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Author | Topic: Rebuttal To Creationists - "Since We Can't Directly Observe Evolution..." | |||||||||||||||||||||||||||||||||||||||
Taq Member Posts: 10073 Joined: Member Rating: 5.2 |
Kleinman writes: You can have multiple evolutionary trajectories occurring in single evolutionary process. But the evolutionary process in one evolutionary trajectory has no mathematical effect on the other trajectories except biological competition. You are still getting the basic maths wrong. You are claiming that the chances of getting an adaptive phenotype is the mutation rate. That is wrong because there can be multiple mutations that are adaptive. It isn't that hard to figure out. If there are 10 possible mutations that confer antibiotic resistance then the rate of adaptive mutations would be 10*mutation rate.
Sexual reproduction doesn't eliminate competition, Desai's experiment demonstrates this by increasing the frequencies of a few beneficial alleles while eliminating some of the less fit alleles. "Desai let his sexual replicators carry out 90 generations of amplification of the beneficial alleles but stopped the competition process at that point by inducing sexual reproduction before fixation occurred."--Kleinman You claimed that sexual reproduction stops competition.
Don't let me stop you, survey of physics expert. Go ahead, compute the entropy change for polymerization of DNA and see what that gets you. Perhaps you can explain how drug resistance evolves and why cancer treatments fail. I doubt it. Thank you for admitting that you can't do the entropy calculations.
Your terminology is as bad as your math and physics. Fixation occurs when an allele is 100% in frequency in the popultation. Moving towards fixation is just increasing in frequency. You claim that multiple alleles can not move towards fixation at the same time, even though they are observed to do so in sexual populations. All you can do is shift the conversations to definitions of fixation.
Do I need to go over the addition rule again with you. You mean the one that allows any number of beneficial alleles to be any frequency whatsoever?
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dwise1 Member Posts: 5949 Joined: Member Rating: 5.2 |
Really, you know where the multiplication rule applies? And more importantly, we also know where it does not apply. Something that you have never been able to figure out; eg, your Message 2466:
Kleinman writes: Every once in a while, someone wins two lotteries of the tens of millions of people that play lotteries and you think that everyone wins two lotteries. You really need to learn how to do the mathematics of biological evolution. First, what you wrote has nothing to do with the mathematics of biological evolution. We all know that, but you don't. Furthermore, despite our attempts to correct you, you obstinately refuse to ever learn. That is a classic definition of stupidity, in this case willful stupidity. Second, if you are claiming that we conclude that "everyone wins two lotteries", then you are a damned liar. Or a completely delusional idiot. Which is it? Third, you just blindly apply the multiplication rule even when it doesn't apply. If you were to develop a math model for evolution, then you could avoid making such a gross mistake. But you refuse to use a math model, so you keep arriving at stupid nonsense. I've been sharing a quote from Marcus Lycus with your circle-jerk buddy, Sledge, but it applies to you too:
quote: Or to steal a line from the Muse, Serendipity:
quote:
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Taq Member Posts: 10073 Joined: Member Rating: 5.2
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Kleinman writes: Really, you know where the multiplication rule applies? It doesn't apply to multiple beneficial alleles in unlinked genes within sexual populations. That's what you have failed to understand from the very beginning. Here is a section from one of my previous posts. By your own admission, the addition rule is: fA + (fB-(fA*fB)) + fC = Tp WherefA = frequency of variant A fB = frequency of variant B fC = frequency of no A or B = ((1-fA)*(1-fB)) Tp = Total population, should be equal to 1 I contend that your made up version of the addition rule, invented to try and explain away your misapplication to variants at different loci, allows A and B to be whatever number I want. In fact, I can have them increasing in frequency in lock stop together, from a frequency of 0.01 on up to a frequency of 1 for each. Your new made up addition (i.e. subtraction) rule allows the very thing you claimed couldn't happen. If I am right then I should be able to increase both A and B and still have the equation equal 1. I will be using the assumption of equal distribution for each variant. I will be using a crude population curve that starts with frequencies of 0.01 to 0.05 in increments of 0.01, and then to speed things up I will change the frequency by 0.05 to 1 by increments of 0.05. I will be using n to represent census numbers and fX to represent the frequencies of each variant. A = n * fAB = n * fB AandB = n * (fA * fB) C = (1-fA) * (1-fB) So for the whole equation: (fA) + (fB - (fA*fB)) + ((1-fA)*(1-fB)) = Tp I will use a population of 100,000 The code looks like this (adding window scroll to reduce size):
n = 100000 for i in range(1, 6): f = i*0.01 fA = f fB = f A = int(n * f) B = int(n * f) AandBn = int(n * (fA * fB)) C = int(n*((1-fA)*(1-fB))) Tp = (fA) + (fB - (fA*fB)) + ((1-fA)*(1-fB)) print(f'{f} = frequency of A and frequency of B') print(f'{A} = number of offspring with A or B') print(f'{AandBn} = number of offspring with AB') print(f'{C} = number of offspring with neither A nor B') print(f'{Tp} = normalized total population') print('\n') for i in range(1, 21): f= round(i*0.05, 2) fA = f fB = f A = int(n * f) B = int(n * f) AandBn = int(n * (fA * fB)) C = int(n*((1-fA)*(1-fB))) Tp = (fA) + (fB - (fA*fB)) + ((1-fA)*(1-fB)) print(f'{f} = frequency of A and frequency of B') print(f'{A} = number of offspring with A or B') print(f'{AandBn} = number of offspring with AB') print(f'{C} = number of offspring with neither A nor B') print(f'{Tp} = normalized total population') print('\n') If you run it on an online python interpreter you will see that the normalized population size (Tp) remains at 1.0 throughout (0.9 repeating is a python glitch). You can also see that the number of AB individuals increases through time. My results:
0.01 = frequency of A and frequency of B 1000 = number of offspring with A or B 10 = number of offspring with AB 98010 = number of offspring with neither A nor B 1.0 = normalized total population 0.02 = frequency of A and frequency of B 2000 = number of offspring with A or B 40 = number of offspring with AB 96039 = number of offspring with neither A nor B 0.9999999999999999 = normalized total population 0.03 = frequency of A and frequency of B 3000 = number of offspring with A or B 90 = number of offspring with AB 94090 = number of offspring with neither A nor B 1.0 = normalized total population 0.04 = frequency of A and frequency of B 4000 = number of offspring with A or B 160 = number of offspring with AB 92160 = number of offspring with neither A nor B 1.0 = normalized total population 0.05 = frequency of A and frequency of B 5000 = number of offspring with A or B 250 = number of offspring with AB 90250 = number of offspring with neither A nor B 1.0 = normalized total population 0.05 = frequency of A and frequency of B 5000 = number of offspring with A or B 250 = number of offspring with AB 90250 = number of offspring with neither A nor B 1.0 = normalized total population 0.1 = frequency of A and frequency of B 10000 = number of offspring with A or B 1000 = number of offspring with AB 81000 = number of offspring with neither A nor B 1.0 = normalized total population 0.15 = frequency of A and frequency of B 15000 = number of offspring with A or B 2250 = number of offspring with AB 72249 = number of offspring with neither A nor B 0.9999999999999999 = normalized total population 0.2 = frequency of A and frequency of B 20000 = number of offspring with A or B 4000 = number of offspring with AB 64000 = number of offspring with neither A nor B 1.0 = normalized total population 0.25 = frequency of A and frequency of B 25000 = number of offspring with A or B 6250 = number of offspring with AB 56250 = number of offspring with neither A nor B 1.0 = normalized total population 0.3 = frequency of A and frequency of B 30000 = number of offspring with A or B 9000 = number of offspring with AB 48999 = number of offspring with neither A nor B 1.0 = normalized total population 0.35 = frequency of A and frequency of B 35000 = number of offspring with A or B 12249 = number of offspring with AB 42250 = number of offspring with neither A nor B 1.0 = normalized total population 0.4 = frequency of A and frequency of B 40000 = number of offspring with A or B 16000 = number of offspring with AB 36000 = number of offspring with neither A nor B 1.0 = normalized total population 0.45 = frequency of A and frequency of B 45000 = number of offspring with A or B 20250 = number of offspring with AB 30250 = number of offspring with neither A nor B 1.0 = normalized total population 0.5 = frequency of A and frequency of B 50000 = number of offspring with A or B 25000 = number of offspring with AB 25000 = number of offspring with neither A nor B 1.0 = normalized total population 0.55 = frequency of A and frequency of B 55000 = number of offspring with A or B 30250 = number of offspring with AB 20249 = number of offspring with neither A nor B 1.0 = normalized total population 0.6 = frequency of A and frequency of B 60000 = number of offspring with A or B 36000 = number of offspring with AB 16000 = number of offspring with neither A nor B 1.0 = normalized total population 0.65 = frequency of A and frequency of B 65000 = number of offspring with A or B 42250 = number of offspring with AB 12249 = number of offspring with neither A nor B 0.9999999999999999 = normalized total population 0.7 = frequency of A and frequency of B 70000 = number of offspring with A or B 48999 = number of offspring with AB 9000 = number of offspring with neither A nor B 1.0 = normalized total population 0.75 = frequency of A and frequency of B 75000 = number of offspring with A or B 56250 = number of offspring with AB 6250 = number of offspring with neither A nor B 1.0 = normalized total population 0.8 = frequency of A and frequency of B 80000 = number of offspring with A or B 64000 = number of offspring with AB 3999 = number of offspring with neither A nor B 1.0 = normalized total population 0.85 = frequency of A and frequency of B 85000 = number of offspring with A or B 72249 = number of offspring with AB 2250 = number of offspring with neither A nor B 1.0 = normalized total population 0.9 = frequency of A and frequency of B 90000 = number of offspring with A or B 81000 = number of offspring with AB 999 = number of offspring with neither A nor B 1.0 = normalized total population 0.95 = frequency of A and frequency of B 95000 = number of offspring with A or B 90250 = number of offspring with AB 250 = number of offspring with neither A nor B 1.0 = normalized total population 1.0 = frequency of A and frequency of B 100000 = number of offspring with A or B 100000 = number of offspring with AB 0 = number of offspring with neither A nor B 1.0 = normalized total population
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Kleinman Member (Idle past 362 days) Posts: 2142 From: United States Joined: |
Kleinman:So, Taq, the expert in probability theory, thinks that if there are 10 possible adaptive mutation then the adaptive mutation rate is 10*mutation rate? You aren't thinking this through. Let's say there are 10 possible evolutionary trajectories that confer antibiotic resistance. You start with a population that is antibiotic sensative and mutations are occurring as the population is growing and there are 10 possible mutations for the first adaptive step. That variant doesn't have 10 possible mutations for the second adaptive step. It needs a specific mutation for the next step on its evolutionary trajectory to drug resistance. Likewise, if another variant gets a different adaptive mutation for its first step on its evolutionary trajectory, there won't be 10 adaptive mutations for the second step. This applies to all variants on the different evolutionary trajectories. Once a variant starts on a particular evolutionary trajectory, it must get specific mutations to improved fitness. This was shown by Weinreich in his paper: Darwinian Evolution Can Follow Only Very Few Mutational Paths to Fitter Proteins Kleinman:It did stop that competition dumb dodo. If it had continued, a single most fit allele would have been fixed whether the population replicated sexually or asexually. Only if the sexual replicators had a successful recombination event (or an adaptive mutation had occurred) would that competition be ended and a new competition started with a new more fit variant in the population. Kleinman:I haven't even tried because I know it is not necessary to do the mathematics of biological evolution. But in your confused state, you think it is necessary to compute that number. Good luck. Kleinman:You must like being stupid, you do it so often. Only one allele can reach fixation. That's why Desai forces sexual reproduction after 90 generations. Kleinman:You do the mathematics like a poorly trained undergraduate.
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Kleinman Member (Idle past 362 days) Posts: 2142 From: United States Joined: |
Kleinman:Yeah, right. You know where to apply the multiplication rule. That's why you can't explain the evolution of drug resistance and why cancer treatments fail. But I'll give the expert in C-- a chance to show where the multiplication rule applies in the use of 3 drug therapy in the treatment of HIV. We won't hold our breath for that explanation.
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Kleinman Member (Idle past 362 days) Posts: 2142 From: United States Joined: |
Kleinman:Taq does a calculation for an imaginary population. Why doesn't he do the mathematics for the tens of millions of successful treatments of HIV with 3 drug therapy? Does the real world make Taqqy Wacky.
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Taq Member Posts: 10073 Joined: Member Rating: 5.2 |
Kleinman writes: Taq does a calculation for an imaginary population. Why doesn't he do the mathematics for the tens of millions of successful treatments of HIV with 3 drug therapy? Does the real world make Taqqy Wacky.
Conman.
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dwise1 Member Posts: 5949 Joined: Member Rating: 5.2
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Yeah, same old stupid bullshit posturing. More than 1550 postings of empty trolling garbage that demonstrate nothing except for what an idiot you are.
And you wonder why everybody laughs at you? Why nobody can take seriously anything you vomit forth? Well, that's mainly because we can all see through your stupid bullshit. And then you just keep pouring out more and more of the exact same stupid garbage. More than 1550 repetitions of the exact same stupid bullshit! More than 1550 instances of a complete waste of bandwidth and server resources! You stupid troll.
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Kleinman Member (Idle past 362 days) Posts: 2142 From: United States Joined: |
Kleinman:What's the matter, in 26 years of research you never tried to model a real example? That explains why you don't understand how biological evolution works. That's why you think there are numerous adaptive mutations when there aren't and that adaptive recombination happens all the time when it doesn't. You are a very sloppy researcher that doesn't understand how stochastic systems behave.
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Kleinman Member (Idle past 362 days) Posts: 2142 From: United States Joined: |
dwise1:What's the matter? Are you having trouble explaining when the multiplication rule applies? I'll give the slow-thinking computer programmer a hint. You apply the multiplication rule every time you compute a joint probability such as two adaptive mutations occurring. It is just like winning two lotteries, you multiply the probability of each happening. You aren't very good at this.
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AZPaul3 Member Posts: 8552 From: Phoenix Joined: Member Rating: 4.9 |
Fraud.
Stop Tzar Vladimir the Condemned!
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Kleinman Member (Idle past 362 days) Posts: 2142 From: United States Joined: |
AZPaul3:He's having a hard time because his bubble has burst.
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Phat Member Posts: 18338 From: Denver,Colorado USA Joined: Member Rating: 1.0 |
I've got one that you might like, Kleinman. Personally I think the guy is full of it, but his presentation goes to great lengths to support the theory of a young earth
Real science, history, geology, population growth, archeology, and common sense prove Earth is only a few thousand years old, NOT billions of years old.
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Kleinman Member (Idle past 362 days) Posts: 2142 From: United States Joined: |
Phat:I watched a couple of minutes of your link. I've never done research on the age of the earth, the problems I have had to deal with are drug resistant infections, so that is what I've studied. The only comment I tend to make on the age of the earth is the presence of soft tissue on fossils and that the half-life of these biomolecules doesn't match with the claimed age that biologists give. I would prefer to talk about how biological evolution actually works. For example, Taq reports that drug resistance of a bacteria can occur by different sets of mutations. He is correct on this point but he either refuses to do the math or doesn't know how to do the math. In addition, Taq appears to think that if there are multiple possible first adaptive mutations, then there must be multiple possible second and additional mutations. This would mean that there are many, many additional evolutionary trajectories to drug resistance but that is not what is measured. When the first adaptive mutation occurs, the remaining additional mutations are specified. This is what Weinreich measured. So, how does this phenomenon of multiple evolutionary trajectories work? How does this change the probabilities for the first adaptive mutation? Since Taq won't do the math, I will have to. How does the probability change when there are multiple possible adaptive mutations at different sites to a given selection pressure? The probability of an adaptive mutation occurring at least once at a single site is given by:P(X)=1-(1-μ)^N where μ is the beneficial mutation rate and N is the total number of replications of that site. If there is more than one possible site in the genome where an adaptive mutation can occur, what is the probability that an adaptive mutation will occur at least once at any of these sites? The following equation gives the "at least one" calculation for the probability of an adaptive mutation occuring at any of multiple possible sites. P(XS)=1-(1-P(X))^S = 1-(1-(1-(1-μ)^N))^S where S is the number of possible sites and P(XS) is the probability that "at least one" adaptive mutation will occur at any one of these sites. Others can talk about the age of the earth. I prefer to talk about the physics and mathematics of biological evolution and how biologists got this science so wrong.
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Taq Member Posts: 10073 Joined: Member Rating: 5.2
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Kleinman writes: What's the matter, in 26 years of research you never tried to model a real example? What's the matter, can't address the math I presented?
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