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Author Topic:   Do you really understand the mathematics of evolution?
Kleinman
Member (Idle past 335 days)
Posts: 2142
From: United States
Joined: 10-06-2016


Message 163 of 239 (878130)
06-25-2020 6:26 PM
Reply to: Message 162 by Taq
06-25-2020 6:03 PM


Re: The mathematics of DNA evolution
Kleinman writes:
What you can't grasp is that these recombination events can occur only under very specific circumstances.
Taq writes:
It happens in every individual in every generation in a sexually reproducing species.
You are having difficulty distinguishing between any recombination event and specific recombination events. It is like the difference between any mutation occurring and a beneficial mutation occurring. You really are having trouble getting a grasp of this math.
Kleinman writes:
People who do weed management aren't dumb enough to use their herbicides as you suggest. Your suggestion is all the more reason that naive school children need to learn the correct mathematics of evolution.
Taq writes:
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?
I have, you just don't have the math skills to recognize it. If you want to start a herbicide-resistant weed breeding program, that's your formula. That's how you deal with the multiplication rule in DNA evolution and the multiplication rule in random recombination to speed up the evolutionary process. You don't even need recombination to do it. Just use single-drug therapy for treating infections. When one drug fails, go on to the next drug and the next drug until you have evolved MRSA. You have demonstrated a brilliant understanding of DNA evolution. Drug-resistant infections will be commonplace for many more years with bright minds like yours in action.

This message is a reply to:
 Message 162 by Taq, posted 06-25-2020 6:03 PM Taq has replied

Replies to this message:
 Message 188 by Taq, posted 06-29-2020 1:30 PM Kleinman has replied

  
Kleinman
Member (Idle past 335 days)
Posts: 2142
From: United States
Joined: 10-06-2016


Message 164 of 239 (878146)
06-26-2020 10:03 AM
Reply to: Message 150 by Kleinman
06-25-2020 1:06 PM


Re: The mathematics of DNA evolution
I must say that I appreciate Taq's arguments about recombination. I hope that Taq understands that the model that I am presenting is the simplest example of DNA evolution for a single selection pressure targeting a single genetic locus. Because of this, recombination has no effect at all on this particular evolutionary example. The reason for this is that when selection is acting at only a single locus, there is no way that recombination can combine mutations.
To illustrate this, let's call the first beneficial mutation on the given evolutionary trajectory "A" and the second beneficial mutation on that evolutionary trajectory "B". For the mutation rate of e-9, 3e9 replications will give us on average one member in that population with mutation A at the genetic locus and another member with the mutation B at the same genetic locus. But recombination without error cannot combine those two alleles to put the A and B mutations on the same member. Perhaps in the very extreme case, some type of chimeric recombination event could occur where the part of the allele with mutation A breaks off and recombines with a broken part of the allele with mutation B, but if you believe that is the basis for evolution, my recommendation to you is stay away from Las Vegas.
With that said, let's get back to the explanation for dwise1 on how the multiplication rule applies to DNA evolution. Yesterday I showed you how to calculate the probability that a beneficial mutation A will occur on a member of a population in a single replication. That calculation is done as follows:
P(A) = P(BeneficialA)
is the mutation rate and P(BeneficialA) is the probability of all the possible mutations that can occur at that site, it is the A mutation. You might be wondering, what is the value of P(BeneficialA). If the only mutations we are considering are base substitutions, P(BeneficialA)=1/3, because there are 3 possible substitutions for the original base. This is also the reason why Taq's computation for the number of replication for the beneficial mutation to occur is 3e9 and not just 1/mutation rate.
So the next question is, what is the probability of A occurring at least once in more than one replication? In order to do that math, you could write out the sample space for all the possible outcomes as each replication occurs but that would become very unwieldy after more than a few replications. What we can use is the "at least one rule" from probability theory. If you are not familiar with that rule, here is a very easy lecture which explains this rule:
https://www.youtube.com/watch?v=a86Vj_7R2aQ
There are shorter videos than this 30 min video but this teacher's explanation is very clear and easy to follow.
I'm going to assume that you understand this rule and continue on with the calculation. The next step is to calculate the probability that mutation A will not occurs in a single, that is the complement of P(A) using the complementary rule.
P(Ac)=1‘P(BeneficialA)
Where P(Ac) means the probability of A not occurring. We can then use the multiplication rule to compute the probability of A not occurring with any number of replications. Set "n" to be the number of replication and that gives the probability of A not occurring with n replications as:
P(Ac)=(1‘P(BeneficialA))^n
What this equation gives us is the probability of mutation A not occurring at all in n replications. What we want to know is the probability of A occurring at least once in n replication. We can compute that number by again applying the complement rule:
P(A)=1-P(Ac)=1-((1‘P(BeneficialA))^n)
To put this into the context of the Kishony experiment, that equation gives the probability of the first beneficial mutation "A" occurring as his colony grows in the drug-free region. A plot of that curve for various mutation rates appears as follows:
To use these curves, select a particular value of P(BeneficialA), then as you go along the x-axis, you can obtain the probability of mutation A occurring for that number of replications.
When that mutation A does occur, you now have a variant that is able to grow in the next higher drug concentration region of the Kishony experiment. Tomorrow, I'll go over the math that shows you how to compute the probability of a second beneficial mutation "B" occurring on some member that already has mutation A and how the multiplication rule applies for this evolutionary transition.

This message is a reply to:
 Message 150 by Kleinman, posted 06-25-2020 1:06 PM Kleinman has not replied

  
Kleinman
Member (Idle past 335 days)
Posts: 2142
From: United States
Joined: 10-06-2016


Message 166 of 239 (878181)
06-26-2020 4:30 PM
Reply to: Message 165 by Straggler
06-26-2020 4:13 PM


Re: Does competition accelerate DNA evolution?
Straggler writes:
You are either due a Nobel prize for revolutionising the whole of biology. Or you are a crackpot.
If I had to put money on which of those two is more probable....
Well - You are the probability whiz - You tell me - Which is more likely?
I really like the Isaac Newton quote: If I have seen further than others, it is by standing upon the shoulders of giants. And there are even some biologists and geneticists in that group. God has given me a gift and I hope it benefits people and reduces suffering in this world.

This message is a reply to:
 Message 165 by Straggler, posted 06-26-2020 4:13 PM Straggler has replied

Replies to this message:
 Message 167 by vimesey, posted 06-26-2020 4:36 PM Kleinman has not replied
 Message 173 by Straggler, posted 06-26-2020 7:03 PM Kleinman has replied

  
Kleinman
Member (Idle past 335 days)
Posts: 2142
From: United States
Joined: 10-06-2016


Message 171 of 239 (878189)
06-26-2020 5:05 PM


Just the mention of God here is like turning on the light in the kitchen.

Replies to this message:
 Message 172 by vimesey, posted 06-26-2020 5:17 PM Kleinman has not replied

  
Kleinman
Member (Idle past 335 days)
Posts: 2142
From: United States
Joined: 10-06-2016


Message 174 of 239 (878193)
06-26-2020 7:10 PM
Reply to: Message 173 by Straggler
06-26-2020 7:03 PM


Re: Does competition accelerate DNA evolution?
Straggler writes:
You are the stats whizz. You tell me objecively speaking which is more likely?
Technically, this is a probabilities model, not a statistical model. And can't you tell a fish story when you hear one?

This message is a reply to:
 Message 173 by Straggler, posted 06-26-2020 7:03 PM Straggler has replied

Replies to this message:
 Message 175 by Straggler, posted 06-27-2020 5:48 AM Kleinman has replied

  
Kleinman
Member (Idle past 335 days)
Posts: 2142
From: United States
Joined: 10-06-2016


(2)
Message 177 of 239 (878210)
06-27-2020 8:51 AM
Reply to: Message 175 by Straggler
06-27-2020 5:48 AM


Re: Does competition accelerate DNA evolution?
Straggler writes:
Have it your way.
Objectively speaking which is more probable?
Kleinman writes:
And can't you tell a fish story when you hear one?
Straggler writes:
I know cod science when I see it...
(Ba ba boom)
On the scale of fish humor, I could go lower than that just for the halibut. But I won't do that because I don't want you to flounder on the mathematics of DNA evolution.

This message is a reply to:
 Message 175 by Straggler, posted 06-27-2020 5:48 AM Straggler has not replied

  
Kleinman
Member (Idle past 335 days)
Posts: 2142
From: United States
Joined: 10-06-2016


Message 178 of 239 (878219)
06-27-2020 12:03 PM


The mathematics of DNA evolution
OK, back to the explanation of the mathematics of DNA evolution. In Message 164, I give the probability equation for some member of the founding colony in the Kishony experiment to get the first beneficial mutation A. That equation again is:
P(A)=1-P(Ac)=1-((1‘P(BeneficialA))^n)
It should be seen when the number of replications in that colony, n, is large enough, mutation A will occur. And for a (beneficial) mutation rate of e-9, that will be on average 1 occurrence for every e9 replications.
When that mutation occurs and that variant with mutation A migrates into the next higher drug-concentration region, it will start a new colony in which members of that colony need another beneficial mutation (call it B) in order to grow in the next higher drug-concentration region of that environment. That mutation B could occur even if it doesn't migrate into the next higher drug-concentration region but later, I will explain why this is much less probable. So, we can calculate the probability of mutation B occurring on some member of this new colony in the same manner as we computed the probability of mutation A occurring in the original founder's colony. We start with the probability of mutation B occurring in a single replication.
P(B) = P(BeneficialB)
Compute the complement:
P(Bc)=1‘P(BeneficialB)
Compute the probability that mutation be won't happen in nA replications where nA is the number of replications of that new colony formed by the A mutant (variant):
P(Bc)=(1‘P(BeneficialB))^nA
Take the complement of that equation to determine the probability of that at least 1 B mutation will occur in nA replications:
P(B)=1-P(Bc)=1-((1‘P(BeneficialB))^nA)
The plots for this equation will be identical to the plots for the P(A) equation. You can go back to Message 164 if you want to review this.
To compute the joint probability of mutation B occurring on some member that already has mutation A can be done using the multiplication rule for joint independent events for this case. That joint probability is written as follows:
P(A)P(B)=(1-((1‘P(BeneficialA))^n))*(1-((1‘P(BeneficialB))^nA))
This same recursion calculation can be done for mutation C for the next higher drug-concentration region, mutation D... and you will obtain an equation that looks like this:
P(A)P(B)P(C)P(D)...=a series of at least 1 probability equations
That's how DNA evolution works for a single selection pressure targeting a single gene where each evolutionary transition requires only a single beneficial mutation to improve fitness. That's how the Kishony experiment works. But the Kishony experiment won't work if the step increase in drug-concentration is too large. That is if two or more mutations are required to increase fitness to that next higher drug concentration. I'll go over that mathematics next week and I will also give the mathematics which explains what will happen if Kishony tries to use two or more drugs in his experiment, even at low concentration increases. After that, I will explain why the Jukes-Cantor/Felsenstein models of DNA evolution are incorrect.
Is this making sense to you dwise1?

  
Kleinman
Member (Idle past 335 days)
Posts: 2142
From: United States
Joined: 10-06-2016


Message 180 of 239 (878314)
06-28-2020 10:12 PM
Reply to: Message 179 by PaulK
06-28-2020 4:46 AM


Re: An important note regarding models of evolution
PaulK writes:
From the Wikipedia article Models of DNA Evolution
The models described on this page describe the evolution of a single site within a set of sequences. They are often used for analyzing the evolution of an entire locus by making the simplifying assumption that different sites evolve independently and are identically distributed. This assumption may be justifiable if the sites can be assumed to be evolving neutrally. If the primary effect of natural selection on the evolution of the sequences is to constrain some sites, then models of among-site rate-heterogeneity can be used.
Let us note that there will be selection constraining some sites in a Kishony experiment.
I'm glad to see that you are starting to think about this. That's part of the problem with these DNA evolution models. But by far, the biggest problem is this assumption:
Wikipedia writes:
For a stationary process, where Q does not depend on time t, this differential equation can be solved.
What that means is that as time goes on the Markov chain goes to equilibrium. That is the frequency of each base at the site considered go to 0.25 and no longer changes. Real evolutionary processes don't do anything like that.
Think about what happens in a competitive environment where fixation occurs, the frequency of the given base at that site will go to 1. Jukes-Cantor and all the derivative models are incorrect for describing real evolutionary processes because these processes don't even get near to approaching equilibrium. The Kishony experiment starts with a founder bacterium in the drug-free region. When that variant does 3e9 replications, if the initial base at that site of the founder was a T, on average 3e9 - 3 members of the population will still have a T at that site, 1 member will have an A base at that site, another member will have a C base at that site, and the third member will have a G at that site. The initial E vector at time 0 will be almost exactly the same as the E vector after 3e9 replications. DNA evolution is a non-stationary Markov process.
Tomorrow I will go over the "at least one method" of DNA evolution for two sites in the genome and I'll show you how to draw the state transition diagram for a second-order Markov chain for DNA and what happens if you make the same assumption as for the first-order Markov chain. And you can try and find a blunder in this math.

This message is a reply to:
 Message 179 by PaulK, posted 06-28-2020 4:46 AM PaulK has replied

Replies to this message:
 Message 181 by PaulK, posted 06-29-2020 12:19 AM Kleinman has replied

  
Kleinman
Member (Idle past 335 days)
Posts: 2142
From: United States
Joined: 10-06-2016


Message 182 of 239 (878337)
06-29-2020 6:08 AM
Reply to: Message 181 by PaulK
06-29-2020 12:19 AM


Re: An important note regarding models of evolution
Kleinman writes:
I'm glad to see that you are starting to think about this.
PaulK writes:
I don’t think that is going to last.
So you are going to stop thinking? Why? Are you afraid you might actually learn how DNA evolution works?
Kleinman writes:
That's part of the problem with these DNA evolution models.
PaulK writes:
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.
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.
Kleinman writes:
What that means is that as time goes on the Markov chain goes to equilibrium. That is the frequency of each base at the site considered go to 0.25 and no longer changes. Real evolutionary processes don't do anything like that.
PaulK writes:
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.
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.

This message is a reply to:
 Message 181 by PaulK, posted 06-29-2020 12:19 AM PaulK has replied

Replies to this message:
 Message 183 by PaulK, posted 06-29-2020 6:49 AM Kleinman has replied

  
Kleinman
Member (Idle past 335 days)
Posts: 2142
From: United States
Joined: 10-06-2016


Message 184 of 239 (878348)
06-29-2020 11:26 AM
Reply to: Message 183 by PaulK
06-29-2020 6:49 AM


Re: An important note regarding models of evolution
Kleinman writes:
So you are going to stop thinking? Why? Are you afraid you might actually learn how DNA evolution works?
PaulK writes:
No, I think you’ll be unhappy because I am thinking and because I do have a good idea of how DNA evolution works.
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.
Kleinman writes:
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.
PaulK writes:
Many bases will be varying neutrally. So a model of neutral evolution will give a reasonable estimate.
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 Jukes-Cantor 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.
Kleinman writes:
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?
PaulK writes:
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.
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 drug-free region that makes it a better replicator for the drug-free 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 drug-sensitive variants.
Kleinman writes:
So, why do you think you can use a model of neutral evolution to estimate divergence times between species?
PaulK writes:
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.
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.
Kleinman writes:
In fact, in your world, not even fixation can occur whether due to drift or selection.
PaulK writes:
And there is another bizarre assertion. The equilibrium would be roughly equal proportions of each base in a single genome.
PaulK, you are really confused here. The Jukes-Cantor/Felsenstein Markov models are only considering a single site in the genome. How did you get so confused on this subject?

This message is a reply to:
 Message 183 by PaulK, posted 06-29-2020 6:49 AM PaulK has replied

Replies to this message:
 Message 185 by PaulK, posted 06-29-2020 11:39 AM Kleinman has replied

  
Kleinman
Member (Idle past 335 days)
Posts: 2142
From: United States
Joined: 10-06-2016


Message 186 of 239 (878356)
06-29-2020 12:18 PM
Reply to: Message 185 by PaulK
06-29-2020 11:39 AM


Re: An important note regarding models of evolution
Kleinman writes:
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?
PaulK writes:
A small portion, certainly not, you’d want a number of genes to get a robust result,
But that's exactly the incorrect approach they are doing with these Markov models. The Jukes-Cantor 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 two-site 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 single-site model the transition diagram is:
And for the second-order 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/...02-03-Probabilities-Markov-HMM-PDF.pdf
Try writing out the transition equations for this case. If you can't, I will show you how to write those equations.
Kleinman writes:
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
PaulK writes:
Oh dear, oh dear, oh dear. This is not at all a rational argument. In fact it is a complete non-sequitur. There are bases which can freely change without deleterious or beneficial effect.
I know, and they show how closely related to a banana you are.
Kleinman writes:
PaulK, you are really confused here. The Jukes-Cantor/Felsenstein Markov models are only considering a single site in the genome. How did you get so confused on this subject?
PaulK writes:
The distribution of the bases in an interbreeding population wouldn’t be random. So how could you apply the probabilities across individuals?
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.
PaulK writes:
Anyway, until you recognise that DNA evolution is driven by neutral drift, you do not understand the mathematics of DNA evolution.
I know, you have correctly shown how closely related to a banana you really are.

This message is a reply to:
 Message 185 by PaulK, posted 06-29-2020 11:39 AM PaulK has replied

Replies to this message:
 Message 187 by PaulK, posted 06-29-2020 1:05 PM Kleinman has replied

  
Kleinman
Member (Idle past 335 days)
Posts: 2142
From: United States
Joined: 10-06-2016


Message 189 of 239 (878367)
06-29-2020 1:53 PM


The mathematics of DNA evolution 2 sites simultaneously
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 drug-concentration 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 e-9, e9 replications will give us 1 variant with the A1 mutation and another variant with the A2 mutation in the initial founder's wild-type 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 e-9, 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 A1-A2 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)
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))^n)
Now, we need to compute the probability that at least a single member of that sub-population 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))^nA1)
Then, the joint conditional probability is:
P(A1)P(A2)=(1-((1‘P(BeneficialA1))^n))*(1-((1‘P(BeneficialA2))^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 e-9, 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 drug-concentration is too large. The carrying capacity of his mega-plate 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 335 days)
Posts: 2142
From: United States
Joined: 10-06-2016


Message 190 of 239 (878368)
06-29-2020 2:08 PM
Reply to: Message 187 by PaulK
06-29-2020 1:05 PM


Re: An important note regarding models of evolution
Kleinman writes:
But that's exactly the incorrect approach they are doing with these Markov models. The Jukes-Cantor and derivative models only consider a single site in the genome.
PaulK writes:
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,
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?
Kleinman writes:
I know, and they show how closely related to a banana you are.
PaulK writes:
You’re only showing that you’re bananas.
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.
Kleinman writes:
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.
PaulK writes:
Apparently I know how DNA evolution works better than you. You do realise that the DNA sequences found in the population are not independent?
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.

This message is a reply to:
 Message 187 by PaulK, posted 06-29-2020 1:05 PM PaulK has replied

Replies to this message:
 Message 192 by PaulK, posted 06-29-2020 3:02 PM Kleinman has replied

  
Kleinman
Member (Idle past 335 days)
Posts: 2142
From: United States
Joined: 10-06-2016


Message 191 of 239 (878370)
06-29-2020 2:20 PM
Reply to: Message 188 by Taq
06-29-2020 1:30 PM


Re: The mathematics of DNA evolution
Kleinman writes:
I have, you just don't have the math skills to recognize it.
Taq writes:
You're a liar. You haven't answered the question.
Taq, we know you are a mathematically incompetent nitwit but do you also have to be a stupid ass as well?
Taq writes:
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?
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.

This message is a reply to:
 Message 188 by Taq, posted 06-29-2020 1:30 PM Taq has replied

Replies to this message:
 Message 201 by Taq, posted 06-30-2020 1:07 PM Kleinman has replied

  
Kleinman
Member (Idle past 335 days)
Posts: 2142
From: United States
Joined: 10-06-2016


Message 193 of 239 (878401)
06-29-2020 4:54 PM
Reply to: Message 192 by PaulK
06-29-2020 3:02 PM


Re: An important note regarding models of evolution
Kleinman writes:
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?
PaulK writes:
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.
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.
Kleinman writes:
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.
PaulK writes:
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.
You are blowing smoke and you know it. Figure out the correct transition matrix for DNA evolution. I have given you plenty of hints.
Kleinman writes:
But mutations are random and that's why DNA evolution works the way I'm showing you
PaulK writes:
Well you’re wrong because it mostly doesn’t.
So, show us the math how it DNA evolution mostly works.
Kleinman writes:
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.
PaulK writes:
Since you don’t know how DNA evolution works, your arguments are worthless.
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 first-order ordinary differential equation by changing t by t. 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 and the are naming that variable .
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.

This message is a reply to:
 Message 192 by PaulK, posted 06-29-2020 3:02 PM PaulK has replied

Replies to this message:
 Message 194 by PaulK, posted 06-29-2020 5:40 PM Kleinman has replied
 Message 195 by Straggler, posted 06-29-2020 6:19 PM Kleinman has replied

  
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