Cell Language Theory, The: Connecting Mind And Matter

Key Terms and Concepts 17

“6x9” b2861 The Cell Language Theory: Connecting Mind and Matter

The unit of information is bits, one bit of information representing the

reduction of uncertainty by a factor of 2. Two bits of information reduce

the uncertainty by a factor of 2^2 = 4, and n bits of information reduce the

uncertainty by a factor of 2n.

Despite the enormous importance of the concept of information in

biology, it is interesting to find that no textbook in biochemistry, molecu­

lar biology, or cell biology that I have consulted discuss any definition of

it. As already mentioned, one of the first quantitative definitions of infor­

mation proposed was that of Shannon given in Eq. (2.6) [38]. H in Eq.

(2.6), called the “Shannon entropy”, is often used interchangeably with

“information” denoted as I. One justification for such a practice is that the

amount of information, I, carried by a message is equal to the amount of

the uncertainty, H, of the message source in a communication system, if

there is no loss of information in the communication channel.

H Kp p=- ∑ iilog, 2 (2.6)

where K is a positive constant which is usually taken to be 1, S is the

summation sign, summing from i = 1 to n, the number of events or choices

available, pi is the probability of the ith event to occur or the ith choice

selected, and log 2 is the binary logarithm (i.e., the logarithm to the base

of 2). As already indicated, the unit of H is “bits” from “binary digits”.

When K = 1, Eq. (2.6) reduces to

H pp=-∑ iilog. 2 (2.7)

Since a dice has six sides with six different numbers on it, from 1 to 6,

we have i = 1, 2, 3, 4, 5, or 6. The probability of the ith number showing

up is pi, which assumes the value of p 1 , p 2 , p 3 , p 4 , p 5 , or p 6. Let us consider

two possible cases:

(1) The dice is fair, so that all the probabilities, pi, are the same, namely

1/6. Then the value of H would be

H = - [1/6 log 2 (1/6) + 1/6 log 2 (1/6) + 1/6 log 2 (1/6) + 1/6 log 2 (1/6)

+ 1/6 log 2 (1/6) + 1/6 log 2 (1/6)]

= - (6) (1/6) log 2 (1/6)

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