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Author | Topic: Rebuttal To Creationists - "Since We Can't Directly Observe Evolution..." | |||||||||||||||||||||||
Kleinman Member (Idle past 335 days) Posts: 2142 From: United States Joined: |
Kleinman:ringo, you are making the same logical inconsistency that Taq makes in his attempt to model human evolutionary fitness improvement. Adaptive mutations are particular mutations, not any mutation. The correct analogy for your raindrop concept would be, what is the probability of two particular raindrops ending up in the same body of water? You might try and argue that raindrops are indistinguishable but mutations are distinguishable. And you must model DNA evolution and the accumulation of adaptive mutations in a lineage using the multiplication rule.
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Kleinman Member (Idle past 335 days) Posts: 2142 From: United States Joined: |
Further response to Percy's Message 175
Kleinman:OK. I'll use the The basic science and mathematics of random mutation and natural selection reference since that is the formulation I used to compute the probabilities in the Lenski paper, Fixation and Adaptation in the Lenski E. coli Long Term Evolution Experiment The Haldane frequency equation gives us a way to compute the subpopulation sizes of the more and less fit variants. We know that the less fit variants will ultimately be driven to extinction so only members of the subpopulation of the more fit variant are candidates for a beneficial mutation "A" that must occur at some site in its genome. We start with the probability that a mutation will occur at that site in a single replication. (Note that this math applies to every site in the genome, not just the site(s) that is/are candidates for adaptive mutations. That is why this is an exhaustive search for every possible mutation). That probability is the mutation rate, call it "mu". But we also need to consider that the mutation that occurs may not be a beneficial mutation. It is possible that the wrong base substitution occurs. In other words, the mutation itself is a random trial with multiple possible outcomes. We can write the set of possible outcomes as follows:P(Ad) + P(Cy) + P(Gu) + P(Th) + P(iAd) + P(iCy) + P(iGu)+ P(iTh) + P(del) + ... = 1 where the first four terms are possible base substitutions, the next four terms represent insertions of bases, the ninth term is the probability of a deletion and the ellipsis represents any other form of mutation you can imagine. Let P(BeneficialA) represent the probability of the mutation that gives improved reproductive fitness. Note that P(BeneficialA) is some number between 0 and 1 and for most cases will have a value of about 1/3 to 1/4. Then, the probability of mutation A occurring at the particular site in a single replication is written: P(A) = P(BeneficialA)*mu (1) One could think of P(BeneficialA)*mu as the "beneficial mutation rate", a probability value slightly lower than the mutation rate "mu". It should be clear that the probability of the A mutation occurring in a single replication is very low. The next step is to compute the probability of that mutation A occurring at least once in "n" replications of the more fit variant. This is done using the "at least one rule". It is a very simple rule to apply and understand. I do a step-by-step derivation in this paper The basic science and mathematics of random mutation and natural selectionfor this case. If there are any questions on how to derive or apply this rule I'll try and answer them. When this rule is applied to equation (1) you get the following probability equation: P(A) = 1 − (1 − P(BeneficialA)*mu)^n (2) Equation (2) is evaluated using the population size generated by the Haldane frequency/fixation equation. I plotted the results for several different fitness parameters in the Lenski fixation/adaptation paper if you are interested. Solving Haldane's frequency/fixation and equation (2) gives the correct mathematical description of the Lenski experiment. It demonstrates mathematically why competition slows the adaptation process. Competition slows the accumulations of replications the more fit variant can do by limiting the energy available to that variant. Note that the more fit variant accumulated replications (the random trial for the next adaptive mutation) most rapidly after the variant fixes in the population. If Lenski used a larger volume in his experiment (increases the carrying capacity), it might not be necessary for that variant to fix in order to achieve the necessary number of replications to give a reasonable probability of the next adaptive mutation to occur on some member of that subset. If the carrying capacity is much larger such as in the Kishony experiment, you can have multiple different lineages taking different evolutionary trajectories to adaptation in the same environment. The math for each of these lineages is the same. If there aren't any questions or comments on this math or how to apply this math to the Lenski experiment, I'll go onto more of Percy's comments, in particular, how to do the mathematics of adaptation using a Markov process random walk model.
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Kleinman Member (Idle past 335 days) Posts: 2142 From: United States Joined: |
Kleinman:Are some of the variants more fit than other variants and are they engaged in biological evolutionary competition? Kleinman:You aren't answering my question about whether all 100,000 individuals are on the same evolutionary trajectory. But, let's get some more detail on your statement that mutations spread through a sexually reproducing population. Is it possible that a beneficial mutation is lost in a sexually reproducing population? What happens if the beneficial allele is heterozygous rather than homozygous? Kleinman:What is the probability of beneficial alleles in a diverse population recombining in the same descendant? Kleinman:Tell us how you compute the probability that a descendant will get beneficial alleles by recombination for the conditions of your model. Kleinman:How do 20 million possible beneficial mutations end up in the lineages of all humans? Recombination? Do all 20 million possible beneficial mutations give an equal improvement in fitness to your different lineages, or is there biological evolutionary competition that causes the loss of some of the less fit variants? Kleinman:Try reading beyond the abstract: quote: Kleinman:OK, Haldane's frequency equation for a sex-linked diploid is: pn^2AA + 2pnqnAa + qn^2aa = 1 Which of the variants fix in the population, AA, Aa, or aa? Kleinman:Taq, you are still confused on this point. Try thinking of it this way. It's the number of replications of a particular allele that determines the probability of an adaptive mutation occurring at some site in that allele. If that allele is homozygous in that diploid then you will get two replications of that allele with every creature replication. If the allele is heterozygous at that site, you will have only one replication of each particular allele.
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Kleinman Member (Idle past 335 days) Posts: 2142 From: United States Joined: |
Kleinman:I get it now! Sexual reproduction works by gravity! I'm impressed by your understanding of the laws of physics. Kleinman:Tell us how you think adaptive alleles are evolved in sexually reproducing populations.
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Kleinman Member (Idle past 335 days) Posts: 2142 From: United States Joined: |
Kleinman:So, in your model of human evolution, the fixation of beneficial alleles does not occur? Kleinman:In your model, does every human lineage have equal reproductive fitness? Do any human lineages go extinct? Kleinman:I don't think you are doing your math correctly on that one. But WRT your model, do all the fathers in your model have beneficial allele A and all the mothers have beneficial alleles B? Kleinman:How do you compute the probability that all these beneficial mutations end up in a single lineage in your model? Kleinman:Try reading this again: quote:How do you compute the rapid fixation of 20 million beneficial mutations in your model? Kleinman:You have yet to show how you do a fixation calculation for your model for a single genetic loci let alone 20 million genetic loci. Are you ever going to get beyond your simple-minded neutral evolutionary model where you assume that 20 million mutations are beneficial? Kleinman:Where did the resistance allele come from that the phage transmits? What if the phage transmitted some other allele besides a resistance allele? Kleinman:Just as a side note to this comment, Kishony's experiment doesn't work when using two drugs simultaneously. Do you understand why? And to your comment here, you have assumed that the two subsets of your sexually reproducing population each have fixed their respective resistance alleles so they exist at frequencies of 1 in their respective populations. If we assume both subsets have equal population sizes when you combine the frequencies of the resistance alleles for each drug will be 0.5. What happens if the resistance alleles are not at high frequency in the population? What is the probability of a recombination event for the beneficial alleles if their frequencies are f1, f2, and where the remaining members of the population have frequency f3, i.e., resistance to neither drug? And BTW, combination therapy works for the treatment of malaria which can do sexual reproduction. And the malaria population size for someone with hyperparasithemia can reach 1 trillion.
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Kleinman Member (Idle past 335 days) Posts: 2142 From: United States Joined: |
Kleinman:The math is way beyond you. 20 million beneficial mutations * 300 generations/fixation = 6 billion generations Kleinman:How much variation? And how many different lineages in your population? Kleinman:So you are claiming that no human lineages have ever gone extinct? What happens to the less fit human lineages when the most fit are fixed in the population? Kleinman:That's still not quite right. What if both the mother and father are homozygous for gene A or homozygous for gene B? Kleinman:Percy likes it when you post from your link. Post the equation you think applies. Kleinman:Fixation isn't rapid and Haldane's mathematical estimate of 300 generations/fixation has been verified experimentally. And if you are concerned that Lenski's experiment uses asexual replicators, don't worry, Haldane's math includes that for diploid sexual replicators. 20 million beneficial mutations * 300 generations/fixation. How many generations are in your model? Kleinman:Plug in a selection coefficient and tell us how many generations to fixation. Then you only have 19,999,999 more fixations to go. Kleinman:So the resistance allele has to evolve in the bacteria and the phage acquires the gene and transmits it to drug-sensitive bacteria. Are you claiming this is the mechanism that gives humans a reproductive advantage over chimps? Kleinman:It appears the selection coefficient is extremely high, all the drug-sensitive variants are killed off in a single generation. Is that how your model for human evolution works? Kleinman:That's the problem, physicians have been taught for years an incorrect way of using antimicrobial agents and it has resulted in multidrug-resistant microbes. Doctors have been taught to use antimicrobial agents as single drug therapy. When one drug fails, go onto the next, and the next, and the next,... Microbiologists need to do a better job teaching physicians how drug resistance evolves. Do you know any microbiologists that know how drug resistance evolves? Kleinman:The probability of that happening depends on the frequency of the different resistance alleles in the population. I'm still waiting for you to figure out that math. I'll even give you a couple of hints. It is the same math as a random card drawing problem except you only have 3 different kinds of cards in the deck. Call one card "A" for the first resistance allele, a second card "B" for the second resistance allele, and the third card "C" for members of the population that have neither the "A" nor "B" resistance alleles. Assume all the members of the population are homozygous at the respective genetic loci to make the math a bit easier and you have nA members in the population, nB members in the population, and nC members so that nA + nB + nC = n, the total population size. What's the probability of drawing an A and B parents to give an AB offspring?
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Kleinman Member (Idle past 335 days) Posts: 2142 From: United States Joined: |
Kleinman:That is hilarious. Kleinman:Taq had no parents, he is a product of a population. Kleinman:All that intermingling of alleles and mutations and still they fix at a rate of greater than 50/generation. The Mexican Salamander has a genome length of 32 billion base pairs. Does that replicator fix 500 mutations/generation? Kleinman:That's fitting, the less fit aren't fit for life anymore. Kleinman:How did all the fathers end up AAbb and all the mothers aaBB? Kleinman:Do you think that Haldane was wrong when he wrote this: Haldane:Haldane posts data from sweet peas in his paper. Sweet peas do sexual replication. And do you have any experimental data that shows that fixation can occur more rapidly in sexual replicators than in asexual replicators? Kleinman:Taq, I've taken multiple biology courses and even 2 years of microbiology. I know how meiosis works. So show us how you use the equations you posted to compute the number of generations to fixation for a single allele. Kleinman:Why should it surprise you when a scientist puts an agent into a population that transfers a resistance allele? This is simply a breeding program with a phage with a known allele and a population of bacteria. Are you claiming that the human reproductive advantage came about due to breeding? Kleinman:Tell us how the Kishony experiment works. And show your math. Kleinman:You still haven't figured out the difference between DNA evolution, biological evolutionary competition, and recombination. Don't worry, I'll be patient with you till you get it.
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Kleinman Member (Idle past 335 days) Posts: 2142 From: United States Joined: |
Further response to Percy's Message 175
Percy, I tried to write the matrix equations using Latex. They appear correctly in the Latex editor I used but don't appear correctly here. Picking up the discussion on computing the probability of a particular mutation occurring using a Markov Process
Kleinman:OK. First, for those readers that don't know what a Markov Chain is, here's a definition from Wikipedia: Markov chain - Wikipedia quote:The first application of a Markov Chain model to DNA evolution was done by Jukes and Cantor and is called the Jukes-Cantor model. The way one does such a calculation is first to draw a state transition diagram of the system of interest. The state transition diagram for a single site in a genome looks as follows: Then, a transition matrix can be written to describe the evolutionary change in time t is = (pij) where the pij gives the probabilities of change from the state Ei to Ej at time t + Δt where Δt is a replication. If we neglect insertions, deletions, transpositions, and other types of mutations (that is substitutions only), the transition matrix would look as follows: P (t)=( pAA pAC pAG pATpCA pCC pCG pCT pGA pGC pGG pGT pTA pTC pTG pTT ) \left[ {\begin{array} p_{AA} & p_{AC} & p_{AG} & p_{AT}\\ p_{CA} & p_{CC} & p_{CG} & p_{CT}\\ p_{GA} & p_{GC} & p_{GG} & p_{GT}\\ p_{TA} & p_{TC} & p_{TG} & p_{TT}\\ \end{array} } \right \][/latex]--> If one assumes that the mutation rates are constant and have the same value for DNA transitions and transversion, we obtain the Jukes-Cantor model. P (t)=[1−μ μ/3 μ/3 μ/3μ/3 1−μ μ/3 μ/3 μ/3 μ/3 1−μ μ/3 μ/3 μ/3 μ/3 1−μ] \left[ {\begin{array}{cccc} 1−μ & μ/3 & μ/3 & μ/3\\ μ/3 & 1−μ & μ/3 & μ/3\\ μ/3 & μ/3 & 1−μ & μ/3\\ μ/3 & μ/3 & μ/3 & 1−μ\\ \end{array} } \right] [/latex]--> The Jukes-Cantor model implicitly assumes a population of one. If one wants to compute the frequency distribution of different variants as a population grows "N", it is done as follows: P (t)=[1−μ/N μ/(3∗N) μ/(3∗N) μ/(3∗N)μ/(3∗N) 1−μ/N μ/(3∗N) μ/(3∗N) μ/(3∗N) μ/(3∗N) 1−μ/N μ/(3∗N) μ/(3∗N) μ/(3∗N) μ/(3∗N) 1−μ/N)] \left[ {\begin{array}{cccc} 1−μ/N & μ/(3∗N) & μ/(3∗N) & μ/(3∗N)\\ μ/(3∗N) & 1−μ/N & μ/(3∗N) & μ/(3∗N)\\ μ/(3∗N) & μ/(3∗N) & 1−μ/N & μ/(3∗N)\\ μ/(3∗N) & μ/(3∗N) & μ/(3∗N) & 1−μ/N\\ \end{array} } \right] [/latex]--> The initial state of the system is written: E0 = (A0, C0, G0, T0) and the state of the system a time ti is: Ei = (Ai, Ci, Gi, Ti) and the state of the system going from state Ei to state Ei+1 is computed by simple matrix multiplication. Ei+1 = Ei [P] For the Jukes-Cantor model one obtains the equations: Ai+1 = Ai(1-μ) + Ci*μ/3 + Gi*μ/3 + Ti*μ/3Ci+1 = Ai*μ/3 + Ci(1-μ) + Gi*μ/3 + Ti*μ/3 Gi+1 = Ai*μ/3 + Ci*μ/3 + Gi(1-μ) + Ti*μ/3 Ti+1 = Ai*μ/3 + Ci*μ/3 + Gi*μ/3 + Ti(1-μ) And for the variable population model: Ai+1 = Ai(1-μ/N) + Ci*μ/(3*N)+ Gi*μ/(3*N) + Ti*μ/(3*N)Ci+1 = Ai*μ/(3*N) + Ci(1-μ/N) + Gi* μ/(3*N)+ Ti*μ/(3*N) Gi+1 = Ai*μ/(3*N) + Ci*μ/(3*N) + Gi(1-μ/N) + Ti*μ/(3*N) Ti+1 = Ai*μ/(3*N) + Ci*μ/(3*N) + Gi*μ/(3*N) + Ti(1-μ/N) Note that A,C,G, and T are frequencies of the particular variants with that particular base at that site. If you multiply any of these frequencies by N, you will get the number of members in the population with that given base at that site. Then, assume that in the initial condition that the base at that site is T but the beneficial mutation is A. The initial condition is written: E0 = (A0, C0, G0, T0) = (0,0,0,1) And do lots of matrix multiplications. For the Jukes-Cantor model and mutation rate 1e-9, one obtains the following frequency curves:
For the variable population transition model and mutation rate 1e-9, one obtains the following frequency curves:
And for comparison, the probability curves for the "at least one" calculation for a beneficial mutation to occur as a function of population size:
Comparing these 3 graphs shows that the Jukes-Cantor model reaches equilibrium at about 3e9 replications. The variable population size model gets an expected number of A variants equal to 1 at about 1.5e8 and the "at least one" calculation for mutation rate 1e-9 gives a rapidly rising probability at about 1.5e8.
Percy:That's right, but ultimately, I think resources are just individual components of the environment that make the energy in the environment available to the replicator. For example, plants need water to convert the energy from the sun to produce sugar. That's why drought and dehydration is a selection pressure. So tell me Percy, is the physics and math that I've presented get us into the ballpark on how biological evolution works?
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Kleinman Member (Idle past 335 days) Posts: 2142 From: United States Joined: |
Taq:It appears that in your microbiology training, none of your instructors taught you what a mutation rate is. If you are able, prepare yourself to be instructed. Mutation rate - Wikipedia quote:And that rate estimated for humans is: quote:The mutation rate is not the ridiculous claim of 50 mutations per generation that you made in your post. You need to put more effort into understanding the equations you use and how you define the variables in these equations. Now put the correct values in your equation and tell us how many generations to fixation for each mutation and then understand why 20,000,000 adaptive mutations are not going to fix in your model.
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Kleinman Member (Idle past 335 days) Posts: 2142 From: United States Joined: |
Taq:The correct value to use in your equation is 1.1x10-8, not 50. Your value is off by about 9 orders of magnitude. There are many reasons why the number you are trying to use is wrong but one of the biggest is that nowhere in your equation is the genome length a variable. That's why I brought up the Mexican Salamander example because its genome is 10x larger than the human genome. That doesn't make the number of fixations 10x larger, 500 in every generation! For neutral evolution, it will take 1/(1.1x10-8) or about 90 million generations for each neutral fixation. Haldane's fixation rate of 1 fixation for every 300 generations under selection is much more generous. How you could imagine that every member of a population ends up with the same 50 neutral (or any) mutations every generation is a mystery.
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Kleinman Member (Idle past 335 days) Posts: 2142 From: United States Joined: |
Kleinman:And 54.5 is not the correct value to use in your equation. Just because the number of mutations that occur in a replication is 54.5 doesn't mean that all 54.5 are fixed. The correct value to use in your equation is 1.1x10-8.
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Kleinman Member (Idle past 335 days) Posts: 2142 From: United States Joined: |
quote:You should watch this video, the derivation of your calculation begins at about the 6:00 minute mark. https://www.youtube.com/watch?v=l2Y8oC6G1us&ab_channel=Kr... You are using a definition of mutation rate based on the entire size of the genome. 2N is the total number of alleles at a given locus and 1/2N is the initial frequency of the first mutation in that allele. The neutral mutation rate being used is just for that genetic locus. If that locus has only a single base, then the neutral mutation rate will be 1.1x10-8. If that genetic locus has 1000 bases, the neutral mutation rate will be about 1.1x10-5 (actually lower if you compute the probability of a mutation occurring at least one site when multiple possible sites are considered). The number of generations to fixation for a single neutral mutation case is about 90,000 generations. Edited by Kleinman, : Correct math
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Kleinman Member (Idle past 335 days) Posts: 2142 From: United States Joined: |
Kleinman:You are making another mathematical blunder here. 1/2N is the initial frequency of the mutant allele. Only a tiny fraction of the genome has mutations. What is the initial frequency for your calculation? It certainly isn't 1/2N. This model only makes sense when considering a single genetic locus because the entire length of the genome and the total number of genetic loci in that genome does not affect the calculation. Not all the mutations in an entire genome are fixed. In fact, some mutations are lost over generations. Perhaps you think that the entire genome is fixed? You should watch the entire video. It gives a very sensible explanation of the equation you are trying to use. You cannot use the entire genome length to compute the mutation rate and do this calculation correctly. It must be done based on the mutations/locus. Mutation rate - Wikipedia quote:I put the boldfacing on the correct definition for mutation rate to be used for this calculation. 90,000 generations/fixation, Haldane's estimate of only 300 generations/fixation but that's with selection. So, when are you going to learn how to do the mathematics of adaptive DNA evolution and give a correct description of the Kishony and Lenski experiments? Don't you think a microbiologist should know how to do that math?
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Kleinman Member (Idle past 335 days) Posts: 2142 From: United States Joined: |
Kleinman:Why are you doing this to yourself? 1/2N is the frequency of the mutant allele. For a diploid population, there are 2N copies of the allele at the particular genetic locus. One of those 2N copies is the first mutant allele. Only 1 member of that population has that mutant allele. The fraction of the population with a mutant allele initially is 1/N. Kleinman:It applies to any mutant allele but only for a single genetic locus. Kleinman:You don't understand this equation. You have even confused the number of copies of alleles with population size. There are 2N copies of an allele in a diploid population size N. And if you somehow want to extrapolate this model to the entire genome means that the entire genome is being fixed. ringo, if you are reading this post, this is GIGO. Kleinman:If you want to do the math for a mutation not in a coding portion of the genome, then the number of bases in that sequence is one and the mutation rate you need to use is 1.1x10-8. That's about 9,000,000 generations to fixation. Post a link to a paper or biology lecture where they do a neutral mutation fixation calculation the way you want to do it. You are just blowing smoke. Nobody does the calculation the way you want because it is nonsense. Kleinman:Fixations aren't adaptation you should understand this by now from the results of the Lenski experiment. Large numbers of replications are what is required for DNA adaptive evolution. Certainly, small populations can have mutations fix more rapidly than large populations. You get all the mutations your parents have plus a few new ones for your own. Populations do exhaustive searches of all possible mutations in order to get just one member with an adaptive mutation. That's why it takes a billion replications for each adaptive mutation in the Kishony and Lenski experiments for a mutation rate of 1e-9. The reason is that "at least one" calculation applies to every site in the genome. The reason why humans have much larger populations than chimps is that humans can do farming on an industrial scale. It is clear that humans had this capability 10,000 years ago. They understood how to irrigate and use animals for labor. You have a population of about 1 billion population with 2 billion chromosome sets replications and use your mutation rate of 1.1x10-8. That doesn't give you many genome replications to work with for adaptive evolution to operate, even if you want to include recombination. Taq:Taq, the math gets orders of magnitude worse if it takes 2 or more mutations to give an improvement in fitness. It introduces another instance of the multiplication rule for each selection condition the population must adapt to. That's really bad for your belief in universal common descent but really good for the fields of medicine and agriculture. It gives a successful treatment of HIV and inhibition of the evolution of herbicide-resistant weeds and pesticide-resistant insects. I don't know if you are ready for that math yet but if you or any other readers of this thread are interested, you can find that paper here: The mathematics of random mutation and natural selection for multiple simultaneous selection pressures and the evolution of antimicrobial drug resistance Don't worry Percy, I'll post equations and quotes in my next few posts if Taq is finished with his neutral fixation model.
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Kleinman Member (Idle past 335 days) Posts: 2142 From: United States Joined: |
Kleinman:You are having trouble doing undergraduate lower division work so I don't know whether I should give you this paper but why not? THE AVERAGE NUMBER OF GENERATIONS UNTIL FIXATION OF A MUTANT GENE IN A FINITE POPULATION Kimura carries out the computation of the fixation of a mutant gene. His calculation doesn't depend on the mutation rate. quote:Now the effective population size can be slightly smaller than the actual population size under certain circumstances that I'm sure you know what they are. Remind me again what the population size you use is. Wasn't it 100,000? That gives the generations to fixation of that mutant gene of 4*100,000=400,000 generations. That really helps. Your lower division equation gave an estimate of 900,000 generations for the fixation of a neutral mutation. How many generations since the divergence of humans and chimps from the common ancestor? Kleinman:You still haven't figured out that different combinations of adaptive mutations give different lineages on different evolutionary trajectories. That math is way over your head. You should stick to trying to figure out an undergraduate lower division equation of neutral fixation.
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