Download Restriction Enzyme - Action of EcoRI

Restriction Enzyme - Action of EcoRI
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Restriction enzymes, also called restriction nucleases (EcoRI in this example) , surrounds
the DNA molecule at the point it seeks(sequence GAATTC). It cuts one strand of the
DNA double helix at one point and the second strand at a different, complementary point
(between the G and the A base). The separated pieces have single stranded "sticky-ends,"
which allow the complementary pieces to combine. The newly joined pieces are
stabilized by DNA ligase.
EcoRI, one of many restriction enzymes, is obtained from the bacteria Escherichia coli.
Site-specific nuclease
This important development came when H.O. Smith, K.W. Wilcox, and T.J. Kelley, working at
Johns Hopkins University in 1968, isolated and characterized the first restriction nuclease whose
functioning depended on a specific DNA nucleotide sequence. Working with Haemophilus
influenzae bacteria, this group isolated an enzyme, called HindII, that always cut DNA molecules
at a particular point within a specific sequence of six base pairs. This sequence is:
5' G T ( pyrimidine: T or C) ( purine: A or G) A C 3'
3' C A ( purine: A or G) ( pyrimidine: T or C) T G 5'
They found that the HindII enzyme always cuts directly in the center of this sequence. Wherever
this particular sequence of six base pairs occurs unmodified in a DNA molecule, HindII will
cleave both DNA backbones between the 3rd and 4th base pairs of the sequence. Moreover,
HindII will only cleave a DNA molecule at this particular site. For this reason, this specific base
sequence is known as the "recognition sequence" for HindII.
HindII is only one example of the class of enzymes known as restriction nucleases. In fact, more
than 900 restriction enzymes, some sequence specific and some not, have been isolated from over
230 strains of bacteria since the initial discovery of HindII. These restriction enzymes generally
have names that reflect their origin--The first letter of the name comes from the genus and the
second two letters come from the species of the prokaryotic cell from which they were isolated.
For example EcoRI comes from Escherichia coli RY13 bacteria, while HindII comes from
Haemophilus influenzae strain Rd. Numbers following the nuclease names indicate the order in
which the enzymes were isolated from single strains of bacteria. Nucleases are further described
by addition of the prefix "endo" or "exo" to the name: The term "endonuclease" applies to
sequence specific nucleases that break nucleic acid chains somewhere in the interior, rather than
at the ends, of the molecule. Nucleases that function by removing nucleotides from the ends of
the molecule are called "exonucleases."
Endonucleases and DNA fragments
A restriction endonuclease functions by "scanning" the length of a DNA molecule. Once it
encounters its particular specific recognition sequence, it will bond to the DNA molecule and
makes one cut in each of the two sugar-phosphate backbones of the double helix. The positions of
these two cuts, both in relation to each other, and to the recognition sequence itself, are
determined by the identity of the restriction endonuclease used to cleave the molecule in the first
place. Different endonucleases yield different sets of cuts, but one endonuclease will always cut a
particular base sequence the same way, no matter what DNA molecule it is acting on. Once the
cuts have been made, the DNA molecule will break into fragments.
Endonucleases and sticky ends
Not all restriction endonucleases cut symmetrically and leave blunt ends like HindII described
above. Many endonucleases cleave the DNA backbones in positions that are not directly opposite
each other. For example, the nuclease EcoRI has the following recognition sequence:
5' G A A T T C 3'
3' C T T A A G 5'
When the enzyme encounters this sequence, it cleaves each backbone between the G and the
closest A base residues. Once the cuts have been made, the resulting fragments are held together
only by the relatively weak hydrogen bonds that hold the complementary bases to each other. The
weakness of these bonds allows the DNA fragments to separate from one each other. Each
resulting fragment has a protruding 5' end composed of unpaired bases. Other enzymes create
cuts in the DNA backbone which result in protruding 3' ends. Protruding ends--both 3' and 5'-are sometimes called "sticky ends" because they tend to bond with complementary sequences of
bases. In other words, if an unpaired length of bases (5' A A T T 3') encounters another unpaired
length with the sequence (3' T T A A5') they will bond to each other--they are "sticky" for each
other. Ligase enzyme is then used to join the phosphate backbones of the two molecules. The
cellular origin, or even the species origin, of the sticky ends does not affect their stickiness. Any
pair of complementary sequences will tend to bond, even if one of the sequences comes from a
length of human DNA, and the other comes from a length of bacterial DNA. In fact, it is this
quality of stickiness that allows production of recombinant DNA molecules, molecules which are
composed of DNA from different sources, and which has given birth to an industry!
Analysis- Please answer in your lab book
1. In your own words, explain how a restriction enzyme cleaves DNA
2. Explain what “sticky ends” are and how they are produced
3. Who isolated the first restriction enzyme? Where and when did they do this
important work?
4. Explain the origin of the name EcoRI.
5. Write a brief summary of why restriction enzymes are used in biotechnology
1. Cut out strips along dotted lines.
2. Tape together top to bottom in any order.
Shaded region = plasmid replication site
Handout 3: Plasmid Base Sequence Strips
Handout 4: DNA Base Sequence Strips
1. Cut out strips along dotted lines.
2. Tape together top to bottom in numerical order.
Shaded region = insulin gene site
Handout 5: Restriction Enzyme Sequence Cards
1. Cut out cards along dotted lines.
2. Compare each enzyme sequence to the base sequences on the plasmid and DNA
strips.
Instructions
1. Cut the plasmid strips along the dotted lines. Tape them together to form a single long
strip. The letters should all be in the same direction when the strips are taped. The two
ends of the strip should then be taped together with the genetic code facing out to form a
circular plasmid.
2. Cut out the DNA base sequence strips, and tape them together to form one long strip. The
pieces must be taped together in the order indicated at the bottom of each strip.
3. Next cut out the restriction enzyme cards.
4. Compare the sequence of base pairs on an enzyme card with the sequences of the plasmid
base pairs. If you find the same sequence of pairs on both the enzyme card and the
plasmid strip, you should mark the location on the plasmid with a pencil, and write the
enzyme number in the marked area. You should do this for each enzyme card. Some
enzyme sequences may not have a corresponding sequence on the plasmid, and that some
enzyme sequences may have more than one corresponding sequence on the plasmid.
5. Once you have identified all corresponding enzyme sequences on the plasmid, identify
those enzymes which cut the plasmid once and only once. Discard any enzymes that cut
the plasmid in the shaded plasmid replication sequence. record their findings in your lab
book
6. Next, compare the enzymes listed against the cell DNA strip. Find any enzymes that will
make two cuts in the DNA, one above the shaded insulin gene sequence and one below
the shaded insulin gene sequence. Mark the areas on the DNA strip that each enzyme will
cut.
7. After you have compared each enzyme with the DNA strip, select one enzyme to use to
make the cuts. The goal is to cut the DNA strand as closely as possible to the insulin gene
sequence without cutting into the gene sequence. Make cuts on both the plasmid and the
DNA strips. Make the cuts in the staggered fashion indicated by the black line on the
enzyme card.
8. Tape the sticky ends (the staggered ends) of the plasmid to the sticky ends of the insulin
gene to create their recombinant DNA.
Discussion Questions
1. Why was it important to find an enzyme that would cut the plasmid at only one site?
What could happen if the plasmid were cut at more than one site? (Cutting at only one
site is important for controlling the variables that will be reproduced. y the restriction
enzyme cut more than one site, then the plasmid might recombine with different
DNAfragments.)
2. Why was it important to discard any enzymes that cut the plasmid at the replication site?
(If the plasmid were cut at the replication site, it would not be able to reproduce and
transfer genetic information to its host bacterial cell.)
3. Why might it be important to cut the DNA strand as closely to the desired gene as
possible? (To make sure that the desired information is transferred to the plasmid without
adding extra unknown or undesirable sequences.)
4. In this activity, you incorporated an insulin gene into the plasmid. How will the new
plasmid DNA be used to produce insulin? (The new plasmid DNA will be introduced into
bacterial cells, where it will reproduce, creating clones of the insulin gene.)