X-Git-Url: https://bilbo.iut-bm.univ-fcomte.fr/and/gitweb/chloroplast13.git/blobdiff_plain/fb06fc26916e64c8e6c452ee7424b6433e255f0f..f8689d1f199221556dc52e80c5237c45addb0218:/classEquiv.tex?ds=inline diff --git a/classEquiv.tex b/classEquiv.tex index b77c51a..41abda4 100644 --- a/classEquiv.tex +++ b/classEquiv.tex @@ -1,55 +1,28 @@ -This step considers as input the set -$\{((g_1,g_2),r_{12}), (g_1,g_3),r_{13}), (g_{n-1},g{n}),r_{n-1.n})\}$ of -$\frac{n(n-1)}{2}$ elements. -Each one $(g_i,g_j),r_{ij})$ where $i < j$, -is a pair that gives the similarity rate $r_{ij}$ between the two genes -$g_{i}$ and $g_{j}$. - -The first step of this stage consists in building the following non-oriented -graph further denoted as to \emph{similarity graph}. -In this one, the vertices are the genes. There is an edge between -$g_{i}$ and $g_{j}$ if the rate $r_{ij}$ is greater than a given similarity -threshold $t$. - -We then define the relation $\sim$ such that -$ x \sim y$ if $x$ and $y$ belong in the same connected component. -Mathematically speaking, it is obvious that this -defines an equivalence relation. -Let $\dot{x}= \{y | x \sim y\}$ -denotes the equivalence class to which $x$ belongs. -All the genes which are equivalent to each other -are also elements of the same equivalence class. -Let us then consider the set of all equivalence classes of the set of genes -by $\sim$, denoted $X/\sim = \{\dot{x} | x \textrm{ is a gene}\}$. -defined by $\pi(x) = \dot{x}$ -which maps each gene into it respective equivalence class by $\sim$. - - - - -For each genome $[g_l,\ldots,g{l+m}]$, the second step computes -the projection of each gene according to $\pi$. -The resulting genome which is -$$ -[\pi(g_l),\ldots,\pi(g{l+m})] -$$ -is again of size $m$. - -Intuitively speaking, for two genes $g_i$ and $g_j$ -in the same equivalence class, there is path from $g_i$ and $g_j$. -It signifies that each evolution step -(represented by an edge in the similarity graph) -has produced a gene s.t. the similarity with the previous one -is greater than $t$. -Genes $g_i$ and $g_j$ may thus have a common ancestor. - - -We compute the core genome as follow. -Each genome is projected according to $\pi$. We then consider the -intersection of all the projected genomes which are considered as sets of genes -and not as sequences of genes. -This results as the set of all the class $\dot{x}$ -such that each genome has an gene $x$ in $\dot{x}$. -The pan genome is computed similarly: the union of all the -projected genomes in computed here. - +Identifying core genes is important to understand evolutionary and +functional phylogenies. Therefore, in this work we present two methods +to build a genes content evolutionary tree. More precisely, we focus +on the following questions considering a collection of +99~chloroplasts annotated from NCBI \cite{Sayers01012011} and Dogma +\cite{RDogma} : how can we identify the best core genome and what +is the evolutionary scenario of these chloroplasts. +Two methods are considered here. The first one is based on NCBI annotation, it is explained below. +We start by the following definition. +\begin{definition} +\label{def1} +Let $A=\{A,T,C,G\}$ be the nucleotides alphabet, and $A^\ast$ be the set of finite words on $A$ (\emph{i.e.}, of DNA sequences). Let $d:A^{\ast}\times A^{\ast}\rightarrow[0,1]$ be a distance on $A^{\ast}$. Consider a given value $T\in[0,1]$ called a threshold. For all $x,y\in A^{\ast}$, we will say that $x\sim_{d,T}y$ if $d(x,y)\leqslant T$. +\end{definition} + +\noindent$\sim_{d,T}$ is obviously an equivalence relation. When $d=1-\Delta$, where $\Delta$ is the similarity scoring function embedded into the emboss package (Needleman-Wunch released by EMBL), we will simply denote $\sim_{d,0.1}$ by $\sim$. The method starts by building an undirected graph based on +the similarity rates $r_{ij}$ between sequences $g_{i}$ and $g_{j}$ (\emph{i.e.}, $r_{ij}=\Delta(g_{i},g_{j})$). +In this latter, nodes are constituted by all the coding sequences of the set of genomes under consideration, and there is an edge between $g_{i}$ and $g_{j}$ if the +similarity rate $r_{ij}$ is +greater than the given similarity threshold. The Connected Components +(CC) of the ``similarity'' graph are thus computed. +This produces an equivalence +relation between sequences in the same CC based on Definition~\ref{def1}. +Any class for this relation is called ``gene'' here, where its representatives (DNA sequences) are the ``alleles'' of this gene. Thus this first method produces for each genome $G$, which is a set $\{g_{1}^G,...,g_{m_G}^G\}$ of $m_{G}$ DNA coding sequences, the projection of each sequence according to $\pi$, where $\pi$ maps each sequence +into its gene (class) according to $\sim$. In other words, $G$ is mapped into $\{\pi(g_{1}^G),...,\pi(g_{m_G}^G)\}$. +Remark that a projected genome has no duplicated gene, as it is a set. The core genome (resp. the pan genome) of $G_{1}$ and $G_{2}$ is defined thus as the intersection (resp. as the union) of these projected genomes.\\ +We then consider the intersection of all the projected genomes, which is the set of all the genes $\dot{x}$ +such that each genome has at least one allele in $\dot{x}$. The pan genome is computed similarly as the union of all the projected genomes. However such approach suffers from producing too small core genomes, +for any chosen similarity threshold, compared to what is usually waited by biologists regarding these chloroplasts. We are then left with the following questions: how can we improve the confidence put in the produced core? Can we thus guess the evolution scenario of these genomes? \ No newline at end of file