Jerry's Calculation of Entropy in Genome

Um...Don't celebrate just yet. Nitwit Ned has not got me silenced just yet. But he's working on it. I fully intend to get to this thread.

I just wonder where that nose has been. ~~~Shudder~~~

Justin:Shouldn't take long to put this whole thread to bed. The math I used simply estimated the entropy increase in the FIRST GENERATION.I calculated S. This math will not calculate continually changing entropies from generation to generation because that it is not S, but deltaS. You are assuming:W = (41469.4 + 1.6)! / (41469.4)!(1.6)! --- 3.66 x 10^173494 / 2.14 x 10^173487W = 1.71 x 10^7Boltzmann's math:S = K log W, S = (1.38 x 10^-23) log(1.71 x 10^7)deltaS = 9.98 x 10^-23,This is not correct!To get deltaS you then have to take that first generation down the lineage:deltaS = S(final) - S(initial)As you can see, since the study showed a steady accumulation of 1.6 mutations per generation, entropy will NEVER begin to decrease, so this should communicate to you that you're probably not doing something right. Paul:

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The figure of 1.6 is NOT directly related to the figure of 41,471. The first is the estimated number of deleterious mutations per generation. The second is the number of nucleotides examined in the study. Jerry's use of the figures has no basis in the study.

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Ok, I just went with nucleotides rather than the triplex. Had you rather I divide the nucleotides into codons and calculate it that way? Doesn't matter to me as you STILL are going to see rising entropy.

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Because the factorial operation (denoted by '!') is only defined for integers.

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I would teach it this way too if I were instructing high schoolers but it ISN'T true and especially so when we get into higher math. I was simply showing whether entropy was positive or negative. When absolute accuracy is necessary, this is not a problem either as there are many excellent programs out there which will accurately calculate fractions of integers using the natural log of a continuous probability distribution. In fact, if I'm not mistaken (couldn't tell with a brief Google) this is what the Windows calculator does.

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If you chose to look at beneficial rather than detrimental mutations you would find that each beneficial mutation increased the entropy.

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You can't show me a half dozen beneficial mutations. Much less enough accumulating to offset the kind of dramatic accumulations of deleterious ones we see in the human genome. Not a factor to consider at all.RAZD:

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as would be the case where you had an identified number of microstates (you don't get half a state)

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Normally true but we often come across fractional states when we average something as was done in that study.That's all that needs addressed on this thread. I think the rest of you are just slinging some mud because you do not like what that study showed. I didn't do it, I just quoted it. Don't kill the messenger.

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Design Dynamics

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Welcome back Jerry!

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Thanks, Limbo. Just here to get caught up so I don't leave people hanging. Then I'm out of here.

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Design Dynamics

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We start off with a W of:

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1. (1000!)/[(1000!)(0!)]=1

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After I mutate a quarter of the nucleotides to deleteriously affect some genes:

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2.) (1000!)/[(750!)(250!)]>1

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After I mutate half of the nuceotides:

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3.) (1000!)/[(500!)(500!)] >>1

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After I mutate all of the nucleotides, so no information is left in the 1000 nucleotide segment:

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4.) (1000!)/[(0!)(1000!)]=1

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As you can see, the change in entropy is always positive except for the last change, from

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(3) to (4), which is:

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5.) Delta W= (1-(>>1))= - N

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I apologize for the abbreviations since I don't have a calculator handy, but you should get the point.

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According to your calculations, if I take a one thousand nucleotide DNA sequence full of genes, and then mutate every nucleotide to deleteriously affect the gene products, then the entropy remains the same. Or, to put it another way, a one thousand nucleotide DNA sequence with half of the nucleotides deleteriously mutated will have a higher entropy than the same sequence with all the nucleotides mutated. The change would be negative if going from the former to the latter using your equation.

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No. This is not my calculation. This is yours and your math is simply incorrect. Please cut and paste where I used a deltaW anywhere.We are calculating entropy, not just the statistical weight. And when you calculate entropy down the lineage you must use deltaS, not just S which will always be the case with this math you are introducing. You KNOW entropy will not magically begin to decrease when the study clearly shows 1.6 harmful mutations continue to accumulate. So why are you mathematically trying to show something mathematically correct that you know to be mathematically incorrect?The combinatorials need only be used once, to calculate the entropic change in an organism with x amount of nucleotides where y mutations are accumulating. Since x and y are always the same, what is there to recalculate using that formula?Please use deltaS to calculate this changing entropy.Considering change in entropy between 500 and 501 descendants:Initial entropy in 500th organism = (500) (9.98 x 10^-23) = S(intial)Final entropy in 501st organism = (501)(9.98 x 10^-23) = S(final)deltaS = S(final) - S(intial)deltaS is positive showing the new accumulated mutations we know occurred in that genome. Now THIS is my math.

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I'm just using the equations you used to show an absurditiy.

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No you're not. You are trying to extrapolate a formula I used in a way I did not use it. This is your math, not mine.

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Please show me exactly where my calculation is in error.

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Um....I think I did. This message has been edited by Jerry Don Bauer, 05-14-2005 10:23 PM

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Design Dynamics