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Author  Topic: Do you really understand the mathematics of evolution?  
PaulK Member Posts: 16775 Joined: Member Rating: 2.8 
quote: I don’t think that is going to last. quote: No, it really isn’t a problem. These are models of neutral evolution. They are used to estimate divergence times between species. Since neutral evolution dominates and since it is not practical to identify which loci were selected in the distant past  except by divergence from these models  that’s obviously a sensible thing. quote: I think you will find that neutral evolution would. The fact that the models don’t describe selection is not an error at all. It’s an intentional feature. Edited by PaulK, : Fixed tag


Kleinman Member (Idle past 183 days) Posts: 528 From: United States Joined: 
So you are going to stop thinking? Why? Are you afraid you might actually learn how DNA evolution works?
Perhaps you think that making assumptions that have no connection with reality will give you any kind of accurate estimate of divergence times between species, but that would just be another example that you don't understand introductory probability theory. If you want to correctly understand DNA evolution, you have to start by making the correct assumptions for your model and these models don't start with the correct assumptions.
You are on a roll, one error after another. Do you think that fish evolve into mammals and reptiles evolve into birds by neutral evolution? How about the Lenski experiment? Do you think it operating by the mathematics of neutral evolution? Of course, not! So, why do you think you can use a model of neutral evolution to estimate divergence times between species? You really have no idea how to apply probability theory to the real world. DNA evolution is nowhere near an equilibrium process in the real world. In fact, in your world, not even fixation can occur whether due to drift or selection. It appears your thinking has already stopped.


PaulK Member Posts: 16775 Joined: Member Rating: 2.8 
quote: No, I think you’ll be unhappy because I am thinking and because I do have a good idea of how DNA evolution works. quote: Many bases will be varying neutrally. So a model of neutral evolution will give a reasonable estimate. quote: No, of course not. But do you think the majority of bases are under selective constraint in the Kishony experiment? It’s only a few that are under selection for antibiotic resistance. Others will be under stabilising selection, but hardly the entire genome. quote: Because many bases will be evolving neutrally. Really it’s not so hard. Of course, looking for bases that change more slowly also helps  neutral change can be too fast. quote: And there is another bizarre assertion. The equilibrium would be roughly equal proportions of each base in a single genome.


Kleinman Member (Idle past 183 days) Posts: 528 From: United States Joined: 
I'm not sure where you get this kind of idea. I like being challenged with good rational thinking. That's the reason I engage in debates like this. If I were playing chess, I would want to play better players than me to press me to improve my game. Your problem is that you are rehashing the same old fish evolve into mammals arguments that I've been hearing for years. If you think that happens, show us the math that would explain this. And if you think I've done the mathematics for the Kishony and Lenski experiment incorrectly, show us all where I made my error(s). Point out the line and shows us the correct math.
Do you think taking a small portion of homologous genome from two different species, ignoring all the other differences in the genomes will give a reasonable estimate of the time separating the two? For example, let's say I find a portion of the genome from a banana that differs by only a single base from a portion of your genome. Can I then plug that into the JukesCantor or Felsenstein model and predict how many generations you have diverged from a banana? I find that a very appeeling idea for what you find as the correct Markov model of DNA evolution.
Sure, every base in the genome is subject to selection. That's why if a detrimental mutation occurs somewhere in the genome, at a minimum, its relative fitness is reduced with respect to the rest of the population. And you might get a variant that gets a beneficial mutation for the drugfree region that makes it a better replicator for the drugfree region. It still won't be able to grow in a region with the drug but it will change the relative fitness of that variant with respect to the other drugsensitive variants.
I like your analysis. Based on your logic and math, I can show that you are closely related to a banana. I bet you didn't know that's what your math is showing how closely related to a banana you are. Depending on the portion of the genome you choose, you might even be more closely related to a banana than a chimp. Think about that the next time you eat a banana and that you might be committing cannibalism.
PaulK, you are really confused here. The JukesCantor/Felsenstein Markov models are only considering a single site in the genome. How did you get so confused on this subject?


PaulK Member Posts: 16775 Joined: Member Rating: 2.8 
quote: A small portion, certainly not, you’d want a number of genes to get a robust result, quote: Oh dear, oh dear, oh dear. This is not at all a rational argument. In fact it is a complete nonsequitur. There are bases which can freely change without deleterious or beneficial effect. quote: The distribution of the bases in an interbreeding population wouldn’t be random. So how could you apply the probabilities across individuals? Anyway, until you recognise that DNA evolution is driven by neutral drift, you do not understand the mathematics of DNA evolution.


Kleinman Member (Idle past 183 days) Posts: 528 From: United States Joined: 
But that's exactly the incorrect approach they are doing with these Markov models. The JukesCantor and derivative models only consider a single site in the genome. You can extend these models to more sites but if you continue to make the assumption that this is a stationary process, when you do the math for those cases, they converge on equilibrium values for the twosite model of 0.0625 (0.25^2). To contrast the one and two site models, consider the Markov transition diagram for both cases. For the singlesite model the transition diagram is: And for the secondorder Markov Chain transition diagram can be found on page 11 of this PowerPoint except not all the possible transitions are included. http://web.mit.edu/...0203ProbabilitiesMarkovHMMPDF.pdf Try writing out the transition equations for this case. If you can't, I will show you how to write those equations.
I know, and they show how closely related to a banana you are.
Oh my, Oh my, now you are going down the Taq trail. Before you toss in recombination, you had better learn how DNA evolution works first.
I know, you have correctly shown how closely related to a banana you really are.


PaulK Member Posts: 16775 Joined: Member Rating: 2.8 
quote: Applying the model to multiple sites is hardly a difficult idea to grasp. Indeed it’s implicit in the part I quoted for you earlier, quote: You’re only showing that you’re bananas. quote: Apparently I know how DNA evolution works better than you. You do realise that the DNA sequences found in the population are not independent?


Taq Member Posts: 8482 Joined: Member Rating: 5.3 
You're a liar. You haven't answered the question. Answer the question. Weed A is exposed to pesticide A in one region and develops resistance. Weed B is exposed to pesticide B in a different region and develops resistance. We bring Weed A and Weed B into the same region. How many generations before you get resistance to pesticides A and B in a single weed? According to your multiplicative rule, it would take just as many generations to develop resistance to both pesticides as it did to one. Is that right?


Kleinman Member (Idle past 183 days) Posts: 528 From: United States Joined: 
On to the mathematics of DNA evolution when more than a single mutation is needed to improve fitness (call those mutations A1 and A2). The full mathematical derivation can be found here:
The mathematics of random mutation and natural selection for multiple simultaneous selection pressures and the evolution of antimicrobial drug resistance And to put this into the context of a real situation, we continue with the Kishony experiment. But in this case, consider what happens if the increase in drugconcentration is so large that only a variant with the A1 and A2 mutations can grow in that region. We have already shown that with a beneficial mutation rate of e9, e9 replications will give us 1 variant with the A1 mutation and another variant with the A2 mutation in the initial founder's wildtype colony. Until one of those A1 variants get an A2 mutation or one of those A2 variants get an A1 mutation, we won't have the variant able to grow in the next higher drug concentration region. Consider the Venn diagram for this circumstance: The area with the A1 label represents the subpopulation with the A1 mutation, the area with the label A2 represents the subpopulation with the A2 mutation the overlap area is the subset that has both the A1 and A2 mutations and the rest of the area consists of variants that have neither the A1 nor the A2 mutations. After e9 replications with a beneficial mutation rate of e9, we will have 1 member in the A1 subset, 1 member in the A2 subset, and (e9)2 members in the rest of the population. What is the probability that there is a member in the intersection of the A1A2 subsets? And how does that probability change as the colony population grows? To do this calculation, start with the mean value of a binomial distribution which gives the number of members in the A1 subset as a function of total population size (call the total population "n". Then the number of A1 mutants (call that number "nA1") is: nA1=n*P(BeneficialA1)μ And we already know from the single selection pressure model the value for the probability of the A1 mutation occurring: P(A1)=1((1−P(BeneficialA1)μ)^n) Now, we need to compute the probability that at least a single member of that subpopulation nA1 will also have mutation A2. This is a conditional probability problem because A2 mutation must occur on a reduced subset (nA1) of the entire population n. P(A2)=1((1−P(BeneficialA2)μ)^nA1) Then, the joint conditional probability is: P(A1)P(A2)=(1((1−P(BeneficialA1)μ)^n))*(1((1−P(BeneficialA2)μ)^nA1)) A plot of this probability equation as a function of n, the total population size looks as follows: From that graph with a beneficial mutation rate of e9, and a population size of n=e12 gives a probability of about 0.6 that a least one A1A2 variant will have occurred. This is why the Kishony experiment will not work if the increase in drugconcentration is too large. The carrying capacity of his megaplate is not sufficient to give the need colony size to give a variant with the A1A2 mutations. This same principle applies if two drugs are used if A is the beneficial mutation for the first drug and B is the beneficial mutation for the second drug. Do you get it dwise1? 

Kleinman Member (Idle past 183 days) Posts: 528 From: United States Joined: 
And if you assume a stationary transition matrix, you will be making your same blunder twice. You know, there's an old saying, there's no education in the second kick of a mule. Are you going to show that you are twice as close to a banana?
It's your mathematics which you claim is correct that shows you are related to a banana! If you do the Markov chain mathematics correctly, you can predict the behavior of the Kishony experiment. That may not be as much fun as your math showing you are related bananas but it is the correct math.
But mutations are random and that's why DNA evolution works the way I'm showing you. And you are not related to bananas. You aren't even related to chimpanzees. You would understand this if you knew how to do the Markov mathematics of DNA evolution correctly. Keep thinking about it, it might come to you.


Kleinman Member (Idle past 183 days) Posts: 528 From: United States Joined: 
Taq, we know you are a mathematically incompetent nitwit but do you also have to be a stupid ass as well?
I've answered the question multiple times and even showed you how to do the mathematics for this situation. You are just too ignorant to understand the math. The answer to your question is yes, that recombination event is possible, but the probability of that event depends on the frequencies of the A and B variants in the population. If A and B are at low frequency then the probability of that recombination event occurring will be low. So don't call me a liar just because you are too stupid to understand the mathematics of recombination. And you hardly do better on the mathematics of DNA evolution.


PaulK Member Posts: 16775 Joined: Member Rating: 2.8 
quote: And you still have no idea. Look, if you can’t understand how these models are used, just admit it rather than inventing silly straw we. quote: It’s the correct math when selection is tightly constraining the viable outcomes at that site, but that’s hardly the normal case for DNA evolution. quote: Well you’re wrong because it mostly doesn’t. quote: Since you don’t know how DNA evolution works, your arguments are worthless.


Kleinman Member (Idle past 183 days) Posts: 528 From: United States Joined: 
I know exactly how fish evolve into mammals clique use these models. I know exactly where they make their error in their assumptions and I know exactly how to correct the error so that the model works correctly to predict the behavior of experiments such as the Kishony and Lenski experiments. I've even told you were the error is and it is in the assumption that DNA evolution is a stationary Markov process. It isn't a stationary Markov process. Now go figure out what I'm telling you. If you can't figure it out, you will have to wait an read my paper when it gets published to get the explanation.
You are blowing smoke and you know it. Figure out the correct transition matrix for DNA evolution. I have given you plenty of hints.
So, show us the math how it DNA evolution mostly works.
Clearly, you need some help in doing this math. Let's go back to this Wikipedia web page and let me show you an alternative way of solving the Markov probability equations as written and perhaps the problem with this model will become more clear: Models of DNA evolution As you read down the page, you come to this line: "Specifically, if E1, E2, E3, E4 are the states, then the transition matrix P(t)=(Pij(t)) where each individual entry, Pij(t) refers to the probability that state Ei will change to state Ej in time t." Note, t here is a replication. And then they write out the transition matrix below that where PAA(t) is the probability that an A base will remain an A base after 1 replication, PAG(t) is the probability that an A base will transition to a G base in 1 replication and so on for the other 14 terms. This is correct so far. Then they go down further and define a Q matrix. This is what they call a rate matrix. I'm going to show you how to do this math without a rate matrix and it is very easy if you know how to write a computer program to do the math. They then go down and formulate a solution by approximating the system as a firstorder ordinary differential equation by changing t by Δt. You don't have to do this to solve this Markov system. Instead, just substitute the mutation rate for the various terms in the P(t) matrix. And you will obtain a transition matrix that looks like this: The only difference is that we have called our mutation rate μ and the are naming that variable α. The Ei vector is simply the frequency of the possible outcomes after the ith replication and you would compute the value for Ej simply by multiplying Ei by the transition matrix shown in the above image. The mathematical equation is written: Ej = Ei * P(t) where this is a matrix multiplication of the vector E times the matrix P(t). Try writing out one replication on this. You will get four very easy to evaluate equations. And if you do lots of recursions of this matrix multiplication (that is j is very large), you will see that E will converge on a value of [0.25,0.25,0.25,0.25], the same value obtained on the Wikipedia page. The reason is that the transition matrix is not a function of time but real DNA evolution processes, the transition matrix is a function of time. See if that helps you figure out what is wrong with this formulation of DNA evolution.


PaulK Member Posts: 16775 Joined: Member Rating: 2.8 
quote: Then why are you misrepresenting it? quote: The models aren’t meant to predict the results of those experiments.


Straggler Member (Idle past 14 days) Posts: 10328 From: London England Joined: 
Are you still here?
Shouldn’t you be lapping up the adulation of the scientific community and preparing your Nobel acceptance speech? I mean you have single handedly and indisputably falsified one of the most accepted and respected scientific theories of all time. The underpinning concept of all biology. You. Standing on the shoulders of giants. The Newton of biology. The Newton of our time. As much as I enjoy basking in your gloriousness and as honoured as we all are by your benevolent presence oh great one, I have to ask  Why are you posting here rather than revoloutionising the whole of biology with your flawless and paradigmshifting proof that descent from common ancestry is a physical impossibility? Why have you chosen us to spread the word rather than  oh I dunno  The biological research community (for example)?



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