"
Why we want to improve it?
According to the experimental record of Freiburg, the success rate is higher than 95%(32/33). However, this result, to some degree, lacks statistical significance.
In the result section, they emphasize that there is a light band at 1200bp, which they believe could indicate that the Golden Gate connection works well. However, after conducting several experiments by ourselves, we find that the key point to indicate whether Golden Gate connection works is not the band at 1200bp. If the band is not clear and specific in the gel, it indicates the experiment doesn’t go well. We can easily find several light bands under the band of 1200bp. Moreover, the second light band is somewhat lighter than the band at 1200bp. Although the Freiburg can explain the results with the repeatability of the TALE sequence, we suppose that the possibility of the mismatch of the sticky ends still can’t be excluded. Frankly speaking, we try to believe that they really made it, but if the success cannot be repeated, there must be something wrong with their system. You can view Detail information in iGEM2012 Freiburg wiki.
- Freiburg's protocol
- Restriction enzyme digestion of plasmid and TAL repeats and gel extraction respectively. By the mole ratio, plasmid to TALE is 1 to 5 and TALE to TALE is 1 to 1. Ligation with T4 ligase in 22 ℃over night.
- The same ratio of plasmid and TALE repeat, but add the TALE repeats one by one and ligation in 22 ℃, 30 minutes
- Every two parts connect at one time, and try to make three intermediates of 400bp, and then mix the plasmid to make the complete TALE.
- The same ratio and ligation with the program of 22℃ 2min, 40℃ 30,25 repeats.
Unfortunately, all of our attempts failed. We didn’t manage to make a complete TALE, or even make two of them together. However, what is important for us is that when we try the 5th protocol, we notice an unexpected result. When we analyze the sequence result, we find that our left adaptor, 1st part and right adaptor connect together. Why do we get this result? We notice that their sticky end is TGAC, GCTC, and ACTC. That is to say, GCTC and ACTC connect with each other by mistake. In another word, if the sticky ends are very similar, they probably connect with each other. Although we failed again, the result gives us confidence to debug 2012 Freiburg's parts.
How do we connect certain monomer?
- Given a sample sequence with repeating amino acids:
What XX means is that it determine the certain kind of base. For one unit of repetition, other amino acids can be identical.
- A fully functional TALE protein contains one sequence, that does not have repetitive units, recognizing base T, and similar sequence but is only half length as its end. That is, one complete TALE protein is able to recognize certain number of repetitive units and two bases.
- The length that can be recognized is not strictly twelve or fourteen. According to the published results, the length and certain sequence are dependent on number and type of monomer.
We can gather 96 bioparts based on Freiburg, and each part has its counterproductive base on certain location(1,2,3,4,5 or 6). By picking two bases on certain location, we are able to design one TALE protein sequence.
The main principles of connection is built upon the idea of Golden Gate Connection.( Sanjana, N. E. et al. A transcription activator-like effector toolbox for genome engineering. Nature Protocols 7, 171–192 (2012).)
The gust of these procedures is more related to one type of restriction enzyme, type II Restriction Enzyme, especially BsmBI enzyme.
The main feature of this enzyme is the recognition sequence is on only one side of cleavage site. It provides the way which can be used to get certain incision without damaging the whole sequence. The sticky end has 4bp base, and it could be designed even for combination of multiple sticky end. That feature is fancy at first, but we cannot regardless its latent shortcomings.
Let’s analyze the example (AA1) provided by Freiburg.
The underlined parts are recognized by BsmBI. Vertical bar(|) is the cutting position. As for this sample, TGAC is one sticky end which can combine with other seven sticky ends.
2012 Freiburg's parts have seven sticky ends:
We all know that certain two parts can combine together, under base-pair rule. However, whether it is possible that unpaired sticky ends can bind together? In fact, the more similar they are, the more possibility that can form new but error base pairs. Spired by BLAST algorithm, we calculate the similarity of each other sticky ends.
The higher score, the higher similarity, and the higher possibility of mismatch. The table shows that more than 30% of pairs’ score is equal to 3, which means that the possibility of mismatch cannot be neglected. Even if we employ the relatively loose rule to calculate the similarity, we can still find that error rates cannot be neglected.
Why did Freiburg insist to use these sticky ends, even the mis-pairing rates are so high? Why not other sticky ends?
Failed to contact the original designers of these sticky ends, what we can do is just to find feasible advantages of these combinations.
Review the TALE repeated amino acids sequence:
The first amino acid is Leu, which is essential for all connection process. There are six different types of base arrangement for Leu, one of the most number of base arrangement.
The counterproductive sticky ends:
The useless of Degeneracy has helped to design seven sticky ends. However, since the codons for identical amino acid are highly similar. This feature, for experimental scientists, is a double-edged sword.
How to improve the Golden Gate sticky ends? A big Table!
Three basic key questions need to be answered:
- Whether it’s possible to find perfect match pair?
- Whether we can find a certain number of sticky ends with least possibility to be mismatched?
- How to make this sticky-end score table?
Loose rule: Match: 1; Mismatch:-1; Gap: 0
Strict rule: Match: 1; Mismatch:0; Gap: -1
The sticky end is composed of four bases, which means that we can design 256 types of sticky ends at most. The forming pair is represented as a 256*256 table.
To solve the TALE parts problem, we need find seven sticky ends, and the similarity score(hereafter referred to as Score) of each pair of them are less than or equal to 1.
When we select Strict Algorithm to find these ends, it is impossible to find seven sticky ends, that each pair of them has score no more than 1. So we have to select Loose Algorithm.
What we are caring about is whether two amino acids can be located on my target sequence, rather than the 4bp bases. So we should convert the sticky ends information to 2 amino acids.
Based on the above table, we are able to calculate the total scores of each combination and find the least one.
Best combination:
Scores Table(Loose rule):
Position in TALE amino acids sequence:
Two main factors to reconstruct DNA sequence:
1.Use the table of best combination and rearrange the sticky ends with your demand.
2.No BsmBI recognition sequence in the reconstruct DNA sequence.
Final DNA Sequence for TALE protein:
1 CTGACCCCGG AACAGGTGGT GGCCATTGCA AGCAACGGTG GTGGCAAGCA GGCCCTGGAG 61 ACAGTCCAAC GGCTGCTTCC GGTTCTGTGT CAGGCCCACG GCCTGACTCC AGAACAAGTG 121 GTTGCTATCG CCAGCCACGA TGGCGGAAAA CAAGCCCTCG AAACCGTGCA GCGCCTGCTT 181 CCGGTGCTGT GTCAGGCCCA CGGGCTCACC CCGGAACAGG TGGTGGCCAT CGCATCTAAC 241 AATGGCGGTA AGCAGGCACT GGAAACAGTG CAGCGCCTGC TTCCGGTCCT GTGTCAGGCT 301 CATGGCCTGA CCCCAGAGCA GGTCGTGGCA ATTGCCTCCA ACATTGGAGG GAAGCAGGCA 361 CTGGAGACCG TGCAGCGGCT GCTGCCGGTG CTGTGTCAGG CCCACGGCTT GACCCCGGAA 421 CAGGTGGTGG CCATCGCCTC CAACGGCGGT GGCAAACAGG CGCTGGAAAC AGTTCAACGC 481 CTCCTTCCGG TCCTGTGCCA GGCCCATGGT CTGACTCCAG AGCAGGTTGT GGCAATTGCA 541 AGCAACATTG GTGGTAAACA AGCTTTGGAA ACCGTCCAGC GCTTGCTGCC AGTACTGTGT 601 CAGGCCCACG GGCTTACCCC GGAACAGGTG GTGGCCATTG CAAGCAACGG TGGTGGCAAG 661 CAGGCCCTGG AGACAGTCCA ACGGCTGCTT CCGGTTCTGT GTCAGGCCCA CGGCCTGACT 721 CCAGAACAAG TGGTTGCTAT CGCCAGCCAC GATGGCGGTA AACAAGCCCT CGAAACCGTG 781 CAGCGCCTGC TTCCGGTGCT CTGTCAGGCC CACGGACTGA CCCCGGAACA GGTGGTGGCC 841 ATCGCCTCCA ACATTGGTGG TAAGCAAGCC CTCGAAACTG TGCAGCGGCT GCTTCCAGTC 901 TTGTGCCAGG CTCACGGCCT GACACCGGAG CAGGTGGTTG CAATCGCGTC TAATATCGGC 961 GGCAAACAGG CACTCGAGAC CGTGCAGCGC TTGCTTCCAG TGCTGTGTCA GGCCCACGGC 1021 CTGACCCCGG AACAGGTGGT GGCCATCGCC TCTAACAATG GCGGCAAACA GGCATTGGAA 1081 ACAGTTCAGC GCCTGCTGCC GGTGTTGTGT CAGGCTCACG GCCTGACTCC GGAGCAGGTT 1141 GTGGCCATCG CAAGCCATGA TGGCGGTAAA CAAGCTCTGG AGACAGTGCA ACGCCTCTTG 1201 CCAGTTTTGT GTCAGGCCCA CGGA
Final Amino acids are remain the same：
1 LTPEQVVAIA SNGGGKQALE TVQRLLPVLC QAHG 35 LTPEQVVAIA SHDGGKQALE TVQRLLPVLC QAHG 69 LTPEQVVAIA SNNGGKQALE TVQRLLPVLC QAHG 103 LTPEQVVAIA SNIGGKQALE TVQRLLPVLC QAHG 137 LTPEQVVAIA SNGGGKQALE TVQRLLPVLC QAHG 171 LTPEQVVAIA SNIGGKQALE TVQRLLPVLC QAHG 205 LTPEQVVAIA SNGGGKQALE TVQRLLPVLC QAHG 239 LTPEQVVAIA SHDGGKQALE TVQRLLPVLC QAHG 273 LTPEQVVAIA SNIGGKQALE TVQRLLPVLC QAHG 307 LTPEQVVAIA SNIGGKQALE TVQRLLPVLC QAHG 341 LTPEQVVAIA SNNGGKQALE TVQRLLPVLC QAHG 375 LTPEQVVAIA SHDGGKQALE TVQRLLPVLC QAHG
Corresponding part：
PART-left: …CTGACCCCGGAGACG
PART1(150bp): CGTCTCGCCCCGGAACAGGTGGTGGCCATTGCAA GCAACGGTGGTGGCAAGCAGGCCCTGGAGACAGT CCAACGGCTGCTTCCGGTTCTGTGTCAGGCCCAC GGCCTGACTCCAGAACAAGTGGTTGCTATCGTGG CGGAAAATGAGACG
How to efficiently connect by using
Golden Gate?
Characterize
and debug TAL effectors parts
designed
by 2012 Freiburg
Why we want to improve it?
Whether the Freiburg's design is efficient or not
According to the experimental record of Freiburg, the success rate is higher than 95%(32/33). However, this result apparently lacks statistical significance.
In the result section, they emphasize that there is a light band at 1200bp, which they believe could indicate that the Golden Gate connection works well. However, after conducting several experiments by ourselves, we find that the key point to indicate whether Golden Gate connection works is not the band at 1200bp. If the band is not clear and specific in the gel, it indicates the experiment doesn’t go well. We can easily find several light bands under the band of 1200bp. Moreover, the second light band is somewhat lighter than the band at 1200bp. Although the Freiburg can explain the results with the repeatability of the TALE sequence, we suppose that the possibility of the mismatch of the sticky ends still can’t be excluded. Frankly speaking, we try to believe that they really made it, but if the success cannot be repeated, there must be something wrong with their system.
The protocol we take to connect the parts of TALE
1. Freiburg's protocol
2. Restriction enzyme digestion of plasmid and TAL repeats and gel extraction respectively. By the mole ratio, plasmid to TALE is 1 to 5 and TALE to TALE is 1 to 1. Ligation with T4 ligase in 22 ℃over night.
3. The same ratio of plasmid and TALE repeat, but add the TALE repeats one by one and ligation in 22 ℃, 30 minutes
4. Every two parts connect at one time, and try to make three intermediates of 400bp, and then mix the plasmid to make the complete TALE.
5. The same ratio and ligation with the program of 22℃ 2min, 40℃ 30,25 repeats.
The motivation to
debug 2012 Freiburg’s parts
Unfortunately, all of our attempts failed. We didn’t manage to make a complete TALE, or even make two of them together. However, what is important for us is that when we try the 5th protocol, we notice an unexpected result. When we analyze the sequence result, we find that our left adaptor, 1st part and right adaptor connect together. Why do we get this result? We notice that their sticky end is TGAC, GCTC, and ACTC. That is to say, GCTC and ACTC connect with each other by mistake. In another word, if the sticky ends are very similar, they probably connect with each other. Although we failed again, the result gives us confidence to debug 2012 Freiburg's parts.
How do we connect certain
monomer?
Some advanced tips for TALE
protein
1) Given
a sample sequence with repeating amino acids:
LTPEQVVAIAS(XX)GGKQALETVQRLLPVLCQAHG
XX=
NGàT
HDàC
NIàA
NNàG or A (NN & NK à G)
What XX means is that it determine the certain kind of base. For one unit of
repetition, other amino acids can be identical.
2) A
fully functional TALE protein contains one sequence, that does not have
repetitive units, recognizing base T, and similar sequence but is only half length as its end. That is, one complete TALE protein
is able to recognize certain number of repetitive units and two bases.
3) The
length that can be recognized is not strictly twelve or fourteen. According to
the published results, the length and certain sequence are dependent on number
and type of monomer.
We can gather 96 bioparts
based on Freiburg, and each part has its counterproductive base on certain
location(1,2,3,4,5 or 6). By picking two bases on certain location, we are able
to design one TALE protein sequence.
Previous Review: Freiburg’s way
of connection
The main principles
of connection is built upon the idea of Golden
Gate Connection.( Sanjana, N. E. et al. A transcription activator-like effector toolbox for genome
engineering. Nature Protocols 7, 171–192 (2012).)
The gust of these procedures is more
related to one type of restriction enzyme, type II Restriction Enzyme,
especially BsmBI enzyme.
5'...C G T
C T C (N)1↓...3'
3'...G C A G A G (N)5↑...5'
The main feature of this enzyme is the
recognition sequence is on only one side of cleavage site. It provides the way which can be used to get certain incision without
damaging the whole sequence. The sticky end has 4bp base, and it could be
designed even for combination of multiple sticky end. That feature is fancy at
first, but we cannot regardless its latent shortcomings.
Let’s analyze the example(AA1) provided
by Freiburg.
CGTCTCA|5’-TGACCCCGGAACAGGTGGTGGCCATCGCCTCCAACATTGGTGGTAAGCAAG
CCCTCGAAACTGTGCAGCGGCTGCTTCCAGTCTTGTGCCAGGCTCACGGCCTGACACCG
GAGCAGGTGGTTGCAATCGCGTCTAATATCGGCGGCAAACAGGCATTGGAGACCGTGCA
GCGCTTGCTTCCAGTGCTGTGTCAGGCCCACGG|GCTCTGAGACG
The underlined parts are recognized by BsmBI. Vertical bar(|) is the cutting position. As
for this sample, TGAC is one sticky end which can
combine with other seven sticky ends.
Evaluate seven sticky ends designed by 2012 Freiburg
2012 Freiburg's parts have seven sticky
ends:
TGAC,GCTC,CTTG,GCTT,ACTG,CCTG,ACTC
We all know that certain two parts can
combine together, under base-pair rule. However, whether it is possible that
unpaired sticky ends can bind together? In fact, the more
similar they are, the more possibility that can form new but error base pairs.
Spired by BLAST algorithm, we calculate
the similarity of each other sticky ends.
|
TGAC |
GCTC |
CTTG |
GCTT |
ACTG |
CCTG |
ACTC |
|
|
TGAC |
\ |
2 |
2 |
2 |
2 |
2 |
2 |
|
GCTC |
2 |
\ |
2 |
3 |
2 |
2 |
3 |
|
CTTG |
2 |
2 |
\ |
3 |
3 |
3 |
2 |
|
GCTT |
2 |
3 |
3 |
\ |
2 |
2 |
2 |
|
ACTG |
2 |
2 |
3 |
2 |
\ |
3 |
3 |
|
CCTG |
2 |
2 |
3 |
2 |
3 |
\ |
2 |
|
ACTC |
2 |
3 |
2 |
2 |
3 |
2 |
\ |
The higher score, the higher
similarity, and the higher possibility of mismatch.
The table shows that more than 30% of
pairs’ score is equal to 3, which means that the possibility of mismatch cannot
be neglected.
Even if we employ the relatively loose
rule to calculate the similarity, we can still find that error rates cannot be
neglected.
|
TGAC |
GCTC |
CTTG |
GCTT |
ACTG |
CCTG |
ACTC |
|
|
TGAC |
\ |
1 |
2 |
1 |
2 |
2 |
2 |
|
GCTC |
1 |
\ |
2 |
3 |
2 |
2 |
3 |
|
CTTG |
2 |
2 |
\ |
3 |
2 |
3 |
2 |
|
GCTT |
1 |
3 |
3 |
\ |
2 |
2 |
2 |
|
ACTG |
2 |
2 |
2 |
2 |
\ |
3 |
3 |
|
CCTG |
2 |
2 |
3 |
2 |
3 |
\ |
2 |
|
ACTC |
2 |
3 |
2 |
2 |
3 |
2 |
\ |
Why did Freiburg insist to
use these sticky ends, even the mis-pairing rates are
so high? Why not other sticky ends?
The Reason why Freiburg used these sticky ends
Failed to contact the original
designers of these sticky ends, what we can do is just to find feasible
advantages of these combinations.
Review the TALE repeated amino acids
sequence:
LTPEQVVAIAS(XX)GGKQALETVQRLLPVLCQAHG（34aa）
The first amino acid is Leu, which is essential for all connection process. There
are six different types of base arrangement for Leu,
one of the most number of base arrangement.
UUA,UUG,CUU,CUC,CUA,CUG
The counterproductive sticky ends:
（C）TGAC,GCTC,CTTG,GCTT,ACTG,CCTG,ACTC
The useless of Degeneracy has helped to
design seven sticky ends. However, since the codons for identical amino acid
are highly similar.
This feature, for experimental
scientists, is a double-edged sword.
How to improve the Golden
Gate sticky ends? A big Table!
Three basic key questions need to be
answered:
1. Whether
it’s possible to find perfect match pair?
2. Whether
we can find a certain number of sticky ends with least possibility to be
mismatched?
3. How
to make this sticky-end score table?
Key algorithms derived from BLAST algorithm
Loose rule: Match: 1; Mismatch:-1; Gap:
0
Strict rule: Match: 1; Mismatch:0; Gap:
-1
The sticky end is composed of four
bases, which means that we can design 256 types of sticky ends at most.
The forming pair is represented as a
256*256 table.
Find target groups of sticky ends
To solve the TALE parts problem, we need find seven sticky ends,
and the similarity score(hereafter referred to as Score) of each pair of them
are less than or equal to 1.
|
Number of Ends |
Algorithm |
|
|
Strict |
Loose |
|
|
2 |
11322 |
15114 |
|
3 |
102844 |
234904 |
|
4 |
169519 |
886369 |
|
5 |
11640 |
624960 |
|
6 |
76 |
61624 |
|
7 |
0 |
640 |
When we select Strict Algorithm to find
these ends, it is impossible to find seven sticky ends, that each pair of them
has score no more than 1. So we have to select Loose
Algorithm.
Four basepair sticky ends convert to
amino acid pair
What
we are caring about is whether two amino acids can be located on my target
sequence, rather than the 4bp bases. So we should convert the sticky ends
information to 2 amino acid.
|
|
U |
C |
A |
G |
|
|
1st position |
FLSYCW |
LPHQR |
IMTNKSR |
VADEG |
|
|
First two positions |
U |
FL |
L |
IM |
V |
|
C |
S |
P |
T |
A |
|
|
A |
Y |
HQ |
NK |
DE |
|
|
G |
W |
R |
SR |
G |
|
|
Last two positions |
U |
FLIV |
SPTA |
THND |
CRSG |
|
C |
FLIV |
SPTA |
YHND |
CRSG |
|
|
A |
LIV |
SPTA |
QKE |
RG |
|
|
G |
LMV |
SPTA |
QKE |
WRG |
|
|
3rd position |
FSYCLPHRITNVADG |
FSYCLPHRITNVADG |
LSPQRITAVAEG |
LSWPQRMTKVAEG |
|
Based on the above table, we are able
to calculate the total scores of each combination and find the least one.
Best choice for seven sticky ends on TALE protein
Best combination:
|
AAAA |
AGGG |
GTAC |
GCTC |
TTTT |
TCGA |
CCCC |
Scores Table(Loose rule):
|
AAAA |
AGGG |
GTAC |
GCTC |
TTTT |
TCGA |
CCCC |
|
|
AAAA |
\ |
1 |
1 |
0 |
0 |
1 |
0 |
|
AGGG |
1 |
\ |
1 |
1 |
0 |
1 |
0 |
|
GTAC |
1 |
1 |
\ |
1 |
1 |
1 |
1 |
|
GCTC |
0 |
1 |
1 |
\ |
1 |
1 |
1 |
|
TTTT |
0 |
0 |
1 |
1 |
\ |
1 |
0 |
|
TCGA |
1 |
1 |
1 |
1 |
1 |
\ |
1 |
|
CCCC |
0 |
0 |
1 |
1 |
0 |
1 |
\ |
Position in TALE amino acids sequence:
|
Sticky
ends |
Amino
Acids |
Sequence
positions |
|
AAAA |
GK* |
LTPEQVVAIAS(XX)GGKQALETVQRLLPVLCQAHG |
|
AGGG |
GG* |
LTPEQVVAIAS(XX)GGKQALETVQRLLPVLCQAHG |
|
GTAC |
VL*
VQ* |
LTPEQVVAIAS(XX)GGKQALETVQRLLPVLCQAHG |
|
GCTG |
LL*
RL* VL* AL* GL* |
LTPEQVVAIAS(XX)GGKQALETVQRLLPVLCQAHG |
|
TTTT |
LL*
VL* |
LTPEQVVAIAS(XX)GGKQALETVQRLLPVLCQAHG |
|
TCGA |
LE* |
LTPEQVVAIAS(XX)GGKQALETVQRLLPVLCQAHG |
|
CCCC |
TP*
LP* |
LTPEQVVAIAS(XX)GGKQALETVQRLLPVLCQAHG |
Reconstruct DNA Sequence
Two main factors to reconstruct DNA
sequence:
1.Use the table of best combination and
rearrange the sticky ends with your demand.
2.No BsmBI
recognition sequence in the reconstruct DNA sequence.
Final DNA Sequence for TALE protein.
1
CTGACCCCGG AACAGGTGGT GGCCATTGCA AGCAACGGTG GTGGCAAGCA GGCCCTGGAG
61 ACAGTCCAAC
GGCTGCTTCC GGTTCTGTGT CAGGCCCACG GCCTGACTCC AGAACAAGTG
121 GTTGCTATCG
CCAGCCACGA TGGCGGAAAA CAAGCCCTCG AAACCGTGCA GCGCCTGCTT
181 CCGGTGCTGT
GTCAGGCCCA CGGGCTCACC CCGGAACAGG TGGTGGCCAT CGCATCTAAC
241 AATGGCGGTA
AGCAGGCACT GGAAACAGTG CAGCGCCTGC TTCCGGTCCT GTGTCAGGCT
301 CATGGCCTGA
CCCCAGAGCA GGTCGTGGCA ATTGCCTCCA ACATTGGAGG GAAGCAGGCA
361 CTGGAGACCG
TGCAGCGGCT GCTGCCGGTG CTGTGTCAGG CCCACGGCTT GACCCCGGAA
421 CAGGTGGTGG
CCATCGCCTC CAACGGCGGT GGCAAACAGG CGCTGGAAAC AGTTCAACGC
481 CTCCTTCCGG
TCCTGTGCCA GGCCCATGGT CTGACTCCAG AGCAGGTTGT GGCAATTGCA
541 AGCAACATTG
GTGGTAAACA AGCTTTGGAA ACCGTCCAGC GCTTGCTGCC AGTACTGTGT
601 CAGGCCCACG
GGCTTACCCC GGAACAGGTG GTGGCCATTG CAAGCAACGG TGGTGGCAAG
661 CAGGCCCTGG
AGACAGTCCA ACGGCTGCTT CCGGTTCTGT GTCAGGCCCA CGGCCTGACT
721 CCAGAACAAG
TGGTTGCTAT CGCCAGCCAC GATGGCGGTA AACAAGCCCT CGAAACCGTG
781 CAGCGCCTGC
TTCCGGTGCT CTGTCAGGCC CACGGACTGA CCCCGGAACA GGTGGTGGCC
841 ATCGCCTCCA
ACATTGGTGG TAAGCAAGCC CTCGAAACTG TGCAGCGGCT GCTTCCAGTC
901 TTGTGCCAGG
CTCACGGCCT GACACCGGAG CAGGTGGTTG CAATCGCGTC TAATATCGGC
961 GGCAAACAGG
CACTCGAGAC CGTGCAGCGC TTGCTTCCAG TGCTGTGTCA GGCCCACGGC
1021 CTGACCCCGG AACAGGTGGT
GGCCATCGCC TCTAACAATG GCGGCAAACA GGCATTGGAA
1081 ACAGTTCAGC GCCTGCTGCC
GGTGTTGTGT CAGGCTCACG GCCTGACTCC GGAGCAGGTT
1141 GTGGCCATCG CAAGCCATGA
TGGCGGTAAA CAAGCTCTGG AGACAGTGCA ACGCCTCTTG
1201 CCAGTTTTGT GTCAGGCCCA
CGGA
Final Amino acids are remain the same：
1
LTPEQVVAIA SNGGGKQALE TVQRLLPVLC QAHG
35
LTPEQVVAIA SHDGGKQALE TVQRLLPVLC QAHG
69
LTPEQVVAIA SNNGGKQALE TVQRLLPVLC QAHG
103 LTPEQVVAIA
SNIGGKQALE TVQRLLPVLC QAHG
137 LTPEQVVAIA
SNGGGKQALE TVQRLLPVLC QAHG
171 LTPEQVVAIA
SNIGGKQALE TVQRLLPVLC QAHG
205 LTPEQVVAIA
SNGGGKQALE TVQRLLPVLC QAHG
239 LTPEQVVAIA
SHDGGKQALE TVQRLLPVLC QAHG
273 LTPEQVVAIA
SNIGGKQALE TVQRLLPVLC QAHG
307 LTPEQVVAIA
SNIGGKQALE TVQRLLPVLC QAHG
341 LTPEQVVAIA
SNNGGKQALE TVQRLLPVLC QAHG
375 LTPEQVVAIA
SHDGGKQALE TVQRLLPVLC QAHG
Corresponding part：
PART-left:
CTGACCCCGGAGACG
PART1(150bp):
CGTCTCGCCCCGGAACAGGTGGTGGCCATTGCAAGCAACGGTGGTGGCAAGCAGGCCCTGGAGACAGTCCAACGGCTGCTTCCGGTTCTGTGTCAGGCCCACGGCCTGACTCCAGAACAAGTGGTTGCTATCGTGGCGGAAAATGAGACG
PART2(219bp):
CGTCTCTAAAACAAGCCCTCGAAACCGTGCAGCGCCTGCTTCCGGTGCTGTGTCAGGCCCACGGGCTCACCCCGGAACAGGTGGTGGCCATCGCATCTAACAATGGCGGTAAGCAGGCACTGGAAACAGTGCAGCGCCTGCTTCCGGTCCTGTGTCAGGCTCATGGCCTGACCCCAGAGCAGGTCGTGGCAATTGCCTCCAACATTGGAGGGCGAGACG
PART3(262bp):
CGTCTCTAGGGAAGCAGGCACTGGAGACCGTGCAGCGGCTGCTGCCGGTGCTGTGTCAGGCCCACGGCTTGACCCCGGAACAGGTGGTGGCCATCGCCTCCAACGGCGGTGGCAAACAGGCGCTGGAAACAGTTCAACGCCTCCTTCCGGTCCTGTGCCAGGCCCATGGTCTGACTCCAGAGCAGGTTGTGGCAATTGCAAGCAACATTGGTGGTAAACAAGCTTTGGAAACCGTCCAGCGCTTGCTGCCAGTACGGAGACG
PART4(224bp):
CGTCTCCGTACTGTGTCAGGCCCACGGGCTTACCCCGGAACAGGTGGTGGCCATTGCAAGCAACGGTGGTGGCAAGCAGGCCCTGGAGACAGTCCAACGGCTGCTTCCGGTTCTGTGTCAGGCCCACGGCCTGACTCCAGAACAAGTGGTTGCTATCGCCAGCCACGATGGCGGTAAACAAGCCCTCGAAACCGTGCAGCGCCTGCTTCCGGTGCTGGGAGACG
PART5(194bp):
CGTCTCCGCTGTGTCAGGCCCACGGACTGACCCCGGAACAGGTGGTGGCCATCGCCTCCAACATTGGTGGTAAGCAAGCCCTCGAAACTGTGCAGCGGCTGCTTCCAGTCTTGTGCCAGGCTCACGGCCTGACACCGGAGCAGGTGGTTGCAATCGCGTCTAATATCGGCGGCAAACAGGCACTCGATGAGACG
PART6(249bp):
CGTCTCATCGAGACCGTGCAGCGCTTGCTTCCAGTGCTGTGTCAGGCCCACGGCCTGACCCCGGAACAGGTGGTGGCCATCGCCTCTAACAATGGCGGCAAACAGGCATTGGAAACAGTTCAGCGCCTGCTGCCGGTGTTGTGTCAGGCTCACGGCCTGACTCCGGAGCAGGTTGTGGCCATCGCAAGCCATGATGGCGGTAAACAAGCTCTGGAGACAGTGCAACGCCTCTTGCCAGTTTTAGAGACG
PART-right:
CGTCTCATTTTGTGTCAGGCCCACGGA...
The recognition sequence of these TALE protein:
TCGATATCAAGC
(End)
All parts are under artificial
synthesis process, so there is few results, which can
prove our changes are useful. However, with the principle of complementary base
pairing, our chioce should be better than original
vision. And if you want our data or use our method to create your own best
sticky ends, just contact us!
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