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Author | Topic: Rebuttal To Creationists - "Since We Can't Directly Observe Evolution..." | |||||||||||||||||||||||||||||||||||||||
Taq Member Posts: 10085 Joined: Member Rating: 5.6
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Dredge writes:
The following article is a bit old (2011), but I thought you might find it interesting nevertheless: Do Shared ERVs Support Common Ancestry? – Evolution News That article is a great example of the disinformation found on creationist sites. From the linked article:
quote: This was written after the chimp genome paper was published which demonstrated that nearly all of the 200,000 ERV's in the chimp and human genomes were shared at the same base in each genome. They are lying to their audience.
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Taq Member Posts: 10085 Joined: Member Rating: 5.6 |
Kleinman writes: What's the probability of a germ line cell being infected hundreds of thousands of times and there not being damage to that cell? Again, we can see this happening in real time in the koala population. They carry ERV's from an active and circulating retrovirus. So the probability is 100%, because it happened. We have the evidence that it happened in the form of ERV's.
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Taq Member Posts: 10085 Joined: Member Rating: 5.6
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Kleinman writes: Since AZPaul3 is more knowledgeable than Jonathan McLatchie, AZPaul3 will now give a mathematical description of descent with modification and recombination. He won't because he is a bubblehead. McLatchie claims that there are less than a dozen shared ERV's between humans and chimps. That's an outright lie. I have even personally notified personnel at the Discovery Institute of this issue and received their acknowledgement that it the article is wrong, and the lie is still on their website.
quote: That's a lie.
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AZPaul3 Member Posts: 8564 From: Phoenix Joined: Member Rating: 5.1 |
Who is going to stop California from taking all of your water? Not your Nazi repugnicans, that's for sure.Stop Tzar Vladimir the Condemned!
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Taq Member Posts: 10085 Joined: Member Rating: 5.6
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Kleinman writes:
I used it right here.
And why doesn't Taq know how to use the addition rule?
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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Taq Member Posts: 10085 Joined: Member Rating: 5.6 |
Kleinman writes: That's interesting, but why should Taq read it, it doesn't fit his narrative. I have read it, many times. It lies about the number of shared ERV's between humans and chimps and about several other characteristics of ERV's. I can talk about the other lies if you like.
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ringo Member (Idle past 442 days) Posts: 20940 From: frozen wasteland Joined:
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Dredge writes:
We've been through that. Science doesn't deal in proof. So, science can't prove that UCD is a fact ... Science can't prove that Coke is better than Pepsi.Science can't prove that pigs can fly. Science can't prove that Taylor Swift can sing. Science can't prove that fill in the blank with ANYTHING you can think of. 1. Science doesn't deal in proof.2. Science doesn't deal in proof. 3. Science doesn't deal in proof. And in case I haven't mentioned it, science doesn't deal in proof. Of course UCD IS a fact. That has been demonstrated by every organism that fits neatly into the nested hierarchy and DNA has confirmed the accuracy of the nested hierarchy.Come all of you cowboys all over this land, I'll teach you the law of the Ranger's Command: To hold a six shooter, and never to run As long as there's bullets in both of your guns. -- Woody Guthrie
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ringo Member (Idle past 442 days) Posts: 20940 From: frozen wasteland Joined: |
Kleinman writes:
I don't need belief. I have science. ringo doesn't need anything but to believe incorrectly.Come all of you cowboys all over this land, I'll teach you the law of the Ranger's Command: To hold a six shooter, and never to run As long as there's bullets in both of your guns. -- Woody Guthrie
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Kleinman Member (Idle past 366 days) Posts: 2142 From: United States Joined: |
Dredge:If you don't agree with the results, it is disinformation. That's why you can't do the mathematics of descent with modification and recombination. You are mathematically incompetent.
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Dredge Member (Idle past 104 days) Posts: 2850 From: Australia Joined: |
ringo writes:
You appear to be contradicting yourself - you're saying UCD is a fact that can't be proven. science doesn't deal in proof. Of course UCD IS a fact. Fascinating.
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Kleinman Member (Idle past 366 days) Posts: 2142 From: United States Joined: |
Kleinman:Yeah, you see hundreds of thousands of viruses invading the germ cell line. You are an idiot.
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Kleinman Member (Idle past 366 days) Posts: 2142 From: United States Joined: |
Kleinman:You believe that a germ cell can be invaded hundreds of thousands of times by viruses and it has zero effect on the cell. That's really smart on your part, if only it was true. You have no idea what a virus does to a cell. And tell us how smart virologists are after the Covid episode.
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Kleinman Member (Idle past 366 days) Posts: 2142 From: United States Joined: |
Kleinman:Who needs republicans, we have democrats to dry you out, drier than a bone. I hope you like it dry, really dry.
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Kleinman Member (Idle past 366 days) Posts: 2142 From: United States Joined: |
Kleinman:Taq is really learning, if only he does real problems.
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Taq Member Posts: 10085 Joined: Member Rating: 5.6 |
Kleinman writes: Taq is really learning, if only he does real problems. I did a real problem. I demonstrated that your subtraction rule allows variants at different loci to be any frequency, contrary to your claims.
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