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# Lab 04 Solution Set by Michael Goldwasser
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# uncomment ONE of the following two lines to pick your source data:
#from human import dna
from guinea_pig import dna


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# Question 0: How often is a base followed immediately by the same base?
# (if completely random, we'd expect this to be 25%)
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count = 0
for k in range(len(dna)-1):    # N.B. stop value
    if dna[k] == dna[k+1]:
        count += 1

percent = count/(len(dna)-1) * 100
print("percent of repeat bases: {:.3f}%".format(percent))
print()

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# Question 1: What percentage of codons across all primary reading frames are ATG?
# (if completely random, we'd expect 1/64)
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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
print("percent of ATG codons: {:.3f}%".format(percent))
print()


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# Question 2: If two consecutive nucleotides match each other,
# how often is the next nucleotide that same nucleotide?
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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
print("percent of identical pairs followed by a third: {:.3f}%".format(percent))
print()


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# Question 3: How many times does pattern CC?AT occur?
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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

print("Pattern CC?AT occurs {} times".format(total))
print()


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# Question 4: For motif of the form CC?AT,
# what is percentage of times that '?' is an A?
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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
print("Percent of CC?AT having an A: {:.3f}%".format(percent))
print()


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# Question 5: What are the relative percentage of bases that immediately follow a 'CCAA' pattern?
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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
print("Following CCAA is")
print("A: {:.3f}%".format(100*ca/total))
print("C: {:.3f}%".format(100*cc/total))
print("G: {:.3f}%".format(100*cg/total))
print("T: {:.3f}%".format(100*ct/total))
print()