 We looked largely at the biology, so there was not a lot of mathematics. We did a couple derivations. No, it's not true. We did do two bona fide derivations. But let me take you through just quickly what we did yesterday. So we're talking about cell division. This is the DNA, the chromosome. It gets replicated. Then the cell divides. We looked at some empirical relations. So basically what is sometimes called the fundamental results of bacterophysiology, these two back-to-back papers by Molas Group in 1958. So in steady state, and this is now balanced exponential growth, where you keep the cells in exponential growth for 10 generations before you take any measurements, they found that the RNA per cell increased basically exponentially with the doubling rate. At a higher rate than the mass per cell, which was, again, at a higher rate than the DNA per cell. They also did shift experiments, and you can rationalize the slope of these lines, at least for the mass per cell and the DNA per cell, by looking at the shift times and the kinetics. So if you go from a poor medium to a rich medium, the time it takes for each of these constituents, if you like, the mass, the DNA, and then the cell number to catch up to the new rate of division will give you the slopes of these lines on a log linear plot, which is surprising, but nonetheless we saw yesterday it was a self-consistency check, if you like. And it comes from the fact that these shifts are so abrupt, they're basically piecewise linear on a log scale. The RNA per cell, on the other hand, we rationalized in a different way, we rationalized it looking at the mechanics. What's really going on here? So that was something to do with the Neidhardt-Magasanic paper, where we saw that the rate of protein production by their assumption was proportional to the number of ribosomes, which then again was proportional to the RNA content in the cell, and that's what gave you this straight-line relationship at fast enough growth rates. With the interpretation that the slope of this line is inversely related to this translation rate, the number of amino acids that you make per second per ribosome. It's nice, it gives you an idea of mechanistic explanation for this increase in the RNA to protein ratio, and then possibly this increased slope in the RNA per cell. This connection between the slopes of these lines in these kinetic times is less clear. And so from about 1958 on, there was a search for some more detailed or mechanistic explanation for where these times come from and what was underlying the regulation of this switching to the new growth rates. So that's what we'll talk about today. It will probably take us till tomorrow as well, because it's an incredible experiment. So it takes some unpacking. But let me pause here. Are there any questions about what we did last lecture? No, I mean last day, so the last two lectures. Is that okay? And so I posted my notes on a link accessible from my home webpage, so if you have any trouble, try that out. Otherwise let me know and I can point you in the direction of the lecture notes and the course notes. Okay. So one of the big problems here with trying to infer mechanism, particularly at the level of single cells, is that all of these results are done on, say, 10 to the 9 cells or 10 to the 8 cells. It was a huge population. And so we're looking at population average results. And so the first question that we want to ask today, which will then lead us into resolving this issue later this lecture or maybe tomorrow's lecture, is when I say population average, precisely what probability density am I using to average? And what does the, for example, age distribution of these cells look like in a population? Okay. And so that's what we'll talk about today. And this is leading towards something called the Helmsteader Cooper experiments, but maybe I'll title this more dramatically as a search for synchrony. And by that I mean that in the 1950s, the technology wasn't yet developed to observe single cells in real time. The technology not yet developed to observe single cells. And so the genius of the biologists at this time was it, well, maybe we don't need to observe a single cell carefully. What if we could synchronize a population of 10 to the 8 cells? Then we wouldn't have to worry about measurement noise or anything like that. We would have this large population, but they would be synchronized. It would be acting as though they were a single cell. So the idea at the time was, can we synchronize the population, 10 to the 9, say, or 10 to the 8 cells? That would be effectively a large population of single cells, if you like, could be acting like an observable. I understand the reasoning here. So if this population is synchronized, everything in the population is doing the same thing at the same time, it's very much like just taking repeated measurements of a single cell. It's like an ergodic theorem, if you like. Yeah. Yeah, yeah, how do you know that they're synchronized without being looking? So one way you could do it is you could look and see if they all divide at the same time. Sure. Okay. Yeah, that's true. So you would have to come up with a criterion for what you mean by synchrony. And it will turn out that the criterion that we'll have is this division. So synchrony and age, they're the same age. Yeah, does that make sense? But then again, there would be an implicit assumption that the cell cycle within each cell is somehow the same. If you're at the same age, you're undergoing the same. And for a eukaryotic cell, that would be problematic for bacteria, it's not so bad. Okay, does everybody, is that any questions about this search? And so you can imagine maybe some tricks that you might try. One trick is that people try temperature shifts. So you would make them hot, cold, hot, cold, hot, cold, hot, cold. In the hopes that they would have, they would basically convulse into some sort of stasis. And then you would grow them normally and everybody would relax at the same time. Another trick was nutrient shift. And in fact, these experiments were designed to synchronize and they did not. Neither do temperature shifts. None of these things synchronize cells. And the reason that they don't synchronize cells is because the age is distributed through the population. Some cells are older and some cells are younger. And if you want to synchronize, that means you need to narrow that distribution to a delta function. And playing with temperature, playing with shifts, all it does is move that distribution around. So you don't have to narrow it in any reasonable sense. So what I want to talk about first is what is this distribution that I keep indicating with my hands. And then second of all, how can we resolve this? Or how did they resolve this? And I'll maybe give you a little foreshadowing, which is that it turns out to be more or less impossible to synchronize bacteria. But what you can do is sample, in a very clever way, sample strips from the population distribution in time, which serves the same purpose. I mean, it wasn't really designed to make sense at this point, but that's where we're headed. All right. So then first, what is the distribution ages in the population? What I mean by that is how close are some cells to dividing, or how close are cells to dividing? What's the distribution of that age, if you like? So by age, I mean, age here, I'm going to denote by A, and it's anything between zero and tau, where this is the doubling time or division time. And so you can have newborn cells, which have age zero by definition. And then you can have very old cells, according to their life cycle, which would be tau minus epsilon. And so we're going to work this out, what this age distribution is. But just intuitively, yeah, Matteo. It wouldn't be also the age of the division time, but in this case, I'm not. And so it could be that A is actually going to go from zero to infinity, but we're going to find that it's compactly supported if I make the assumption that tau is a delta function. But we don't need that. So in the derivation, I'm going to leave the distribution as the doubling times or division times as a distribution. And then when we get to the end and we want to normalize the distribution, that's when we're going to put in some additional assumptions just to make it a little bit easier. But for now, I mean this conceptually, you're right. You're free to go further, and some will. Is that okay with today? Yeah, sorry. Yeah, yeah. Yeah. Yeah. Oh, it ceases to be the mother at that point. No, no, no. I mean, it's not, it's not semantic. It means, I mean that it really does reset. So the two that come off at the end go back to the beginning. I mean, you recycle cells back again. You're right. And Matteo's point is that it's not a hard wall. I put it up here just to, so you have a sense of what I mean by age, but think of this tau as a sort of a fuzzy ending to their doubling cycle. Or if you like, you can think of it as a delta distributed. Everybody divides exactly 40 minutes after birth. That's conceptually fine as well. But is that okay? So this modularity is important, and we'll come back to it in one second. But you had a question? No. That was it. Okay, now my question is, given this modularity that he suggested, what can you tell me qualitatively about this population distribution? Anything? Uniform. Uniform, okay. Can it be uniform? Okay, Gaussian and uniform are perfectly legitimate, knee-jerk suggestions. Which is of course, as a physicist, that's what you would, I mean, that's what I would say. If it's not uniform, it's either Poisson or Gaussian. And that's totally legitimate. But there's something that we know about this process that's, if you like, unique to this system, which is exactly what you were saying, right, that mothers become two daughters. What does that imply about the population distribution? Pardon me? It's sort of power law, but then, maybe I'm not framing it well. Do you expect more young cells or more old cells? More young cells. Young cells. And how many more? Twice as many. There's always twice as many young cells in your population as there are old cells. Because all the time, the old cells are having, they're dividing into two. Which is very weird. I mean, now you can rationalize it backwards, but it's odd that no matter what distribution you have for these doubling times or whatever it is, you are going to have twice as many young cells as you have old cells. And what we'll do is analytically derive what this distribution needs to be, and you'll see that that is indeed true. But just qualitatively, I hope it makes sense, that because these cells grow, grow, grow, then divide and sort of recycle again, you're always dumping in from the end of the distribution, you're dumping back in again to the front twice as many. So it's like a periodic boundary condition with amplification, if you like. I don't know if that's helpful. Yeah? All right. And so we know qualitatively, twice as many newborn cells, cells about to divide. Divided. But what precisely is the distribution? And so let me introduce some notation and then let's manipulate this distribution until we can solve for it. So I'll give you a sort of, what would you call it, a synopsis of what we're about to do. I'm going to define these probability distributions, both for the division time and for the age distribution. And then we're going to use properties of statistical distributions, namely that you can look at cumulative probabilities to end up with a difference equation for that distribution, which is going to be horrible. But then we can use Taylor series to turn that into a differential equation which we can solve, lickety-split. And once we have that, of course, end normalization becomes an issue. And so we'll normalize after making an assumption about this distribution for these doubling times. It's not necessary, but for us it's fine. For what we're going to use this distribution for, it'll be sufficient. So let me first define what I mean by these distributions or what notation I'm going to use and then let's talk about it. So the cell age distribution or the age distribution is denoted by, well, I will denote it by phi A. And what I mean is that it's a continuous distribution which represents a probability that A bacterium aged between A plus DA, however you want to do this, A minus DA over 2, however, in some interval of width DA. To divide. Because the newborn cells will take pretty much the same time as their mothers for dividing. Yeah, more that mothers are dumping, every time they divide, they dump in two newborns. So the instant that they divide... Why are you calling them able to divide? They are as able to divide as the newborns. Cells about to divide. Oh, I mean at the threshold of division, their age is tau. Sorry, is that not clear? So cells about to divide, i.e. their age is about tau. So A approximately equal to tau is what I mean. I mean they're at the end of their career as a single cell. Is that better? No? Is it the about or is it what do I mean? Okay, what do we talk about? Okay, is this okay, this distribution? What do I mean by this? Okay, let me talk then about the distribution for the doubling times. So also, we have distribution for the doubling times, which I'm going to call f tau, lowercase tau, where here I mean that f is a continuous probability distribution that denotes the probability, the a bacterium, a somewhere between tau and tau plus d tau. Let me pause it. So these are going to be the two, the two sort of things that we're going to be playing with. This one we're going to be, we're going to assume it's a given. This is our input, if you like, and we want to find this output. So we need to find an equation that's going to transform our given distribution of doubling times to some probability of ages, if that makes sense, I hope. Is that sensible? Let's take a look. So I'm going to move this down so that I can use the board, but hopefully a between tau and tau plus d tau. All right, so the first step is going to be to look at the cumulative distribution of this doubling time. So I'm going to integrate it from some time up to infinity. I'll do that in one moment. But one thing that I want to call your attention to is that I'm going to be thinking of this as a steady-state distribution. And so there'll be no time dependence in this distribution for now. Well, forever in this course, but... All right, so this guy's steady-state. Now let's look at the cumulative distribution. So we'll introduce cumulative density. This is going to be called F greater than, which is the fraction of cells with doubling time. Okay, so you can think of it as the complement of the... Sure, I could call it A cumulative density then, rather than D. And so by definition, we'll have this. Well, let me put it... Let me put primes here, sorry. There we go. And so this is really the extent of the probability theory that we need. And now what we're going to do is manipulate this and try... Oh, sorry. And try to express a certain fraction in a second in two different ways. And that'll give us our difference equation. But let me pause. Is the notation comfortable for everybody? Anyone have any questions about notation? Yeah. This guy? Yeah, to the quantity of the age and the probability of the whole duration, like for example, a person can live 100 years, but then you're calculating how many... the probability of having age between, for example, 5 and 10. Exactly, exactly. But how can you calculate the mean division? Yeah, so this is an input. We either need to measure this or just guess what it might be. And here a Gaussian distribution would be maybe a sensible guess. Or something like this. Yeah, you would have to either have a guess or some empirical reason for your choice. You might even try a uniform between two bounds. But the idea here is that given this distribution, we want to chug it through the biology to get this distribution. Yeah, possibly. I mean within reason. But here what we're thinking of is steady state at a single growth rate. And that'll be important. If you want to do shifts and things like that, then you'll need to modify this derivation. But you should have enough tools to do that if you want it to. It's okay? Alright, let's take a look. Alright, so suppose that we have a bacterium that reaches age A. So then what is the probability that it will reach age A plus T without dividing? And this is kind of one of those classic assignment question in probability and statistics. But let's work through it. So is that comfortable? So we've got them, we know that we're at age A, whatever that age might be. We have some finite interval, A plus T, and I haven't said what, you know, T is just some finite interval number. And we want to know what's the probability of getting that far without dividing? Okay, so here my delta T is just T. So you can think of this as delta T. Is that okay? No? Okay, alright. So let's look at it. This is age A, this is age T, this is age infinity. Alright, and so this, we go like this, F greater than A is the fraction with doubling time greater than A. By definition, that makes sense. And in this one, this will be F greater than A plus T, which is the fraction with doubling time greater than T, or sorry, greater than A plus T, if that makes sense. Is that alright? That's by definition. Okay, now we want the fraction right like this. So given the bacterium reaches age A, probability, no division in this interval A to A plus T is going to be equal to one. And you can either tell me directly or you can, we can argue pictorially and you can tell me what element of this picture I'm talking about. Can you say one more time? Yeah, I agree. So what is this fraction? This fraction is probability that no division occurs between 0 and A plus T. Is that true? So the point here is that we've got only this rectangle to worry about. We've said because it's a conditional probability we're at all this stuff with probability one we've reached age A. Okay, so this dark rectangle is going to be the fraction with doubling time between A and A plus T. Exactly. Can I pause for one second? I'll just pause. Take a look at that and tell me if you believe it. Now given that rectangle what's the fraction of that rectangle? Of course it's problematic because one of the sides of the rectangle is infinitely far away but you can just bend yourself there. The doubling time and the division time same thing, sorry. So a doubling time is the division time. They're synonymous. Same. So they divide the numbers double. So I want the light part of this I want to know what fraction of that rectangle is not hatched. Does not have lines on it. Yeah, that's true. That's true that this might, this is optimistic that it's probably starts to be one after that but it still stands if T is sort of small enough then you would think that there is an appreciable fraction that is here and appreciable fraction that's here. It doesn't really matter for us if it's appreciable or not. Just abstractly what would be the fraction of this rectangle. It's not this but that was a good guess. That would be if A is 0. So I'm going to write it. There are a couple of suggestions up here but it's the ratio of this to this because what we're interested in is this piece. We want to know the relative area of that piece relative to the whole rectangle. Let me write it up and you tell me if you believe it. This is going to be f greater than A plus T divided by f greater than A. And that's going to be the area of the part of the rectangle that's kind of next through it. So two questions. One, does that match your intuition for what these functions are representing? And two, does everybody see what I'm trying to calculate? Anybody Nazi? Anybody want to back up a bit? Is that okay? I guess maybe not okay. Okay is not the word. Does that seem reasonable? Yeah? Alright. Okay. So here what I'm asking is given that you start here, what's the probability that you make it here rather than dividing here? Okay and the way that we do that is we'll ask what the probability is for this whole rectangle and then look at the probability that you missed the hatched part of it. And undoubtedly you can manipulate that in different ways. Right? But what I want to do is look at it as this fraction because I want to express it then in terms of this age distribution at the moment. Alright so if we have that, let's start fiddling with it. So now suppose we start N bacteria with age zero. So we have a perfectly synchronized initial population. Now I want to ask how we write that fraction in terms of this exponentially growing population. So the number of bacteria at age a a plus d a will be so if I start with N bacteria at age zero and then I have this distribution what's the I shouldn't have put them at age zero. Sorry. Start with N bacteria. Suppose we have N bacteria. What's the number of bacteria that have an age between a and a a plus d a? Given that we have a probability distribution for the age. Yes. So we'll get N multiplied by phi a d a. Does that make sense? So this instant right now now suppose we wait a time t what's going to happen? So now let me write one more thing and let's talk about it. Number undivided a plus t is going to be this. So this will be number undivided age a plus t will now be what? So suppose I have that many bacteria and if that fraction I can multiply the two and so I'll write this up and let's talk about it. By a d a times this fraction I promise you we're getting somewhere for right at the threshold of getting somewhere. So far so good. This is the number that are at this age so that's the number that are here and then we have this fraction that makes it for a finite step delta t or t in this case. So far so good. So does this match then with your intuition for what these functions mean? It's okay? Alright. Here's where the biology comes in. So over this interval a plus t the population increases. So suppose I'm imagining he's exponentially growing cells in a steady state what's going to happen over in finite interval or non-zero interval t will the population increase too? Did I give you enough to go on? What's that? So we have n cells right now these cells are in balanced exponential growth. We wait to this time so at time a so t is equal to 0 we have n cells at time t is equal to t we have how many cells? After a period t what happens? What's the population? So t is a finite amount of time what's lurking behind here is we're in balanced exponential growth so what can you tell me about this population? Think of n as being so large that our population is continuous I feel like I'm not giving enough information I'm going to write it up and you tell me if this is true wait maybe that's what you were saying were you using a base 2? So you were saying 2 to the doubling rate times t is that what you're saying? Because this is also the same thing as n 2t divided by some doubling rate t times mu some average doubling rate What's that? Yeah we need to we're eventually going to get f back again Oh I see Okay I think we should pause and kind of step through this okay so your cells are growing exponentially balanced exponential growth we have this idea that we have a distribution for their ages we don't know what that is we have a distribution for their doubling times we also don't really know what that is but empirically we do know that the population itself is growing exponentially and so then the question is given that I know this is true so this is going to be my empirical input or constraint that I need to satisfy that I have some unknown function for the doubling times and for the cell age what can I do with it how can I find this from this and this right but this is an empirical constraint that we have to deal with the cells are doubling exponentially obviously irrespective of I mean if we were free to choose any distribution of doubling times imaginable probably we could no we couldn't we can't break that we don't know what the distribution of doubling times is precisely but we do know that the cells are growing exponentially is that does everybody is that okay do you agree that this end over this interval will have to increase exponentially okay so then if we want to look at let me go one more step and then let's talk about it so then the fraction undivided cells is a filing it will be then fraction undivided a plus t will be this big mess and then let's talk about it greater than a plus t f greater than a t a divided by n e lambda t so this is the number that are undivided at a plus t but this is the total population size at time a plus t and so the fraction is dividing both of these and so if I carry one more step I will get let me do it here I will get by a f greater than a plus t f greater than a by a and disappears by e minus t okay but then I want to express this as a fraction I'm assuming all of these cells are independent so this is also a probability what probability is it which is the same as maybe it's best if I write it up here i.e. the probability a cell has age a plus t to a plus t plus d so let's pause the threshold now of actually being able to derive an equation for that phi irrespective of what the distribution of doubling times is okay so let's go back to this guy is that one let's go way back okay and in this empirical constraint that our cells are increasing exponentially and we have a fraction of undivided is going to be that number divided by this total population and then we tidy it up a bit and then we just cancel this exponential I'm putting in the numerator but then I look at that this is the fraction that are undivided at a plus t so then that tells me that this is a probability that a cell has an age a plus t an a plus t plus d a is that okay and now my question to you is is there another way to write that you exactly right so in terms of phi what would it be exactly phi a plus t d a and so this where we've advanced the probability distribution ahead by a finite increment is equal to this and that gives us our difference equation so let me leave that up and then let's let me rewrite it over here phi an a plus t and I'll pause here and then we can talk about it equals phi greater than a plus t over greater than a e minus t and so the problem then becomes you give me a distribution of doubling times and you obviously also need to tell me the growth rate of the population those are my inputs and then it's incumbent upon me to solve this difference equation for phi it's subject of course to normalization things like that and it's horrible you don't want to solve this thing but anyway let's talk about and make sure that it's okay and what we'll do is take a Taylor series for small t and solve the differential equation and then what I ask you to do is an exercise at the end of those lecture notes is prove that the distribution you get from the Taylor series also satisfies the difference equation but for now let's just take the Taylor series but before I do that is this equivalence comfortable anybody have any questions about this equivalence and so I wouldn't expect you to reproduce this argument it's far too many steps for a person to keep in their head so don't worry about memorizing this it's more important to me that you can you can see that it's baby step baby step baby step baby step gives you what crazy solution in the end alright and I'm actually even more interested that you see the distribution later even qualitatively but I want to take you through the calculation once so that if you felt like it in future you could go through it without any trouble with a minimal amount of labor alright so that said is this equivalence comfortable and so then we have this difference equation how do we solve it so how do we solve this alright so we will use a perturbation expansion so we will use a Taylor series for t very small ok so maybe we'll do that we'll take a break for maybe 5-10 minutes or something and we'll come back and solve this guy but before we go any questions before you go alright they might develop so yeah I'll see you in a bit to hand this out again yeah for sure give you before which is this Bayes theorem ok and that's that's his derivation and so just to remind you so I'm going back to the rectangle that I had which I think was not so great maybe so this is f greater than a plus t and then I had this guy which is f greater than a alright and what we were looking for was again the probability so given the age age what is the probability no division between a and a plus t i.e. it's a survival probability that you reach the age a plus t ok and so just to remind you this is Bayes theorem we've got this conditional probability is expressible as a joint divided by the initial probability if you like and so in terms of what we're talking about we're looking for the conditional probability that given that the doubling time is greater than my present age what is the probability that the doubling time is also greater than my age plus t alright so this conditional probability is precisely what we're after is that clear clearer I hope probably than this rectangle but then we can express that as this joint probability that the doubling time is bigger than a plus t and the doubling time is bigger than a divided by the probability that the doubling time is bigger than a so far so good we pause alright now it's lovely inside is that this is a redundant restriction that if the doubling time is greater than a plus t t is non-zero and positive then this is automatically true it's redundant the first condition implies the second and so then you end up with this probability that the doubling time is greater than my age plus t divided by the probability that the doubling time is greater than my age but then that's just expressible in terms of a cumulative function f greater than a plus t divided by f greater than and so that rectangle that I showed earlier was a misguided attempt at circumventing phase is that okay? I hope that's better for those for whom this the rectangle is not good hopefully this brings this all together does it? good so that's the new way that I will teach this, thanks alright so let's go to the Taylor series I'll leave that up if you want and we'll come back to it okay so if we take this use a Taylor series with t going to zero then we can re-express these finite increments as infinitesimal changes so we'll Taylor expand this we'll also Taylor expand this okay so we have that f greater than a plus t and we have phi a plus t and we have e negative lambda t only here am I going to keep the order symbol so I'm only going to take it up to first order and then I'm going to throw out all the order t squared stuff okay so what I end up with is f greater than a plus t f prime greater than a plus t phi prime of a and then here I get 1 minus lambda t and all these are plus some order t squared okay and I'm going to truncate here and ignore the rest is that okay everybody's okay with this Taylor series so the idea is that I want to get rid of this finite increment and so now if I plug that into here what do I get the first step is a mess with a little bit of algebraic manipulation we can pull it together maybe I'll pass this around I always forget to pass this around alright so then if we put this into our difference equation it becomes this we'll end up with phi prime of a over phi of a is equal to f greater than a prime f greater than this is nice and we get this ratio of derivatives over their functions and so this immediately suggests to us that we start using natural logarithms the next step is less obvious but you'll be able to see in retrospect that it's okay so let me write that up and then let's talk about it this is da times the natural logarithm of phi of a is equal to d by da the natural logarithm of and normally we would just be looking at the natural logarithm of this guy but that lambda is going to give us a little bit of trouble except that we can group everybody together e negative lambda a so if I've done everything correctly I think that that will give you the same expression and so now what we have is a separable differential equation before it didn't look like it it looked horrible I guess you could see that it was a little bit separable but anyway here's a better form for it and so we can integrate let me pause though from here to here via this is I hope okay any questions though it's okay so from here to here okay at least in retrospect I mean I don't think any same person would rewrite this like this without knowing what the answer was alright so if we integrate we'll get phi of a is equal to some normalization constant which we don't know yet e negative lambda a f greater than a which we can now rewrite in terms of our our distribution for doubling times so now we have phi zero which is this normalization constant and I'll come to that in a second times this integral from a to infinity of f of tau prime and that in in principle is our distribution for ages which is contingent or relies upon this assumption that my population is growing exponentially and it's in some balanced state of growth alright and so we haven't said anything about f yet this distribution of doubling times nor have I said anything about the normalization condition but we'll come to that are there any questions about the solution of this differential equation first off does this seem reasonable alright even without looking at a particular value of this distribution function or this normalization qualitatively can you say anything even without an explicit what can you say about two term and this term as a gets larger so phi at zero is newborn cells and now as a function of a what can you say about the remaining two terms you got it they're monotonically decreasing the exponential I think is okay this guy's monotonically decreasing with a this guy is two though it doesn't matter what this probability distribution is this one minus accumulative is always going to be monotonically decreasing and so we know that this guy is as big as the distribution gets there are always more newborn cells it doesn't matter what your distribution is um phi zero is always the max e negative lambda a and this integral monotone decreasing and so now this comes from a derivation by Powell from 1957 to derive the normalization condition for arbitrary f but we don't need to do that I mean if you're interested you can look at how he does it but for us what I'm going to look at is a delta distributed distribution or a delta distributed um doubling times so it's possible to solve for the normalization arbitrarily but it's a lot of work well not a lot but it's work alright so for us it is, it will be sufficient to consider delta distributed doubling times or division times so this f tau prime is going to just be a delta distribution a direct delta distribution around some macroscopic or fixed doubling time so this will then give us that phi a phi a is going to be phi at 0 e negative lambda a or 0 so this is for a 0 and tau and this is otherwise if I did that right sorry which part this part ok so suppose now we put a delta function so the delta function is 0 except for one point and then if we integrate across that point then we get 1 and so if a is above tau this delta function will be over here in the interval we'll miss it, we won't integrate across it does that make sense but then if the a is going to be less than tau then we'll have a delta function that will integrate over and we'll get 1 and so this thing is either going to be 1 depending if a is greater or less than tau ok let me write that up and then let's talk about it for one more second i.e. we have that this integral delta t prime minus tau dt prime is equal to 1 if a is less than tau and 0 if a is greater than tau then your definition of how to integrate the delta function you could put an equal to sign on the top one or the bottom but let's I'll just leave it, unless it drives you crazy in which case we can decide what we want to do with it so everybody understand what I'm saying so it's not well defined when I integrate with this tau right at one of the boundaries is what I'm saying so I'm going to ignore the boundary and I'm going to pass now is that distribution is this function now sensible given this definition for the delta function it's okay any concerns good so now we'll look at the normalization and then we'll be we'll be able to plot this guy and then we'll be done so after normalization i.e. enforcing this condition the continuity of phi a d a is equal to 1 we end up with the following solution for phi 0 we get phi of a is equal to and you can do this if you like or you can take my word for it e negative lambda a right so long as a is less than tau but I'm going to make one more change in a second but then we know that I mean in this case where the tau is a macroscopic doubling time that that means we're just rewriting the natural base e in a base 2 if you like and so this lambda which is my exponential growth rate is related to tau which is my base 2 doubling rate by the natural logarithm of 2 so this thing let me write one more thing which is that lambda is equal to long 2 over tau if we had a general distribution for the doubling times that wouldn't be true and so that's another hassle about dealing with generalized distributions for the doubling time but for delta distributed that's by definition true okay this is the time it takes me to increment by a factor of 2 this is the time it takes me to increment by a factor of 2.7 and so then I end up with let me write one more line and let's talk about it 2 long 2 divided by tau 2 minus a over tau for a up to 1 and so let me pause first of all is that okay? does it mix with the definitions we've been using so far is basically my point anybody feel okay and so normally or often or often we we use a relative h say a hat which I'm going to divide by tau in which case this distribution becomes the following and I do this just because I want to be able to write it in a very compact way so then we get phi of a hat is going to be equal to 2 long 2 of 2 raised to the negative a hat for a hat going between 0 and 1 now my question is okay let me pause there's been a lot of algebra hidden underneath the sink here but this is the this in the end is what I wanted to get at today I wanted to get at this because for two reasons well for one main reason is because when we look at the experiments that were done on trying to synchronize cells this distribution is going to show up and it's important to see where it came from I think but before that what does this distribution look like this is phi this guy so this is only for this f of tau prime equal to t minus t prime that's part of the problem with the general distribution is not only do you need to normalize you also need to relate the empirical doubling time to this distribution to some moment of that distribution does that make sense and so that derivation I'll post that paper and you can look through it so it's not part of the papers that I have on the website right now but I'll make it part of that web those papers do you know what I mean if you're interested in the general derivation I'll put it up on the website for this part it's okay so if you are interested in the more general argument I'll make that available to you let me pause though any questions it's okay alright what's this distribution look like that there's a question this guy minus oh wait no because it's in the basement I know you're right negative land yeah thanks yeah very nice okay what does this function look like yeah perfect it just looks like this and so this thing is two long two and this thing is one and so the relative abundance of babies is twice as much as their mothers if you like or not that's not quite right they're twice as much as those just about to have babies or just about to divide alright let me pause any questions about any of these okay alright so this distribution is where we're gonna start I don't know when this one ends does this one end what's the time that these things end at 11 10 oh okay okay alright so maybe I'll say ah you know this is a good place to stop so next next time so that will be tomorrow at 11 we'll we'll start with this distribution and so the genius of a guy named Helmstetter is that he's gonna come up with an apparatus that samples as a function of time tiny strips from this distribution and so essentially he has a time series of synchronized cells and it took him 10 years to come up with this idea and um maybe I'll give you alright I'll give you just one sort of um what do you call it foreshadowing that he imagined himself lying in bed and a bunch of chickens on his ceiling dropping eggs on his head and that was the sort of the epiphany for him so with that in mind imagine how you would sample this distribution and we'll talk about that tomorrow but let me pause again any questions about this so the details are in the lecture notes and in the course notes and as I say I'll make the more general derivation available through the website it's okay? alright ok see you guys tomorrow