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  Topic: Creating CSI with NS, H T T H H H T H T T H H H H T T T< Next Oldest | Next Newest >  
RumraketR



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Joined: Nov. 2012

(Permalink) Posted: Nov. 20 2012,16:18   

Quote (Jerry Don Bauer @ Nov. 20 2012,15:34)
Quote (Jerry Don Bauer @ Nov. 20 2012,15:51)
We are discussing Complex Specified Information and what makes certain information complex, or not and/or specified or not.

This has little to do with the length of anything or the amount of loci it harbors.

Quote
You refer to the example of a 6 billion base-pair diploid genome, divided by the number of possibilities pr site (4):

610^9 base pairs/diploid genome x 1 byte/4 base pairs = 1.510^9 bytes or 1.5 Gigabytes, about 2 CDs worth of space!

In other words, the information content of a sequence of DNA, for example 12 base-pairs in length, AUGAATAUGTTA, is equal to 12 base pairs x 1 byte/4 base pairs = 3 bytes.

Am I correct in my understanding here?

You are referring to a link I referrenced. The purpose of that link was to show that even a genome contains much more information than the 500 bits upper probability boundary. Therefore, an entire organism most certainly would be over 500 bits and therefore CSI.....


That was all I was pointing out.....I certainly did not want to get into genomic entropy and the like at this point.

Quote (Jerry Don Bauer @ Nov. 20 2012,15:51)
We are discussing Complex Specified Information and what makes certain information complex, or not and/or specified or not.

This has little to do with the length of anything or the amount of loci it harbors.

Quote
You refer to the example of a 6 billion base-pair diploid genome, divided by the number of possibilities pr site (4):

610^9 base pairs/diploid genome x 1 byte/4 base pairs = 1.510^9 bytes or 1.5 Gigabytes, about 2 CDs worth of space!

In other words, the information content of a sequence of DNA, for example 12 base-pairs in length, AUGAATAUGTTA, is equal to 12 base pairs x 1 byte/4 base pairs = 3 bytes.

Am I correct in my understanding here?

You are referring to a link I referrenced. The purpose of that link was to show that even a genome contains much more information than the 500 bits upper probability boundary. Therefore, an entire organism most certainly would be over 500 bits and therefore CSI.....

Okay, fair enough, I think I understand. But just to be sure, you agree the quantity of information in the genome there is 1.5 gigabytes? Not CSI, not Entropy, just 1.5 gigabytes of information, and 1.5 gigabytes is more than 500 bits(and 500 bits would be the bound above with the quantity of information would qualify as being CSI). Right?

If my understanding is not correct, could you clarify:
A): How to calculate the quantity of information in an arbitrary string of DNA, for example?

You can use any stretch of DNA you want, like a real world promoter sequence(or mRNA transcript or whatever you like), or just use a small random string for the purpose, like the one I supplied. Anything is fine with me, I just want to make sure that we agree on how to calculate the quantity of information in a string of symbols, like DNA, in bytes.

I understand that the quantity itself is not what makes it Complex or Specified. I just want to make sure we agree on how to calculate the quantity.

Quote (Jerry Don Bauer @ Nov. 20 2012,15:51)
That was all I was pointing out.....I certainly did not want to get into genomic entropy and the like at this point.

That's absolutely fine with me, we don't have to delve into entropy or anything. I just want to reach an agreement on the basics, like how to calculate information quantity in stretches of DNA.

That's why I brought up the example you quoted earlier, because you seemed to be using a method that corresponded to length of string divided by number of symbols and reporting the result in bytes.

If this is not how you would calculate information content in a string of symbols, how else? Give an example and I would be most grateful.

Thanks again for your time :)

  
  128 replies since Oct. 06 2012,18:57 < Next Oldest | Next Newest >  

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