# Dinucleotide Circular Codes

Authors: Christian J. Michel; Giuseppe Pirillo
Journal: ISRN Biomathematics
 (2013)

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## Abstract

We begin here a combinatorial study of dinucleotide circular codes. A word written on a circle is called circular. A set of dinucleotides is a circular code if all circular words constructed with this set have a unique decomposition. Propositions based on a letter necklace allow to determine the 24 maximum dinucleotide circular codes (of 6 elements). A partition property is also identified with eight self-complementary maximum dinucleotide circular codes and two classes of eight maximum dinucleotide circular codes in bijective correspondence by the complementarity map.


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## Body

## 1. Introduction

We continue our study of the combinatorial properties of circular codes in genes, that is, on the nucleotide alphabet𝒜4={A,C,G,T}. A dinucleotide is a word of two letters (diletter) on 𝒜4. A trinucleotide is a word of three letters (triletter) on 𝒜4. The two sets of 16 dinucleotides and 64 trinucleotides are codes in the sense of language theory but not circular codes [1, 2]. In order to have an intuitive meaning of these notions, codes are written on a straight line, while circular codes are written on a circle, but, in both cases, unique decipherability is required.Trinucleotide comma-free codes, a very particular case of trinucleotide circular codes, have been studied for a long time, see for example, [3–5]. After the discovery of a trinucleotide circular code in genes with strong mathematical properties [6], circular codes are mathematical objects studied in combinatorics, theoretical computer science, and theoretical biology. This theory underwent a rapid development, see for example, [7–27].Trinucleotides are the fundamental words for genes, that is, the DNA sequences coding the amino acids constituting the protein sequences. However, dinucleotides are also words with important biological functions in genomes. Dinucleotides are involved in some genome sites, for example, the splice sites of introns in eukaryotic genomes are based on the dinucleotidesGT and AT [28, 29]. Dinucleotides are also involved in some genome regions, for example, the dinucleotide CG in animal and plant genomes allows a positive or negative control over gene expression [30], and the dinucleotides CA [31, 32], CT [33], and TG [34] in eukaryotic genomes occur as concatenated words (l1l2)+, l1,l2∈𝒜4 (called tandem repeats in biology).We begin here a new combinatorial study concerning the dinucleotide circular codes. Their number, their list, and a partition according to the complementarity map are determined with propositions based on a letter necklace.


## 2. Preliminaries

The following definitions and propositions are classical for any finite set of words on any finite alphabet [1]. We recall them for dinucleotides, that is, words of length 2 on a 4-letter alphabet. Let 𝒜4={A,C,G,T} denote the genetic alphabet, lexicographically ordered by A<C<G<T. The set of nonempty words (resp., words) on 𝒜4 is denoted by 𝒜4+ (resp.,𝒜4*). The set of the 16 words of length 2 (dinucleotides or diletters) over 𝒜4 is denoted by 𝒜42. The set of the 64 words of length 3 (trinucleotides or triletters) over 𝒜4 is denoted by 𝒜43.Definition 1.
A setX of words in 𝒜42 is a dinucleotide code if, for each x1,…,xn,x1′,…,xm′∈X,  n,m≥1, the condition x1⋯xn=x1′⋯xm′ implies n=m and xi=xi′ for  i=1,…,n.Dinucleotide codes are read on a straight line.Definition 2.
A dinucleotide codeX in 𝒜42 is circular if, for each x1,…,xn,x1′,…,xm′∈X,  n,m≥1, p∈𝒜4*, s∈𝒜4+, the conditions sx2⋯xnp=x1′⋯xm′ and x1=ps imply n=m, p=ε (empty word), and xi=xi′ for i=1,…,n.Dinucleotide circular codes are read on a circle.Remark 3.
The set𝒜42 is a code but not a circular code.Definition 4.
Two dinucleotidesx and y are conjugate if there exist two letters l1 and l2, l1,l2∈𝒜4, such that x=l1l2 and y=l2l1.Proposition 5 (see [1]).
A dinucleotide circular code cannot contain a word of the form
u
2
with
u
≠
ε
.The periodic dinucleotidesAA, CC, GG, and TT cannot be in a dinucleotide circular code.Proposition 6 (see [1]).
A dinucleotide circular code cannot contain conjugate dinucleotides.Example 7.
The dinucleotidesAC and CA cannot be in the same circular code.The set operations complementarity𝒞, permutation 𝒫, and mirror image  ̃  defined later are involutions.Definition 8.
The nucleotide complementarity map𝒞:𝒜4→𝒜4 is defined by 𝒞(A)=T,𝒞(T)=A,𝒞(C)=G, and 𝒞(G)=C.Definition 9.
The dinucleotide complementarity map𝒞:𝒜42→𝒜42  is defined by 𝒞(l1l2)=𝒞(l2)𝒞(l1)   for all  l1,l2∈𝒜4.Example 10.
𝒞
(
A
C
)
=
G
T.Definition 11.
The complementary dinucleotide set𝒞(X) of a dinucleotide set X is the set obtained by applying the dinucleotide complementarity map 𝒞 to all the dinucleotides of X.Remark 12.
𝒞
2
(
X
)
=
𝒞
(
𝒞
(
X
)
)
=
X.Definition 13.
A dinucleotide circular codeX is self-complementary if, for each x∈X,𝒞(x)∈X.Definition 14.
The (left) dinucleotide circular permutation map𝒫:𝒜42→𝒜42 permutes circularly each dinucleotide 𝒫(l1l2)=l2l1,l1,l2∈𝒜4.Definition 15.
The permuted dinucleotide set𝒫(X) of a dinucleotide set X is the set obtained by applying the circular permutation map 𝒫 to all the dinucleotides of X.Remark 16.
𝒫
2
(
X
)
=
𝒫
(
𝒫
(
X
)
)
=
X.Definition 17.
The mirror image of a dinucleotidex=l1l2 is x~=l2l1, l1,l2∈𝒜4.Definition 18.
The mirror imageX~ of a dinucleotide set X is the set of the mirror images of all the dinucleotides of X.Remark 19.
X
~
~
=
X.Remark 20.
For a dinucleotidex and for a dinucleotide set X, we have 𝒫(x)=x~ and 𝒫(X)=X~.Proposition 21 (see [27]).
A dinucleotide codeX is circular if and only if the dinucleotide code X~ is circular.Proposition 22.
A dinucleotide code
X
is circular if and only if the permuted dinucleotide code
𝒫
(
X
)
is circular.Proof.
By Proposition21 and Remark 20.Remark 23.
Proposition22 is not true with trinucleotides [6].


## 3. Results

In this paper, we identify the subsets of𝒜42 which are circular codes. Based on a letter necklace, we prove a necessary and sufficient condition for a set of dinucleotides to be a circular code.Definition 24.
Let
l
1
,
l
2
,
l
3
,
…
,
l
n
,
l
n
+
1 be  letters in 𝒜4. One says that the ordered sequence l1,l2,l3,…,ln,ln+1is a (n+1)-necklace for a subset X⊂𝒜42 if each dinucleotide  l1l2,l2l3,…,lnln+1 belongs to X.Proposition 25.
Let
X
be a subset of
𝒜
4
2
. The following conditions are equivalent:
(1)
X
is circular code,
(2)
X
has no 5-necklace.Proof.
(1)→ (2). Let X be a circular code. We have to prove that X has no 5-necklace. Suppose, by way of contradiction, that l1,l2,l3,l4,l5 is a 5-necklace for X. As 𝒜4 contains four letters, for some i,j∈{1,2,3,4,5}, i<j, we have that li=lj. Remark that the maximum value of j-i is 4. (i)
Ifj-i=1, then X has a periodic dinucleotide αα. (Contradiction with Proposition 5.)(ii)
Ifj-i=2, then X has two conjugated dinucleotides αβ and βα. Contradiction with Proposition 6.(iii)
Ifj-i=3, then either l1=l4 or l2=l5.(iiia)
Ifl1=l4, the 5-necklace l1,l2,l3,l4,l5 is l1,l2,l3,l1,l5. So, l1l2,l2l3,l3l1,l1l5∈X. Consider the sequence l1,l2,l3,l1,l2,l3. Put x1=l1l2, x2=l3l1, x3=l2l3, x1′=l2l3, x2′=l1l2, x3′=l3l1. Note that x1,x2,x3,x1′,x2′,x3′ belong to X. Now, the following relations hold: l2l3l1l2l3l1=l2·l3l1·l2l3·l1=l2x2x3l1=x1′x2′x3′ and l1l2=x1. (Contradiction with the assumption that X is a circular code.)(iiib)
The case ofl2=l5 is analogous.(iv)
Ifj-i=4, then l1=l5, and the 5-necklace l1,l2,l3,l4,l5 is l1,l2,l3,l4,l1. So, l1l2,l2l3,l3l4,l4l1∈X. Consider the sequence l1,l2,l3,l4,l1. Put x1=l1l2, x2=l3l4, x1′=l2l3, x2′=l4l1. Note that x1,x2,x1′,x2′ belong to X. Now, the following relations hold: l2l3l4l1=l2·l3l4·l1=l2x2l1=x1′x2′ and l1l2=x1. Contradiction with the assumption that X is a circular code.
(2)→ (1) Let X be without 5-necklace and suppose, by way of contradiction, that X is not a circular code. As X is a uniform code, there exist x1,…,xn,x1′,…,xn′∈X, n≥1, p1,s1∈𝒜4+, such that s1x2⋯xnp1=x1′⋯xn′ and x1=p1s1. Moreover, as all the elements of X have a length of 2, there exist p2,…,pn,s2,…,sn∈𝒜, such that p2s2=x2,…, pn-1sn-1=xn-1, pnsn=xn and s1p2=x1′, s2p3=x2′,…, sn-1pn=xn-1′, snp1=xn′. Nowifn≥3, then p1,s1,p2,s2,p3 is a 5-necklace,ifn=2, then p1,s1,p2,s2,p1 is a 5-necklace,ifn=1, then p1,s1,p1,s1,p1 is a 5-necklace.In any case, there is a5-necklace. Contradiction.If a dinucleotide setX is a circular code, then there exists no word of X with two different decompositions of their products written on a circle.Example 26.
Consider the setX containing only the dinucleotides AC and GT.Let x1x2x3,…,xn be any sequence with xi=AC or xi=GT.As X does not contain CA,CG,TA and TG,the sequence x1x2x3,…,xn cannot have a double decomposition on a circle. The set X has no 5-necklace as, if l5 is C (resp., T) then l4 must be  A (resp., G), but A  (resp., G) is never a suffix in X.There are sets with six dinucleotides which are circular codes. For example, any sequence with the set of dinucleotides x1=AC, x2=AG, x3=AT,x1′=CG, x2′=CT,x3′=GT has no double decomposition on a circle.More generally, write the sequencel1,l2,l3,l4,l5,l6 of 𝒜4 on a circle. If x1=l1l2, x2=l3l4, x3=l5l6, x1′=l2l3, x2′=l4l5, x3′=l6l1 belong to a set Y, then Y cannot be a circular code because the sequence l1l2l3l4l5l6 can be read in two ways: x1x2x3 (with l1 as the first letter) and x1′x2′x3′ (with l2 as the first letter). There is a double reading of the sequence l1,l2,l3,l4,l5,l6 (corresponding to a double reading of the sequence l2,l3,l4,l5,l6,l1). In this case, l1,l2,l3,l4,l5 is a 5-necklace for Y.Example 27.
Ifx1=l1l2=AC,x2=l3l4=GT,x3=l5l6=AG,x1′=l2l3=CG,x2′=l4l5=TA,x3′=l5l6=GA are dinucleotides of Y,we have the following relations: ACGTAG=AC·GT·AG=x1x2x3,CGTAGA=CG·TA·GA=x1′x2′x3′=Cx2x3A and AC=x1.So, Y is not a dinucleotide circular code (also a consequence of the fact that Y contains two conjugate dinucleotides AG and GA).Proposition 28.
A dinucleotide circular code has at most 6 elements.Proof.
There are 16 dinucleotides. Four dinucleotides are periodic:AA, CC, GG, and TT. The remaining 12 dinucleotides are partitioned in six conjugation classes: {AC,CA}, {AG,GA}, {AT,TA}, {CG,GC}, {CT,TC}, and {GT,TG}. By Proposition 6, a dinucleotide circular code has at most one dinucleotide in each of these conjugation classes. So, a dinucleotide circular code has at most 6 elements.Proposition 29.
Let
(
i
,
j
,
h
,
k
)
be a permutation of
(
A
,
C
,
G
,
T
)
. If
(1)
X
=
{
i
j
,
i
h
,
i
k
,
j
h
,
j
k
,
h
k
}
,
then
X
is a dinucleotide circular code.Proof.
Suppose, by way of contradiction, thatX is not a dinucleotide circular code, and let l1,l2,l3,l4,l5 be a 5-necklace of X. Note that, with the exception of l1, the other letters l2,l3,l4,l5 composing the necklace must be a suffix of a dinucleotide of X.
Claim1. Forα∈{2,3,4,5}, lα≠i.
Proof of Claim1. By inspection, i is never a suffix of a dinucleotide of X.
Claim2. Forα∈{3,4,5}, lα≠j.
Proof of Claim2. By inspection, j is a suffix only of ij. For α∈{3,4,5}, if lα=j, then lα-1=i which is impossible by Claim 1.
Claim3. For eachα∈{4,5}, lα≠h.
Proof of Claim3. By inspection, h is a suffix only of ih and jh. Suppose, by way of contradiction, that l5=h. Then, l4=i or l4=j. If l4=i, we are in contradiction with Claim 1 and if l4=j, we are in contradiction with Claim 2. Suppose, by way of contradiction, that l4=h. Then, l3=i or l3=j. If l3=i, we are in contradiction with Claim 1 and if l3=j, we are in contradiction with Claim 2.
Claim4.  l5≠k.
Proof of Claim4. By inspection, k is a suffix only of ik, jk, and hk. Suppose, by way of contradiction, that l5=k. Then, l4=i,  l4=j, or l4=h. In the first case, we are in contradiction with Claim 1; in the second case, we are in contradiction with Claim 2; and in the third case, we are in contradiction with Claim 3.
By Claims1,2,3, and 4, we have l5≠i,l5≠j,l5≠h,l5≠k, and so, X has no 5-necklace. Consequently, X is a dinucleotide circular code.Definition 30.
A maximum dinucleotide circular code is a dinucleotide circular code having 6 elements.Remark 31.
In Proposition29, we have considered an arbitrary permutation (i,j,h,k) of (A,C,G,T), and we have proved that a maximum dinucleotide circular code corresponds to it. As the number of possible permutations is 24, the number of maximum dinucleotide circular codes is at least 24, and we will prove hereafter that it is exactly 24.In the maximum dinucleotide circular codeX={ij,ih,ik,jh,jk,hk} (Proposition 29), the letter i has three occurrences in prefix of dinucleotides of X (shortly in prefix of X), the letter j has two occurrences in prefix of X, and h has one occurrence in prefix of X. The letter k never occurs in prefix of X. This is a general fact, in the sense that in each maximum dinucleotide circular code X there is a letter, say i, with three occurrence in prefix of X, and a letter, say j, with two occurrences in prefix of X, and a letter, say h, with one occurrence in prefix of X, while the remaining letter, say k, never occurs in prefix of X.We will prove formally this general fact. In the sequel, a set of4 nonnegative numbers having a sum equal to 6 is called a 4-partition of 6. By “set” we rather mean a “multiset” as some numbers can be equal,forexample,{2,2,2,0}. Define lA (resp., lC, lG, lT) as the number of occurrences of A (resp., C, G, T) in prefix of a maximum dinucleotide circular code X.Lemma 32.
IfX is a maximum dinucleotide circular code, then {lA,lC,lG,lT} is a 4-partition of 6.Proof .
By Proposition28.Lemma 33.
In any dinucleotide circular code, one haslA≤3,lC≤3,lG≤3,lT≤3.Proof.
The alphabet𝒜4 contains four letters, and a dinucleotide circular code cannot contain periodic dinucleotides.Example 34.
ForX={ij,ih,ik,jh,jk,hk}, the 4-partition {li,lj,lh,lk}={lA,lC,lG,lT} of 6 is {3,2,1,0}.The following lemma will prove that the unique possible 4-partition{lA,lC,lG,lT} for a maximum dinucleotide circular code is {3,2,1,0}.Lemma 35.
For each maximum dinucleotide circular codeX, there exists a permutation π of {A,C,G,T} such that π(A) has three occurrences in prefix of X,π(C) has two occurrences in prefix of X,π(G) has one occurrence in prefix of X, and π(T) has no occurrence in prefix of X.Proof.
Putting the valueslA,lC,lG,lT in nonincreasing order, by Lemma 33, we have to consider only the following cases: {3,3,0,0}, {3,2,1,0}, {3,1,1,1}, {2,2,2,0}, and {2,2,1,1}.
Case
{
3,3
,
0,0
}
. Let i be the letter with three occurrences in prefix of X. Let j,h,k be the three other letters of {A,C,G,T}. We have ij,ih,ik∈X. Without loss of generality, suppose that j has three occurrences in prefix of X. Necessarily one of the two dinucleotides ji and jj must be in X. But, in the first case, we are in contradiction with Proposition 6, and in the second case, we are in contradiction with Proposition 5.
So, the case{3,3,0,0} is impossible.
Case
{
3,1
,
1,1
}. Let i be the letter with three occurrences in prefix of X. Let j,h,k be the three other letters of {A,C,G,T}. Then, we have ij,ih,ik∈X and ii,ji,hi,ki∉X; otherwise, we are in contradiction with Propositions 5 and 6.
Now, suppose that inX-{ij,ih,ik}, the same letter, say j without loss of generality, has two occurrences in suffix of X,thatis,hj,kj∈X. The letter j cannot be a prefix of X. Indeed, ji∉X, jh,jk∈X are in contradiction with Proposition 6, and jj∈X is in contradiction with Proposition 5.
So, inX-{ij,ih,ik}, the letters j,h,k must have only one occurrence in suffix of X. Without loss of generality, we have jh,hk,kj∈X. But, jhkjh is a 5-necklace for X. By Proposition 25, we are in contradiction.
So, the case{3,1,1,1} is impossible.
Case
{
2,2
,
2,0
}. Let i be one of the three letters with two occurrences in prefix of X. Let j,h,k the three other letters of {A,C,G,T}. Without loss of generality, we have ij,ih∈X. With the two other letters having two occurrences in prefix of X, we have three possibilities {j,h}, {j,k}, {h,k}.
Case
{
j
,
h
}. By Propositions 6 and 5, ji,jj,hi,hh∉X, but ij,ih,jh,jk,hj,hk∈X. As jh,hj are conjugate, we are in contradiction with Proposition 6.
Case
{
j
,
k
}. By Propositions 6 and 5, ji,jj∉X, but jh,jk∈X. As kj∉X (otherwise, we are in contradiction with Proposition 6), the two dinucleotides with k in prefix of X must be ki and kh and, consequently, ij,ih,jh,jk,ki,kh∈X. But, kijki is a 5-necklace for X. By Proposition 25, we are in contradiction.
Case
{
h
,
k
}. By Propositions 6 and 5, hi,hh∉X, but hj,hk∈X. As kh∉X (otherwise, we are in contradiction with Proposition 6), the two dinucleotides with k in prefix of X must be ki and kj and, consequently, ij,ih,hj,hk,ki,kj∈X. But, ihkih is a 5-necklace for X. By Proposition 25, we are in contradiction.
So, the case{2,2,2,0} is impossible.
Case
{
2,2
,
1,1
}
.Let i be one of the two letters with two occurrences in prefix of X. Let j,h,k be the three other letters of {A,C,G,T}. Without loss of generality, we have ij,ih∈X. Consider the following cases:(i)
j has two occurrences in prefix of X, and h and k have one occurrence in prefix of X. By Propositions 6 and 5, ij,ih,jh,jk∈X, and hk is the unique possible dinucleotide of X with h in prefix of X. By Propositions 6 and 5, kj,kh,kk∉X. If ki∈X then ijkij is a 5-necklace of X, and by Proposition 25, we are in contradiction. So, k cannot be a prefix of X. Contradiction.(ii)
h has two occurrences in prefix of X, and j and k have one occurrence in prefix of X. By Propositions 6 and 5, ij,ih,hj,hk∈X, and jk is the unique possible dinucleotide of X with j in prefix of X. By Propositions 6 and 5, kj,kh,kk∉X and ki∉X (otherwise, ijkij is a 5-necklace for X, and by Proposition 25, we are in contradiction). So, k cannot be a prefix in X. Contradiction.(iii)
k has two occurrences in prefix of X, and j and h have one occurrence in prefix of X. By Propositions 6 and 5, we have three possible cases ki,kj∈X, ki,kh∈X, and kj,kh∈X.(iiia)
ki,kj∈X. So, ij,ih,ki,kj∈X. By Propositions 6 and 5, ji,jj,jk∉X, but jh∈X. By Propositions 6 and 5, hj,hk∈X. In the first case X={ij,ih,ki,kj,jh,hj}, and by Proposition 6, we are in contradiction. In the second case, X={ij,ih,ki,kj,jh,hk}. But, ihkih is a 5-necklace for X. By Proposition 25, we are in contradiction.(iiib)
ki,kh∈X. So, ij,ih,ki,kh∈X. By Propositions 6 and 5, hi,hh,hk∉X, but hj∈X. By Propositions 6 and 5, jk is the unique possible dinucleotide of X with prefix j. So, X={ij,ih,ki,kh,hj,jk}. But, ihjki is a 5-necklace for X. By Proposition 25, we are in contradiction.(iiic)
kj,kh∈X. So, ij,ih,kj,kh∈X. By Propositions 6 and 5, ji,jj,jk∉X, but jh∈X. By Propositions 6 and 5, hi,hh,hk∉X, but hj∈X. As hj,jh∈X are conjugate, we are in contradiction with Proposition 6.So, the case{2,2,1,1} is also impossible.
Only the 4-partition{3,2,1,0} is realized by {ij,ih,ik,jh,jk,hk}. It corresponds to the permutation (i,j,h,k) of (A,C,G,T). In other words, the permutation, whose existence is proved, is π(A)=i, π(C)=j, π(G)=k, and π(T)=k.Proposition 36.
There are 24 maximum dinucleotide circular codes.Proof.
By Proposition29, each permutation (i,j,h,k) of (A,C,G,T) is associated with a maximum dinucleotide circular code {ij,ih,ik,jh,jk,hk}. As there are 24 permutations, the number of maximum dinucleotide circular codes is at least 24.
Now, letX be a maximum dinucleotide circular code. By Lemma 35, its 4-partition must be {3,2,1,0}. Let i (resp., j, h, k) be the letter of 𝒜4 having 3 (resp., 2, 1, 0) occurrences in prefix of X. As i has three occurrences in prefix of X, we must have ij,ih,ik∈X. As j has two occurrences in prefix of X, and as ij is already in X, we must also have jh,jk∈X. Finally, as h has only one occurrence in prefix of X, and as ih and jh are already in X, we must have hk∈X. Consequently, X={ij,ih,ik,jh,jk,hk}, and X is one of the 24 maximum circular codes already considered. Thus, the number of maximum dinucleotide circular codes is exactly 24.A computer calculus confirms that there are exactly24 maximum dinucleotide circular codes (Table 1).Table 1
The24 maximum dinucleotide circular codes and their properties.
Symbol
Dinucleotide circular code
𝒞
𝒫
𝒫
𝒞
X
1
{
A
C
,
A
G
,
A
T
,
C
G
,
C
T
,
G
T
}
𝒞
(
X
1
)
=
X
1
𝒫
(
X
1
)
=
X
24
𝒫
(
𝒞
(
X
1
)
)
=
X
24
X
2
{
A
C
,
A
G
,
A
T
,
C
G
,
C
T
,
T
G
}
𝒞
(
X
2
)
=
X
13
𝒫
(
X
2
)
=
X
23
𝒫
(
𝒞
(
X
2
)
)
=
X
12
X
3
{
A
C
,
A
G
,
A
T
,
C
G
,
T
C
,
T
G
}
𝒞
(
X
3
)
=
X
17
𝒫
(
X
3
)
=
X
22
𝒫
(
𝒞
(
X
3
)
)
=
X
8
X
4
{
A
C
,
A
G
,
A
T
,
C
T
,
G
C
,
G
T
}
𝒞
(
X
4
)
=
X
4
𝒫
(
X
4
)
=
X
21
𝒫
(
𝒞
(
X
4
)
)
=
X
21
X
5
{
A
C
,
A
G
,
A
T
,
G
C
,
G
T
,
T
C
}
𝒞
(
X
5
)
=
X
9
𝒫
(
X
5
)
=
X
20
𝒫
(
𝒞
(
X
5
)
)
=
X
16
X
6
{
A
C
,
A
G
,
A
T
,
G
C
,
T
C
,
T
G
}
𝒞
(
X
6
)
=
X
18
𝒫
(
X
6
)
=
X
19
𝒫
(
𝒞
(
X
6
)
)
=
X
7
X
7
{
A
C
,
A
G
,
C
G
,
T
A
,
T
C
,
T
G
}
𝒞
(
X
7
)
=
X
19
𝒫
(
X
7
)
=
X
18
𝒫
(
𝒞
(
X
7
)
)
=
X
6
X
8
{
A
C
,
A
G
,
G
C
,
T
A
,
T
C
,
T
G
}
𝒞
(
X
8
)
=
X
22
𝒫
(
X
8
)
=
X
17
𝒫
(
𝒞
(
X
8
)
)
=
X
3
X
9
{
A
C
,
A
T
,
C
T
,
G
A
,
G
C
,
G
T
}
𝒞
(
X
9
)
=
X
5
𝒫
(
X
9
)
=
X
16
𝒫
(
𝒞
(
X
9
)
)
=
X
20
X
10
{
A
C
,
A
T
,
G
A
,
G
C
,
G
T
,
T
C
}
𝒞
(
X
10
)
=
X
10
𝒫
(
X
10
)
=
X
15
𝒫
(
𝒞
(
X
10
)
)
=
X
15
X
11
{
A
C
,
G
A
,
G
C
,
G
T
,
T
A
,
T
C
}
𝒞
(
X
11
)
=
X
11
𝒫
(
X
11
)
=
X
14
𝒫
(
𝒞
(
X
11
)
)
=
X
14
X
12
{
A
C
,
G
A
,
G
C
,
T
A
,
T
C
,
T
G
}
𝒞
(
X
12
)
=
X
23
𝒫
(
X
12
)
=
X
13
𝒫
(
𝒞
(
X
12
)
)
=
X
2
X
13
{
A
G
,
A
T
,
C
A
,
C
G
,
C
T
,
G
T
}
𝒞
(
X
13
)
=
X
2
𝒫
(
X
13
)
=
X
12
𝒫
(
𝒞
(
X
13
)
)
=
X
23
X
14
{
A
G
,
A
T
,
C
A
,
C
G
,
C
T
,
T
G
}
𝒞
(
X
14
)
=
X
14
𝒫
(
X
14
)
=
X
11
𝒫
(
𝒞
(
X
14
)
)
=
X
11
X
15
{
A
G
,
C
A
,
C
G
,
C
T
,
T
A
,
T
G
}
𝒞
(
X
15
)
=
X
15
𝒫
(
X
15
)
=
X
10
𝒫
(
𝒞
(
X
15
)
)
=
X
10
X
16
{
A
G
,
C
A
,
C
G
,
T
A
,
T
C
,
T
G
}
𝒞
(
X
16
)
=
X
20
𝒫
(
X
16
)
=
X
9
𝒫
(
𝒞
(
X
16
)
)
=
X
5
X
17
{
A
T
,
C
A
,
C
G
,
C
T
,
G
A
,
G
T
}
𝒞
(
X
17
)
=
X
3
𝒫
(
X
17
)
=
X
8
𝒫
(
𝒞
(
X
17
)
)
=
X
22
X
18
{
A
T
,
C
A
,
C
T
,
G
A
,
G
C
,
G
T
}
𝒞
(
X
18
)
=
X
6
𝒫
(
X
18
)
=
X
7
𝒫
(
𝒞
(
X
18
)
)
=
X
19
X
19
{
C
A
,
C
G
,
C
T
,
G
A
,
G
T
,
T
A
}
𝒞
(
X
19
)
=
X
7
𝒫
(
X
19
)
=
X
6
𝒫
(
𝒞
(
X
19
)
)
=
X
18
X
20
{
C
A
,
C
G
,
C
T
,
G
A
,
T
A
,
T
G
}
𝒞
(
X
20
)
=
X
16
𝒫
(
X
20
)
=
X
5
𝒫
(
𝒞
(
X
20
)
)
=
X
9
X
21
{
C
A
,
C
G
,
G
A
,
T
A
,
T
C
,
T
G
}
𝒞
(
X
21
)
=
X
21
𝒫
(
X
21
)
=
X
4
𝒫
(
𝒞
(
X
21
)
)
=
X
4
X
22
{
C
A
,
C
T
,
G
A
,
G
C
,
G
T
,
T
A
}
𝒞
(
X
22
)
=
X
8
𝒫
(
X
22
)
=
X
3
𝒫
(
𝒞
(
X
22
)
)
=
X
17
X
23
{
C
A
,
G
A
,
G
C
,
G
T
,
T
A
,
T
C
}
𝒞
(
X
23
)
=
X
12
𝒫
(
X
23
)
=
X
2
𝒫
(
𝒞
(
X
23
)
)
=
X
13
X
24
{
C
A
,
G
A
,
G
C
,
T
A
,
T
C
,
T
G
}
𝒞
(
X
24
)
=
X
24
𝒫
(
X
24
)
=
X
1
𝒫
(
𝒞
(
X
24
)
)
=
X
1There are eight self-complementary maximum dinucleotide circular codes:X1, X4, X10, X11, X14, X15, X21, and X24 (Table 1). The 16 remaining ones are partitioned in two classes of eight maximum dinucleotide circular codes in bijective correspondence by the complementarity map (Table 1).Proposition 37.
IfX is a maximum dinucleotide circular code, then 𝒞(X) is also a maximum dinucleotide circular code.Proof .
By inspection (Table1).Proposition 38.
IfX is a maximum dinucleotide circular code, then
(2)𝒫(𝒞(X))=𝒞(𝒫(X)).Proof.
By inspection (Table1).This proposition is not true with maximum trinucleotide circular codes, see for example, [6].


## 4. Conclusion

This new combinatorial study of circular codes in genes has proved that there are exactly24 maximum dinucleotide circular codes on the 4-letter genetic alphabet {A,C,G,T}. They are listed in Table 1. Propositions 22, 37, and 38 lead to interesting properties with dinucleotide circular codes in DNA. Indeed, they ensure that several maximum dinucleotide circular codes can exist in the two strands of the DNA double helix simultaneously. Indeed, a maximum dinucleotide circular code X in a given strand s of DNA implies that its complementary set 𝒞(X) in the complementary strand 𝒞(s) of DNA is also a maximum dinucleotide circular code (Proposition 37) and according to two possibilities: 𝒞(X)=X or 𝒞(X)=Y with Y≠X (Table 1). Furthermore, its permuted set 𝒫(X) in s, obtained by a frameshift of one letter of X in s, is also a maximum dinucleotide circular code (Proposition 22). Finally, its complementary permuted set 𝒞(𝒫(X)) in 𝒞(s) is also a maximum dinucleotide circular code (Proposition 38).Chemical modification of nucleotides is ubiquitous in RNA and DNA. So far, a total of 107 modified nucleotides, for which chemical structures have been assigned, have been reported in RNA (see the RNA Modification Database athttp://rna-mdb.cas.albany.edu/RNAmods/ [35]). The largest number, that is, 81, with the greatest structural diversity, is found in tRNA, with 30 in rRNA, 12 in mRNA, and 13 in other RNA species, most notably snRNA. The four nucleotides can be chemically modified, for example, methyladenosine, dimethyladenosine, trimethyladenosine, methylcytidine, dimethylcytidine, thiocytidine, methylguanosine, dimethylguanosine, trimethylguanosine, methyluridine, dimethyluridine, thiouridine, pseudouridine, dihydrouridine, but also inosine, lysidine, wybutosine, wyosine, queuosine, and archaeosine. In DNA, the cytosine in the CG dinucleotide, involved in gene regulation, can have two chemical forms (methylcytosine, hydroxymethylcytosine). This chemical change allows to store additional information, thus expanding the alphabet {A,C,G,T} by two letters. Thus, the generalization of dinucleotide circular code propositions over larger alphabets is very interesting and should be investigated.Dinucleotide circular codes may be involved in retrieval of the modulo2 frame in genomes, for example, in the dinucleotide repeats.Dinucleotide circular codes may also have a biological function in the coding process of amino acids. In the standard genetic code, eight amino acidsAla (A), Arg (R), Gly (G), Leu (L), Pro (P), Thr (T), Ser (S), and Val (V) are coded by sets of trinucleotides involving dinucleotides. Indeed, for each of these eight amino acids, there exists a dinucleotide d∈𝒜42 such that all the trinucleotides of the form dl (where l is any letter of 𝒜4) code the same amino acid (Table 2).Table 2
Amino acids coded by the trinucleotidesdl in the the standard genetic code where  d∈𝒜42 and  l  being any letter of  𝒜4.
Amino acid
Trinucleotidesdl
Ala (A)
G
C
l
Arg (R)
C
G
l
Gly (G)
G
G
l
Leu (L)
C
T
l
Pro (P)
C
C
l
Thr (T)
A
C
l
Ser (S)
T
C
l
Val (V)
G
T
lNow,Gly (G) and Pro (P) cannot be coded by a dinucleotide circular code as their dinucleotides are periodic {CC,GG}. Moreover, Ala (A) and Arg (R) cannot be coded simultaneously by a dinucleotide circular code as their dinucleotides are conjugate {CG,GC} and similarly for Leu (L) and Ser (S) with the conjugate dinucleotides {CT,TC}. On the other hand, as any subset of a maximum dinucleotide circular code is also a dinucleotide circular code, the following properties exist.(i)
The four amino acidsArg(R), Leu(L), Thr(T), and Val(V) can be coded by the dinucleotide circular code {AC,CG,CT,GT} which is a proper subset of the maximum dinucleotide circular code X1 (Table 1).(ii)
The four amino acidsAla(A), Leu(L), Thr(T), and Val(V) can be coded by the dinucleotide circular code {AC,CT,GC,GT} which is a proper subset of the maximum dinucleotide circular code X4 (Table 1).(iii)
The four amino acidsAla(A), Ser(S), Thr(T), and Val(V) can be coded by the dinucleotide circular code {AC,GC,GT,TC} which is a proper subset of the maximum dinucleotide circular code X5 (Table 1).These results contribute to the research field analysing the mathematical properties of genetic codes.



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*Source: 538631-2013-02-12.xml*