You can download all of our solutions as lab04.py

1. What percentage of codons across all primary reading frames are ATG?
   If this were completely random, we'd expect 1/64=1.5625% of the triples. We observe 1.196% for guinea pig and 0.978% for human.

   count = 0
   for k in range(len(dna)-2):    # N.B. stop value
       if dna[k:k+3] == 'ATG':
           count += 1
   
   percent = count/(len(dna)-2) * 100
         

2. If two consecutive nucleotides match each other, how often is the next nucleotide that same nucleotide?
   If nucleotides were completely random, we’d expect 25%;
   We observe 28.392% in guinea pig and 30.620% in human.

   doubles = 0
   triples = 0
   for k in range(len(dna)-2):    # N.B. stop value
       if dna[k] == dna[k+1]:     # neighbors match
           doubles += 1
           if dna[k] == dna[k+2]: # the third of the triple matches as well
               triples += 1
   
   percent = 100*triples/doubles
         

3. How many times does a motif of the form CC?AT occur within the sequence? (where ? could be anything)
   For guinea pig, 111 times; for humans, 132 times.

   total = 0
   for k in range(len(dna)-4):    # N.B. stop value
       if dna[k:k+2] == 'CC' and dna[k+3:k+5] == 'AT':
           total += 1
         

4. When the motif CC?AT does occur, what percentage of the time is the middle nucleotide an A? (A so-called cat box CCAAT)
   For guinea pig, 27.027%; for humans, 21.212%.

   motifs = 0
   catbox = 0
   for k in range(len(dna)-4):    # N.B. stop value
       if dna[k:k+2] == 'CC' and dna[k+3:k+5] == 'AT':
           motifs += 1
           if dna[k+2] == 'A':
               catbox += 1
   
   percent = 100*catbox/motifs
         

5. The pattern CCAAT is known as a "cat" box. What are the relative percentage of bases immediately following the pattern CCAA in the dna?
   

   Guinea Pig
   A: 28.431%	C: 31.373%	G: 10.784%	T: 29.412%
   Human
   A: 39.416%	C: 29.197%	G: 10.949%	T: 20.438%

   ca = 0   # count for A
   cc = 0   # count for C
   cg = 0   # count for G
   ct = 0   # count for T
   for k in range(len(dna)-4):    # N.B. stop value
       if dna[k:k+4] == 'CCAA':
           if dna[k+4] == 'A':
               ca += 1
           elif dna[k+4] == 'C':
               cc += 1
           elif dna[k+4] == 'G':
               cg += 1
           elif dna[k+4] == 'T':
               ct += 1
   
   total = ca+cc+cg+ct
   # ... can then display ca/total, cc/total, and so on