Frequency of BstE II cutting?
Mikhail Alexeyev
malexeyev at biost1.thi.tmc.edu
Mon Jun 24 13:46:18 EST 1996
In article <DtIG58.8wK.B.midge at bath.ac.uk>, bspwrb at bath.ac.uk (W R
BENNETT) wrote:
> OK,
>
> If BstE II has a restriction site of GGTNACC, does it cut at the same
> frequency as a six-cutter (i.e. an average 1 in 4096 disregarding sequence
> distribution considerations), which is the "intuitive" answer, or does it
> cut with reduced frequency (which is what I'd like!). Promega's technical
> department felt that it definitely cut at reduced frequency, but couldn't
> really say why, or what the frequency was....
>
Since 4096= 4^6 is a number of DIFFERENT 6-nucleotide combinations that
can be composed out of 4 bases (A,T,G and C), the frequency of 7-base
cutter (with non-degenerated recognition sequence) should be 1 in 4^7=
4096 x 4 =16384. However, since Bst EII recognition sequence is
degenerated (redundant) at one position, there should be 4 recognition
sequences (ggtAacc, ggtTacc, ggtCacc AND ggtGacc) for BstEII in every
16384 base pairs that makes it 1 in 4096. To put it in other words, there
are only 6 positions that matter for recognition sequence, therefore, the
frequency should be the same as for 6-base cutter. Same should be true for
(hypothetical) enzymes with recognition sequences of GGT(N)xACC were x
could be any (reasonable) number. However, for an enzyme with recognition
sequence GGT(G/C)ACC the frequency of cutting should be 1 in 4096 x 2
=8192.
Yet another way to put it is in terms of probability to encounter a
specific nucleotide at a specific position. For BstEII it should be:
G G T N A C C
1/4 1/4 1/4 4/4 1/4 1/4 1/4
Probability is: (1/4)^6 x 4/4= 4^-6= 1 in 4096
I hope, that written above is more helpful than it is confusing :).
Regards,
M. Alexeyev
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