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At the transcription start site of a gene. any of the four nucleotides can occur with equal probability p. the shannon entropy s. given by s = - 2=p;ln p;. for
At the transcription start site of a gene. any of the four nucleotides can occur with equal probability p. The Shannon Entropy S. given by S = - 2=P;ln p;. for this start site is Given data: In(2) = 0.69
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Narayan singh
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Given the fact that there is equal Probability for all 4 nucleotides, so p=0.25 or 1/4. 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...
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Abhijeet Gaurav Best Answer
Given the fact that there is equal Probability for all 4 nucleotides, so p=0.25 or 1/4. 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)] As 4 nucleotide occur equally so; = -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 So the Shanon entropy for start site is 1.38.