: 3`-CTTAAG-5`Recognition sequence: 5`-GAATTC-3`

Palindromic

words

you

can read forward and backword and they are the same like the name Otto!

Palindromic

DNA

sequences

you

can

read

forward

and

backword

and

they

have

the

same

base

sequences

!

Specifity

of Restriction EnzymesLike all enzymes also restriction enzymes are effect specific. they can cut their recognition

sequence

in

a

specific

way: sticky or blunt

Specifity

of EcoRIEcoRI cuts sticky ends: sticky end

DNA

Eco

RI

Specifity

of HaeIIIHaeIII cuts blunt ends: recognition sequence

is

: 5`- GGCC-3`

Hae

III

blunt

ends5`- GGCC-3`

Statistical

Calculation

of the Number of CuttingsThe length of human DNA is 3 billion basepairs (3 x 109 Bp)If the recognition sequence would

be

only

2

neighbar

baseslike GA by the 4 DNA bases (A,T,G,C)  42 = 16 That means: Statisticly the probability to find exactly this GA sequence again is each 16 bases.

Statistical

Number

of CuttingsIf the recognition sequence of a restriction enzyme would be only 2 baseslike GA

by

4 different

bases

(

A,T,G,C

)

 42 = 16 16 possibilities (42) are:AG TA CA GGAT TT CT GTAC TC CC GCAA TG CG GA

Statistical

Number

of CuttingsIf the recognition sequence would be 3 neighbar bases e.g. GAA by the 4 different DNA bases: A,T,G,C



4

3

= 64

That

means: Statisticly after each 64 bases the recognition sequence of GAA should occur again on the DNA and a cut should happen done by the restriction enzyme..

Statistical

Number

of CuttingsIf the recognition sequence is 6 neighbar bases e.g. GAATTC by the 4 different DNA bases: A,T,G,C

4

6

= 4096

That

means: Statisticly after each 4096 bases the recognition sequence of GAATTC should occur again on the DNA and a cut should happen done by the restriction enzyme..

Statistical

Number

of CuttingsThe length of human DNA is 3 Billion Basepairs (3 x 109 Bp)Statisticly after each 4096 bases the recognition sequence

of

GAATTC

occurs

on

the

DNA again3 x 109 bases : 4096 bases = ? We will have statistically 732 421 cuts732 422 fragments (linear DNA)These fragments are polymorph, they all have different lengths (RFLP).

Restriction

Enzyme

ClaI*Substrate specifity: Recognition sequence is 5‘ ATCGAT 3‘ 3‘ TAGCTA 5‘  It`s a palindromic sequenceEffect specifity of

Cla

I

:

It

cuts sticky ends 5‘ AT CGAT 3‘ 3‘ TAGC TA 5‘Enzyme is blocked by CpG methylation * ClaI is the first restriction enzyme isolated out of the bacterium Cladosporium spec.

Restriction

of

CpG-methylated

and

non-

methylated

λ-DNA with ClaI CpG-methylatedNon-methylated5‘ … ACAACATCGATAATGAATCAT … 3‘Recognition sequence for

Cla

I

Restriction

enzyme

Cla

I

Methyl

group

5‘ … ACAAC

AT

CGAT

AATGAATCAT … 3‘

Restriktion

enzyme

Cla

I

Recognition

sequence

for

Cla

I