Analytical Derivation of One-Locus Two-Allele Selection Dynamics

[00:00:02][Music] Hi. So now last discussion we simulated using Microsoft Excel uh how one locus 2 al selection case uh behaves and uh before we go forward I would like to say that some of you might have found it a little difficult to follow me and uh create the excel sheet the way we uh did it. So for those people uh you can actually you know go to this uh website.

[00:00:49]This is the one where we had uh gone previously to simulate Hardy Weineberg.

[00:00:54]So I'll quickly take you over there.

[00:00:57]So you can go here and you can say start and then yeah uh go to individual simulations.

[00:01:09]Remember we are assuming that all other conditions of Hardy Weineberg are operative. So we will take population size as infinite number of generations actually you know as we saw 200 or so is good enough. So we'll keep that here and then you have to go to additional settings and go to selection and then you are going to have all these fitness coefficients for A1 A1 A1 A2 and A2 A2 set these and you will be good to go. So let's say we'll put at 6 and we will put this at uh.9 and let's say we put this as uh whatever four and just say run simulation and you will be able to see you know the frequency of the alals as well as the frequency of the genotypes.

[00:01:58]Okay. So everything that we did with our Microsoft Excel simulation you'll be able to do over here. uh I would still recommend that if possible try to do the Excel thing because that way you actually understand step by step how the process is happening and uh I mean while this kind of simulators are very convenient from a teaching perspective they don't really give you the same level of intimacy with the subject the way uh you know handcoded thing like the one we did in Microsoft Excel has. So this is only for those people who are finding it difficult to work on uh a spreadsheet. They can look at this one.

[00:02:44]Okay. So let's get back to our PowerPoint. So in our last discussion we said that you know we made a bunch of observations from our Excel simulations and we said that in order to understand those observations in greater detail we will need to analytically figure out what's really happening in the one locus two al selection case. So I'm going to analytically derive the expression and I'm going to know draw various kinds of graphs and mathematically show what the dynamics is likely to be under different situations. Now quite frankly this particular discussion that we are having today is probably mathematically speaking the most involved discussion in the whole you know 30 hours that we are going to spend together.

[00:03:40]uh this particular derivation is slightly advanced in the sense that most regular textbooks of evolutionary biology don't carry it. It's only the slightly more theoretically oriented books and then also not all of them who show this particular derivation. Uh it would have been very easy for me to just give you the formula and then request you to take it on faith and go from there. But my personal experience of teaching this for many many years at Isa Pune is that until and unless you actually see where the derivation is coming from and what kind of uh assumptions are being made at various places, you don't really get the feel for the whole thing. So, uh I'm going to do the derivation. This is going to require one or two very basic concepts uh in uh algebra and uh a little bit of calculus, a very little bit of understanding of probability.

[00:04:39]Uh I will try to explain it in as much of detail as simply as possible. I'm trying to aim for those people who might have taken a course in maths in class 12th or maybe they took their last course in class maths course in class 10th but they are still not afraid of the maths thingy and therefore you know they would like to follow what I'm trying to do just in case this derivation you know it seems somewhere in the middle that it's a little too tough for you or too many concepts you are not really figuring out how I'm getting there. Uh I would still request you to go ahead and see what are the main results that are coming and in case even that becomes too difficult for you don't bother too much. Okay. In the next discussion, I will actually start that discussion with the result that I'm getting today and I'm going to discuss the implications of those results without going into the maths part of it.

[00:05:50]But as I said, evolution is one of the most highly formalized subject you know mathematically formalized subject in whole of biology. So understanding this particular derivation and where it is coming from is actually going to be very very useful and uh which is the reason for which I'm going to great length to present it over here. So in order to do this we will head to a different software. This is uh Microsoft whiteboard and huh okay so before we even start doing the derivation there's one little concept that I have found people often you know students they have a little bit of an issue in comprehending that people who have done a course in algebra for them it's not really an issue but for others sometimes it is an issue so I'll just present this thing for you over here.

[00:06:50]So suppose you have uh okay let me slightly increase the yeah this is good let's say you have five numbers four four five and five okay so four three times five two times and I tell you that please take the average of this what will you do you are going to do 4 + 4 + 4 + 5 + 5 and since there are five numbers over here in total you're going to divide the whole thing by five and that is going to be your average right now I can write this thing as 3 * 4 + 2 * sorry 2 * 5 divided by 5 right 3 * 4 + 2 * which I can write as 3x 5 into 4 + 2x 5 into 5, right? Which I can further write, you know, think about it.

[00:08:07]What is 3x5? 3x5 is the number of times four is there in this entire sample. In other words, 3x 5 is the frequency of four and similarly 2x 5 is the frequency of five. Therefore, I can write this as frequency of a value multiplied by that value itself.

[00:08:38]And then this entire thing this is you know the summation sign this entire thing summed over all values right? In other words, if you have a bunch of numbers and you need to take their average, all you need to do is to figure out the frequency of each number and then multiply that number with the corresponding frequency and add them all up. Add you know all these products and that is going to be your average.

[00:09:17]Right? very very simple concept but this is a concept that's worth keeping in mind before we start this entire thing.

[00:09:26]Okay. So the main derivation starts now.

[00:09:30]So we are talking about the one locus sorry two alil selection.

[00:09:42]Okay. And if you remember yesterday that I said that we are explicitly talking about the viability selection model viability selection right so what are our two alles let's assume that the alles are a1 and a2 and these alles let's further assume that their frequencies are p and q where q is 1 minus p. So we'll be going back and forth between Q and 1 minus P. Okay.

[00:10:15]So this obviously leads to three different genotypes. A1 A1 A1 A2 and A2 A2. Right? So these are my genotypes.

[00:10:30]And remember these genotypes before selection their frequencies are P square 2pq and Q ² right so these are genotypic frequencies before selection okay now we have selection happening and we have explicitly viability selection happening What does it mean? That means that each of these three genotypes they have a different uh fitness. Let's call them W11, W12, W22 and let me slightly increase the point of the pen here. Let's W11 W12 W22 and these are the fraction of the genotype that is surviving right so the no progeny has been born and now they're surviving while they're going to adulthood and each genotype has a slightly different survivorship and those are the ones that are being given by W11 W12 and W22 so these are our genotypic fit uh sorry genotypic fitnesses.

[00:11:53]Right?

[00:11:55]Now, if this be the case, what are the frequencies after the selection? So, after selection, the frequencies are P ² W1, 2 PQ, W12 and Q² W22.

[00:12:12]Right? So these are genotypic frequencies post selection.

[00:12:20]Right?

[00:12:23]Now as we saw yesterday during the simulation these things are not going to sum to one and therefore we need to do some scaling so that all these three frequencies they sum to one. What is that scaling? that scaling is essentially dividing each one of them by the by this sum. Now this sum we are going to what is this sum? This is p ^ 2 w11 + 2 pq w12 + q ² w22 right now what is this? As you can see, these are the values of the genotypic fitnesses multiplied by the corresponding frequencies. Right? So if that be the case, these by what we discussed just now, these are the average genotypic fitness in the population. In other words, w bar which is this sum is equal to p ^ 2 w11 + 2 ppq12 + q q w22 and that is the average sorry average genotypic frequent uh fitness right.

[00:13:46]So therefore this is unscaled what we have over here.

[00:13:55]And the scaled fitnesses these are P² W1 sorry W11 by W bar 2 PQ 2 PQ by W bar and Q ² W22 by W bar. Right now before we proceed forward, it's actually going to be very useful for us to derive a certain quantity. Okay, a certain concept and that concept is the concept of the marginal alilic fitness.

[00:14:43]Now what the hell do I mean by that? Now think about it. Fitnesses are actually never derived at the alic level.

[00:14:52]Fitnesses are defined at the genotypic level. Okay. A1 A1's fitness, A2 A2's fitness, A1 A2's fitness and so on and so forth. So what do I even mean by saying that I want to look at the fitness of an alil. So here we borrow a concept from the you know subject of economics and the concept is that of a marginal value. What essentially means is the marginal alytic fitness is the fitness of all the average fitness of all the individuals in the population who contain at least one alil A1. Okay.

[00:15:35]In other words, uh let's see. So marginalic fitness marginal alic fitness is the average over average fitness of all individuals who have at least at least one A1 LE. This is marginal fitness for A1. Okay, at least one A1 LE.

[00:16:21]Now, how do I get to this?

[00:16:25]Which are the genotypes which have at least one A1 L?

[00:16:29]A A1 A1 and A1 A2. Right? A2 A2 has no A1 L. So, forget about it. Now what is the genotypic fitness of A1 A1 that is W11. What is the genotypic fitness of A1 A2? That is W12.

[00:16:46]Right? Now we have to do the average fitness. For this we have to somehow scale it with some measure of frequency.

[00:16:54]How do we get there? Now think about the definition. The definition says average fitness of all individuals who have at least one A1 L. Right? So let's say that this A1 al is fixed right now if one al is fixed as A1 we have already picked it up what is the probability assuming random meeting of course what is the probability that the other al that you pick is also going to be A1 under the Hardy Weineber conditions which is where we are operating in over here that is equal to the frequency of the al A1 in other words that is equal to P.

[00:17:36]Similarly, assuming that one AL already has been picked up as A1, what is the probability that the other al is going to be A2? And that probability is simply Q. So W1 star which is the marginal quitness of al A1 is equal to P W11 + Q W12.

[00:18:04]Okay. Similarly let's think about the marginal liquidness of al A2 which is W2 star equal to what is going to happen?

[00:18:15]Which are the two genotypes which have A2? One is A1 A2 the other is A2 A2 right? So in this the genotypic fitness of this is W12 genotypic fitness of this is W22 right now given that one al has been fixed as A2 what is the probability that the other al is A1 that is equal to the frequency of A1 which is P. Similarly, given that one AL is A2, what is the probability that the other AL is also A2? That is equal to the frequency of A2 which is Q. Right? So the sum in other words W2* is equal to PW12 plus QW22.

[00:19:05]Okay, these two quantities although it's not entirely clear to you as of this moment why we need these two quantities it will become clear in a few minutes. So in order to figure out what really is happening we come back over here and we start looking at this W bar.

[00:19:27]Okay. So we said this W bar is equal to P ² W1 + 2 PQ W12 + Q² W1 22. Now I can write this as W bar is equal to P ² W1 plus the 2 PQW12 I will break it up as PQw12 plus Q you know another PQ W12 okay just writing it in two parts plus Q ^2 W22 right equal to this is there's a P Here there's a P here. I'll take the P as a common. So P common PW11 + QW12 plus here I'll take Q common P W12 plus QW22 right and if you just look at what we derived 1 second earlier PW11 plus QW12 is W1 star and PW12 plus QW22 is W2 star. So if we come back over here, this simply becomes equal to P W1* plus Q W2 star. Right? This is my W bar.

[00:20:55]This is another thing that we need to keep in mind.

[00:21:00]Right?

[00:21:01]Now let's keep a third thing in mind. We again go back to this definition of uh w bar and now let's start writing it you know in all in terms of p. So what are we going to do? So w bar is equal to p² w11 + uh this is 2pq w12 right so 2 p and q I'll write it as 1 - p okay w12 + q ² w22 so q ² I'll write it as 1 - p² w22 easy equal I'll expand it. P² W11 + 2 P W12 2 P W12 + rather minus 2 P² W12 + so 1 - P whole square you know is 1 - 2 P + P² so 1 into W22 is + W22 minus 2 P from inside into W22 - 2 P into W22 + P² W22 P² into W22 simple algebraic uh expansion nothing else equal to [Music] yeah okay actually this is fine now I want to know how W bar changes is with change in P. Okay. Now in order to know that from calculus we know that we need to look at the quantity DW bar by DP right. So we are going to differentiate this with respect to P. Now if you have forgotten differentiation just you know take it from me that hopefully what I'll do is correct. So 2 P W11 plus 2 W12 minus 4 P W12.

[00:23:29]This W22 goes away. minus uh 2 W 222 plus 2P W22 right now if you remember your calculus you can see that there should be a few extra terms which I'm not putting over here why am I not putting them because I am explicitely ly assuming that the W11's W12s and W22s all these are constant. In other words, I'm explicitly assuming that these things do not vary with P.

[00:24:16]If I did not make that assumption, then I would have to add three extra terms over here. We'll come to that towards the end of today's discussion. But as of this moment we are going to assume that all these w11 w12 etc these are frequency independent.

[00:24:37]So what we are explicitly assuming is that there is frequency independent selection happening.

[00:24:50]Okay, frequency independent means that w bars the values of uh sorry the values of w11 w12 w22 they do not depend on p.

[00:25:00]So that be the case let's go forward this is my dw by dp. Now I can write this dwar dp is equal to see I have a two everywhere right? So I'll just take this two outside and this becomes pw11 plus uh w12 minus so I have taken two outside so I have 2 pw12 inside so this 2pw12 I'll write this as minus pw12 and I will again write it as minus pw12 okay 2 has gone outside 2 PW12 remains.

[00:25:50]I have split it into minus pw1 to - pw12 minus pw12 and then 2 has gone outside so I have - w22 has gone outside so I have + pw22 right equal to two outside PW11 is fine plus so this is w12 and w2 one2 over here. So I can take the w12 common w12 1 minus p right so this term is done this term is done this term is done I'll take the minus outside and this will become pw12 uh + w22 minus pw22 okay now Let's equal to 2.

[00:26:58]This is pw11 and 1 minus p is q. So this is + qw12 minus I have pw12 over here and I have w22us pw22. I can take w22 common over here. Then this will become W22 into 1 minus P which basically means QW22. In other words, I have PW12 plus QW22 right equal to I think you can already see where this is going. PW11 + QW12 remember is W1 star. So equal to 2 W1* minus W2* right this is the third relationship that we need to develop that DW bar DP is equal to 2 W1 star minus W2 star assuming that our genotypic fitnesses are frequency independent again I cannot cannot stress this enough. This is assuming frequency independent selection.

[00:28:19]Great. So now that we have these things, let's come back to our main stuff.

[00:28:27]Remember we started with trying to figure out the uh recussion of P due to selection across generations. So these are the things that we have got. So now al frequency in next generation okay in the among the offspring that is what is that we'll call that as p hat uh sorry p prime is equal to how are you going to get it remember this is my capital p this is my capital q then p prime is simply p plus/ q in other words P prime is equal to P ² W1 by W bar plus half of this 2pq by W bar in other words I'll write it as PQ by W bar okay now equal to W bar is the common denominator so W bar is my common denominator equal do uh I will take the P outside PW11 plus sorry there is a PQ12 over here. Okay.

[00:29:55]PW11 + QW12 by W bar. Again the stuff inside over here you can see this is W1 star. So equal to P into W1* by W bar. This is my next generation al frequency.

[00:30:16]Now suppose I want to look at the change in L frequency. In other words, I want to look at delta P which is defined here as Pdash minus P. Okay. By how much has selection been able to change the al frequency over one generation? Great. So what is that value going to be? So we just saw that p prime is this. So delta P is equal to P W1* by W bar minus P equal to P W1* minus PW bar divided by W bar or I can take the P outside P by W bar W1* R minus W bar.

[00:31:20]Okay.

[00:31:22]Now if you remember we at some point when we were looking at you know our expressions we saw that W bar is equal to PW1 star plus QW2 star right we have already kept this one over here. So we will import that one over here equal to P by W bar W1* minus minus P W1* minus Q W2 star right equal to let's take it a bit forward equal to uh p by w bar.

[00:32:16]This is w1 star minus pw1 star. I can take the w1 star common. In other words, this one become will become w1 star into 1 minus p which is qw1 star. So equal to QW1* minus Q W2 star equal to I can take the PQ uh I can take the Q outside equal to PQ by W bar W1* minus W2 star.

[00:32:55]Now if you remember when we were talking about dwar dp we derived that dwar dp equal to 2 w1 star minus w2 star under the frequency independent condition. In other words w from this we can see that w1* minus w2* is equal to half of dwar dp.

[00:33:23]Right? So now I can take this relationship over here and I can put it over here and say that delta P is equal to P Q by W bar there's a half so I'll put a two over here into DW bar DP right This is one of the most famous equations in evolutionary biology known as civil rights equation for fitness landscape.

[00:34:16]Okay. Sometimes people take it one step forward and what they do is okay let me do it over here. Uh they write this as delta p is equal to pq by 2 uh d of lawn where lawn is a natural uh logarithm uh lawn war dp. Okay. The reason this works is because don w bar is equal to 1x w bar dw bar. So this is also another form in which the equation is often talked about. Although this is the form in which we are going to analyze it. Okay.

[00:35:04]So why is this equation so important?

[00:35:08]Why is this? You know, I'm saying it's one of the most famous equations.

[00:35:13]To understand the implications of this equation, what we first need to do is to just look at its structure. So see that it has P, it has Q, it has W bar over here. What is P? P is an al frequency.

[00:35:30]So we know that it will lie between 0 and 1, which means it's positive.

[00:35:35]Similarly, Q 1 minus P will also lie between 0 and 1, which means it's positive. W bar is the average genotypic fitness, which basically means this is average of lots of values, each one of which is between 0 and one. So, average of lots of positive values means itself it's going to be positive. Therefore, this stuff over here, whatever is, you know, inside this circle, this thing is always positive.

[00:36:04]That implies that the sign of delta p is going to be equal to the sign of dwarp p. Now what is delta p? Remember delta p is p dash minus p which means by how much is the al frequency changing in one generation due to selection. So what this is saying is that the sign of that is going to be the same as the sign of DW by DP. Now what exactly do I mean by that statement?

[00:36:41]What I mean by that statement is that if you end up having P on X-axis and W bar on Yaxis then you're going to get some kind of a graph and we'll talk about it in a few minutes. what kind of a graph you will get but that graph let you know it's uh let's say you will get it something like this or let's assume that you end up getting something like this whatever you can get all kinds of shapes okay but whatever you get for any value of p the sign of delta p in other words whether p will increase in this direction or p will decrease in this direction that is going to depend on the slope of this particular curve at that point. Okay. If the slope is positive, delta P is positive, P will increase. If the slope is negative, delta P is negative, P will decrease.

[00:37:42]Okay. Now, let us further examine this W bar. It's a fantastic, you know, thing.

[00:37:51]So we started by saying that this is our W bar right? P² W1 plus let me reduce the size a bit. Uh yeah P q W11 + 2PQW12 + Q ^2 W22. Now as you can see this is quadratic in P. What do I mean by that?

[00:38:15]What I mean is that the powers of P, the max power that P can have is two, right?

[00:38:23]It can't go beyond a square. Now, we also know that if you have a quadratic, then that quadratic is going to have at max one maxima or one minima, right? And that maxima or minima is going to happen when DW bar DP is equal to zero.

[00:38:43]Now what is the meaning of DW by DP is equal to zero. The meaning of DW by DP is equal to 0 is that the AL frequency is not changing. It has gone to an equilibrium. Right? Now what are the ways in which this equation can go to an equilibrium. So if you think about it this equation can go to an equilibrium.

[00:39:05]So this is P right? So P you know is going from 0 to 1, right? So when P is equal to0 that means that P =0 means that Q is equal to 1 which means that the entire population is full of A2 A2 individuals right there are no A1 A1 individual there are no A1 A2 individual so if that be the case at this point the average genotypic fitness of the population is basically just W22 because there are no other indiv individuals.

[00:39:42]Similarly, when P is equal to 1 and Q is equal to zero, what will happen? You only have A1 A1 individuals in the population. There are no A1 A2. There are no A2 A2. In other words, at this point W bar is going to be equal to W11.

[00:40:01]Here W bar is going to be equal to W22.

[00:40:04]Right? So that point on the this axis can lie anywhere. Similarly over here you know it can lie anywhere. These are the two fixed points. Once the population has gone here you know either fixed for L A1 or fixed for L A2 nothing can change. But barring those two points, what is going to happen in the middle for intrinsic values of P. Now suppose you have a scenario like this where the slope of this curve is always positive.

[00:40:43]Then as I said delta P is always going to be positive. In other words, P will always increase and you will get a fixed point over here at P you know P equal to 1. Similarly, if you have a scenario which looks something like this, you are going to have you know uh equilibrium here at P is equal to zero. It will go like this. Okay. But I mean these are the things that we saw yesterday, right?

[00:41:12]We saw that for many many cases the al frequencies were either at equilibrium were either going to zero or going to one depending on who had a higher fitness. However, we did end up seeing certain situations wherein the al frequency was going to an equilibrium at an intermediate value of p. In other words, some p which was between 0 and 1. So at these positions dwar dp is going to be equal to zero.

[00:41:45]Now if dwar dp is going to be equal to zero. We have already figured out that DW bar DP is equal to 2 W1 star minus W2 star. Which means that at equilibrium, you know, 2 W1* minus W2* will be equal to zero. In other words, W1* will be equal to W2 star.

[00:42:14]Now, what is the formula for W1 star and W2 star? So remember uh let me just get this above. Huh? Remember P W11 plus QW12 that's my W2 star will become equal to PW12 + Q W22 that's my W2 star. So let me collect all the P's and the Q's on one side. So P W11 minus P W12 is going to be equal to Q W22 minus Q W12.

[00:42:55]So I take P as common W11 minus W12 is equal to I take Q as common W22 minus W12 right at equilibrium. Remember that's the condition we're dealing here. Now think about it. P as we just discussed is an al frequency. So it's always positive. Q as we just discussed is an al frequency. So always positive.

[00:43:27]Therefore forget about the magnitude.

[00:43:29]Just think in terms of the sign. The sign has to be equal on both sides.

[00:43:32]Right? So this relationship in terms of its sign can be true if and only if w11 and w1 uh uh sorry w1 and 1 and w22 either both of them are greater than w12 or w11 and w22 both of them are less than w12. Why am I saying that? Because let's take the first condition. W11 is greater than W12. So this is positive.

[00:44:06]W22 is greater than W12. So this is positive. Positive positive. Great. But suppose that's not the case then you'll have positive on one side and negative on the other side which cannot happen.

[00:44:18]Right? Therefore this W11 W22 both being greater than 1 12 W12 will ensure that you are having a positive sign on both sides. Similarly, if W11 is less than W12, you're going to get a negative sign over here. W22 is less than W12, you you'll get a negative sign over here.

[00:44:38]However, if you have a scenario where W11 is greater than W12 is greater than W22 or the other way around, then this is not going to work out. You are not going to get an internal equilibrium.

[00:44:51]Right?

[00:44:53]Now, great. So we now know the conditions that will lead to this equilibriums. But remember we are dealing with a quadratic form. In other words, our equation or you know graph is going to be a parabola. Now a parabola we know can either have a maxima or it can have a minima. Now knowing this doesn't allow us to tell which one is which one. Right? So in which case we are going to get a maxima which we are going to get a minima that's not entirely clear. So in order to figure that out what do we need to do? We simply need to take a double differentiation right. So we'll start from this point. Okay. So we will differentiate this again sorry into W bar by DP² equal to we'll take the keep the two uh aside. Okay. Uh this is going to be equal to W11.

[00:45:56]This W12 will go remember this is 2p W12. So this is - 2 W12 and this W22 will go plus PW22 means plus W22.

[00:46:11]Right? So this is my double differential. So we know that when this double differential is negative that's when we are going to get a maxima and when this double differential is positive that's when we are going to get a minima. This is what calculus tells us right? When is this thing going to be positive? This thing is going to be positive when both W11 and W22 when both of them are going to be greater than W12. So this case right. So when W11 and W22 both are greater than W12 then we expect a minima.

[00:47:02]And similarly when W11 and W22 both are less than W12 we expect a maxima.

[00:47:13]Right?

[00:47:15]So with this knowledge now let's start plotting.

[00:47:20]So as I said you can have three kinds of graphs over here.

[00:47:26]Okay.

[00:47:29]In all these graphs we have P the al frequency of al A1 on X axis and we have W bar on the Y-axis.

[00:47:40]Okay.

[00:47:41]You can have a situation where the thing is you know posit uh negative or positive throughout. The slope is negative or positive throughout. And in all those cases irrespective of where you start your P the DW by DP is always going to be positive or always going to be negative. And therefore P will either you know go in this direction and get fixed at one or it will go in this direction and get fixed at zero. Okay zero. So this is zero and this is one.

[00:48:19]If you have a situation which is like this w11 w22 is greater than w12 then you will get a graph which will look like this. Okay. the minimum will be somewhere in the middle. And this situation where W11 W22 is greater than uh W12.

[00:48:42]This means that the hetererozygous form is less fit than both homozygous forms.

[00:48:48]This is what is known as under dominance. Under dominance.

[00:48:59]Okay. This is the case where w11 w22 is greater than w12.

[00:49:08]And this situation where the hetererozygous has the maximum fitness and compared to the two homozygous conditions. This is the case where you will get something like this.

[00:49:23]Okay? Where you are going to get a maxima somewhere in the middle.

[00:49:30]Now obviously these points these are the points at which DW bar dp is equal to zero. In other words, these are the points at which P has stopped changing.

[00:49:41]These are equilibrium points. However, there is a qualitative difference between the nature of these two equilibrium points. Why that is so?

[00:49:50]Let's see. Let's first start with this one. In this particular case, what is happening on the left side? The slope of this curve is positive at all the points.

[00:50:03]Right? So slope means dw by dp. So dw by dp is positive to the left uh is greater than zero to the left and it is less than zero. dw by dp is less than zero to the right. Therefore, what did I say about delta P? Delta P will have the same sign as DW by DP. So on this side, delta P is positive. Which means if the AL frequency is anywhere in this zone, it will go here. Okay?

[00:50:37]Whereas this side it is negative. So if the AL frequency is anywhere in this zone, it will go in this direction.

[00:50:45]Right? Now what happens when the AL frequency goes like this? This is W bar.

[00:50:49]Right? So the population the ali average genotypic fitness of the population will go like this for this side and on this side it will go like this like this right now what exactly happens over here as we said this is an equilibrium point but once you have reached this point if you slightly perturb the system if you slightly change the al frequencies it will come back to this stating that this is a uh no stable equilibrium point.

[00:51:24]Okay. Uh I forgot to tell you that the situation where the heterrozygos uh condition is the most fit this is what is called overdominance.

[00:51:33]These are technical technical terms over dominance and underdominance.

[00:51:39]Great. So in other words in this situation the fitness is always increasing. Now what is going to happen over here? In this case again this is an equilibrium point but the slope of dwp is positive on the right hand side. So if you slightly perturb the system the system is going to go in this direction over here till it hits one or if you start perturb it in this on this side of this equilibrium point it will go in this direction and it will hit zero.

[00:52:13]In both cases, what happens to the average fitness? The average fitness in both cases goes up, right? Even in this case, the average fitness is going to go up. Okay? What happens here? In this particular case, here if you see that the slope is positive over here. So, any point where you start the average fitness will go like this, right? Because delta P is always positive. So P will go in this direction. So it will go like this. And for the other one, this is negative. And therefore it will the average fitness will go like this. Delta P will you know keep on reducing and the average fitness will go like this.

[00:52:58]In other words, in all the graphs that we are seeing, we see that the average fitness of the population is always going to go up. Okay? And this is what leads to the thought process in the minds of most people that the beauty of selection is that it is always going to increase the average fitness of the population.

[00:53:28]That's what selection does. Okay. Note that that intuition is mathematically well supported but it is well supported only in this scenario that we are talking about the one locus 2 al scenario under the frequency independent selection case.

[00:53:45]Now as you can guess if I'm stressing on this harping on this so much then obviously that there has to be a twist in the tail. What is the twist in the tail? The twist in the tail is related to this thing that we did over here. Okay, we said that because this is frequency independent. Therefore, we are dropping three terms. Okay, what are the three terms that we dropped? So, let's now go to that scenario.

[00:54:14]Let's go to the scenario that where you have the frequency dependent case.

[00:54:21]Okay, frequency dependent genotypic fitnesses fitness. Now we are looking at dwar dp. Okay. And all the terms that we did over here each one of these terms they will all exist. And therefore we are going to get the final thing that we got which is equal to 2 W1* minus W2 star. This part is constant. I mean the similar however on top of that there will be three other terms. What are those three terms?

[00:55:05]Remember now we are explicitly assuming that these things these three genotypic fitnesses are function of P themselves.

[00:55:12]And therefore the three terms are going to be p² dw11 dp + 2 p q d w1 2 dp + uh q² dw 222 dp. Okay, this is the rule of uh differentiation.

[00:55:39]Now look at these three terms. What are they? These three terms are three differentials. The differentials of the corresponding genotypic fitnesses multiplied by their corresponding frequencies.

[00:55:56]Right? Therefore this itself these three terms together they form an average. An average of what? An average of the genotypic fitnesses. So this or rather the derivatives average of the derivatives of the genotypic fitnesses.

[00:56:11]So this can be written as two W1* minus W2* plus average can in statistical sense is also known as expectation. Expectation of DWDP.

[00:56:30]Note this is not DW bar. expectation of DWDP which essentially means the average over these three derivatives of these three genotypic witnesses right so if this be the case then then remember we need this stuff stuff right so then w 1* minus w2* is going to be equal to half of this goes this side right. So this will become half of DW bar DP minus expectation of DWDP.

[00:57:23]Okay. And this now we have to you know superimpose this stuff on uh okay where are we just one sec.

[00:57:41]Ouch.

[00:57:44]Okay. So yeah. So this is where this gets super imposed. So delta P so let me write it properly properly. So for the frequency dependent selection delta P is equal to PQ by 2. This uh part stays the same. PQ by 2 uh expectation PQ by 2 W bar sorry uh DW bar DP minus expectation of DW DP.

[00:58:47]Right?

[00:58:49]Now here comes the first big major problem. What is the problem? The problem is that it's not at all clear that these two terms will go to zero at the same point. In other words, when DW bar DP will go to equal to zero, it's absolutely not certain that DWDP should also go to zero. In fact, in many many cases, it will not. Therefore, dw by dp going to zero doesn't necessarily mean that your delta p has gone to zero.

[00:59:25]In other words, you have a situation where it's not entirely clear that you know you will definitely get a internal equilibrium. In many many cases, what you actually end up getting is all kinds of complicated dynamics including oscilly dynamics. In other words, it doesn't settle to an internal equilibrium. The fitness doesn't always increase. Sometimes it increases, sometimes it decreases. And under this situation, you can also have scenarios wherein due to selection, the average fitness of the population will keep on decreasing finally leading to extinction. A phenomena which is technically known as Darwinian extinction. Right? Now just think about what we are saying.

[01:00:12]We are saying that when you have the one locus 12 in selection case then under frequency independent selection frequ sorry frequency independent genotypic fitnesses selection can increase the population fitness average population fitness. However, even under the simplistic one locus 2 LL case, the moment the fitness itself, genotypic fitness itself becomes dependent on frequencies, this is not going to happen and you can get a situation where selection leads to way more complicated dynamics, oscilly dynamics or even extinction and this is with one locus.

[01:00:52]The moment you go to two losi or more than two loi the interaction of selection actually becomes very very complicated.

[01:01:02]This is why I wanted to show you guys this particular derivation because if you talk to any normal person even biologists forget about you know non-scientists even scientists we have this implicit notion in our head that selection will always improve the species selection or the population selection will always increase fitness that entirely comes from you know this simplistic IC rights uh equation which is this one. The moment you go out even slightly of that simplistic highly restrictive scenario, you are no longer guaranteed that selection is going to increase fitness.

[01:01:48]Selection can do many many things. And people like us who actually work with selection and know experimental evolution in our lab, we actually know how complicated a thing selection can be. Okay. So it's only under the simplest scenarios that selection improves fitness. Under even slightly more complicated scenarios, selection doesn't really improve fitness all the time. It improves fitness times. It or at least some of the times, but definitely not all the times. So we are going to stop over here and when we come back for our next discussion we are going to ask what are the implications of this and I'm going to start that discussion with some of the insights that we are getting over here and we'll probably let me see if we have time we'll quickly do some Excel simulations to drive home some of uh these points that we are making and uh we will explicitly answer the three or four observations that we had last time and we'll explicitly answer why those observations are the way they are. In fact, if you just take today's discussion and go back to those observations, the answers should become very clear to you. But for the sake of those people who are finding the mass a little tough, I will end up discussing those explicitly in the next discussion.

[01:03:11]See you. Bye.