This is R programming.

(1) Given a phonelist with one number per entry such as phonelist =2" and returns a vector with all DNA subsequences (as strings) that start with triplet ("codon") "AAA" or "GAA" and end with triplet ("codon") "AGT" and have at least 2 and at most "n" other triplets ("codons") between the start and the end. Eg., for "GAACCCACTAGTATAAAATTTGGGAGTCCCAAACCCTTTGGGAGT" and for n=2, the answer is c= (GAACCCACTAGT", "AAATTTGGGAGT"). For e.g, n=10 the answer is: c("GAACCCACTAGTATAAAATTTGGGAGT", "AAACCCTTTGGGAGT") (ii) Modify the function in (i) so that it returns a matrix containing the starting position of each match along with the actual number of codons between the start and the end for each match. E.g. for "GAACCCACTAGTATAAAATTTGGGAGTCCCAAACCCTTTGGGAGT" and n=10, the answer is 1 7 31 3 Apply the functions for several values of n of your choice on the DNA sequence in file DNAseq.txt (posted in D2L) and state what you observe