On Sun, 28 Mar 1999 23:37:39, Lars Anders Carlson <lars.anders.carlson at telia.com>
wrote:
> Could anyone give a reasonable explanation why class II restriction
> enzymes have come to recognise palindromic sequences?
There are two reasons:
1) The enzymes are homodimers. Each subunit binds to and cleaves a specific sequence
(usually 4 to 8 bases) of target DNA. Since the 2 subunits are identical, the
recognition sequences must be identical; if the enzyme is to cut both strands of DNA
close to each other, the recognition sequences must be on opposite strands. The
simplest location is for the recognition sequences to overlap perfectly, in which
case they must be palindromes. (Actually, the recognition site isn't a palindrome but
a two-fold symmetrical sequence.)
The reason the enzymes cut opposite strands is that they are _restriction_ enzymes:
their biological function is to prevent foreign DNA (usually viral) from gaining a
foothold in the cell, so they need to fragment the DNA.
2) Having two subunits greatly increases the accuracy and the DNA-binding affinity of
restriction endonucleases. Suppose the Kd for DNA of a single subunit is 1x10(-4)
[0.1 millimolar, not especially tight]. (This is not an experimental value.) Then the
affinity of two subunits binding together is the product of the affinity of each
subunit, or [1x10(-4)]x[1x10(-4)]. This is 1 x 10(-8), or 10 nanomolar, which is very
tight.
Also, suppose the probability that a single subunit will bind to the wrong hexameric
sequence is 1 x 10(-4). (I just made up this number for illustration!) This is 1 in
10,000 hexamers, or 1 in 60,000 bases [someone should check my reasoning], which is
unacceptably high; for a bacterial genome of 2-6 x 10(6) bases, we'd get a several
hundred erroneous (and lethal) cleavages! Now suppose that two such subunits must
bind simultaneously to cleave their target: the probability that they will both bind
to the _same_ wrong sequence is [1x10(-4)]x [1x10(-4)], or 1 x 10(-8). This is 1 in
10(8) hexamers, or 1 in 6x10(8) bases, so the likelihood of incorrectly cleaving a 6
x 10(6) base genome is only 1x10(-2).
Hope this helps. It would be great if someone could plug in actual data into the
above calculations.
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| Dr. Peter Gegenheimer | Vox: 785-864-3939 FAX: 785-864-5321 |
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