Problem 3. The deoxyribonucleic acid (DNA) is a molecule that contains the genet

Question

Problem 3. The deoxyribonucleic acid (DNA) is a molecule that contains the genetic instructions required for the development and functioning of all known living organisms. The basic double-helix structure of the DNA was co-discovered by Prof. Franci ick, a long-time faculty member at UCSD. The DNA molecule consists of a long sequence of four nucleotide bases: adenine (A), cytosine (C), gua- nine (G) and thymine (T). Since this molecule contains all the genetic information of a living organism, geneticists are interested in understanding the roles of the variuos DNA sequence patterns that are con- tinuously being discovered worldwide. One of the most common methods to identify the role of a DNA sequence is to compare it with other DNA sequences, whose functionality is already known. The more similar such DNA sequences are, the more likely it is that they will function similarly Your taslk is to write a C program, called dna.c, that reads three DNA sequences from a file called dna-input.dat and prints the results of a comparison between each pair of sequences to the file dna output.dat. The input file dna input.dat consists of three lines. Each line is a single se- quence of characters from the set (A, C. G. T). that appear without spaces in some order, terminated by the end of line character In. You can assume that the three lines contain the san MacBook Air

Explanation / Answer

SOurceCode:

#include <stdio.h>

/*=============================================================================
Constants and definitions
=============================================================================*/

#define MAX_IN_LENGTH 241
#define OUT_LENGTH 60

/*=============================================================================
Global variables
=============================================================================*/

FILE *input;
FILE *output;
char DNAseq1[MAX_IN_LENGTH], DNAseq2[MAX_IN_LENGTH], DNAseq3[MAX_IN_LENGTH];
char DNAcomp12[MAX_IN_LENGTH],DNAcomp13[MAX_IN_LENGTH],DNAcomp23[MAX_IN_LENGTH];

/*=============================================================================
Forward function declarations
=============================================================================*/

int read_DNA(char sequence[]);
double compare_DNA(char seq1[], char seq2[], char seq3[], int n);
void print_DNA(char seq1[], char seq2[], char seq3[], int n);

/*@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

The main function:
The program compares pairs of dna sequences in the file dna_input.dat.
(There are 3 dna sequence in the dna_input file.)
In each comparison, the program prints the two DNA sequences with their
comparison in between. In the comparison line, there is a space if one of DNA
characters does not matche the DNA character in the ther sequence. Otherwise,
it prints the matching letter if the letters match in the DNA sequence. The
program then prints the overlap percentage of the pair of DNA sequences.

@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@*/

int main()
{
   int char_count1, char_count2, char_count3;
   double overlap_percent1, overlap_percent2, overlap_percent3;

   input = fopen("dna_input.dat","r");
output = fopen("dna_output.dat", "w");

/* Reading DNA sequences */
//each of char counts should have the same value.
char_count1 = read_DNA(DNAseq1);
char_count2 = read_DNA(DNAseq2);
char_count3 = read_DNA(DNAseq3);

/* Comparing DNA sequences and computing their overlap percentage */
overlap_percent1 = compare_DNA(DNAseq1, DNAseq2, DNAcomp12, char_count1);
overlap_percent2 = compare_DNA(DNAseq1, DNAseq3, DNAcomp13, char_count2);
overlap_percent3 = compare_DNA(DNAseq2, DNAseq3, DNAcomp23, char_count3);

/* Prints comparison between sequence 1 and sequence 2 */
fprintf(output, "Comparison between sequence #1 and sequence #2: ");
print_DNA(DNAseq1, DNAseq2, DNAcomp12, char_count1);
fprintf(output, "The overlap percentage is %.1lf%% ", overlap_percent1);

/* Prints comparison between sequence 1 and sequence 3 */
fprintf(output, "Comparison between sequence #1 and sequence #3: ");
print_DNA(DNAseq1, DNAseq3, DNAcomp13, char_count2);
fprintf(output, "The overlap percentage is %.1lf%% ", overlap_percent2);

/* Prints comparison between sequence 2 and sequence 3 */
fprintf(output, "Comparison between sequence #2 and sequence #3: ");
print_DNA(DNAseq2, DNAseq3, DNAcomp23, char_count3);
fprintf(output, "The overlap percentage is %.1lf%% ", overlap_percent3);

fclose(input); fclose(output);

return 0;
}

/*=============================================================================
Function definitions
=============================================================================*/

/* Reads a DNA sequence from the file and puts it into an array.
* It then returns the number of letters. */
int read_DNA(char sequence[])
{
   int letter_count = 0, i = 0;

   //Puts characters from input into array and counts number of letters
   do{
      fscanf(input,"%c",&sequence[i]);
      letter_count++;
      i++;
   } while (sequence[i-1] != ' ');

   return letter_count-1; //Disregards the newline character
} /* END OF read_DNA */

/* Compares a pair of DNA sequences and stores their comparison into
* another array. The function also returns the percentage of overlap. */
double compare_DNA(char seq1[], char seq2[], char seq3[], int n)
{
   int i, matches = 0;

   //Compares each character in the sequences and counts the number of matches
   for(i = 0; i < n; i++)
   {
      if (seq1[i] == seq2[i])
      {
         seq3[i] = seq1[i]; //Puts the matching letter into the comparison array
         matches++;
      }
      else seq3[i] = ' '; //Puts blank space into comparison array
   }

   //Subtracting 1 disregards the newline character in calculations
   return (double) (matches)/(n)*100;
} /* END OF compare_DNA */

/* Prints the DNA sequences with their comparison with a maximum of
* OUT_LENGTH characters per line. */
void print_DNA(char seq1[], char seq2[], char seq3[], int n)
{
   int i,j;

   //Prints the sequences with a maximum of OUT_LENGTH
   //characters per line.
   for(i = 0; i < n; i+= OUT_LENGTH)
   {
      //Prints first DNA sequence
      for(j = 0; j < OUT_LENGTH && j+i < n; j++)
      {
         fprintf(output, "%c", seq1[i+j]);
      }
      fprintf(output, " ");
    
      //Prints comparison between the two sequences
      for(j = 0; j < OUT_LENGTH && j+i < n; j++)
      {
         fprintf(output, "%c", seq3[i+j]);
      }
      fprintf(output, " ");

      //Prints second DNA sequence
      for(j = 0; j < OUT_LENGTH && j+i < n; j++)
      {
         fprintf(output, "%c", seq2[i+j]);
      }
      fprintf(output, " ");
   }
} /* END OF print_DNA */

dna_input.dat

ACGTTTTAAGGGCTGAGCTAGTCAGTTCATCGCGCGCGTATATCCTCGATCGATCATTCTCTCTAGACGTTTTAAGGGCTGAGCTAGTCAGTTC
ACGTTTTAAGGGCTTAGAGCTTATGCTAATCGCGCGCGTATATCCTCGATCGATCATTCTCTCTAGACGTTTTAAGGGCTAAGGCGCGTAATTA
TCGTTTGAAGGGCTTAGTTAGTTAGTTCATCGGCGGCGTATATCCTCGATCGATCATTCTCTCTAGACGTTTTAAGGGCTGAGCCGGTCAGTTA

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