In this problem, we think of a DNA molecule as a long string formed from an alph

Question

In this problem, we think of a DNA molecule as a long string formed from an alphabet of four characters. In certain methods for finding the sequence of a piece of DNA, one first finds the position in the DNA of many smaller, overlapping line segments (as in the figure below). Then one wants to select a subset of those segments so that every part of the original DNA string is covered by at least one of the selected segments, and so that the total length of the selected segments is as small as possible. For example in the figure below one could select segments denoted 1,2,3,4,5,6 to cover all the original DNA - but this is not the selection that minimizes the total length of the selected segments. We want an algorithm to solve this problem. One way to do this is to cast the problem as a shortest path problem in a directed graph derived from the figure. Show how to do this and explain why this approach is correct. Hint: each line segment in the input becomes a node in the graph.

1 3 5

--------------- --------------- ------- --------------

2

------- -------------- ------------ ---------- ---------------

------------- ------------- --------------- ------- ------ ----

4 6

------------------- ------------------ ----- ---

DNA

------------------------------------------------------------------

Explanation / Answer

DNA stores organic data. The DNA spine is impervious to cleavage, and both strands of the twofold stranded structure store the same natural data. Organic data is duplicated as the two strands are isolated. A noteworthy bit of DNA (more than 98% for people) is non-coding, implying that these segments don't serve as examples for protein arrangements.

The two strands of DNA keep running in inverse headings to one another and are accordingly against parallel. Connected to every sugar is one of four sorts of nucleobases (casually, bases). It is the arrangement of these four nucleobases along the spine that encodes natural data. Under the hereditary code, RNA strands are meant determine the grouping of amino acids inside of proteins. These RNA strands are at first made utilizing DNA strands as a layout in a procedure called translation.

Inside of cells, DNA is sorted out into long structures called chromosomes. Amid cell division these chromosomes are copied during the time spent DNA replication, giving every phone its own complete arrangement of chromosomes. Eukaryotic living beings (creatures, plants, parasites, and protists) store the greater part of their DNA inside the cell core and some of their DNA in organelles, for example, mitochondria or chloroplasts.[4] conversely, prokaryotes (microbes and archaea) store their DNA just in the cytoplasm. Inside of the chromosomes, chromatin proteins, for example, histones conservative and arrange DNA. These minimized structures direct the cooperations in the middle of DNA and different proteins, controlling which parts of the DNA are deciphered.

The DNA twofold helix is balanced out essentially by two strengths: hydrogen bonds in the middle of nucleotides and base-stacking communications among fragrant nucleobases.[17] In the watery environment of the cell, the conjugated obligations of nucleotide bases adjust opposite to the hub of the DNA atom, minimizing their association with the solvation shell and in this manner, the Gibbs free vitality. The four bases found in DNA are adenine (truncated A), cytosine (C), guanine (G) and thymine (T). These four bases are connected to the sugar/phosphate to shape the complete nucleotide, as appeared for adenosine monophosphate. Adenine sets with thymine and guanine sets with cytosine. It was spoken to by A-T base sets and G-C base sets.

We instate the arrangement network same as the information diagram matrix.Then we upgrade the arrangement lattice by considering all vertices as a middle of the road vertex. The thought is to pick the vertices one by one and redesign all most brief ways which incorporate the picked vertex as a middle vertex in the most limited way. When we pick vertex number k as a middle of the road vertex, we as of now have considered vertices {0, 1, 2, .. k-1} as transitional vertices. For each pair (i, j) of

source and destination vertices separately, there are two conceivable cases.

1) k is not a transitional vertex in most brief way from i to j. We

keep the estimation of dist[i][j] as it seems to be.

2) k is a transitional vertex in most brief way from i to j. We redesign

the estimation of dist[i][j] as dist[i][k] + dist[k][j].Given a weighted digraph with a weight capacity w : E R, R is the arrangement of genuine numbers that decide the length of the most brief way between all pair of vertices in G. Given an information, n*n grid, "W" speak to the edge weights of n vertices; i.e., W= (wij), wher

wij=0 if(i=j)

=w(i,j)if (!=j)and(i,j)belongs to e

infinite if(i!=j)and (i,j)!=e

Read n; //number of vertex
For(int i=1 ;i<=n;i++)
For(int j=1 ;j<=n;i++)
{
D[i][j]=”0”; // initialize the weighted matrix
P[i][j]=0 ; // P matrix is used to store the intermediate
node value
}
For(int i=1 ;i<=n;i++)
{
For(int j=1 ;j<=n;i++)
{
If(i != j) //check source and distination not
same
{
Read D[i][j]; //read the weight of
graph in the form of //genome DNA sequence
}
}
}
//Convert Genome DNA sequence into decimal number
For(int i=1 ;i<=n;i++)
For(int j=1 ;j<=n;i++)
If(i != j)
{
I[i][j] =BTD (D[i][j]) ;
}
// Calculation
For(int k=1 ;k<=n; k++)
For(int i=1 ;i<=n;i++)
For(int j=1 ;j<=n;j++)
{
//This condition used to check source
,destination and //intermediate node should not
be same
If((i != j) && (j !=k) && (j !=k))
{
If(I [i][j] >( I[i][k] + I [k][j]))
{
I [i][j] =( I[i][k] + I [k][j]);
P[i][k] = k; // P matrix used to store the intermediate
node
}
}
}
}
}
}

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