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Author  Topic: Jerry's Calculation of Entropy in Genome  
Limbo Inactive Member 
Welcome back Jerry! :)
 
RAZD Member Posts: 19757 From: the other end of the sidewalk Joined: Member Rating: 5.8 
still waiting for evidence in the fossil record that shows a {just created} state in the {human\hominid} ancestry and clear evidence of the superior height from which we have since "devolved" among others we are limited in our ability to understand by our ability to understand RebelAAmerican.Zen[Deist
 
Jerry Don Bauer Inactive Member 
quote: Thanks, Limbo. Just here to get caught up so I don't leave people hanging. Then I'm out of here. ;) Design Dynamics
 
JustinC Member (Idle past 2921 days) Posts: 624 From: Pittsburgh, PA, USA Joined: 
quote: I honestly have a hard time following this. I'll just calculate W since it is directly proportional to entropy. Also, I know I was only calculating W, not delta W. But, delta W could be found by comparing the different generations. Here is the calculation again. I will assume ancestral nucleotides to begin with. We start off with a W of: 1. (1000!)/[(1000!)(0!)]=1 After I mutate a quarter of the nucleotides to deleteriously affect some genes: 2.) (1000!)/[(750!)(250!)]>1 After I mutate half of the nuceotides: 3.) (1000!)/[(500!)(500!)] >>1 After I mutate all of the nucleotides, so no information is left in the 1000 nucleotide segment: 4.) (1000!)/[(0!)(1000!)]=1 As you can see, the change in entropy is always positive except for the last change, from (3) to (4), which is: 5.) Delta W= (1(>>1))=  N I apologize for the abbreviations since I don't have a calculator handy, but you should get the point. 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. quote: I'm just using the equations you used to show an absurditiy. That should communicate to you that your equations aren't sound. Please show me exactly where my calculation is in error. This message has been edited by JustinC, 05142005 08:40 PM
 
Jerry Don Bauer Inactive Member 
quote: 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. quote: 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. quote: Um....I think I did. ;) This message has been edited by Jerry Don Bauer, 05142005 10:23 PM Design Dynamics
 
PaulK Member Posts: 14751 Joined: Member Rating: 1.7 
Shaded quotes are from my previous post. quote: Your reply is just confused. I made no reference to the genetic code. I simply pointed out that the "1.6" applied to the ENTIRE GENOME not the smaller figure of 41,471 nucleotides referred to in the study. quote: It IS true that the factorial operation '!' is only defined for positive integers. And if you think about it 0.6 of a point mutation makes no sense whatsoever. You could have just used 1. quote: Which shows how useless your argument is. My variation of your argument relies only on a simple definitional change  not affecting the validity or changing the premises. And you completely reject it and start talking about evidence instead. So now we know that even you don't consider your argument valid.
 
JustinC Member (Idle past 2921 days) Posts: 624 From: Pittsburgh, PA, USA Joined: 
quote: It doesn't matter if I use W or S, or Delta W or Delta S, W and S are directly proportional. All I would be doing is taking the ln of the of the number and multiplying it by a constant. You can do the extra math if you want, but the results will turn out the same. quote: Yes, I know I am using W. They are directly proportional. Do the extra math if you would like, the answer will be the same. For instance, look at my last generation, with all the nucleotides deleteriously mutated: W= (1000!)/(0!)(1000!)=1 S= k ln W= 0 So delta S going from equation (3) to (4) would be: Delta S= SfSi= 0(k ln (>>1))=N That final entropy would be zero, the entropy before that would be a positive number, giving a negative delta S. A decrease in entropy as more deleteriously mutations accumulate. quote: You were the one trying to equate information loss (in the sense of changing the ancestral state of the genome) with entropy increase. You used that equation to show it. I used that equation to show an absurdity, which (if correct) calls into question your whole calculation. Go from (3) to (4) in my calculation, calculate S's for both, and then find delta S. It will be a decrease in entropy as more deleterious mutations occur. quote: I don't see what's so hard to understand. I am increasing y, i.e., mutating more than half of the genes. When I do this, your equation says the entropy will decrease. This calls into question your calculation, since I don't think you want to be saying this. quote: I did above, but I'll do it again for the hell of it. Equation (4) says W=1, so the entropy will be: Sf= k ln 1= 0 Equation (3) says the entropy is more than 0 (I'll use W=1.3 as an example), so: Si= k ln (1.3)>0 So Delta S, going from (3) to (4), would be: Delta S= SfSi= 0 (>0)= N. It will be negative still. quote: I have absolutely no idea what that is supposed to show. We are talking about the entropy of a genome if we dichotomize into ancestral and deleteriously mutated nucleotides, using the equation for statistical weight (N1+N2)!/ ((N1!)(N2!)). What does the generation of the organism have to do with that, and how would that factor into the calculation. The only way I can see it factoring into the equation would be if we write the number of deleterious mutations as a function of the generation. The result will be the same, once we get to a certain point the more delteriously mutations that accumulate the entropy will decrease. quote: The formula should be consistent with the point you are trying to prove. You are trying to prove that deleterious mutations in the genome constitute and increase in entropy. You equation says it does this to a point, and then the entropy will decrease. quote: I don't. Calculate the entropies. Entropy will decrease after a certain point as more deleterious mutations accumulate. The reason I am going through the trouble of this is because I think you just pulled that equation of the internet and tried to use it without understanding it. It works great with the gas in a box, but you can't just extrapolate it to any binary system, as PaulK was saying quote: That's the crux. This message has been edited by JustinC, 05172005 07:45 PM
 
derwood Member Posts: 1457 Joined: 
I wonder how Jerry Don's amazin' math takes things like gene duplication, chromosomal segmental duplications, etc. into consideration. I have a suspicion that it simply does not. 


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