Do you really understand the mathematics of evolution?

Do you think that for every single adaptation there is only one possible mutation in the entire genome that can produce that adaptation?Do you think that for every single environmental challenge there is only one single adaptation that will work?Why do you keep claiming that mutations have to wait for one another when sexual reproduction can combine them into a single offspring?

That's not a valid model for sexually reproducing populations in real environmental settings.

Yes indeedy it is. DNA evolution works the same for all replicators. You can superimpose recombination on the DNA evolution process and it changes the math slightly but you had better start with the simpler case without recombination until you learn the mathematics of DNA evolution in the non-recombination situation. In addition, do you think that the Jukes-Cantor and Felsenstein models don't apply to sexually reproducing organisms? Because when you write the correct transition matrix for these models, you will get consistent results with this model.Now, if you said this model doesn't apply to competition, that would be correct. But you can impose competition on a DNA evolutionary process such as the Lenski experiment. And you also can impose recombination on a DNA evolutionary process. You need to learn how to distinguish the different evolutionary transformation processes and then learn how to superimpose the different solutions depending on the replicator and environmental conditions in order to predict the behavior of the particular system.

DNA evolution works differently in sexual and asexual species. Until you learn this lesson you will continue to fail at understanding this topic.

Oh really? DNA evolution works differently in sexual and asexual species? How so? Does one type of replicator get mutations and the other doesn't? Does the multiplication rule disappear for one type of replicator and it doesn't apply for the other? What's the difference? And show your math.

Diploid v. haploid. Look it up. The multiplication rule does disappear for diploid organisms. The fact you can't understand this says a lot.

Where's that link that shows that DNA evolution works differently for sexual and asexual species? You are confusing the method of reproduction with DNA evolution. Whoops, a daisy! Are you telling us that combination herbicides don't impair the evolution of herbicide resistance and combination pesticides don't impair the evolution of pesticide resistance? Look it up, those links exist and there are lots of them.

Here:Sexual reproduction - Wikipedia Go back to your Kishony experiment. The multiplicative rule in that case requires evolution of resistance to one drug and then evolution of resistance to the other drug after that. This isn't the case for sexually reproducing species. The evolution of resistance can occur separately in two individuals, and then their descendants can mate and have offspring with resistance to both drugs. I have explained this so many times now that you would have to be an idiot not to understand it.

Where in that link do they talk about DNA evolution? Are you telling us that the DNA in sexual reproduction is not replicated? So there are never any errors in that non-existent replication process? Each additional selection pressure causes additional instances of the multiplication rule to be applied to the DNA evolutionary process. And the reason why combination herbicides and pesticides are used is to prevent the increase in the frequency of any variant that might have resistance to one drug or another. If you use single herbicides for a long enough period, you will increase the frequency of variants resistance to that particular herbicide. When that herbicide fails and you switch to another herbicide, you may have high enough frequency of the resistant variants to get that recombination event. But you can also get DNA evolution giving a doubly resistant variant simply by mutation. And in case you didn't notice, the multiplication rule applies to recombination events.

Go back to your Kishony experiment. The multiplicative rule in that case requires evolution of resistance to one drug and then evolution of resistance to the other drug after that. This isn't the case for sexually reproducing species. The evolution of resistance can occur separately in two individuals, and then their descendants can mate and have offspring with resistance to both drugs. I have explained this so many times now that you would have to be an idiot not to understand it. Weed A is exposed to pesticide A in one region and develops resistance. Weed B is exposed to pesticide B in a different region and develops resistance.We bring Weed A and Weed B into the same region. How many generations before you get resistance to pesticides A and B in a single weed? According to your multiplicative rule, it would take just as many generations to develop resistance to both pesticides as it did to one. Is that right?Edited by Taq, : No reason given.

What you can't grasp is that these recombination events can occur only under very specific circumstances. That's why combination selection pressures impair the evolution of drug-resistant hiv, herbicide-resistant weeds and pesticide-resistant insects. You have to be an idiot not to understand it. People who do weed management aren't dumb enough to use their herbicides as you suggest. Your suggestion is all the more reason that naive school children need to learn the correct mathematics of evolution.

It happens in every individual in every generation in a sexually reproducing species. Answer the question.Weed A is exposed to pesticide A in one region and develops resistance. Weed B is exposed to pesticide B in a different region and develops resistance.We bring Weed A and Weed B into the same region. How many generations before you get resistance to pesticides A and B in a single weed? According to your multiplicative rule, it would take just as many generations to develop resistance to both pesticides as it did to one. Is that right?

You are having difficulty distinguishing between any recombination event and specific recombination events. It is like the difference between any mutation occurring and a beneficial mutation occurring. You really are having trouble getting a grasp of this math. I have, you just don't have the math skills to recognize it. If you want to start a herbicide-resistant weed breeding program, that's your formula. That's how you deal with the multiplication rule in DNA evolution and the multiplication rule in random recombination to speed up the evolutionary process. You don't even need recombination to do it. Just use single-drug therapy for treating infections. When one drug fails, go on to the next drug and the next drug until you have evolved MRSA. You have demonstrated a brilliant understanding of DNA evolution. Drug-resistant infections will be commonplace for many more years with bright minds like yours in action.

I must say that I appreciate Taq's arguments about recombination. I hope that Taq understands that the model that I am presenting is the simplest example of DNA evolution for a single selection pressure targeting a single genetic locus. Because of this, recombination has no effect at all on this particular evolutionary example. The reason for this is that when selection is acting at only a single locus, there is no way that recombination can combine mutations.To illustrate this, let's call the first beneficial mutation on the given evolutionary trajectory "A" and the second beneficial mutation on that evolutionary trajectory "B". For the mutation rate of e-9, 3e9 replications will give us on average one member in that population with mutation A at the genetic locus and another member with the mutation B at the same genetic locus. But recombination without error cannot combine those two alleles to put the A and B mutations on the same member. Perhaps in the very extreme case, some type of chimeric recombination event could occur where the part of the allele with mutation A breaks off and recombines with a broken part of the allele with mutation B, but if you believe that is the basis for evolution, my recommendation to you is stay away from Las Vegas.With that said, let's get back to the explanation for dwise1 on how the multiplication rule applies to DNA evolution. Yesterday I showed you how to calculate the probability that a beneficial mutation A will occur on a member of a population in a single replication. That calculation is done as follows:P(A) = P(BeneficialA) is the mutation rate and P(BeneficialA) is the probability of all the possible mutations that can occur at that site, it is the A mutation. You might be wondering, what is the value of P(BeneficialA). If the only mutations we are considering are base substitutions, P(BeneficialA)=1/3, because there are 3 possible substitutions for the original base. This is also the reason why Taq's computation for the number of replication for the beneficial mutation to occur is 3e9 and not just 1/mutation rate.So the next question is, what is the probability of A occurring at least once in more than one replication? In order to do that math, you could write out the sample space for all the possible outcomes as each replication occurs but that would become very unwieldy after more than a few replications. What we can use is the "at least one rule" from probability theory. If you are not familiar with that rule, here is a very easy lecture which explains this rule:
https://www.youtube.com/watch?v=a86Vj_7R2aQ
There are shorter videos than this 30 min video but this teacher's explanation is very clear and easy to follow.I'm going to assume that you understand this rule and continue on with the calculation. The next step is to calculate the probability that mutation A will not occurs in a single, that is the complement of P(A) using the complementary rule.P(Ac)=1‘P(BeneficialA)Where P(Ac) means the probability of A not occurring. We can then use the multiplication rule to compute the probability of A not occurring with any number of replications. Set "n" to be the number of replication and that gives the probability of A not occurring with n replications as:P(Ac)=(1‘P(BeneficialA))^nWhat this equation gives us is the probability of mutation A not occurring at all in n replications. What we want to know is the probability of A occurring at least once in n replication. We can compute that number by again applying the complement rule:P(A)=1-P(Ac)=1-((1‘P(BeneficialA))^n)To put this into the context of the Kishony experiment, that equation gives the probability of the first beneficial mutation "A" occurring as his colony grows in the drug-free region. A plot of that curve for various mutation rates appears as follows: To use these curves, select a particular value of P(BeneficialA), then as you go along the x-axis, you can obtain the probability of mutation A occurring for that number of replications.When that mutation A does occur, you now have a variant that is able to grow in the next higher drug concentration region of the Kishony experiment. Tomorrow, I'll go over the math that shows you how to compute the probability of a second beneficial mutation "B" occurring on some member that already has mutation A and how the multiplication rule applies for this evolutionary transition.

You are either due a Nobel prize for revolutionising the whole of biology. Or you are a crackpot.If I had to put money on which of those two is more probable....Well - You are the probability whiz - You tell me - Which is more likely?