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Author | Topic: Do you really understand the mathematics of evolution? | ||||||||||||||||||||||||
Kleinman Member (Idle past 363 days) Posts: 2142 From: United States Joined: |
It seems that nwr is not the only one with a vague understanding of DNA evolution, Taq doesn't do much better. So let's see if we can help.
Taq writes:
So, if the genome length is 4.6E6 and if mu = 1E-9, it would take about 218 replications of the original founder bacterium of Kishony's drug sensative lineage before we would expect to see the first mutation in one of those 218 members, somewhere it its genome. The remaining 217 members should be exact clones.
mu is mu. In a 4.6 million base genome there are 13.8 million possible substitution mutations, so if you are calculating the number of replications needed to produce all possible mutations, one of which is the one of interest, then you need to use the 13.8 million number. |
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Taq Member Posts: 10084 Joined: Member Rating: 5.1 |
Kleinman writes: Why won't you tell us how man replications it takes for a beneficial mutation to occur for a single step in the Kishony experiment? I think you already saw it: "The Poisson distribution is ok." If this is a throw away argument, I really don't want to go through the effort of digging through all the formulae and doing the math. Is there a point you want to make with all of this?
Why? It's a very rare thing in life when you understand something exactly. Sometimes, all you can do is give a ballpark estimate. You still need valid assumptions to get ballpark estimates.
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Kleinman Member (Idle past 363 days) Posts: 2142 From: United States Joined: |
Kleinman writes:
The reason why the Poisson distribution is ok but not correct because the probability of success (the mutation rate) is very small and population sizes are very large (the number of trials). Under these circumstances, the Poisson distribution gives similar results as the binomial distribution (the correct distribution function for this probability problem).
Why won't you tell us how man replications it takes for a beneficial mutation to occur for a single step in the Kishony experiment?Taq writes: I think you already saw it: "The Poisson distribution is ok."Taq writes:
There isn't a lot of work to go through to answer the question of how many replication necessary for the Kishony experiment to get a variant to adapt to the next higher drug-concentration region. You can even use the Poisson distribution to do your calculation. I've already given you a link that explains how to do the math. If you can't do the math for this simplest of evolutionary experiments, how are you going to do the math for the more complex evolutionary experiments?
If this is a throw away argument, I really don't want to go through the effort of digging through all the formulae and doing the math. Is there a point you want to make with all of this?Kleinman writes:
Quite correct. So, do the math for the simplest case. Assume there is only one possible beneficial mutation at a single site in the genome that will give the necessary improvement in fitness for that new variant to grow in the next higher drug-concentration. Assume you start with a single wild-type bacterium in the drug-free region. What is the number of replications of that wild-type variant needed for that improved fitness variant expected to appear? Please show your math.
Why? It's a very rare thing in life when you understand something exactly. Sometimes, all you can do is give a ballpark estimate.Taq writes: You still need valid assumptions to get ballpark estimates.
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Taq Member Posts: 10084 Joined: Member Rating: 5.1 |
Kleinman writes: The reason why the Poisson distribution is ok but not correct because the probability of success (the mutation rate) is very small and population sizes are very large (the number of trials). Under these circumstances, the Poisson distribution gives similar results as the binomial distribution (the correct distribution function for this probability problem). A Poisson distribution is a binomial distribution: "A familiar nongenetics example of a binominal probability distribution is a coin flip: heads and tails are two discrete outcomes, and the probability of each is 0.5 on any single flip. One type of binomial distribution is the Poisson distribution, which expresses the probability of a given number of occurrences of an event that occurs during a fixed time period."https://www.genetics.org/content/genetics/202/2/371.full.pdf If you can't do the math for this simplest of evolutionary experiments, how are you going to do the math for the more complex evolutionary experiments? So what is the ultimate point you are trying to make? Explain to us why it is worth our time to do the calculations? If we do the calculations, will you admit that people do understand the calculations and stop posting in the thread?
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Kleinman Member (Idle past 363 days) Posts: 2142 From: United States Joined: |
Kleinman writes:
Use whatever distribution you think is appropriate. Tell us how many replications are needed to give a reasonable probability of a drug-resistant variant occurring able to grow in the next higher drug-concentration region in the Kishony experiment and show your math.
The reason why the Poisson distribution is ok but not correct because the probability of success (the mutation rate) is very small and population sizes are very large (the number of trials). Under these circumstances, the Poisson distribution gives similar results as the binomial distribution (the correct distribution function for this probability problem).Taq writes: A Poisson distribution is a binomial distribution: "A familiar nongenetics example of a binominal probability distribution is a coin flip: heads and tails are two discrete outcomes, and the probability of each is 0.5 on any single flip. One type of binomial distribution is the Poisson distribution, which expresses the probability of a given number of occurrences of an event that occurs during a fixed time period."https://www.genetics.org/content/genetics/202/2/371.full.pdf Kleinman writes:
Don't you think that correctly describing the evolution of drug-resistance is important? And this subject is important in understanding how to treat cancer, especially as targeted cancer therapies are developed. And why is it so important to you to stop this thread? What is so terrible about correctly describing the physics and mathematics of evolution? My understanding of evolution has been helped by discussions like this. If you can't do the math for this simplest of evolutionary experiments, how are you going to do the math for the more complex evolutionary experiments?Taq writes: So what is the ultimate point you are trying to make? Explain to us why it is worth our time to do the calculations? If we do the calculations, will you admit that people do understand the calculations and stop posting in the thread? I think that Darwinian evolution is qualitatively correct. What's wrong with quantitating his theory?
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Taq Member Posts: 10084 Joined: Member Rating: 5.1 |
Kleinman writes: Tell us how many replications are needed to give a reasonable probability of a drug-resistant variant occurring able to grow in the next higher drug-concentration region in the Kishony experiment and show your math. Going with a per base mutation rate of 1E-9 and a genome size of 4.6E6 bases, that would be one mutation per 217 replications. If we are looking for a specific substitution mutation that would be 1 mutation out of 4.6E6 bases and 3 possible mutations per base for a total of 1.38E7 mutations. multiply the number of replications per mutation by the number of possible mutations and you get 3E9 replications. So you would need about 3 billion replications to have a reasonable chance of getting any specific substitution mutation in E. coli with that specific mutation rate and genome size. E. coli in saturated liquid culture is about 1E6 to 1E7 per ml. If we added 1 bacterium to 100 ml to 1 liter of culture and let it reach stationary phase we should get our mutant. Now what?
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Kleinman Member (Idle past 363 days) Posts: 2142 From: United States Joined: |
Kleinman writes:
Very good! I hope AZPaul3 is following this discussion because his estimate was between 1 and infinity. 3 billion is definitely in the ballpark. There are several directions this discussion can take but let's stick with the line we are on and try to give the mathematical explanation of why the Kishony experiment won't work with two drugs instead of one. And to do this, consider the 100ml to 1 liter of culture that you described with the bacterium with the beneficial mutation in that population of 3 billion. How large would the growth solution have to be if you took those 3 billion bacteria (with the one member with the resistance mutation) in order to get a variant with the first 2 beneficial mutations? In other words, how large must the population be for a variant to occur with both mutations necessary to grow in the next higher drug-concentration region with two drugs? Assume that you need only a single mutation for each drug. Tell us how many replications are needed to give a reasonable probability of a drug-resistant variant occurring able to grow in the next higher drug-concentration region in the Kishony experiment and show your math.Taq writes: Going with a per base mutation rate of 1E-9 and a genome size of 4.6E6 bases, that would be one mutation per 217 replications. If we are looking for a specific substitution mutation that would be 1 mutation out of 4.6E6 bases and 3 possible mutations per base for a total of 1.38E7 mutations. multiply the number of replications per mutation by the number of possible mutations and you get 3E9 replications. So you would need about 3 billion replications to have a reasonable chance of getting any specific substitution mutation in E. coli with that specific mutation rate and genome size. E. coli in saturated liquid culture is about 1E6 to 1E7 per ml. If we added 1 bacterium to 100 ml to 1 liter of culture and let it reach stationary phase we should get our mutant. Now what? And by the way, the size of the genome is actually not a factor. When you have a mutation rate of 1e-9 and 3 billion replications of that genome, you will have on average, every substitution possible at every site in the genome no matter how long the genome in some member of that population. Edited by Kleinman, : Correct the title
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Taq Member Posts: 10084 Joined: Member Rating: 5.1 |
Kleinman writes: There are several directions this discussion can take but let's stick with the line we are on and try to give the mathematical explanation of why the Kishony experiment won't work with two drugs instead of one. I think we can all agree that needing two mutations at once is going to make adaptation nearly impossible if you start with zero variation. Is there a point beyond that? What would this look like if you start out with a certain amount of variation?
When you have a mutation rate of 1e-9 and 3 billion replications of that genome, you will have on average, every substitution possible at every site in the genome no matter how long the genome in some member of that population. I knew as much when I started, but sometimes it is easier to follow a train of thought instead of employing algebra to eliminate redundancies. In the same vein, the neutral fixation rate is equal to the mutation rate no matter the size of the population. Edited by Taq, : No reason given.
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Kleinman Member (Idle past 363 days) Posts: 2142 From: United States Joined: |
Kleinman writes:
We are not talking about two simultaneous mutations. We are talking about a second particular mutation occurring on some member of the population where a first particular mutation has already occurred. In the context of the Kishony 2 drug experiment, in your 3 billion population, you already have a variant with a mutation for the first drug. You also have a variant with a mutation for the second drug as well. How large must the total population size be for a mutation for the second drug occur on a member that already has a mutation for the first drug or for a mutation for the first drug occur on a member that already has a mutation for the second drug? This is a classical conditional probability problem. I gave you a Venn diagram earlier in the thread to help you understand the problem:
There are several directions this discussion can take but let's stick with the line we are on and try to give the mathematical explanation of why the Kishony experiment won't work with two drugs instead of one.Taq writes: I think we can all agree that needing two mutations at once is going to make adaptation nearly impossible if you start with zero variation. Is there a point beyond that?Taq writes:
We know what the variation is for the Kishony experiment, you are starting with a population size of 3e9. Of that 3e9 population, 13.8 million are variants with each of these members having one of the possible mutations but the vast majority of the starting population is still "wild-type" that is they are exact clones of the original founder bacterium. And we have only 1 member with a beneficial mutation for the first drug and only 1 member with a beneficial mutation for the second drug. Now, the 3e9 bacteria start doubling, how large must that population be for there to be a reasonable probability of getting a variant with both the A1 and A2 mutations. I'll give you a hint here to help you understand this problem. In the single drug experiment, that single drug-resistant mutant moves into the next higher drug concentration region and must achieve 3e9 replication for the next beneficial mutation to occur.
What would this look like if you start out with a certain amount of variation?Kleinman writes:
I don't agree with that. If you can't do the mathematics, the best you can have is a vague understanding of the phenomenon. We are trying to give nwr more than a vague understanding of evolution.
When you have a mutation rate of 1e-9 and 3 billion replications of that genome, you will have on average, every substitution possible at every site in the genome no matter how long the genome in some member of that population. Taq writes: I knew as much when I started, but sometimes it is easier to follow a train of thought instead of employing algebra to eliminate redundancies.Taq writes:
That's not correct but since fixation does not pertain to the Kishony experiment, it is not worth discussing this subject at this time. In the same vein, the neutral fixation rate is equal to the mutation rate no matter the size of the population. Try to figure out how large the population size has to be for a variant to occur with 2 beneficial mutations for 2 drugs to make one evolutionary step in the Kishony experiment if 2 drugs are used.
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Taq Member Posts: 10084 Joined: Member Rating: 5.1 |
Kleinman writes: We are talking about a second particular mutation occurring on some member of the population where a first particular mutation has already occurred. The same calculations would apply, starting with the first bacterium to gain resistance to a single drug. You would need another 3 billion descendants from that first resistant bacterium to get the mutation for resistance to the 2nd drug.
We know what the variation is for the Kishony experiment, you are starting with a population size of 3e9. That doesn't tell us what the genetic variation is. A population of 3 billion that recently descended from a common ancestor is going to have less genetic variation than a sample of 3 billion individuals from a population that has been dividing for a long time period. It is similar to the genetic variation in 10 close family members vs. the genetic variation of 10 individuals randomly sampled from across the globe.
If you can't do the mathematics, the best you can have is a vague understanding of the phenomenon. I did the math. Your only complaint seems to be that I didn't do the math the way you wanted me to.
That's not correct but since fixation does not pertain to the Kishony experiment, it is not worth discussing this subject at this time. Just in case you were curious.
quote:
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Kleinman Member (Idle past 363 days) Posts: 2142 From: United States Joined: |
Kleinman writes:
That's correct for the single drug experiment. How does the math work for the 2 drug experiment where some variant has to appear in the drug-free region with a beneficial mutation for each of the drugs before it can grow in the next higher drug-concentration region? How large does the founding colony have to be for there to be a reasonable probability of that variant appearing?
We are talking about a second particular mutation occurring on some member of the population where a first particular mutation has already occurred.Taq writes: The same calculations would apply, starting with the first bacterium to gain resistance to a single drug. You would need another 3 billion descendants from that first resistant bacterium to get the mutation for resistance to the 2nd drug.Kleinman writes:
The math tells us what the variation is. You start with a single wild-type bacterium without any resistance alleles. When we have 3e9 replication with a mutation rate of 1e-9, that gives us on average 1 member in the population with any of the 13.8 million possible substitution mutations, The remaining members will be wild-type clones of the original founder. Of course, somewhere else in the world they may be other variants but try to understand the Kishony experiment first.
We know what the variation is for the Kishony experiment, you are starting with a population size of 3e9.
Taq writes: That doesn't tell us what the genetic variation is. A population of 3 billion that recently descended from a common ancestor is going to have less genetic variation than a sample of 3 billion individuals from a population that has been dividing for a long time period. It is similar to the genetic variation in 10 close family members vs. the genetic variation of 10 individuals randomly sampled from across the globe.Kleinman writes:
You have started to do the math but then you said you can explain your train of thought without algebra. That is vague. And your approach to the math was slightly different than my approach but your approach is correct as well. The problem with your approach becomes obvious when you try to do the math that would predict the behavior of the Kishony experiment when using two drugs instead of one. As you try to do the mathematics for the two-drug experiment,you will see why your approach doesn't work. The problem is that you are working with two linked binomial probability problems. Perhaps you can get your approach to work but I don't think so.
If you can't do the mathematics, the best you can have is a vague understanding of the phenomenon.
Taq writes: I did the math. Your only complaint seems to be that I didn't do the math the way you wanted me to.Kleinman writes:
Fixation is not applicable to the Kishony experiment. When you finally figure out the math for the two-drug Kishony model, we can discuss fixation in the Lenski experiment (except that is fixation for a haploid and that fixation is selective). This will give you practice in applying evolutionary mathematics to real, measurable, and repeatable experimental examples of evolution. Just in case you were curious.Taq writes: Just in case you were curious.
quote: So, try to figure out how large the population size has to be for a variant to occur with 2 beneficial mutations for 2 drugs to make one evolutionary step in the Kishony experiment if 2 drugs are used.
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Taq Member Posts: 10084 Joined: Member Rating: 5.1 |
Kleinman writes: How does the math work for the 2 drug experiment where some variant has to appear in the drug-free region with a beneficial mutation for each of the drugs before it can grow in the next higher drug-concentration region? I would assume it would be the multiplicative, so 3E9^2.
The math tells us what the variation is. You start with a single wild-type bacterium without any resistance alleles. Genetic variation is an observation. The Kishoni experiment does start with a single bacterium, but I am asking a different question. What if we start with a wild population that already has genetic variation?
You have started to do the math but then you said you can explain your train of thought without algebra. I did the math. I started it AND FINISHED IT. What I didn't do is construct all of the equations and cancel out variables using algebra.
Fixation is not applicable to the Kishony experiment. It is applicable to molecular evolution. If you ask for the math of evolution part of that math is the rate of fixation of neutral mutations. You will also get neutral mutations moving towards fixation in the Kishony experiment.
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Kleinman Member (Idle past 363 days) Posts: 2142 From: United States Joined: |
Kleinman writes:
That is incorrect. Do you understand the difference between joint probabilities when the individual probabilities are independent and when they are conditional? Because what you are assuming here is that the mutations in this circumstance are independent when they are actually conditional (dependent). If you want to understand that with words rather than with algebra, what that means is what is the probability that mutation A2 occurs on some member after mutation A1 has already occurred. Your calculation assumes that the mutations have to occur simultaneously.
How does the math work for the 2 drug experiment where some variant has to appear in the drug-free region with a beneficial mutation for each of the drugs before it can grow in the next higher drug-concentration region?Taq writes: I would assume it would be the multiplicative, so 3E9^2.Kleinman writes:
That's the point of correctly understanding the mathematics of evolution. If you go back far enough in time, that variant doesn't exist anywhere in the universe.
The math tells us what the variation is. You start with a single wild-type bacterium without any resistance alleles.Taq writes: Genetic variation is an observation. The Kishoni experiment does start with a single bacterium, but I am asking a different question. What if we start with a wild population that already has genetic variation?Kleinman writes:
You have just started doing the mathematics of evolution and that's just for one step of DNA evolution to a single selection pressure. And for a mutation rate of 1e-9, it takes 3e9 replications of a variant. And your math grossly overestimates the number of replications necessary for DNA evolution to 2 simultaneous selection pressures. The number of replications necessary for those evolutionary conditions is about 4-5 orders of magnitude smaller.
You have started to do the math but then you said you can explain your train of thought without algebra.Taq writes: I did the math. I started it AND FINISHED IT. What I didn't do is construct all of the equations and cancel out variables using algebra.Kleinman writes:
Fixation is part of molecular evolution but any fixation of neutral mutations in the Kishony experiment will be due to those neutral mutations hitchhiking on the variant with the beneficial mutation. But there is no fixation occurring in the Kishony experiment. The less fit variants (drug-sensitive variants) are still happily growing in the lower drug concentration regions while DNA evolution is proceeding. You need to understand DNA evolution in a non-competitive environment (the Kishony experiment) before you attempt to do the mathematics of DNA evolution in a competitive environment (the Lenski experiment). You should start by learning how to do the mathematics of conditional probabilities because that is the correct mathematics to use when considering 2 or more simultaneous selection pressures. Fixation is not applicable to the Kishony experiment.Taq writes: It is applicable to molecular evolution. If you ask for the math of evolution part of that math is the rate of fixation of neutral mutations. You will also get neutral mutations moving towards fixation in the Kishony experiment. If you want a simple example that demonstrates the difference between joint independent probabilities and joint conditional dependent probabilities, study the example of random card drawing, with and without card replacement. Here's a simple explanation of the difference between independent and dependent probabilities.https://www.youtube.com/watch?v=aBvWUr9hTLE When considering the application of these principles to DNA evolution and the Kishony experiment, the single drug experiment gives rise to independent joint probabilities while the multiple drug experiment gives rise to dependent joint probabilities. So, do you want to try to redo the math where you estimate 3E9^2 replications for the 2-drug Kishony experiment?
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Taq Member Posts: 10084 Joined: Member Rating: 5.1 |
Kleinman writes: That is incorrect. Do you understand the difference between joint probabilities when the individual probabilities are independent and when they are conditional? How is it incorrect? How are they conditional and not independent? Let's see your math.
That's the point of correctly understanding the mathematics of evolution. If you go back far enough in time, that variant doesn't exist anywhere in the universe. We aren't going back in time. I am talking about right now. What if we gathered E. coli from around the world and put them in a single population? What then?
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Kleinman Member (Idle past 363 days) Posts: 2142 From: United States Joined: |
Kleinman writes:
In the single selection pressure case, the evolutionary steps are independent. Each step is a new binomial probability problem independent of the previous step and a new sample space occurs for each step. You can see this in the Kishony experiment, mutations A1 and A2 occur in separate drug-concentration regions and these regions correspond to the mathematical sample spaces. This paper shows how you do the math in for this DNA evolutionary process: That is incorrect. Do you understand the difference between joint probabilities when the individual probabilities are independent and when they are conditional?Taq writes: How is it incorrect? How are they conditional and not independent? Let's see your math.Just a moment... For the multiple selection pressure situation, mutations A1 and A2 must occur in the same region if two drugs are used. So A1 and A2 are occurring in the same mathematical sample space as well. What makes this a conditional probability situation is that A2 must occur on the reduced sample space of those variants that already have the A1 mutation (or vice versa if mutation A1 occurs on some variant that already has mutation A2). Mutations A1 and A2 occurring on some member that has neither mutation do not give the drug-resistant variant but they will contribute to their respective subsets of either A1 or A2 which is why the joint probability of the A1-A2 variant is much better than 3E9^2 (but still requires about 1e13 replications for that variant to appear). Here's the paper which explains how to do the math for multiple simultaneous selection pressures:Just a moment... Kleinman writes:
You can do that but then you are not talking about evolution. You are talking about migration. What I'm trying to get across to you is how DNA evolution by common descent works. When DNA replicates with error, you get the diversification of the population. And when adaptation requires the accumulation of specific sets of mutations, that can be achieved by lineages in the population only if a sufficient number of replications at each evolutionary step occur. That process works most efficiently when only a single selection pressure is acting on the population at a time but for a mutation rate of 1e-9, it takes about 3e9 replications for each evolutionary step. That's the point of correctly understanding the mathematics of evolution. If you go back far enough in time, that variant doesn't exist anywhere in the universe.Taq writes: We aren't going back in time. I am talking about right now. What if we gathered E. coli from around the world and put them in a single population? What then? The Kishony experiment gives the most ideal environment to demonstrate this because his lineages only face minimal competition and don't require fixation at each evolutionary step. On the other hand, the Lenski experiment is carried out in a highly competitive environment (with bottlenecks every 6-7 generations). This slows the accumulation of the 3e9 replications of the most fit variants to get their next beneficial mutation. Kishony's lineages accumulate their 5 adaptive mutations in a matter of days, Lenski's lineages have taken more than 30 years to accumulate their 100 or so beneficial mutations.
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