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

Do you?Let's say we have three adaptive mutations: lactase persistance, lower melanin production in higher latitudes, and malarial resistance.If these adaptive mutations were in three different unlinked genes, would they be dependent or independent? Could all three mutations start moving towards fixation at the same time, or would they have to move towards fixation one at a time? Why?

The problem is how Kleinman tries to apply probabilities to genetics. He seems to think that if one beneficial mutation is at a frequency of 0.5 then all other beneficial mutations throughout the genome can not add up to more than a frequency of 0.5. I hope you can see how ridiculous that is.

Show me where I said that. It's called sexual reproduction. Mommy has A and Daddy has B.

Then let's look at achondroplasia which is caused by mutations in the FGFR3 gene and cystic fibrosis which is caused by mutations in the CFTR gene. More than 99% of people have the healthy allele for both and only an extreme few have both cystic fibrosis and dwarfism. So let's do the math:0.99A + 0.99B + 0.0000001C != 1Your math doesn't work. You just said they were mutually exclusive. Can you give me a single example of two mutations in two different unlinked genes that are mutually exclusive?

Diploidy, sexual reproduction, and meiosis also exist. What you still can't seem to understand is that the mathematics of asexual reproduction don't always apply to sexual reproduction. Which of the genetic differences between humans and chimps do you think microevolution could not produce, and why?

Unlinked. What would this mean for how each allele moves through population now that they are not tied together?In the Kishony and Lenski experiments all of the genes are linked because they are on the same chromosome, the organisms are haploid, and there is no crossover between copies of genomes. In the case of diploid, sexually reproducing organisms this isn't the case. Therefore, you no longer have the case where the fittest allele in the entire genome drives natural selection. Does that ever happen in the Lenski and Kishony experiments? Do you think this process may have an affect on what alleles can be selected for?

Why would I want to do that? The mutation for lactase persistence can occur in one individual and the mutation for lower melanin production can occur in another individual. Later, their descendants can mate and have offspring that have both mutations. In fact, we see this all of the time in the human population.Why do you still not understand how sexual reproduction works?

You still don't understand that a baby gets a copy of Mom's and Dad's genome. Both. If Mom has an adaptive mutation A in gene I and Dad has an adaptive mutation B in gene J, guess what can happen? Baby can get both mutations. You don't have to have mutation A happen in someone who already carries mutation B, or vice versa.Also, the frequency for A and B can exist at a frequency higher than 0.5 within the population if they are unlinked. You still can't seem to understand that. Reality insists differently. Heterozygosity applies to alleles for the same gene, not alleles for unlinked genes. You are the one who is applying the addition rule to unlinked genes, not I. If A and B are on separate chromosomes why would you even need a recombination event?You do realize that there is more than one chromosome, more than one gene, and that recombination events are extremely common within a chromosome, right? This is why genes that are on distant parts of the same chromosome are considered unlinked because there are about 1 to 3 crossover events per chromosome per generation which gives a very high chance of alleles switching chromosome copies if they are distant from each other on the same chromosome.There is no reason to do any math until you learn the basic concepts of sexual reproduction, diploidy, and meiosis. It works because only genes located close to each other on a chromosome are linked to one another. The vast majority of genes are independent in diploid organisms. I understand both asexual and sexual reproduction just fine which is why I understand where they differ. You don't seem to understand how each differs from each other. You keep applying the concepts from asexual reproduction when they don't apply to sexual reproduction.In bacteria, there is no crossover between copies of bacterial genomes because they are asexual and haploid. HGT does occur between bacteria on occasion and there are plasmids, but let's not complicate matters at this point. Let's say mutation I in gene X occurs in one bacteria in the population and mutation J happens in gene Y in a different bacteria in the same population. What would need to happen to get both mutations in the same individual in this case? We would have to have a repeat of the mutation in each of the lineages. However, the fittest of I and J could drive the other mutation to extinction in the mean time. All of this is very true in asexual populations.Is this the case in sexual populations? NO, not in the case of unlinked genes. Do you know why? Because descendants of those carrying mutation I and J can mate and have offspring with BOTH MUTATIONS. The mutations doesn't have to happen again. We don't even need the math. It is all explained in the pictures I have given you for combining mom's and dad's dna, linked/unlinked genes, and cross-overs.

You need to understand the system before you can apply maths (nod to our UK particpants) to it. It is your lack of understanding of diploidy, multiple chromosomes, gene linkage, meiosis, and sexual reproduction that causes you to misapply the maths.

Your equation doesn't have the subtraction of the intersection. It just has the frequencies of both mutations.Define the following variables:
n – is the total population size.
nA – is the number of members in the population with beneficial allele A.
nB – is the number of members in the population with beneficial allele B.
nC – is the number of members in the population that have neither beneficial allele A nor beneficial allele B.
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In addition, we have the following condition: nA + nB + nC = n.
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And the frequency of each of the variants are:
f_A = nA/n
f_B = nB/n
f_C = nC/nWhere's the subtraction? It's not there. Then why do you keep insisting that the frequency of mutations at unlinked genes must add up to 1? Will you now admit that they don't have to add up to one? Why would I apply the addition rule when you yourself are saying that it doesn't apply?

Kleinman (and to whomever),Look at the picture below. Look at the last example of genes on separate chromosomes. Even though recombination will occur within each chromosome, is that even needed in order to get independent association of these alleles? No. They are on separate chromosomes. You can have 6 possible combinations in a given gamete if the carrier is heterozygous at both genes:5A/6A
5A/6B
5B/6A
5B/6BHow many combinations can you get with two gametes with the same mixture of alleles at two unlinked genes?What would this look like if one parent is 5AA and 6BB and the other parent is 5BB and 6AA? Could their offspring have a mixture of alleles at both genes without a recombination event needed? (The answer is yes, in case you were wondering)

I am saying that we can't apply your addition rule to unlinked genes in diploid organisms. You are saying that we can. Who do you think is right? I will repeat:In bacteria, there is no crossover between copies of bacterial genomes because they are asexual and haploid. HGT does occur between bacteria on occasion and there are plasmids, but let's not complicate matters at this point. Let's say mutation I in gene X occurs in one bacteria in the population and mutation J happens in gene Y in a different bacteria in the same population. What would need to happen to get both mutations in the same individual in this case? We would have to have a repeat of the mutation in each of the lineages. However, the fittest of I and J could drive the other mutation to extinction in the mean time. All of this is very true in asexual populations.
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Is this the case in sexual populations? NO, not in the case of unlinked genes. Do you know why? Because descendants of those carrying mutation I and J can mate and have offspring with BOTH MUTATIONS. The mutations doesn't have to happen again. See the picture above on linked and unlinked genes.

You keep asking how we get two beneficial mutations in the same individual. Those pictures demonstrate how it is done in diploid organisms who are sexually reproducing.Notice the first case. You get a new allele combination between two different genes on the same chromosome. Notice the second case of unlinked genes. You get a new combination of chromosomes for different chromosome pairs.That's how diploidy, multiple chromosomes, sexual recombination, and meiosis affects DNA evolution.

I already did that:Then let's look at achondroplasia which is caused by mutations in the FGFR3 gene and cystic fibrosis which is caused by mutations in the CFTR gene. More than 99% of people have the healthy allele for both and only an extreme few have both cystic fibrosis and dwarfism. So let's do the math:
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0.99A + 0.99B + 0.0000001C != 1
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Your math doesn't work. I just showed that it doesn't.

Why don't you show us the experiment where life was supernaturally created?