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shadowpool
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Posted at 5:37 pm on July 1, 2008 |
| Someone with a good computer science education should know how given the formula used in the program--qalc. |
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GodsFireAngel
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Posted at 5:24 pm on July 1, 2008 |
| It would be intresting to calculate how long such a calclation would take, anyone know how? |
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Bobman21
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Posted at 8:16 pm on June 26, 2008 |
| Don't trick yourself into thinking your answer will be even approximately close to the truth. There are 64 codons, nearly 5% of which are stop codons. Take into consideration that a protein is defined as a polypeptide of >100 amino acids and therefore 3000 a minimum of 3000 codons, that there are at least 10,000 genes coding for proteins, and that therefore nowhere in the reading frame of these 10,000 proteins can you find a stop codon. That's only one of the problems... assuming you care that such problems exist. My guess is that you already know they exist but dont care. Oh well, my two cents. |
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Event Horizon
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Posted at 4:11 pm on June 26, 2008 |
| well now you know how many. the only problem is that our genome is specific in many places, so that the number of possible combinations will decrease as certain spots are locked in place. |
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shadowpool
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Posted at 3:59 pm on June 26, 2008 |
Quote: from Event Horizon at 7:37 pm on June 26, 2008
i knew he was missing a zero. you want to calculate the possible combinations of the human genome on your home computer? 
On my laptop. I think I might give up. I just wanted an idea of the number of possibilities. Now that I know how many digits the number has, calculating it isn't necessary. Except I then I will have wasted 15 CPU hours calculating 4^187500000. But for all I know, it could take months or years. |
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Event Horizon
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Posted at 3:37 pm on June 26, 2008 |
| i knew he was missing a zero. you want to calculate the possible combinations of the human genome on your home computer? |
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GodsFireAngel
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Posted at 3:36 pm on June 26, 2008 |
| yah I am missing a zero, I didn't count the zeros in the question he asked properly 112,880,000- to 112,890,000 no forula, just guesswork. what formula did you use? |
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shadowpool
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Posted at 3:33 pm on June 26, 2008 |
| Log(4)*187500000=112,886,248.374 So I'm looking at 107.7MB of digits. Not too bad! Except I want to raise that result to the 16th power. For a total of 4^3000000000 which represents the total number of possible combinations of DNA in the human genome. And that's 1.6GB of digits. I'm not sure I have enough to ram to calculate that one out. I'd have to write a specialized program to write the intermediate results to the disk. :( |
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Event Horizon
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Posted at 3:30 pm on June 26, 2008 |
| what formula did you use, I think you are missing a zero |
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GodsFireAngel
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Posted at 3:16 pm on June 26, 2008 |
| probably closer to 11,290,000 in fact |
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GodsFireAngel
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Posted at 3:15 pm on June 26, 2008 |
| around 11,280,000 digits |
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Event Horizon
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Posted at 3:08 pm on June 26, 2008 |
| should be to the magnitude of 10^112500000 i believe basically, a 1 with 112,500,000 zeros after it lol |
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xenchantedx
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Posted at 3:06 pm on June 26, 2008 |
| I have no idea, but why are you trying to figure this out? |
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matto
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Posted at 3:02 pm on June 26, 2008 |
| There probably is. I haven't learned it, though. :( If you find out, let me know, I'd be interested. |
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shadowpool
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Posted at 2:57 pm on June 26, 2008 |
| I know there's a way to do this. |
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