02-15-2017, 02:02 AM
(02-15-2017, 01:56 AM)SunilNagpal Wrote:Thank you Sunil sir.....My answer was also 1.38(02-14-2017, 09:45 PM)CHANDA7 Wrote: Can anyone answer this question?Given the fact that there is equal Probability for all 4 nucleotides, so p=0.25 or 1/4.
At the transcription start site of a gene, any of the four nucleotides can occur with equal probability p.
The Shannon Entropy S for this site would be? (Details in the image)
Now, all we have to do is fill the same into the given equation:
S= -[0.25*ln(0.25) + 0.25*ln(0.25) + 0.25*ln(0.25) + 0.25*ln (0.25)]
= -0.25[4ln(0.25)]
=-1*ln(0.25)
=-ln(0.25)
=-ln(1/4)
=-[ln(1) - ln(4)]
= -[0 - ln (2*2)]
= ln(2) +ln (2)
=0.69 + 0.69
= 1.38
I hope this is the way to go about it...