JP2000353156A - Method for obtaining solution of np-complete problem - Google Patents

Abstract

PROBLEM TO BE SOLVED: To solve a combination problem (NP-complete problem) for finding out a satisfactory solution from a huge number of combinations by a smaller number of processes by utilizing a condition that a DNA molecule forms a secondary structure. SOLUTION: In a node type Boolean expression (e.g. an expression for selecting each node from three variables) which is expressed by m variables (e.g. (a) to (f)) and n nodes (e.g. 1 to 10) and in which each of n nodes includes only m variables, each variable is expressed by a DNA molecule having 30 residues, for instance. Each node is peculiarly expressed by a base sequence of linkers connected to both end parts or one end part of a base sequence which represents variables. When DNA molecules to which linkers are attached are mixed with each other, these DNA molecules are mutually connected by oxygen and a product in which n DNA molecules are connected like a chain is generated. A solution can be obtained by removing DNA molecules forming a hairpin structure and not corresponding to the solution from the product.

Description

DETAILED DESCRIPTION OF THE INVENTION

[0001]

The present invention relates to an NP complete problem (for example, a satisfiability problem of a nodal Boolean expression) in which the number of combinations is enormous and it is difficult to find a solution. It relates to a method of solving by a biological technique.

[0002]

2. Description of the Related Art Solving combinatorial problems using DNA molecules is described, for example, in "Molecular Computation of Solutions of Combinatorial Problems" (Leonard M. Adleman, Science, Vol. 266, pp. 1021 to 1024, November 1994). On March 11). In this paper, the Hamiltonian path problem is taken as an example of the combination problem (NP complete problem). Specifically, point (0)-
There are seven points of point (6), and it is determined that each point can pass in one direction or both directions. Point (0) is the starting point, and points (1) to (5) The problem is whether it is possible to reach the end point (6) via each of them only once.

[0003] In order to solve this problem, each of the points (0) to (6) is represented as a single-stranded DNA molecule in which 20 bases unique to each point are connected. In addition, a traversable route between each point is defined as a half (10 bases) of a single-stranded DNA molecule representing a point on the starting point side of the route and a half (10 bases) of a single-stranded DNA molecule representing a point on the ending point side of the route. Single-stranded DNA complementary to the single-stranded DNA molecule linked to
Expressed as a molecule. Then, after preparing single-stranded DNA molecules representing these points and routes, and mixing all the prepared single-stranded DNA molecules, some of the single-stranded DNA molecules representing the points represent the routes between those points. Single-stranded DNA
The molecules combine as battens to produce various DNA molecules that represent traversable pathways.

[0004] Of these DNA molecules, the one whose starting point is point (0) and whose ending point is point (6) is
After amplification and selection by the PCR method, those that have passed through five points between the starting point and the ending point (in other words, those having a length of 140 base pairs) were separated by gel electrophoresis. And purify. Further, those passing through all of the points (1) to (5) are sequentially selected by hybridization with an oligomer having a sequence complementary to the DNA molecule representing the points (1) to (5). . At this time, by binding each oligomer to magnetic beads, a DNA molecule that has formed a hybrid can be easily selected. In this way, a satisfactory solution has been obtained.

[0005]

In the technique described in the above article, only some of the conditions (points and routes) constituting the Hamiltonian pathway problem are encoded in the form of the base sequence of a DNA molecule. Not all problems are coded. In other words, the conditions for setting the starting point at the point (0) and the ending point at the point (6), the condition of passing through each point only once, and the like are not coded. For this reason, after mixing DNA molecules representing points and routes, it is necessary to perform various operations for each condition in order to select one satisfying each condition in question. An object of the present invention is to solve a combination problem (NP complete problem) for finding a satisfactory solution from an enormous number of combinations using fewer steps by utilizing the fact that a DNA molecule forms a secondary structure.

[0006]

According to the present invention, there is provided a method for obtaining a satisfactory solution to the NP complete problem, comprising the steps of encoding and representing the NP complete problem in a plurality of types of DNA molecules having different base sequences from each other; Mixing the DNA molecules and causing the DNA molecules to react with each other to perform the operation autonomously; and, from among the products of the reaction between the DNA molecules,
A step of separating a product corresponding to the solution of the problem and a product not corresponding to the solution of the problem (claim 1). By expressing the problem formula in a DNA molecule in this way, a product corresponding to the solution of the problem can be obtained without requiring other operations other than the mixing of a plurality of types of DNA molecules,
A product that does not correspond to the solution of the problem can be obtained. A product that does not correspond to the solution of the above problem can be produced, for example, as having a hairpin structure composed of several base pairs (claim 2).

[0007] Examples of specific embodiments of the present invention include:
"Represented by m variables and n clauses (where m and n are each a positive integer of 2 or more), and each of the n clauses includes only the m variables. In the clause-type Boolean satisfiability problem, a DNA molecule is used as a means for expressing each of the m variables, and the DNA molecule has a different base sequence for each variable, and a variable between the variables or the variable itself. It is a base sequence that does not form a hairpin structure in it, and as a means for representing each of the n nodes, a linker connected to both ends or one end of the DNA molecule representing the variable, and The linker is configured to have complementarity only to the linker of the adjacent node, and after connecting the linker to the DNA molecule to obtain a “variable-linker” DNA molecule, -Linker " A mixture of NA molecules, n pieces of DN
A product in which the A molecules are linked in a chain is obtained, and then, by eliminating the products that have formed a hairpin structure from the products, only the products corresponding to the satisfaction of the above-mentioned Boolean expression are obtained. To obtain a satisfactory solution of the above clause-form Boolean satisfiability problem. (Claim 3). Here, the removal of the product having formed the hairpin structure can be performed by a method comprising a step of cutting the hairpin structure with a restriction enzyme and a step of amplifying only the product having no hairpin structure by a PCR method. (Claim 4).

[0008] The method of the present invention can also be applied to Boolean expressions having a large number of clauses. That is,
m variables and n × p clauses (where m, n, and p are each a positive integer of 2 or more), and each of the n × p clauses is the m After dividing the clause-form Boolean expression into p groups including n clauses and defining the first to p-th groups, the above NP complete problem The method for obtaining the solution is applied sequentially from the first group to the p-th group in such a manner that the product corresponding to the satisfactory solution of the previous group is added to the latter group, and finally the node form in the p-th group It is possible to obtain a Boolean satisfying solution (claim 5).

[0009]

DESCRIPTION OF THE PREFERRED EMBODIMENTS Hereinafter, the present invention will be described based on specific examples. First, a clause type Boolean expression which is an example of the NP complete problem will be described. As a simple clause Boolean expression:
Take F = (a or b) and (! A or! C) as an example. In the above formula, “! A” and “! C” represent negation of “a” and “c”, respectively. a! A variable or negation of a variable, such as a, is called a literal. Also, (a or b) and (! A or! C) that are part of the expression are called clauses. The above equation is simplified and expressed as F = (a + b) (! A +! C). Candidates for satisfying solutions of clause type Boolean expressions are ones in which literals are selected and arranged one by one from each clause. a, a! c 、 b ・! a, b ...! c. Among them, a! a is a (affirmative),
! It represents an inconsistent combination that is also a (negative), and is not a satisfactory solution. Therefore, the satisfaction solution is a!
c 、 b ・! a, b ...! c.

The present invention solves a clause type Boolean satisfiability problem having m variables and n clauses using a DNA molecule in the following manner. First, each of the m variables is represented by a specific base sequence that differs for each variable. In addition, each of the n nodes is defined by both ends (sections 2 to (n-1)) or one end of a base sequence (also referred to as “variable sequence” in the present specification) representing a variable. Sections 1 and n
Section). Here, each of the n nodes is represented by a specific base sequence that differs for each node. The base sequence corresponding to the linker is determined so that the base sequence is not rotationally symmetric so as not to bond with the same linker.

A base sequence representing a negation of a variable is represented by a sequence complementary to the variable sequence. There are as many combinations of clauses and variables as there are Boolean expressions. For example, if each clause contains three variables, there are 3 × n combinations. The combination of these variables and clauses is covered by the combination of the variable sequence and the linker connected to both ends or one end thereof. here,
The variable selected from the first section has a base sequence of a PCR primer binding site instead of a linker at the 5 ′ end. The variable selected from section n has the base sequence of the PCR primer binding site instead of the linker at the 3 ′ end.

After the base sequences corresponding to the variables and the nodes are determined in this manner, the double-stranded DN consisting of the variable sequence and its complementary sequence
Prepare A. The 5 'end of the DNA molecule is extended to form a single-stranded portion. This single-stranded portion is filled with a linker sequence. Here, the combination of the variable sequence and the linker is indicated by a given Boolean expression as described above. The 5 'end of the DNA molecule has a phosphate group. The 3 'end has a hydroxyl group. Such a DNA molecule can be easily obtained by annealing two oligonucleotides.

When the above-mentioned DNA molecules are mixed and subjected to a ligation reaction, a product is obtained in which n DNA molecules are linked in a line by linking them at a linker portion. What is produced here is a mixture of DNA molecules having a plurality of different base sequences. For example, if each clause contains three variables, 3 n powers of DNA molecules are obtained. These D
In each of the NA molecules, one variable from each of the first to nth sections is connected in this order.

The products of the above ligation reactions include all cases of selection from each clause, expressed by Boolean expressions, and do not include selection methods that contradict Boolean expressions. The product contains multiple types of DNA
The molecules are amplified by a PCR reaction in a mixed state. At this time, one of the primers is 5 ′
One having a chemical functional group bonded to the terminal is used. This functional group has a property of strongly binding to the solid support. The PCR-amplified DNA molecule is converted to a single-stranded form by binding to a solid phase via either 5 ′ end and subsequent denaturation. Since the single-stranded DNA molecule is firmly bound to the solid phase, there is no interaction between the molecules.

[0015] Within a single-stranded DNA molecule, one that contains both positive and negative for a variable forms a hairpin structure by intramolecular interaction. In this hairpin structure, a certain variable sequence (base sequence representing a variable) and its negative sequence are annealed to form a double strand. In order to prevent annealing of unrelated variable sequences, a base sequence suitable for the purpose must be used as the variable sequence. Tools for designing such base sequences are:
It has already been developed and reported by several research groups.

In order to remove the DNA molecules forming the hairpin structure, the following two operations are performed. First, a few residues in the middle of the variable sequence are reserved for restriction enzyme sites so that the duplex formed by the variable sequence and its complementary sequence can be cleaved by the restriction enzyme. The DNA molecule cleaved by the restriction enzyme is not amplified by a PCR reaction using the above primers.

The second is a method utilizing the properties of DNA polymerase. Since this enzyme cannot penetrate into the double strand of the hairpin structure and synthesize DNA, synthesis stops before the double strand. Therefore, a DNA molecule having a hairpin structure is not amplified by PCR.

By combining the above two methods, a DNA molecule having a hairpin structure can be removed and a linear DNA molecule can be concentrated. By cloning the DNA molecule thus concentrated and determining the base sequence, it is possible to obtain a method of selecting a variable that does not include contradiction of affirmative / negative for any variable. In these selection methods, a satisfactory solution can be obtained by reading whether each variable is selected as positive or negative. This operation is very easy because it can be performed with a time proportional to the number of nodes.

[0019]

The present invention will be described below based on examples (experimental examples). In the embodiment, an object is to obtain a solution of the following clause type Boolean expression G. G = (a + b +! C) (a + c + d) (a +! C +!)
d) (! a +! c + d) (a +! c + e) (a + d +!)
f) (! a + c + d) (a + c +! d) (! a +! c +
! d) (! a + c +! d) Here, the alphabets of “a” to “f” represent variables, and those with “!” added to the left side of the alphabets of “a” to “f” are “!”. Represents the negation of what is not attached. "()" Indicates a clause, and "+" in the clause is "or"
And there is a “and” relationship between clauses. Each variable can take a value of “0” or “1”. When a certain variable is “1”, the negation is “0”, and when the variable is “0”, the negation is “1”.

For example, from each section, "! Ca-! D
-! ca-a-d-! d-! c-! d ”, a = 1, c = 0, d = 1, and what the value of the Boolean expression will be. When G = 1 × 1 × 0 × 1 × 1 × 1 × 1 × 0 × 1 × 0 = 0 ”. Even if d = 0, G = 0.

"A" to "1" so that the value of the Boolean expression becomes 1.
Whether the value of “f” can be determined is called a Boolean satisfaction problem, and such values of “a” to “f” are called a satisfaction solution. Whether this Boolean expression is satisfiable is equivalent to the question, "Can we choose variables from each clause without choosing both positive and negative for any variable?"

There are three 10th powers (= 59,049) in the case where variables are selected from each clause of the equation G. In this problem, only 24 of them correspond to correct selection. These 24 ways all correspond to the same satisfactory solution. Although this example is not a difficult problem, it is not easy to understand at a glance what the satisfaction solution of the G expression is, so it is an appropriate problem to verify the effectiveness of the present invention. Based on the property of producing 59,049 cases with a DNA molecule and further forming a secondary structure of the DNA molecule, all of these cases were checked at the same time, and finally a solution was obtained. Hereinafter, the method of the experiment will be specifically described.

Each of the variables (af) is represented by a sequence of 30 bases. a. TTGGTTGATAGACCCAGGATCGAGTGCCAT b. TTGGCATAAGTTGCCAGGCAGGAACTCCAT c. TTGGAACGTAGTACCAGGAGTCTCCTCCAT d. TTGGTGATACGGACCAGGCCTTCTTACCAT e. TTGGTTGCCCTCTCCAGGTGACTAATCCAT f. TTGGACTGCTCATCCAGGACTGAAGACCAT

A linker connected to both ends or one end of these base sequences representing variables is represented by a four-base sequence as follows. The number n (n = 1 to 9) displayed on the left side of the base sequence is a linker number, and indicates that the linker is between the nth section and the (n + 1) th section. 1. CAGA 2. CTTC 3. GTGT 4. TCCT 5. ACTC 6. ATGC 7. TCAC8. TTCG 9. GAAC

At the 5 'end of the variable selected from section 1,
Primer binding site (pbs
1) and a primer binding site (pbs2) to be connected instead of a linker at the 3 'end of a variable selected from section 10
Is a sequence of 20 bases as follows. pbs1: AGCGAGTGTTAACCCTGCCT pbs2: CGACTACGATATTCGCGCAG

The above coding is shown in FIG. In FIG. 1, the first section includes a base sequence representing any of the variables “a”, “b”, and “! C”, and a linker 1 (“3 ′ end”) linked to one end (3 ′ end side) of the base sequence. (4 bases). The second section includes a base sequence representing any of the variables “a”, “c”, and “d”, a linker 1 ′ linked to one end (5′-end side) of the base sequence, It is represented as a base sequence having a linker 2 linked to the other end (3 ′ end side) of the sequence. Hereinafter, Sections 3 to 10
The same applies to clauses. In FIG. 1, the primer binding site (pbs1) attached to one end of the first section and the primer binding site (pbs2) attached to one end of the tenth section are not shown.

As described above, after coding the variables and clauses by the base sequence, a DNA molecule having a linker connected to both ends or one end of the variable sequence (base sequence representing the variable) (referred to as “variable -Linker "DNA molecule). To facilitate the production of these DNA molecules, a variable sequence selected from three to four consecutive nodes is
It was made as a series of DNA molecules and then cut off later. The sequences of nine such DNA molecules are shown below. In practice, this DNA molecule is a double-stranded DNA, which is designed to be cleaved by the enzyme at the linker sequence site so that the linker becomes a single-stranded portion of the 5 'overhang.

Sequence of DNA molecule DNA1-1 (pbs1-a-1-c-2-! D-3) AGCGAGTGTTAACCCTGCCT-TTGGTTGATAGACCCAGGATCGAGTGCCAT-CAGA-TTGGAACGTAGTACC AGGAGTCTCCTCCAT-CTTC-ATGGTAAGAAGGCCTGGTCCGTATCACCAA-GTGT-TTGGACACAC DNA1-2! -4-! C-5-a-6-d-7) ACACACCCAT-GTGT-ATGGCACTCGATCCTGGGTCTATCAACCAA-TCCT-ATGGAGGAGACTCCTGGTAC TACGTTCCAA-ACTC-TTGGTTGATAGACCCAGGATCGAGTGCCAT-ATGC-TTGGTGATACGGACCAGGCC TTCT-CATAC-TCAC-GG d-9-! a-pbs2) ACACACCCAT-TCAC-TTGGTTGATAGACCCAGGATCGAGTGCCAT-TTCG-ATGGTAAGAAGGCCTGGTCC GTATCACCAA-GAAC-ATGGCACTCGATCCTGGGTCTATCAACCAA-CGACTACGATATTCGCGCAG

DNA2-1 (pbs1-b-1-a-2-! C-3) AGCGAGTGTTAACCCTGCCT-TTGGCATAAGTTGCCAGGCAGGAACTCCAT-CAGA-TTGGTTGATAGACCC AGGATCGAGTGCCAT-CTTC-ATGGAGGAGACTCCTGGTACTACGTTCCAA-GTGT-TTGGACACAC DNA2-2 (3-! C-4-e -5-d-6-! A-7) ACACACCCATGTGT-ATGGAGGAGACTCCTGGTACTACGTTCCAA-TCCT-TTGGTTGCCCTCTCCAGGTGA CTAATCCAT-ACTC-TTGGTGATACGGACCAGGCCTTCTTACCAT-ATGC-ATGGCACTCGATCCTGGGTCT ATCAACCAA-TCAC-TTGGACAC-8 -pbs2) ACACACCCAT-TCAC-TTGGAACGTAGTACCAGGAGTCTCCTCCAT-TTCG-ATGGAGGAGACTCCTGGTAC TACGTTCCAA-GAAC-ATGGTAAGAAGGCCTGGTCCGTATCACCAA-CGACTACGATATTCGCGCAG

DNA3-1 (pbs1-! C-1-d-2-a-3) AGCGAGTGTTAACCCTGCCT-ATGGAGGAGACTCCTGGTACTACGTTCCAA-CAGA-TTGGTGATACGGACC AGGCCTTCTTACCAT-CTTC-TTGGTTGATAGACCCAGGATCGAGTGCCAT-GTGT-TTGGACACAC DNA3-2 (3-d-4-a- 5-! F-6-c-7) ACACACCCATGTGT-TTGGTGATACGGACCAGGCCTTCTTACCAT-TCCT-TTGGTTGATAGACCCAGGATC GAGTGCCAT-ACTC-ATGGTCTTCAGTCCTGGATGAGCAGTCCAA-ATGC-TTGGAACGTAGTACCAGGAGT CTCCTCCAT-TCAC-TTGGACAC-d-7-a pbs2) ACACACCCATTCAC-ATGGTAAGAAGGCCTGGTCCGTATCACCAA-TTCG-ATGGCACTCGATCCTGGGTCT ATCAACCAA-GAAC-TTGGAACGTAGTACCAGGAGTCTCCTCCAT-CGACTACGATATTCGCGCAG (Note)
The base sequence of DNA encoding a variable selected from three to four consecutive nodes is shown. Hyphens in the base sequence indicate boundaries between linkers and variables. The notation in parentheses described on the right side such as “DNA1-1” is obtained by translating the base sequence into the arrangement of variables and linkers. The numbers in parentheses are the linker numbers.

In order to cut off the nucleotide sequence at the linker site, a restriction enzyme BstXI (manufactured by Toyobo) was used. Since this enzyme cleaves double-stranded DNA at the CCANNNNNNTGG sequence portion (N may be any of A, C, G, and T), the CCAN and NTGG sequences are connected before and after the linker, respectively. Each variable array has already been designed to start with NTGG and end with CCAN. The reaction of the restriction enzyme was carried out continuously at 45 ° C. for 18 hours using a buffer attached to the restriction enzyme.

In this way, 30 "variable-linker" DNA molecules were obtained. Next, these DNA molecules (10 pmol in total) were put together and a ligation reaction (capacity: 0.006 mL) was performed. The composition of the reaction buffer is 66 mM Tris-HCl (pH 7.6), 6.6 mM MgCl ₂ , 10 mM
DTT, 1 mM ATP, the enzyme used was T4 DNA ligase (18 units) (Takara Shuzo), and the reaction conditions were 1
6 ° C., 90 minutes. 10 "variable-linker"
The product to which the DNA molecule is bound is the desired product,
In practice, shorter DNA molecules are also produced, so they are separated by polyacrylamide gel electrophoresis,
Only molecules were cut out of the gel and extracted. In this way, the pool of DNA molecules expressing the selection method when selecting variables one by one from all the clauses of the Boolean expression G is 1n
g was obtained. Since the DNA of 1ng include 10 ⁹ molecules,
On average, more than 10,000 DNAs for each selection
It will contain molecules. Therefore, it can be said that all possible selection methods are included in this pool.

To find a solution from this pool,
The DNA molecules forming the hairpin structure may be removed, and the DNA molecules corresponding to the solution may be concentrated. For this purpose, first, a DN having a hairpin structure is
Decompose A. For this purpose, the middle five residues of each variable sequence are designed as recognition sites for the restriction enzyme BstNI.

Here, the method for removing the DNA molecules forming the hairpin structure will be briefly described with reference to the drawings. In FIG. 2, a single-stranded DNA molecule 1 is a double-stranded region 2 having a hairpin structure. And one end is attached to the dynabeads 3. When the restriction enzyme BstNI acts thereon, the double-stranded region 2 is cleaved at the cleavage site by the restriction enzyme BstNI, as shown in FIG. In addition, as shown in FIG. 4, in the case of having a hairpin structure, amplification is started from one end of the DNA molecule by the DNA polymerase, but is inhibited at the end of the double-stranded region 2.

[0035]Confirmation of usefulness of digestion by restriction enzymes  Use restriction enzymes to remove hairpin DNA
In some methods, DNA with a hairpin structure is double-stranded
Because it has a part, it can be expected that it will be cut.
On the other hand, DNA molecules that should not form a hairpin structure
In some cases, an unexpected secondary structure causes
It could be done. Therefore, for preliminary experiments
Therefore, a DNA molecule without a hairpin structure is a restriction enzyme
It is especially important to make sure that
is there. Therefore, the following experiment was performed.

The oligomer A has a base sequence in which the variable a, the linker 1, and the variable b are arranged in this order. The base sequence is as follows. Oligomer A: TTGGTTGATAGACCCAGGATCGAGTGCCATCAGA
TTGGCATAAGTTGCCAGGCAGGAACTCCAT Oligomer B has a base sequence in which variable a, linker 1, and negation of variable a (! A) are arranged in this order. The base sequence is as follows. Oligomer B: TTGGTTGATAGACCCAGGATCGAGTGCCATCAGA
ATGGCACTCGATCCTGGGTCTATCAACCAA

That is, the oligomer B forms a hairpin structure, and the oligomer A does not form a hairpin structure.
FIG. 5 shows the results before and after the enzymatic reaction analyzed by electrophoresis using a polyacrylamide gel (8 M urea, 6% acrylamide). In the figure, lanes 1 and 2 show oligomer A, and lanes 3 and 4 show oligomer B. Lanes 1 and 3 are before cleavage by the enzyme, and lanes 2 and 4 are after cleavage by the enzyme. The sample before enzyme digestion was kept at 94 ° C. for 1 minute, cooled to room temperature, and analyzed. As can be seen in lane 3, oligomer B migrates faster than expected from its length (same as oligomer A), indicating a hairpin structure. The enzyme reaction is performed by heating the sample to 94 ° C.
After cooling for 1 minute, cool to 60 ° C and use the restriction enzyme BstNI
(New England Biolabs) was added, and the mixture was incubated for 1 hour and 30 minutes. As a reaction buffer, a buffer attached to a commercially available enzyme was used. In Lane 4, although some oligomer B remains after being partially cleaved, most is degraded, and a band appears at a position where the migration distance is long. On the other hand, oligomer A
As is clear when lane 1 and lane 2 are compared, no change is seen at all. From the above experiments, it was shown that the degradation by an enzyme is effective for specifically removing DNA having a hairpin structure.

[0038]Degradation by restriction enzymes  For the above pool, perform PCR between pbs1 and pbs2
Increased the amount. The PCR reaction was performed using a commercially available Ex T
Perform using aq DNA polymerase (Takara Shuzo).
Reaction buffer and nucleotides
An acid mixture was used. At this time, the plug that binds to pbs1
The 5 'end of the immer is biotinylated. Thus obtained
The PCR product obtained is treated with streptavidin via biotin.
Bead (Dynal). This
The experiment using the beads was based on the theory attached to the commercially available beads.
It was performed based on the written instructions. However, denature with alkali
After that, 0.1N NaOH, 0.3M sodium acetate buffer (pH
5.2), B & W buffer (5 mM Tris-HCl (pH 7.5), 0.5 mM
EDTA, 1.0 mM NaCl, TE buffer (5 mM Tris-HCl (pH 7.
5), 0.5 mM EDTA) and then neutralized.

DNA bound to beads via biotin
The molecule is denatured with an alkali and converted into a single strand, and the liquid phase is subjected to an enzyme reaction buffer (50 mM Tris-HCl (pH 7.5), 10 mM MgCl 2).
_2. Replace with 1 mM DTT). Thereafter, the temperature is kept at 94 ° C. for 30 seconds and cooled to 60 ° C. over 2 minutes. 1 at 60 ° C
After holding for 1 min, the restriction enzyme BstNI is added at 1
The unit was added, and the temperature was kept at 60 ° C. for 30 minutes. afterwards,
After washing with an alkali, an enzymatic reaction was performed again in the same procedure. After washing again with alkali, a part of the beads was sampled and subjected to PCR, and only the DNA molecules which were not cut and remained at two sites, pbs1 and pbs2, were collected. Since the DNA molecules thus obtained are mixed in various lengths, only the molecules corresponding to the length of 10 variables were purified and used for the next reaction.

As described above, when manipulating a small amount of a DNA molecule, it is essential to increase the amount by PCR, and thus, mutation may occur in the DNA molecule during PCR. When such a displacement occurs at the restriction enzyme site,
The above method cannot remove DNA having a hairpin structure. For this reason, a second method is prepared complementarily, and the DNA molecule having the hairpin structure is removed by this method.

[0041]Has a hairpin structure using PCR reaction
Removal of DNA molecules  The second approach is that DNA polymerase has a hairpin structure.
Utilizing the property of not being able to successfully replicate DNA molecules
ing. However, DNA molecules that form a hairpin structure
However, it always takes a hairpin structure in any state
However, in normal PCR, a hairpin structure is formed.
Such DNA molecules are efficiently amplified. for that reason,
It is necessary to devise PCR conditions.

The DNA molecules recovered after the enzymatic reaction contained 20
Add only ng or less to 0.025 mL of PCR reaction. Initially, heat at 94 ° C. for 30 seconds without adding enzymes or primers and cool to 60 ° C. over 2 minutes. And 60
0.025 mL PCR reaction solution (in solution, 5 pm
ol primer and 0.5 units of Taq DNA polymerase. ), Followed by nine PCR cycles. Each cycle was maintained at 60 ° C. for 1 minute, 9
A process of incubation at 9 ° C. for 3 minutes and cooling to 60 ° C. for 2 minutes is included in this order. Finally, in the tenth cycle, only the temperature is kept at 60 ° C. for 1 minute. Immediately before each cycle, half of the reaction solution (0.025 mL) is discarded, and the same amount of the reaction solution at 60 ° C (2.5 pmol of primer and 0.25 mol
Unit enzyme is included. ).
Thereby, DNA molecules that have not been replicated due to the formation of the hairpin structure are diluted by half in each cycle. Only DNA molecules that do not form a hairpin structure will be present in the reaction at the initial concentration. After the completion of the reaction in the 10th cycle, normal PCR amplification is performed. However, the amplification is limited to 8 to 10 cycles.

The DNA molecule thus obtained was cloned, and the nucleotide sequence of each DNA molecule was analyzed. Analysis of the four clones resulted in DNA molecules corresponding to the solution. The nucleotide sequences of these four clones are shown in order of the variables. 1. bd-! d-! ae-! f-! aa-! d-! d 2. bdadadda-! dc 3. bc-! d-! ae-! fcc-! d-! a baaded-! ac-! ac It is apparent at first glance that the third clone contains only positive or negative for any variable, and thus the satisfactory solution (a, b, c, d, e) , F) = (0, 1,
1,0,1,0).

[0044]Method with a large number of clauses and variables  Next, a case where the number of clauses and variables is large will be described. This
In such a case, the full length of the Boolean expression is generated at once.
It is not possible. Because the number of cases can be very large,
DNA content that can be accommodated in a normal size test tube
This is because the number of children exceeds the upper limit. 100 variables
And 400 clauses, so that each clause has three variables
The method will be described by taking a simple Boolean solution as an example.

The satisfiable number of the Boolean expression is to assign “0” or “1” to each variable. By assigning “0” to the negation of the variable and “1” to the affirmative, (a,
b, c) = (0, 1, 1) It is the same value as how to select the variable abc. In this way, a variable that is selected once and each of the variables is selected and determined as either positive or negative can be regarded as a “solution candidate”.

In the above experimental example, the "solution" was read from the Boolean expression in a satisfied state. However, for a large problem, "solution candidates" were also prepared together with the DNA molecule representing the Boolean expression.
As the calculation progresses, the "solution candidates" are narrowed down so that "solutions" remain. First, a Boolean expression consisting of 400 clauses is represented by a DNA molecule every 10 clauses.
This is done as described above. The first ten clauses contain at most 30 variables. All “solution candidates” containing these variables were created and the first 10 DNs
Connect to A molecule. The types of DNA molecules thus obtained are such that they can be contained in a normal test tube.

Here, the calculation based on the formation of the hairpin structure described above is performed to remove the DNA molecules having the hairpin structure.
For each of the remaining DNA molecules, a "solution candidate"
Is recovered by PCR. The number of “solution candidates” has been reduced by this operation. The above is the first stage of the calculation. A “solution candidate” including a variable newly appearing in the next section 10 is created with a DNA molecule, and this is connected to the “solution candidate” remaining in the first-stage operation.
Further, it is connected to a DNA molecule expressing a Boolean expression for the ten clauses, and performs an operation. When the remaining “solution candidates” are collected, the second stage of the operation is completed. In this way,
The operation proceeds sequentially to the last clause. At this time, the “solution candidates” are narrowed down by reducing the types, and sequentially include many variables, and finally, “solutions” including either positive or negative for all variables Is obtained.

[0048]

According to the method of the present invention, since the NP complete problem is encoded in a plurality of types of DNA molecules, the operation can be performed autonomously simply by mixing these types of DNA molecules. Done. In the clause-type Boolean expression, a DNA molecule that does not correspond to a satisfactory solution is generated as having a hairpin structure. Therefore, using a restriction enzyme and a PCR method, the material having a hairpin structure is removed to represent a satisfactory solution. Only DNA molecules can be obtained. Then, by analyzing the base sequence of the obtained DNA molecule, a satisfactory solution can be obtained. As described above, the method of the present invention provides a principle of an autonomous calculation method of a combination problem using DNA molecules, and is very promising as a solution to a problem that cannot be solved by a current computer. .

[Brief description of the drawings]

FIG. 1 is a diagram showing a clause type Boolean expression encoded.

FIG. 2 is a diagram showing a state in which a DNA molecule has formed a hairpin structure.

FIG. 3 is a diagram showing a state in which a DNA molecule having a hairpin structure is cleaved by a restriction enzyme in a double-stranded region.

FIG. 4 is a diagram for explaining that DNA amplification is interrupted at the end of a double-stranded region in a DNA molecule having a hairpin structure.

FIG. 5 is a diagram showing the results of electrophoresis of oligomer A and oligomer B.

[Explanation of symbols]

 1 DNA molecule 2 Double-strand forming region 3 Dynabeads

Claims (5)

[Claims] 1. An autonomous NP-complete problem by encoding a plurality of types of DNA molecules having different base sequences from each other and expressing the NP complete problem by mixing the plurality of types of DNA molecules and causing the DNA molecules to react with each other. And a step of separating, from the products of the reaction between the DNA molecules, a product corresponding to the solution of the problem and a product not corresponding to the solution of the problem. To obtain a solution to the complete NP problem. 2. The method according to claim 1, wherein the products not corresponding to the solution of the problem have a hairpin structure. 3. m variables and n clauses (where m, n
Is a positive integer of 2 or more. ),
In the clause type Boolean satisfiability problem in which each of the n clauses includes only the m variables, a DNA molecule is used as a means for representing each of the m variables, and the DNA molecule is The base sequence is different for each variable, and the base sequence is such that a hairpin structure is not formed between the variables or within the variable itself. As means for representing each of the n nodes, DNA representing the variable A linker connected to both ends or one end of the molecule is used, and the linker is configured to have complementarity only to a linker of an adjacent node, and the linker is connected to the DNA molecule. "Variable-
After obtaining the “linker” DNA molecule, the “variable-linker” DNA molecule is mixed to obtain a product in which n DNA molecules are linked in a chain, and then, among the products,
The method for obtaining a solution to the NP-complete problem according to claim 2, wherein only products corresponding to the satisfying solution of the above-mentioned clause type Boolean expression are obtained by eliminating the products that have formed a hairpin structure. 4. The method according to claim 3, wherein the removal of the product having the hairpin structure comprises the steps of cutting the hairpin structure with a restriction enzyme, and amplifying only the product having no hairpin structure by PCR. NP described in
How to get a complete solution. 5. m variables and n × p clauses (where,
m, n, and p are each a positive integer of 2 or more. ), Where each of the n × p clauses is divided into p groups of n clauses in the clause Boolean expression containing only the m variables. , After defining the first to p-th groups, the method for obtaining the solution of the NP complete problem according to claim 3 or 4 corresponds to the first to p-th groups corresponding to the satisfactory solution of the previous group. A method of obtaining a solution of the NP-complete problem, in which the products are sequentially applied as they are added to the subsequent groups, and finally a satisfactory solution of a clause-type Boolean expression is obtained in the p-th group.