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To: Southack
If you feel that you can mathematically demonstrate that...

I'll try. Let's stick to coding sections of the genome. IIRC current estimates are that there are ~30K genes. Let's say the average gene codes directly for a 100 amino acid protein (I'm sure this is conservative). That gives 3*10^6 codons.

As you know, there is significant redundancy in DNA coding. Every codon can code 64 values but there are only 20 amino acids. Let's go wth the average and say that every codon can have three variations.

Varying all the codons independently gives 3^(3*10^6) functionally equivalent (hence viable) humane genome variants. That's ~10^954242.

695 posted on 04/09/2002 12:04:41 PM PDT by edsheppa
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To: edsheppa
"Varying all the codons independently gives 3^(3*10^6) functionally equivalent (hence viable) humane genome variants. That's ~10^954242."

I don't think that anyone is arguing over the quantity of potential permutations, but rather, the quantity of viable permutations...

698 posted on 04/09/2002 12:32:09 PM PDT by Southack
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