From: simon@dcs.warwick.ac.uk (Simon Clippingdale)
Subject: Re: Information in DNA  (was: Creationist CRAPOLA!)
Organization: Department of Computer Science, Warwick University, England
Date: Wed, 30 Sep 1992 19:21:52 GMT

In article <1992Sep28.154730.428@aurora.com> isaak@aurora.com (Mark Isaak) writes:
>In article <1992Sep28.100020.27769@hsr.no> onar@hsr.no (Onar Aam) writes:
>>In junk-DNA (j-DNA) there are Four amino acids, A,T,C & G, which can give four
>>different combinations: A-T, T-A, C-G, G-C. . . .
>>You can easily see that describing such chains, no matter content, requires
>>INFORMATION.
>
>Just because an alphabet has four letters doesn't mean any arrangement
>of those letters carries information, any more than the fact that English
>has 26 letters implies that uow vzlihlwgvdl pm cxsafhudhkec lnwlqmciyovf.

Paradoxically enough, in information theory terms, a string of symbols/letters
each randomly selected from some alphabet carries *more* information than does
a sequence in which correlations exist (the correlations equate to redundancy
or reduction of information: thet is wyy you kam uderstand thas if yuo try).
In a totally random splurge with no intersymbol correlation, you can't
correct errors because there is no redundancy. In info theory terms,
information doesn't imply comprehensibility as you seem to imply above;
it implies a reduction of uncertainty about the message which occurs when
the message is received.

>A question to those who know more about the subject than me:  Information
>theory has some simple formulas (which I can't recall offhand) to
>measure how much information a message carries.  Someone has undoubtedly
>applied this to DNA.  How much information is there in a length of DNA?
>Does the amount of information in junk-DNA differ from the rest of it?

Dunno about the last bit, but the zeroth-order entropy (loosely, the
*average* information content assuming no intersymbol correlation, so
receiving one symbol tells you nothing about what the next will be) of
a message consisting of N symbols a[i], 1 <= i<= N, with respective
probabilities p[i], is

                   N
	H   =   - sum ( p[i] log  p[i] ) bits per symbol.
                  i=1           2

If you have four nucleotides A,T,C,G each of which has a probability of
0.25 (I'm not saying this is actually the case in nature, BTW) and there are
no intersymbol correlations (i.e. knowing the neighbours of some base position
doesn't tell you anything about which base is more likely to appear there),
then the entropy is

	H   =   - (0.25 * -2) + (0.25 * -2) + (0.25 * -2) + (0.25 * -2)

	    = 2 bits per symbol, which is what you would expect since
you need two bits to specify one of four equally-probable outcomes.

Any intersymbol correlations (e.g. A follows T more often than does C) or
deviation from the equal-probability assumption (e.g. A occurs more frequently
than T) reduces the entropy and thus the information content of the message
(when p[i1] = 1.0 and p[any other symbol] = 0.0, you have H = 0: there is
no information in a string of ....AAAAAAAAAAAAAAAAAAAAAAAAA..... for instance).

Ball park time: a mammalian genome of 10^10 base pairs per haploid cell
(Futuyma p.49) would have a maximum information content of 2 (bits per
symbol) * 10^10 = 20 gigabits.

>Mark Isaak           "Every generation thinks it has the answers, and every
>isaak@aurora.com      generation is humbled by nature." - Philip Lubin

Cheers

Simon
-- 
Simon Clippingdale                simon@dcs.warwick.ac.uk
Department of Computer Science    Tel (+44) 203 523296
University of Warwick             FAX (+44) 203 525714
Coventry CV4 7AL, U.K.
Newsgroups: talk.origins