User talk:Kelly Brock

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Personal/Lab Info

Assignment 1. Python Epicness

I organized the Python code by questions 1,2,3, and 4 as indicated in the comments. I redirected the output stream into a txt file, which is what I turned in on Thursday for my answers. This was by far my favorite problem set this week! I like python as a language because it seems like a really good mix of C, Scheme (especially the lists and dictionaries), and Matlab. As far as the experiment itself goes, part 4 was the most interesting to me because it modeled actual mutations instead of providing intrinsic data about the sequence.

<syntax = python>

1. Kelly Brock
2. File: BiophysAsst3P1.py
3. Answer four parts of Assignment 3

import random

1. For part 4, when we have to do multiple experiments

TRIALS = 6

1. Input genetic sequence into memory

sequence = "cggagcagctcactattcacccgatgagaggggaggagagagagagaaaatgtcctttag" sequence += "gccggttcctcttacttggcagagggaggctgctattctccgcctgcatttctttttctg" sequence += "gattacttagttatggcctttgcaaaggcaggggtatttgttttgatgcaaacctcaatc" sequence += "cctccccttctttgaatggtgtgccccaccccccgggtcgcctgcaacctaggcggacgc" sequence += "taccatggcgtagacagggagggaaagaagtgtgcagaaggcaagcccggaggcactttc" sequence += "aagaatgagcatatctcatcttcccggagaaaaaaaaaaaagaatggtacgtctgagaat" sequence += "gaaattttgaaagagtgcaatgatgggtcgtttgataatttgtcgggaaaaacaatctac" sequence += "ctgttatctagctttgggctaggccattccagttccagacgcaggctgaacgtcgtgaag" sequence += "cggaaggggcgggcccgcaggcgtccgtgtggtcctccgtgcagccctcggcccgagccg" sequence += "gttcttcctggtaggaggcggaactcgaattcatttctcccgctgccccatctcttagct" sequence += "cgcggttgtttcattccgcagtttcttcccatgcacctgccgcgtaccggccactttgtg" sequence += "ccgtacttacgtcatctttttcctaaatcgaggtggcatttacacacagcgccagtgcac" sequence += "acagcaagtgcacaggaagatgagttttggcccctaaccgctccgtgatgcctaccaagt" sequence += "cacagacccttttcatcgtcccagaaacgtttcatcacgtctcttcccagtcgattcccg" sequence += "accccacctttattttgatctccataaccattttgcctgttggagaacttcatatagaat" sequence += "ggaatcaggatgggcgctgtggctcacgcctgcactttggctcacgcctgcactttggga" sequence += "ggccgaggcgggcggattacttgaggataggagttccagaccagcgtggccaacgtggtg"

1. Part 1 - CG content
2. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

print "Kelly Brock\n" print "Biophysics 101 Asst 3\n" print "Part I\n\n"

1. Variable to keep track of how many 'cg's we've encountered

count = 0

1. Check each character in our genetic string

for i in range(0, len(sequence)): if (sequence[i] == 'g') | (sequence[i] == 'c'): count += 1

1. Compute fraction of total characters equal to c or g

answer = count*1.0/len(sequence) print "CG fraction is: " + str(answer) + "\n"

1. Part 2 - Find reverse complement
2. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

print "\nPart II\n"

1. Make list to hold reversed sequence

RevSeq = list((sequence[::-1]))

1. Change all values to their complements

for i in range(0,len(RevSeq)): if RevSeq[i] == 'c': RevSeq[i] = 'g' elif RevSeq[i] == 'g': RevSeq[i] = 'c' elif RevSeq[i] == 't': RevSeq[i] = 'a'; elif RevSeq[i] == 'a': RevSeq[i] = 't'

1. Recast our sequence back into a string

RevSeq = "".join(RevSeq) print "Reverse Complement Sequence" print RevSeq

1. Part 3 - Determining protein sequence
2. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

print "\nPart III\n"

1. Hardcode the protein dictionary

standard = { 'ttt': 'F', 'tct': 'S', 'tat': 'Y', 'tgt': 'C', 'ttc': 'F', 'tcc': 'S', 'tac': 'Y', 'tgc': 'C', 'tta': 'L', 'tca': 'S', 'taa': '*', 'tga': '*', 'ttg': 'L', 'tcg': 'S', 'tag': '*', 'tgg': 'W',

'ctt': 'L', 'cct': 'P', 'cat': 'H', 'cgt': 'R', 'ctc': 'L', 'ccc': 'P', 'cac': 'H', 'cgc': 'R', 'cta': 'L', 'cca': 'P', 'caa': 'Q', 'cga': 'R', 'ctg': 'L', 'ccg': 'P', 'cag': 'Q', 'cgg': 'R',

		'att': 'I', 'act': 'T', 'aat': 'N', 'agt': 'S',
		'atc': 'I', 'acc': 'T', 'aac': 'N', 'agc': 'S',

'ata': 'I', 'aca': 'T', 'aaa': 'K', 'aga': 'R',

 		'atg': 'M', 'acg': 'T', 'aag': 'K', 'agg': 'R',

'gtt': 'V', 'gct': 'A', 'gat': 'D', 'ggt': 'G', 'gtc': 'V', 'gcc': 'A', 'gac': 'D', 'ggc': 'G', 'gta': 'V', 'gca': 'A', 'gaa': 'E', 'gga': 'G', 'gtg': 'V', 'gcg': 'A', 'gag': 'E', 'ggg': 'G' }

1. Make function to find protein abbreviations
2. sequence is forward genetic seq, RevSeq is reverse
3. complement, and posneg indicates whether we want to
4. do all frames (>1) or just the positive ones (1)

def proteinabbr(sequence, RevSeq, posneg):

# Will hold the list of one-letter abbreviations for the proteins # encoded by p53 in different frames protein = list() totalprot = list()

# Top loop chooses + open frame (0) or - open frame (1) for l in range(0,posneg):

# Use + open frame with normal sequence if l == 0: sign = " + " seq = sequence

# The second time, use reverse complement sequence else: sign = " - " seq = RevSeq

# There are 3 possible reading frames for both normal and # reverse complement sequences for m in range(0,3): print "\nFrame" + sign + str(m+1)

# Go through each triple in our frame for i in range(m,len(seq),3):

# Prevents error if not evenly divisible by 3 if (i+2) < len(seq):

# Lookup protein value in dictionary and add it to the # protein list protein.append(standard[seq[i:(i+3)]])

# Prints result as string and clears list print "".join(protein) totalprot.append(protein) protein = list() return totalprot

1. Do all that we just defined and store as original, unmutated sequence

original = proteinabbr(sequence, RevSeq, 2)

1. Part 4
2. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

print "\nPart IV\n"

1. Have to do this for a certain number of trials

for j in range(0,TRIALS):

# Make list to hold the random numbers - there should be 1% of total mutspot = random.sample(range(0,(len(sequence)-1)),len(sequence)/100)

# Make string of sequence mutable mutseq = list(sequence)

# Possibilities for mutation for each character A = ['c','g','t'] C = ['a','g','t'] G = ['a','c','t'] T = ['a','c','g']

# Choose another nucleotide for each spot where you assigned a mutation for i in mutspot: if mutseq[i] == 'a': mutseq[i] = random.choice(A) elif mutseq[i] == 'c': mutseq[i] = random.choice(C) elif mutseq[i] == 'g': mutseq[i] = random.choice(G) else: mutseq[i] = random.choice(T)

print "\nMutated Protein Sequence for Frames +1,2,3 in Trial " + str(j) print "\nMUTATED SEQUENCE" print "".join(mutseq)

# Translate mutated string into 3-frame protein abbreviations mutprotseq = proteinabbr("".join(mutseq), RevSeq, 1)

print "\nNumber of immature stop codons: "

# Go through each ORF we computed for k in range(0,3):

# We also want to see how many changes were introduced countmut = 0

# We want to count how many times a new stop codon is introduced # into the code, compared to the original sequence. * = stop countstop = 0

# Find each protein within each ORF for l in range(0,len(mutprotseq[k])):

# Is it a mutation? if mutprotseq[k][l] != original[k][l]: countmut += 1

# Did you introduce a new stop codon? if mutprotseq[k][l] == '*': countstop += 1

print "\nThe total number of protein mutations was " + str(countmut) print "Of these, " + str(countstop) + " were incorrect stop codons." </syntax>

Assignment 0. Python and Excel I'm currently having technical difficulties getting Python to run - it doesn't want to recognize the matplotlib or numpy libraries. However, I did complete the Excel graphs - with increasing k for the first equation, the function values also increased as expected, resulting in different endpoints of the curve. For the second equation, the curve decreased to very negative numbers, like a reflection of a normal exponential curve. For the third graph, I got that all values were zero since (k* e^x * (1-e^x)) would always be <= 0, and the max would automatically choose zero.