DNA Sequencing Project

1. Start with this code

public class DNA {

  public static void main(String[] args) {

    //  -. .-.   .-. .-.   .
    //    \   \ /   \   \ / 
    //   / \   \   / \   \  
    //  ~   `-~ `-`   `-~ `-

  }

}

2. Write a comment near the top of the program that describe what the program does.

3. Here are the three DNA strands that you are going to use to test your program:

   • "ATGCGATACGCTTGA"
   • "ATGCGATACGTGA"
   • "ATTAATATGTACTGA"

   Store them in different strings: dna1, dna2, and dna3.

4. Create a generic String variable called dna that can be set to any DNA sequence (dna1, dna2, dna3).

5. To warm up, find the length of the dna string.

6. Remember that a protein has the following qualities:

   1. It begins with a start codon ATG.
   2. It ends with a stop codon TGA.
   3. In between, the number of nucleotides is divisible by 3.

   First, let’s start with the first condition. Does the DNA strand have the start codon ATG within it?

   Find the index where ATG begins using indexOf().

7. Next, does the DNA strand have the stop codon TGA?

   Find the index where TGA begins.

8. Lastly, you’ll find out whether or not there is a protein!

   Let’s start with an if statement that checks for a start codon and a stop codon using the && operator.

   Remember that the indexOf() string method will return -1 if the substring doesn’t exist within a String.

9. Add a third condition that checks whether or not that the number of nucleotides in between the start codon and the stop condon is a multiple of 3.

   Remember that the modolo operator % returns the remainder of a division.

10. Inside the if statement, create a String variable named protein.

    And find this protein in the dna by using the substring() string method. Think about where you want the substring to begin and where you want the substring to end.

    Remember that a codon is 3 nucleotides long.

11. Add an else clause that print out No protein..

12. You are all done!

    Let’s test your code with each DNA strand. These should be the results:

    • dna1: Contains a protein.
    • dna2: Does not contain a protein.
    • dna3: Contains a protein.