Rebuttal To Creationists - "Since We Can't Directly Observe Evolution..."

That article is a great example of the disinformation found on creationist sites.From the linked article:

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Out of tens of thousands of ERV elements in the human genome, roughly how many are known to occupy the same sites in humans and chimpanzees? According to this Talk-Origins article, at least seven. Let’s call it less than a dozen. Given the sheer number of these retroviruses in our genome (literally tens of thousands), and accounting for the evidence of integration preferences and site biases which I have documented above, what are the odds of finding a handful of ERV elements which have independently inserted themselves into the same locus?

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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.

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.

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.

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Out of tens of thousands of ERV elements in the human genome, roughly how many are known to occupy the same sites in humans and chimpanzees? According to this Talk-Origins article, at least seven. Let’s call it less than a dozen. Given the sheer number of these retroviruses in our genome (literally tens of thousands), and accounting for the evidence of integration preferences and site biases which I have documented above, what are the odds of finding a handful of ERV elements which have independently inserted themselves into the same locus?

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That's a lie.

Not your Nazi repugnicans, that's for sure.

I used it right here.By your own admission, the addition rule is:fA + (fB-(fA*fB)) + fC = TpWhere
fA = 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 1I 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 * fA
B = n * fB
AandB = n * (fA * fB)
C = (1-fA) * (1-fB)So for the whole equation:(fA) + (fB - (fA*fB)) + ((1-fA)*(1-fB)) = TpI will use a population of 100,000The 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

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.

We've been through that. Science doesn't deal in proof.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.

I don't need belief. I have science.

Dredge:

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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

Taq:

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That article is a great example of the disinformation found on creationist sites.

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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.

You appear to be contradicting yourself - you're saying UCD is a fact that can't be proven.Fascinating.

Kleinman:

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What's the probability of a germ line cell being infected hundreds of thousands of times and there not being damage to that cell?

Taq:

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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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Yeah, you see hundreds of thousands of viruses invading the germ cell line. You are an idiot.

Kleinman:

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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.

Taq:

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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.

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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.

Kleinman:

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Who is going to stop California from taking all of your water?

AZPaul3:

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Not your Nazi repugnicans, that's for sure.

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Who needs republicans, we have democrats to dry you out, drier than a bone. I hope you like it dry, really dry.

Kleinman:

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And why doesn't Taq know how to use the addition rule?

Taq:

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I used it right here.

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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.