User talk:Ridhi Tariyal

Assignments:
October 29, 2009

I am not sure where we landed on the discussion last week, but it looks like there is an oppportunity for overlap between the projects? If that is the case and since I don't have strong programming skills, I would like to develop some methodology for aggregating and analyzing adverse events by pharmaceutical agent. (This could involve mining the Adverse Events Reporting System, speaking with major hospitals, primary clinician interviews...)Ideally, I would like to provide a data set that is organized by some characteristics (yet to be defined) such as frequency of events, any patterns in reported data, etc. which would provide focus and direction to basic research about genetic links which drive therapy selection and dosing decisions.

Check out the following links:

20% of patients who receive warfarin experience bleeding events. Directly related to a genetic predisposition which effects metabolism of the drug. 

Example output of companies already serving this market:

September 24, 2009 Assignment 3

1)

#first I established that p53 was a string that contained the p53 dna segment as provided by the hw   p53_list = list(p53) print len (p53) gc=0 for x in range(len(p53_list)): if p53_list[x]=='g' or p53_list[x]=='c': gc +=1 print gc/float(len(p53_list))*100

2)

#first I established that p53 was a string that contained the p53 dna segment as provided by the hw   p53_list = list(p53) p53_list.reverse #print p53_list for x in range(len(p53_list)): if p53_list[x]=='g': p53_list[x] ='c'       elif p53_list[x]=='c': p53_list[x] ='g'       elif p53_list[x]=='a': p53_list[x] ='t'       elif p53_list[x]=='t': p53_list[x]='a'   print p53_list

3)

#first I established that p53 was a string that contained the p53 dna segment as provided by the hw   counter = 0 p53_list= [] for x in range(0,(len(p53))/3): p53_list.append(p53[counter:counter+3]) counter+=3 print p53_list print len(p53_list) codon = [] for y in range(len(p53_list)): codon.append(standard[p53_list[y]]) print codon

second frame:

p53_new=p53[1:len(p53)]+p53[0] counter = 0 p53_list= [] for x in range(0,(len(p53))/3): p53_list.append(p53[counter:counter+3]) counter+=3 codon = [] for y in range(len(p53_list)): codon.append(standard[p53_list[y]]) print codon

third frame:

p53_new=p53[2:len(p53)]+p53[0]+p53[1] counter = 0 p53_list= [] for x in range(0,(len(p53))/3): p53_list.append(p53[counter:counter+3]) counter+=3 codon = [] for y in range(len(p53_list)): codon.append(standard[p53_list[y]]) print codo

for the reverse sequences, I only added one procedure to the three pieces of code above:

def reverse(b): a = list(b) a.reverse return''.join(a) p53_r = reverse(p53)

4)

p53_list = list(p53) alt_list = list(p53) print p53_list dna_insert = ['a','t','g','c'] for i in range(int(round(len(p53)*.1))): r1 = random.randint(0,len(p53)) p53_list[r1]=random.choice(dna_insert) print p53_list d = [] for a in range(len(alt_list)): if p53_list[a] != alt_list[a]: d.append(a) print d   print len (d) print len(alt_list) print float(len(d))/float(len(alt_list))*100

September 15, 2009

- It seems that if k > 1, the function describes a phenomenon of exponential decay and if k > 1, the equation describes exponential growth.



- If you hold the initial value constant, say at 1, and you use the multiple values given for k, it seems that after k = 3, the shape of the curve is unaffected even though the scale of the axis changes.

- Increasing the initial value has a considerable impact on the time it takes for the curve to reach the exponential vertical portion of the curve. It is much slower to be characterized by a swift, almost vertical increase at lower initial values.