But it is related in a sense because the conference is mainly concentrated on various random structures, but this will be about quasi-periodic quasi-periodic matrix, which is sort of a transition between deterministic things and random things, so I think it's sort of relevant in the sense Okay, so So and this will be the most well-known quasi-periodic operator because it's the simplest one and it models electron on a two-dimensional square lattice in perpendicular magnetic field So it's sort of lattice potential and the magnetic field on top of it and that creates possibility for quasi-periodic phenomenon Okay, so this is the we just look at this the following operator This is H as an operator on L2 of Z It's sort of given by the given by the following expression To lambda cosine Okay, so this is a okay. So as I said, it's models 2d electron So on the square lattice in perpendicular magnetic field so The parameter alpha Comes from the magnetic field and the parameter lambda is a hopin parameter on the lattice So if lambda is equal to 1 we have uniform lattice Square lattice and if it's not one then one direction is preferable to the another direction Okay, so this leads to some interesting Phenomenon this competition between magnetic field and the lattice So namely, we would like to ask the question. What is the spectrum of this operator? How it looks like and then we observe the following that for Irrational irrational alpha we have Following cases so first lambda larger than one then it turns out that This corresponds to the so-called localization which means that the spectrum is pure point The pure point spectrum with exponentially decreasing eigenfunctions And this is true for almost every alpha and theta But not for all So if lambda equals one we have a so-called critical critical regime Critical critical regime So here we have a singular continuous spectrum and this also as I said lambda equals one corresponds to the Uniform uniform lattice that is where this interactions in both directions are the same Sort of this is the most interesting therefore case. I think Okay, and lambda less than one so lambda between zero and one So this is a delocalization Delocalization It corresponds to absolutely continuous spectrum or all alpha and that So this is a kind of understandable in a sense that if lambda is very large then this part dominates And so it's obviously pure point and if lambda is zero it is sort of a cosine as eigenfunctions So it's a capitol of the continuous, but the transition is exactly at the point lambda equals one in this sense Okay, so but this is a So the next thing is that we will be so I will be kind of explain you what happens To the spectrum as a set So well, I just look at the spectrum of H So we will denote it simply Sigma of H So as a set this is bounded set on the real line and So how does it look like right? So so first of all if alpha is rational Then it is a standard standard result for Periodic operators that it consists of Q intervals or P over Q theta This Q intervals General they can touch but in this case they normally don't touch so if Alpha is a rational I will just so far remark that that this Sigma of H alpha theta is independent of the phase theta Yes, and in fact looking back here physically significant is not the spectrum Not this spectrum, but rather the union of all set us because the the the proper parameters alpha lambda and this This is a kind of Auxiliary so we would like it to be we're interested in the union of all over set up but for alpha irrational doesn't matter because It is as I said doesn't it does not depend sort of adjust their Godicity property of this Of the same, okay So we will also denote denote Billy back measure of this Set the following way so now we can formulate a theorem which says that for all Irrational alpha The real Theta and Lambda no not equal to zero the spectrum is Is a counter set of measure one minus lambda? One minus long four times one minus long Actually, we can restrict our attention to positive lambda because it's very simple transformation relating positive and negative Okay, so This Serum is a very very long history. It started somewhere in In the 60s already. I think in the 60s and so now this kind of it used to be known as conjecture So the measure conjecture used to be called Aubrey Aubrey and andre injecture And the counter set conjecture became known as a 10 martini problem because Ten martini was apparently offered by mark cuts for anyone could to solve this conjecture and and Barry Simon then Propagated this this name and so it became now it's sort of Called ten martini problem He's a problem and now I will give you some history of the of the proof of this Some history of the proof Okay, but first I would like to make a kind of a note for the critical case if lambda is equal to one So then the Lebesgue measure is zero and it follows from here immediately that The spectrum is a counter set. So because the counter set by definition is Closed no, but then said to that isolated points I Absence of isolated points for this kind of from the general or Gothic series Can be reduced and any spectrum is of course closed and the only thing which remains is this nowhere dense So basically the statement is that it is nowhere dense Okay, so and if it's measure zero then immediately it's clear that in nowhere dense. Otherwise, it's cannot be Okay, so but if lambda is not equal to one of then it is not clear because it's positive measure So it can be a counter set that may not And can happen Okay, so the history of the of the proofs Maybe one of the first rigorous result was to show that the sigma is Contour set for a dense set of pairs Lambda alpha was done by Billy Sardin Simon. I think it was 1982 okay, so then Then it was shown the next that sigma is a counter set For alphas which are very quickly approximable by rational Approximations so they they are they are called I call I call them sort of vaguely willion alpha So for you know alpha there they have a measure zero the typically this is which are proximal by rationals Faster than exponential exponentially quickly fast so So so this this was done By joy Elliott and and you in 1990 it was very kind of important development Okay, so then then there were some some works on the Continuity of the spectrum with the spectrum of irrational is approximated by spectrum of rational Approximates from here it came it came a very nice result that sigma That the that the measure of the of the sigma Get a measure of sigma is equal for minus minus lambda or almost every For almost every alpha and I can be even more precise and say which which alpha Actually such that that the limit when you approximate by by the rational Approximates your alpha So then then this limit should be zero So these are the irrational switch have coefficients of a continued fraction expansion from an unbounded sequence and They form a set of full measure and for that for that last I proved this formula in 1993-94 here's the two papers. So in particular if lambda is One If lambda equals one then it follows that there's zero measure Okay, but but then there was there was there was a problem in his proof because well Not in the proof, but he couldn't prove it for for all alpha because you know They didn't have enough Strong enough continuity for for the spectrum and so that remained for a while unsolved Then then actually that's what that's how I started with with this Well, it's not started, but but then we were now paper was Lana Zhitomirskaya We actually could show that this the measure is Given by this formula for lambda not equal to one like this and Dollar-Rational for the rational alpha. So this is our paper was in 2002 I think Done, okay, so Yes, so that that then we use the fact that In sort of supercritical regime we have exponentially decreasing Eigenfunctions that allowed us to improve continuity. Okay, but then it wasn't clear how to how to Do it for the case for the critical case first of all and then it wasn't clear How to answer the question about the counter structure and Here the the next step was to show that that sigma For lambda not equal one is indeed a counter set or for some diaphanthine Irrationals, so this is done by push in 19 in 2004. Yes So this diaphanthine irrationals are those which are poorly approximated by rational So as I'm with the golden mean is a standard Representation says standard irrational this time But it's so they form this this form set of measure zero Okay, so So then then and still it didn't didn't didn't cover the case lambda equals one so then then The fact that the measure of this sigma is equal to zero for lambda equals one and And all the rational alpha the rational alpha so was done in the work of Avala and Gregorium 2006 so here it has some Significance for the for the continuation of the talk. So sort of here they used they used The theory of dynamical systems in some sense so dynamical dynamical systems It is rather lone paper since 30 30 pages 40 pages So to to extend so to extend the result the result of last diaphanthine diaphanthine Alpha so the argument works there for for diaphanthine alpha and So together with the with this kind of result of last It settles the settles of the case of lambda equals one and so what remains what remains here is just to to show that it is a counter set for for the rest of Irrational alphas when it is not equal to one right so then it will be and this was completed that Sigma or Lambda not equal to one is Country set for all for all rational alpha so and this was a joint work of Avala and Gtomerska in 2009 I think Okay, so this is a this is a brief history of The proof of this of this property So now so we have now a new work with sweet Lana Where we found a much easier argument for the critical case So so that that I would like to present a little bit. So from now on we could see the only Cedar the case Lambda equals one Everything which I will be telling you from now on will be about lambda equals one and I remove your lamp Okay, so and in fact what we what we do that in this case we we actually give two elementary proofs of The serum for this case lambda equals one. So Namely, we show that Sigma absolute value of Sigma is equal to So measure of Sigma is zero for all irrational alpha. And so our argument work at the same time for all irrational alphas Rather than separating actually works so argument works and so the the idea of the of the proof is just to Use a special unit retransformation Which transforms this operating in a different form and this Kind of utilizes the fact that the problem is two-dimensional and you can the right the magnetic field in different gauges And this is sort of corresponds to the change of the gauge of magnetic field But we will write a couple of moments later on the the new representation Okay, but so what what new can we do this is kind of reproof of the known result, but we could Also prove something new and it's also an interesting Interesting question It has also long history From the 80s and so the question because it was conjectured for a long time that for this critical case the spectrum is Has a zero measure. So the natural question, of course is what is the what is this set? How can I characterize a set and the the most? Obvious question is what is the house dwarf dimension of the set, right? So so question question What is house dwarf dimension of the sigma? Okay, so So here I Maybe I will remind you about the well just the definition of the house dwarf dimension To make it clear So first we first we define the so-called house dwarf measure Responding to parameter t of a set a so a is a subset of the real line So and we define measure in the following way, so that's a limit on delta tends to zero of the infimum of the sums what value E so a is covered by this intervals and This is a delta cover Take a infimum of all delta covers and then you take the limit when delta tends to to zero And so what is a delta cover that it's just that here why these are the interval intervals of length Less so equal than delta, okay So if you look at this object, you will see that very often it will be there is a zero infinity and in fact If t is very small then it will be infinity And if t is close to one it may be maybe zero And so there there is there is a certain point where it switches from zero to infinity and this point is known as a house dwarf dimension so house dwarf dimension of the set Can be written as an infimum over t such that this house dwarf measure is is finite some point it will Start becoming infinite and this is it. So if we have a set on the real line this Dimension will be somewhere between zero zero and one zero and one right, so this is the the question So now So thawless 1980s So by by numerical numerical computations So observed and as he observed that actually measure of the spectrum on the in the rational case p over q and union over all factors This measures it behaves some constant divided by q when as Alpha is a as p over q approach itself some irrational So it is Rather small and this led him of course to the conjecture again that it is which was Slightly earlier that it is kind of zero that for alpha irrationalities as zero measure But of course, you know, there is a question of Proving that and a question of proving continuity and there are two questions so So if if we assume that if all intervals are of the same size If all q intervals Of the spectrum of this rational operator The were Of same size Which is in reality not not so But if they were all at the same size The size of one the size of one would be or is some constant over q q n squared or q squared Or q of them q squared and So this suggests That the household dimension is one half Because if you take it to the power one half and multiply by q the number of integrals, it will be a finite number And if you take something larger than one half, uh, then It will it will be small So currently zero and Smaller it will be infinity or it will be something it will be Something growing with q Right. So sort of intuitively it it would follow that So So one Injections that House dork dimension of this sigma is one half For irrational alpha So this is Unfortunately, it's not true In general, it may still be true in some sense Maybe for almost all but in general it is not true because there are exceptional cases. So, uh, so they exist Leovillian alpha Such that this house dork dimension Is actually zero It's very very thin set So this was, uh shown by by last And uh shamis Rather recently in 16 But on the other hand, uh, it is also known that they exist. They exist, uh some Daffentine alpha Such that this dimension is positive positive So this was uh done in the work of uh, elfer With uh collaborators I think you and Joe If I'm not I might miss a name here. Okay, so So in principle We have this a little bit mixed picture on the other hand. It was shown by last that that actually Or Some uh Daffentine alpha this dimension Uh is less Is less or equal than one half It was done by last in 1994 Okay, so The apparent there is also there also a conjecture apparently due to billy sar that actually Uh, it should be, you know, the same number for almost all irrational alphas Okay, but uh, what what did we prove there here is that we prove the following theorem So So we say that for all Irrational alphas A real theta So this dimension Is uh, actually less or equal than half For all irrationals So, uh, as I said, we do not expect it to be equal to one half It might be that it is equal to certain number less than one half or equal to one half for almost all that would be reasonable Okay, so So now, uh, maybe I will give you the main element of the proof Because there are several elements Give you the one which sort of gave us the right idea Uh, how to proceed Proof is the following We know that there exists a unitary transformation Which is uh, which is explicit That that shows the following that uh, the the union of the spectrum of h Two alpha theta is equal then the union of Of a set of spectrum of h with head alpha theta Where the operator with head acts the following way So it it's a gives you two sine two pi Alpha n minus one plus zeta Times v of n minus one plus two sine Pi alpha n plus zeta Times v of n plus one And that's it. There is no diagonal. There is no diagonal term and uh, so why uh, so this first of all physically This corresponds to the change of the gauge of magnetic field. So it sort of utilizes the two dimensional structure of the of the model And why is it uh, why is it good it is good because sometimes this, uh Objects become very small for some n This becomes very very small sometimes And here uh, here there is no chance because it's one and one here. So diagonal elements sometimes become very small And so if you look at this matrix Like this For example, this is small and this is symmetrically small And if this is very small you can see that it's practically splits into finite dimensional blocks So since uh Of diagonal elements can be small So this h hat of let's say quasi separates So which allows to obtain a better continuity statement for the spectrum So which allows to improve continuity result on the spectrum a result Uh on the spectrum on the spectrum as h p over q Sort of approximates approximates h alpha And from here we after that after that one can After this statement is improved sufficiently one can use the the argument used by last before in 1994 So and from here from here So the the result follows basically from here the fact that absolute value that Measure is zero and dimension sigma Less than equal one half follows So in fact, I should say that This is much easier statement the measure zero for that only a tiny improvement of continuity Was sufficient But in order to prove this much it's a much stronger improvement improvement of continuity Yes, so this is a I think uh should I finish already or I have Uh-huh. So maybe I just tell you what about about this continuity a little bit because it sounds a bit mysterious So Yes So for that I just tell you how last obtained his result So last On that for Rational case or the rational case The spectrum the union over theta Is between something bounded some constant over q some other constant over q is explicit constants Something like eight times e some pi times something in this constant So which combined was a combined which which is which which has combined them with continuity statement following that For every point in the spectrum of the alpha Irrational There exists a pointy prime in the spectrum of Rational approximate union over theta Such that the distance between e and d prime Is less than some constant times the square root alpha minus p over q That is a continuity statement, which was obtained by simon van moosh And avran previously So what it means is that So suppose you have the your points of the spectrum of your presumable Contour set of irrational Irrational situation irrational h alpha And so it says that they cannot be very far away from the from the intervals of the rational approximate So we have intervals of rational approximate and must be somewhere over here And and this and this boundary where they can deviate The size of this deviation is of order square root alpha minus p over q This is what it says simply And so then it gives a simple bound for the for the measure So sigma of h alpha then Is bounded by uh sigma of h p over q Over union over theta Absolute uh yes Over union over theta of this Plus uh the error. What is the error? The error is is this interval Times the number of edges The q interval to q edges. So so it's to q Some constant and the root alpha minus p over q And and that is it because This thing He showed us one over q. So it if you increase q it it becomes smaller and smaller And this thing will will tend to zero As p over q tends to alpha and q q increases for almost every Alpha that's why I wrote your condition before but not for all So in other words the right hand side of this inequality can be made arbitrary small From which is follows that the measure this one should be zero So it then follows that the measure is zero. So it follows that measure of sigma of alpha theta Actually zero For almost every Alpha so this is an argument of last for the household dimension one can Use a bit somewhat similar construction Okay, so if we if we can improve this Then then of course We will we will have that it will be true for for all for all alpha Because even a slight improvement of this root there is one half plus epsilon This is then true for all the rationals All the rationals and so but we we actually we show that Certain sense often We have this e minus c prime Where so equal some constant and the first power of alpha minus p over q slightly worse with some logarithmic Badness here. Yes. Not not for all points, but and not for all Thetas but often And so that of course immediately Sort of if you q is multiplied by absolute value that it is goes to zero for all Irrational and also it allows us to To have this continued to household dimension statement as well as continuity and this continuity We could only obtain because we had this nice structure of the matrix with matrix elements being very small Which means that uh, you can So technically here you need to cut off the test function and it is very nice to cut off Exactly at these points where you know, it was a very little error Because in a previous representation, uh, it wouldn't matter if it's always one Okay, so that's why so it works. Okay. Thank you very much


Hello guys, I am Nazrin Naseer. Here are 4 things you need to know for building a successful restaurant. As a restaurant, one must make sure business appeals to the audience it is targeting. They will only update your menu and ensure that the food introduced is top notch. The people that make up your team will have just as much to do with the success of your restaurant as your food will. Through the forgetful host or waiters can cause the experience of a guest to go from great to poor in a sentence. It is advisable to invest in high quality, comfortable furniture. Make sure your furniture choices are complimented by the rest of the room from paint color to wall decorations and lighting. Make sure you are marketing your restaurant effectively. It's a great idea to claim your business on local search pages like Google Places, Google Maps, Yelp, Swarm and Facebook. Thank you and have a good day.


TRED, Traffic Root Extraction and Anomaly Detection is a machine learning algorithm designed to extract and analyze maritime traffic patterns from raw AIS data. It is able to process varying levels of intermittency, sensor coverage and performance, as well as time lags between successive observations. This article was authored by Karna Bryan, Juliana Pallana, and Michelle Vesp.


The challenge of the Yukon. The Wonder Dog King, swiftest and strongest of Esmoli dogs, blazes the trail through storm and snow for Sergeant Preston as he meets the challenge of the Yukon. Sergeant Preston was typical of the small band of northwest-mounted police who preserved law and order in the new northwest country where the greed for wealth and power led to frequent violence and bloodshed. But in spite of the odds against them, Sergeant Preston and his Wonder Dog King met that challenge and justice ruled triumphant. The story behind the fabulous Valera diamond was one mixed with good and bad fortune for those who possessed it. It seems strange indeed that this diamond, resting safely in the United States in 1897, was destined to be the cause of death in the Yukon, a territory as incredible as the almost legendary story of the precious stone itself. In the old Barlow mansion in New York City, there was a deep quiet. The servants talked in hushed whispers as nurses in crisply starched uniforms went back and forth through the echoing halls to Mrs. Barlow's bedroom. Doctors held lengthy consultations. There were consultations too in the kitchen as the cook brewed an early morning cup of tea. Here now, Adams, just settle yourself comfortable. What you ever need of it is a good cup of tea. You too, Mary. Well, so Mrs. Barlow will hardly last the day. Yeah, she's pretty far gone. Young Mr. George is a mighty sad man. Is it he didn't have enough on his mind? End it. Just what are you meaning by that? Why, that girl, Emily Quinn, her and her father left for Seattle yesterday. From there, she says they're going to the Yukon. The note he got this morning was a bad blow to him, I can tell you. Oh, the better she's gone, if you're asking my opinion. That forsaken Yukon country is the place for the lights of her. No family background at all. His mother would never hear of the match. His mother will have nothing to say about it. She's a dying woman. That's the only reason young George didn't take the next teamer up with the Yukon himself. Well, he's up with his mother now. Poor soul. She knows there's not much time. George, dear. Yes, mother? I want you to promise me something. Try to rest, please. No, listen to me. It's the diamond. What about it, mother? Promise me you'll never give it up. I'm an old woman and perhaps a foolish one. But the stone has brought misfortune to anyone who's ever gotten it through wickedness. I promise, mother. Remember, dear, it's the only thing I ask of you. It was a few days after Mrs. Barlow's death. In the library of the old house, George Barlow spent a hurried few minutes going through miscellaneous papers. His cousin, Henry Drake, walked slowly around the room peering at the titles of the books along the shelves. Well, that's that. All finished. At last. George, about the diamond. Huh? Aunt Mary told me she wanted you to keep it with you. As a matter of fact, she made me promise I always would. Why? Taking it with you. I gave my promise, Henry. Though I don't like the idea of carrying around the stone that size. Since you were determined to go off to the Yukon, I thought you might take it with you. Then after you've had enough of that place, go over to Amsterdam. I haven't made up my mind yet what I'll do. I want to thank you for standing by as you did. That's all right. I thought a lot of Aunt Mary, you know. This is goodbye then for a while. Take care of yourself, Henry. Maybe we'll run across each other. Maybe we will. Goodbye. We'll run across each other all right, my dear cousin. Going to the Yukon isn't exactly my idea of a pleasant trip, but the valor of diamond is quite an inducement. An expression of bitterness crossed the young man's face. Litched inside his coat from a pocket, he drew out a slender sheath. Pulling out this stiletto, he held a point expertly between his thumb and index finger. Then stepping back, he faced the wall paneling. An old Mexican knife trick. A very worthwhile accomplice. From Skagway, it was only 600 miles to Dawson by way of Chilcute Pass, or White Pass and Arctic Lakes and the wild White Horse Rapids. This was the shortcut for the coarse gold and nuggets countless hundreds of men sought. George Barlow hired a guide and took the shortcut because of a clue he picked up in Skagway. Somewhere in that desolate country, he would find Emily Quinn and her father. Several days later, King ran far ahead of Sergeant Preston's dogs breaking a trail in the snow for the sled. The sharpness of the air exhilarated the great Huskies and he set a fast pace for the rest of the pack. Okay, hold you, Huskies! Here we are, boy. This is where we'll stop. A short time later, King ran through the timber over the crest of a snow-covered hill. He tilted his nose in the air and stopped a moment as he caught the scent of another camp off in a distance. His curiosity aroused, he went forward quickly. King approached the two men unobserved. He knew he'd never seen either of them before, so he stopped at a prudent distance and watched. One of them might have been an Indian as far as the dog could make out. He was the one who carried a rifle as he walked. As King watched him, he was conscious of another scent. The two men had paused for a moment so that the lynx coming suddenly into view was unaware of its danger. Two ladies saw them. The lynx! Your rifle! See! See! He felt rather than saw the pain of the wounded animal. For a fraction of a second, the lynx hesitated. Then ran in leaps, zigzagging confusedly, desperately. Then the man who was apparently unarmed did his surprising thing. He had a slender knife in his hand, calculated for a moment, and at the next frenzied leap of the lynx, he threw it. The knife caught the animal in the chest. How's that for accuracy, Jose? It was all over. King heard the cruel edge in the man's laughter, and he was contemptuous of them.   And they use no sheath. What is danger? I use it better than any Mexican, so I might as well carry it as they do. Only one sheath and I would have to leave in that confounded steamer.  It was early evening, two days later. A light snow was still falling, a sergeant Preston called... Okay, oh, you huskies! A campfire was burning back from the trail, and as the mountain approached, a young girl and an elderly man rose to meet him. Welcome, stranger. Mind if my dogs and I share your campfire? I'm Sergeant Preston in the Northwest Mountain Police. Pleased to meet you, sergeant. I'm Jack Quinn, and this is my daughter, Emily. Emily?  What's his name? Well, this is King. Well, I haven't been up here long, but I can tell from the looks of him, he's a mighty fine husky. Say, Emily, seems like we're going to have more company. There's another sled coming. Yes, you're right, Mr. Quinn. This is more people than we've seen since we left Skagway, Dad. Eh, Jesus, Dad. Sure is a lonesome country. I don't think Dad's as happy up here as he thought he'd be, Sergeant. You see, I... I'm Miss Quinn. Is there something wrong? Emily Child, you're as wide as a sheet. What's the matter? George! Emily! Emily, darling! Well, I'll be if it isn't young Barlow. As the two young people talked happily, Sergeant Preston got this story behind George Barlow's coming to the Yukon. The young man told the Mountie how the previous night his pack had mysteriously been stolen. Meanwhile, another sled approached the camp, where Preston talked to George Barlow, Emily Quinn, and her father. I should have known it wouldn't be in the pack. I was a fool not to think of the pouch before. Must be around his neck. This time, Lux on my side. With the snow to cover my tracks, it's a perfect setup. Marshal, men and lutes! We're joining my cousin, Jose. Tonight we'll all bunker on the same campfire. King watched Sergeant Preston's new friends greet the man they called Henry. He watched, too, for his master's reaction to the newcomer he had seen a few days ago. And he sensed that because these people seemed to know each other, the Mountie accepted Henry Drake as one of them. At Preston's feet several hours after supper, a great dog carefully weighed the feeling of goodwill everyone showed Henry against his own dislike for him. The two were not to be reconciled. His instinct told him the man with the hard laugh was an anime. Not to be trusted. It was much later that night. Curled up beside Preston, King looked from one sleeper to the other. Everyone slept heavily after a long day on the trail, but as the dog listened to Henry Drake's breathing, he realized the man was awake. Henry got up, cautiously made his way over to where George Barlow lay sleeping. He hesitated a moment. Then bending over his cousin, he listened to his breathing. King waited. Henry had slowly and carefully lifted something from around Barlow's neck. Torn between the desire to jump him or keep quiet, King watched. He saw him pull out the knife he had thrown at the links when the dog repaired the jump. It was only a chord Henry held in his fingers and he cut it quickly. When he stood up, the satisfaction crossed his face. He put the knife inside his mackinac and stepped to a fire. At last, a valorant diamond. King glanced at the sleeping man from whom Henry had taken this thing that caught the light and glittered. He had waited tensely, giving this man the benefit of the trust his friends felt for him. But when he made a movement to slip the pouch inside his mackinac, King jumped. The dog's weight knocked Henry to his knees. What's wrong? What's wrong? The pouch is gone in his teeth. Come back here! Gury swept over Henry's grave like a wave of insanity. He started after the great dog and the dog, remembering the fate of the links, dodged from one side to the other.  So he ran from the camp out into the darkness. Henry ran like a madman. Everything forgotten but the diamond. A narrow gully lay across King's path and the dog jumped it easily. But his pursuer, plunging blindly after him, fell headlong into it. Come here, boy! What have you thought about? It's Henry. Pull that lantern up a bit, will you, Mr. Quinn? Here comes that dog of yours, Sergeant. King, King fella. Looks like he's got something in his teeth. The pouch. He wants you to take it, George. How did he get it? Where's Henry? Lying down there in that gully. Well, I have to carry him back to camp. It was his gelato. He falled. He falled in the gelato. I tell him this is dangerous to carry without the sheath. That's what happened, all right. He fell on his own knife. That pouch, King, gave you, George. What's in it? Perhaps you've heard of the Ballora Diamond, Sergeant. This is it. I've been wearing it in the pouch around my neck. You can see where the cord was cut. Mm-hmm. Oh, look here by your sleeping bag. Those footprints. Yes. Henry came over here and took it while I slept. Well, King's the only one who knows what really happened. But it's my guess he jumped Henry and took the diamond from him, planning to return it to you as he did. No one knew I brought it with me except Henry.  Anyone who's tried to steal it has always come to a bad end. Well, I'm not a superstitious man myself, George. It's my own opinion. Your very lucky King was here to guard it for you. Yes, fella. You've done a good job. These copyrighted dramas originate in the studios of WXYZ Detroit, and all characters, names, places, and incidents used are fictitious. They're sent to you each week at the same time.


All right, there we go. All right, so let me take you back to where we were yesterday. Yesterday we were talking, we spent quite a good deal of time talking about this age distribution. The idea was that the experiments that we'd seen previously, the experiments that we'd seen previously were population average. And it's going to be important, and I'll show you why, in a moment, that we have some sense of what that probability over which we're averaging looks like. And some way, if we can, to deconvolve these experimental results from this averaging over the population. And so I'll tell you about that in a second. What I'm looking toward is explaining if we can these experiments by molar that were now, at the time that I'm about to talk about, a decade in the past. So these are from 1958. The experiments that we'll talk about today are from 1968. And just to remind you, the idea was that the faster cells grow, the faster the bacteria grow, the more mass per cell they have, that is to say, the bigger they are, and the more DNA per cell they have. And that's puzzling, as you say. We want to resolve that puzzle. So where is that DNA coming from? And in second of all, maybe not why are they bigger, but why is this scale here one doubling per hour? And why is that a characteristic time? OK, well, let me pause. Any questions about what we did last time? OK, one other piece of administration, periodically through the lectures, people have asked questions that I've deferred. Like for example, what's the average age of a cell? I think somebody asked. So I've put papers onto that website that hopefully address those questions. And so there's a paper that addresses a question, what if we don't have a delta distributed doubling time? There's a paper that addresses, do cells age, do bacteria age? And there was another question that is, can we experimentally ascertain what distribution we should give for the doubling time? That's also on the page. If you have any other questions, ask me as we go, ask me after the lectures, and I'll post relevant literature if you want to go deeper or farther, OK? All right, so then maybe I'll pause here. Yeah, could you say one more time? No. And that's because they're immortal. So the question is, is there another way to define age? And there could be, right? There could be, what would you call it? This would be the age relative to the bacteria. And then there could be age relative to the laboratory. Like these are four days old or something like that. But because these are in balanced exponential growth, they're sort of outside of time. The only time that matters is their time between birth and division, if that makes sense. And so this comes back to a question you asked, I think two lectures ago, which is, do the mother, or is there any noticeable difference between, say, the great-grandmother and the progeny of these cells? And for E. coli, none that we can tell. They're immortal. It's strange. I mean, from our point of view, it's humans. Yeah, yeah. So can you trace back? Can you say, if I gave you this cell and this cell, could you tell me which one was the grandmother? No. They're totally the same. It just happens that in the laboratory time, one came before the other. But chemically identical. Oh, we have to be careful, though. This is not a laboratory time. So this is, say, the experiment would only take maybe six hours. So there's no evolution or anything working on this. They're genetically identical to the zeroth order approximation. And so that allows you to say, that's what I'm using to tell you that they're indistinguishable. If you let these guys evolve for hundreds of years or something like that, then you can tell just from the genetic drift you could order them in terms of their genealogy, if that makes sense. So they were breaking outside of species rather than just looking at progeny, grandmother, mother, daughter. I don't know if I did. Was that? Yeah? Good. Any other questions? So what I want to talk about today is, like I say, basically DNA replication. So it's not obvious that that's all tied together, but that's where we're headed. And what I really like about this experiment is that it's, again, it's very much like a mathematical proof. You have tiny steps that at each point take a little bit of thinking, but don't break your brain. But then at the end, you end up with something that's absolutely incredible. It's like a proof by our comedies or something, except it's an experiment. And you say, how in the devil did he ever think this up? But this is Helmstetter. And then Cooper as well, although Helmstetter is the originator of the experiment that I'm about to talk about. This is Helmstetter 1968, and Cooper as well, who is his co-pilot. All right, and so the idea is if we could synchronize these cells, we would take this age distribution, which is this sort of exponential, if you like, distributed across the possible ages of our bacteria. And again, there are more newborns in our cells about to divide. And if we were going to synchronize, we would somehow chemically or biologically force this distribution to narrow. So we would turn this into some delta distribution. It turns out for bacteria, that's seemingly impossible. All right, but the next best thing, this is very difficult, let's not say impossible. No one has ever done it before. The next best thing to sample a narrow strip.  So something like this. And then you know that all the cells in that strip have the same age. OK, so you don't sample them in age. You sample them in time, if you like. Yeah? Can you look at the skews and make sure that you feel everybody's ready? I think on paper that would work, right? And for eukaryotic cells, it's much easier because they have checkpoints. And so these guys are everything simultaneous. And so you would have to have some kind of very accurate killing mechanism that's cued to something that's not a checkpoint.  But you would have to cue it to, say, kill the cells at some certain point of replication in DNA or something, or no, allow those to survive. So we're going to do something similar here. What's going to happen is we're going to hold back all the ones that are not at a particular age and let go ones that aren't. Yeah, exactly, exactly. So we have no analogous cycling in the system. Yeah. OK. And the reason that Helmstetter wanted to sample now age distributions is because they didn't understand how DNA replicated. So I'll tell you what they knew at the time. So Helmstetter's goal was to understand DNA replication. OK, what was known at the time, I sort of moved myself out of board space, but I'll do it here. So what was known at the time was that when these chromosomes, so they knew that the E. coli had a circular chromosome, and when it replicated, there was only a single point of replication. So a single origin of replication. And so it starts, it's like a zipper, if you like. So it starts replicating at this closing point of the zipper, then it unzips like this, and then you have two pieces, and then the cell divides. Is that clear? OK, I said that. I mean, you don't have to tell me, but tell me why it's not clear. OK, then you do have to tell me. Yeah. OK, so this is the DNA molecule that contains all the genetic information. And it looks like a circle, but say imagine a ribbon. Is that OK? A ribbon. And so you've got a circular ribbon, and now you cut it at one point, and then you tear it into two circles. Is that an OK image? Do you have the image that I'm talking about? Yeah, yeah. So what happens here is, so let me first tell you why I contrast this with another model.  Contrast that with perforating it at, say, 50 different places, and then just pulling it apart. So there's only a single point of replication. And what happens is it some machinery crawls along, so this is indicated by these triangles, crawls along this tear point. So this is called a replication fork. Makes new DNA here. So it pulls it apart. It's a double-stranded molecule. Pulls it apart into single strands, and then stitches on new DNA. So it replicates it. And then as it goes down, this is the old DNA, and this is the new DNA. Does that make sense? So this is one double-strand becomes two double-strands. In principle, they're perfectly replicated. I mean, it's true that occasionally errors in replication occur, and those are what we call mutations. OK, is this OK? So this Y is a blow-up of this triangle area. And so what you have is two forks of replication that are coming down the ring of DNA. And in the end, so you'll go from one ring to two rings, like this. So these are the forks. These are the final replicated chromosomes. Is that sensible? All right, because logically, it's going to start to get jumbled in a second. I mean, not too jumbled, I hope. So this was known. And so far so good. Then Mola postulated that the speed that these replication forks proceed at is constant, irrespective of growth rate. He had no evidence for that, really, but it's a simplest possible assumption. And Mola always went for that. So Mola proposed, let's say, the speed of DNA replication was growth rate independent. And to a good approximation, it is. So let's say that the time it takes to go from initiation of this replication event to separation of the sister chromosomes, so call time it takes to replicate. So you start here. A second later or an instant later, you have this system. Then a little bit later, you have something that looks more like that. I mean, this should be coming out toward you. Just not a very good draw. And then it will be like this. And then finally, you'll have two. So fork keeps coming down. And then they separate. So call this time tau cycle. That's a cell cycle time, if you like. And that's the time it takes not only to terminate this replication process, but to separate the chromosomes and then to divide this out. Is that sensible? So the table's set now. So two questions. One is, if your doubling time is much, much longer than this tau cycle. So suppose this is, I don't know, 60 minutes. But your doubling time is 90 minutes. And that speed is constant. Then what would be one of the predictions? No, no, keep going. Or unless you don't want to. Some cells have more DNA per cell than another and this depends on the new. So now you, OK, I'm not sure. Yeah, yeah, OK. So I'm going to paraphrase what she said, but it's not exactly what you said. But if this thing took, say, 60 minutes and you're doubling every 90 minutes, then you replicated one and a half times per life, then we can't possibly have a balanced state of growth. Because the first time that you grew, you would have 1.5 times your chromosomes. Then again, you would have 1.5 times 1.5 number of chromosomes. Then 1.5 times, 1.5 times, and so on. But you'd have a number greater than one raised to the power n, which is your generation. You would just explode with DNA eventually. So how logically would you have balanced growth if this thing was, say, 60 minutes and you were doubling at 90 minutes? So you would just stop. You'd have a gap of 30 minutes where nothing happens. Then it starts. You have 60 minutes of replication, then you have division. Does that make sense? So if the doubling time tau is greater than tau cycle, there should be a gap in DNA replication. But there should be a period where there's no DNA replication. Does that make sense? So there'd be a period where nothing's happening in the cell for, say, 30 minutes. Otherwise, you would have too much DNA. And it would keep multiplying, multiplying, multiplying. Let me pause. That's important. Does that make sense? All right, two questions then. One is how, yeah, say one more time. You could think of it that way. You could think of it as a maturation time. Something's got to happen until it decides to replicate its chromosome. What that Q is, you know, is debatable, but you're right. You could think of it as a maturation. OK, two questions. One would be, how would you ever see this gap?  You've got it only occurring in, say, young cells. Well, fine. But you have the old cells where there is DNA replication, and they wash that out. So the only way that you could see this gap is by somehow isolating cells at different stages of their growth. I mean, there might be another way to do it, but that's what Helmstetter was thinking. Does that make sense? So his motivation is to, in some sense, validate Mola's proposal. But then I think even more importantly, ask what would happen if your doubling time was smaller than this cycle time? If it takes 60 minutes to double this thing and you're growing at 20 minutes, what the devil's going on? This is crazy. All right, so that's what we're going to do today. Does everybody see the problem? Sure. Sure. Because if not complicated, the thing would not be able to divide further. Yeah. If I do that, if the distribution was uniform, then I would have a fraction which would not divide, which would be equivalent to the one set, average finished division, the leftover is taken in my order of the gap. Yeah, that's a good way. So his suggestion is, kill anything that has one of these triangles, and then count how many survived. You could do that. So if you could find something that specifically killed something with a replication triangle, then you'd be done. How about if tau is less than tau cycle, then we're in trouble, and then everybody's going to try. All right, so Helmstetter wanted to validate this picture and then probe the case where tau is less than tau cycle. How did he do it? And so as I said, he wanted to sample narrow distribution from the age distribution. So Helmstetter, and this took 10 years, what I'm going to show you seems like, yeah, OK, in hindsight, obviously. But so do a lot of our committee's proofs. So it's got that same quality. So Helmstetter hit upon a very simple, in retrospect, simple way to sample from the age distribution. So what he did was he took a funnel like this, and he had a filter paper here, and he poured his bacteria through the filter. So he took a test tube full of bacteria, and he dumped it in. Pour bacteria through the filter, and then he flipped it over. And you end up with, so again, the same filter paper. The funnel's going up now, though, like this. And then he ran media through the back end of the filter. So now this is a poor drawing here. Now fresh media starts dumping through like this. And as luck would have it, some of the bacteria would stick to the filter paper. So you'd end up with some, let's see, this guy's halfway done. This one just had a baby. So you had some cells stick to the filter. That's these guys. These are called the mother cells.  And so as they divide, their daughters or babies washed away. But the babies who fall off all have the same age. They're freshly newborn. Does that make sense? And this came from a dream that he had of chickens on the roof dropping eggs on his head. It's brilliant. And so this came to be called the baby machine. It's super cheap. It's super easy. And you'll see in a second that it cracks this question wide open in a way that was impossible beforehand. Okay, but let me pause. Someone had a question? Yeah. Yeah, yeah, yeah. Yeah. Are they stick to the chipper and also stick to the filter? Yeah, you'll see that. So her suggestion is that this is highly idealized. Not all of them are going to be stuck at the poles. And that's absolutely true. Some of them will be stuck sideways. Some of them may eventually fall off. That's quite rare. But the sideways one, you'll see that quite a few of them do stick sideways so that their progeny remains fixed to the funnel. But you'll see where that comes in. Well, actually, let me tell you what this looks like and then imagine what's going to happen. Is the experiment okay? Do you guys see what's happening?  Is it with the mothers? So when they get stuck to the filter, they have this, it's just a billion cells that follow this distribution. So this is in some sense the number distribution of ages in the cells that get stuck to the filter paper. And so then as time goes on, so here's tau. Here's two tau. Here's three tau. This is the babies. Baby cells. And I'm going to call this flow the effluent. So it's sort of the waste media, if you like. So what's happening? These guys fall off first. So let's call this T1. Then later on, the babies from these cells fall off. Does that make sense? So you've got them all stuck. They're very big and about to have a baby drop their baby right away. The ones that are newborn, they have to grow and then they drop their baby. And so what's that distribution going to look like? So again, remember that this was a probability distribution. Now I'm multiplying it by the number of mothers that get stuck to my filter. So now it's a number distribution. What's a number distribution going to look like here? It's probably easiest if we think of zero to tau first, like this. One possibility is it looks like this. What do you think? One more time, the mother cells. No, no, no, exactly right. So you also have to assume that the sticking probability is uniform across the population. And to a good approximation it is. So there's no bias. This guy, yeah? And so then what will you get? Who falls off first is maybe the best question. It's a little image. So it's going to look like this. So where this is, if I say, you know, relative concentration, say, where I don't have to worry about normalization, this would be two. This would be one. And then boom, boom, boom, boom. And then this would go up.  Because these guys are the T1 cells. And these guys are the T2 cells. Does that make sense? So in the babies, it's like time is reversed. That's small step, but sort of, you know, disorienting step number one. Okay, so in the babies, age is reversed. I don't know if that's the best way to say it, but hopefully you see what I'm saying. The ones to fall off first are coming from the oldest cells stuck to the filter paper. Yeah. Oh, yeah. So this is after there is a transient. You're right. So this is say, you know, wait for the transient, wait for the first division, then start. You're right. I mean, it'll go like this. Is that okay? So ignore the transient. Does that make sense? What I want to orient you to is this inversion. So do you mean why is it decreasing here? Oh, oh, oh, should go. Yeah. What do you think? So two things here. One is that it's flipped in time. Is that okay? Now, why is it not an escalator? So it's exactly. So he's exactly where you guys. So this is probably the image to keep in mind. Once this event happens, that age resets itself basically. It grows, dumps its baby, but the baby has left the system, left the filter paper. So it's as it was a generation ago. Ideally, but your previous question is going to come back. Okay. So is everybody. So first of all, the time is okay, I think. But now how about the drops? Drop. Drop. Is that okay? Yeah. Exactly. Exactly. That's the problem. If they were synchronized, we wouldn't need to go through all this because we would have, we would have, we would have zero peak, zero peak, zero peak. This dropping is because, it's because once the baby is born, she's washed away and only one of the mother's remains. And so it's, it's as though you reset again. So this is, this point here is where the newborn cells have their babies and they're them because there were more newborn cells to begin with stuck to the filter. But now suddenly it's a periodic boundary condition. You end up now looking at the, the oldest cells who are now ready again to, to have babies again. So you keep moving through this distribution backward in time with a periodicity. Does that make sense? And the reason it's not cumulative is because the babies don't stick to the filter. If the baby's stuck to the filter, wow, this, this wouldn't make sense. It would just be zero. And you would end up with that and that. Yeah? Okay. Yeah, yeah, yeah. Let's look. Yeah. You would, right? But what turns out is that you have more of what she suggested, which is that the baby stick. And so here's the picture that I have, or this is the data. So ignore the kind of transients. Who was worried about transients? So ignore these transients. But this periodicity comes from the cell division cycles. This slow upward creep comes from babies getting stuck to the filter. And so there are, there are, you know, many non-idealized factors here. But the overall sort of tup, tup, tup is what they're after. Yeah. So, so this, this thing is kind of going up like this. What's, what's also brilliant about this experiment is that it doesn't matter that these things are going up. That there's some slowly creeping offset. It doesn't matter because what he's going to be measuring is delta changes in a secondary quantity that I'll talk about in a second. But I just want, I just want conceptually people to see that this, this oscillation comes from that babies getting born. The slow creep up comes from what she was suggesting, which is that babies gets, you know, don't, in an ideal world you would have just a flat baseline, but you have inevitable creep. Yeah, exactly. So, so as she says, this is not particularly because of this creep. This is not a sustainable, I mean, this is his first run at it. People got decidedly better at it and developed filters that were decidedly less prone to this.  And you lose a signal in the noise, right? And then you also have these problems as she suggested. If the density on the filter becomes too high, then it's not, you can't guarantee that they're growing exponentially in some balanced state. If that makes sense. And so at best you can maybe get, especially in his first run at this experiment, you can maybe get three or four or five generations reliably. But it will turn out that that's all that he needed. Okay. Yeah. Yeah. Yeah. Yeah. Yeah. Yeah. Yeah. Yeah. Yeah. Yeah. Yeah. Well, he's going to want it. So there's, there's three things here. One is that this is a finite amount of time. Okay. Fine. But then you want to sample the population that washes out to be high enough that you can measure reliably. So you want something of the order of say millions or tens of millions of cells. So you have to either really ramp up the amount of cells that you stick to your filter. But then what if they're not in balanced growth?  And so there's optimization that you need to fiddle with, right? And then this question of, of growth rate is something that he wants to toggle independently. So at each growth rate, you would need to, you know, fiddle with these things to make sure that you're not incurring any sort of problematic artifacts. But in the end, what he's going to do is run this experiment. In fact, these are all different growth rates. And his sample period, I think it's for about a minute. So he, you know, holds us big and open for a minute and samples us. All right. So this, this is not yet the experiment. I mean, this is the setup. This is the baby machine. This is the narrow sort of ticking out of the population. But we haven't yet said why this has anything to do with DNA replication. And as it stands, it does not. So, but we'll get to that in one moment. All right. But I want this to be clear because otherwise it just starts to get foggy and foggy and then it's just incomprehensible. But is this okay?  So here is more or less a transient as things. So you've got a big puddle of cells at the beginning. You dump them and they wash out and then you have to let the system equilibrate a bit. And so it sort of takes a little bit of bouncing around. Is that, I don't know if I answered. Yeah. If my faith is a prediction as I go into the patients right. Yeah. And that's because the babies are sticking to the filter. Which is, which is, you know, you want that not to be there. The other thing too is that these are not sharp, I mean they should be vertical, but they have some, what would you call it, slope, non-infinite slope if you like, right? And that's because the age distribution in practice is not delta distributed, and so you have sort of a curve here, all right. But again, these are all small points that mercifully don't impact the experiment that we'll talk about in a moment. But again, let me pause, anyone want to clarify this? Times flipped, yeah, yeah, yeah. So this, think of this as a relative concentration. So this might be six billion, and this would be three billion. And it would go, there would be, there are half as many about to have babies as there are who were just born. And that's because when they're born they make two here. So these guys are falling off of this part. And there are, there are half as many, so here, there are twice as many here as there are here. Does that make sense? Okay, all right, I should have labeled both axes. Any other questions? Okay, let's get to the weird part. So the way that Helmstetter wants to look at DNA synthesis is the following. So he will, he has his cells growing in a flask, and then he's going to dump them into this filter. But before he does that, so to, I should turn this off if it's not relevant. To measure DNA synthesis rate, he does something that's called pulse labeling. So he grows his cells for one minute in radioactive thymidine. So, I'll write it up. And this is one of the DNA bases. So it only appears in DNA. I mean, there's free thymidine, but not much. It doesn't appear in RNA, it doesn't appear in proteins. This is a chemical that's specific to DNA. And you can get it, in the world it's just regular. It's made up of carbon-13, but he's going to get especially made carbon-14 thymidine. And so then he, for one minute, lets his cells grow. And then he flips them and pours them through the filter paper. Okay? Then flips or pours them through the filter. And then he measures the radioactivity in the cells that come out. So the new media that he pours through has no radioactivity. The media that he grew them in before had no radioactivity. And so what he's trying to measure is the number of forks, basically. Because the amount of radioactivity that a cell takes on in this one minute is proportional to the number of forks, replication forks that it has going on. Okay, so his idea is the following, and then let's talk about it. So his idea is the level of radioactivity in a cell is proportional to the number of forks, the number replication forks. Okay, so you have them exposed to this radioactivity for one minute. If it has two forks, then it takes in twice as well, two is okay. So if it has two forks, it takes in a certain amount of radioactivity. If it happened to have four forks, for example, it would take in twice as much radioactivity, if that makes sense. Or if it had no forks, it would take in no radioactivity. And then suddenly when it had two forks, it would take twice as much radioactivity, if that makes sense. So everybody see what I'm saying? So by reading the radioactivity in the baby cells, then he should get some sense for the number of forks that were going on at that age in the parents. It's like three levels of inference, but hopefully they're okay. Oh, you just put it into the media. So you grow them in regular media. Probably there would have been, I wonder if he supplied the thymidine. He probably does. So what he would do is grow them cells that can't make their own thymidine. He would supply it in a medium non-radioactive. So they're bringing it in and incorporating it. And then he would add radioactive thymidine in huge excess, like say 500 times more than was in the media to begin with. Then you're guaranteed that the thymidine that you're taking in is radioactive. You've like less than 1% of a chance to take in a non-radioactive thymidine. Does that make sense? You have to make sure they can't produce their own thymidine. You have to make sure they can't produce their own thymidine, otherwise you have to label all the carbon sources. Does that make sense? Yeah. Is it lethal? Oh, I see. So this was, although less so now because of safety regulations all over the place, but it still remains a fairly standard procedure in biology, is if you want to label something, and now we do with fluorescent proteins primarily, but labeling with radioactivity is a well-established practice. It turns out that radioactivity is usually non-hazardous to us and to the bacterium. Is that what you mean? Is that what you're worried about? Or that the chemicals are distinct enough that the cell will treat them differently? Yeah, yeah, okay. So they don't seem to incur mutations. They're not high enough energy radiation to worry too much about that. Any other questions? Okay, so his idea is this. Suppose you have the mothers. So now we're gonna talk about the mothers, and then in a moment we're gonna talk about their daughters.  Suppose something like this happens. So this is now radioactive, radioactivity per cell. Suppose something like this happens. What would you suppose was going on? So this is now relative radioactivity. So it was one, and it jumps to two. What would be your interpretation? It could be, yeah, okay, even easier would be the gap one. So suppose the gap goes like this. Now it's gonna be in absolute units. But how would you interpret that? What does a change represent? Division happens at the end. What do you mean? Division happens at the end. So division happens at the end here. So this age at which you see a jump corresponds to, so a jump, let me write it out and let's talk about it. A jump in the relative radioactivity per cell, per cell corresponds to the initiation of DNA replication. All of a sudden on the scene appears two replication forks that can pull in radioactivity that weren't there before. And this line that I've drawn then is what, I mean, I think could be reasonably denoted the age of initiation or initiation age. Now let's say age of initiation. And so for cells that are growing much, much longer than the cycle time that it takes to replicate the DNA, that age of initiation would make up the gap. And it turns out that it's actually like this. It's a gap at the beginning of the cell, cells lifetime. And then DNA starts to replicate and then terminates its cell division. Does everybody see the logic? Yeah. No, OK, so I have in mind the case where the cell is not replicating its DNA. So this is a cell doubling, say, every 100 minutes. And yeah, but it doesn't suck in this radioactive thymidine if it's not replicating DNA. That's the key. That's the brilliant part of this. It's only going to take in this radioactivity if it's replicating DNA. It's a logically compelling argument. I mean, it has to be clear otherwise then it's not compelling. But is it clear? Exactly. So this would be the case if this replication time was much, much longer than the time it takes to replicate DNA. Yes, yes, exactly, exactly, exactly. Yep, exactly. OK, all right. And so what? Oh, that's not what I wanted to do, sorry. This is what I wanted to do. OK, so I'm going to tell you about the experiment he did in 1968. In 1967, he looked at slow growing cells, saw this gap, and he was very pleased, very excited. Because the benefit of this experimental setup, in addition to being cheap and super easy, is that you don't touch the cells after they've been labeled. You label them when they're growing exponentially in a flask.  And everybody's doing it, so nobody's going to get mad about that part. Then you dump it through his filter paper. You haven't touched it. It's still growing exponentially in balanced growth. It's a minimal perturbation. It's no surprise that Helmsfeder spent a lot of time in Mola's lab. So they're very sympathetic in their approach to these systems. And Helmsfeder was trying to find a way to do this without perturbing the cell, without your observation disturbing the information that you get out of it. So in 1967, I want you to get the data because it might be tricky. It's better if I show you the new data from 1968. So this is now, Helmsfeder is growing cells that are much more rapidly growing. So they have doubling times of, say, 30 minutes, 40 minutes. And he's joined by Cooper. And together they look at this data. And what he sees is, so this is radioactivity in the mother. And so in the baby cell, so this is now radioactivity. And I'm going to put this on a log base 2 scale. Log base 2 of the radioactivity per baby cell in the effluent. And it's going to look like this. So here is 2. Here is 1. Here is 0.5. Here is now 0.25. So these are equally spaced because I'm on a log base 2. And I want to show you what he sees. So he'll see what's going to happen here. Zoop, zoop, zoop.  I'll get out of the way in one second. I just want to do that. He saw this. And so the interpretation is that you have something that goes like this in the mother. So we won't yet talk about why this thing looks so weird. We will, however, talk about what he measured. So then he measured this. So he said the first time that you see a rise in the DNA replication rate corresponds to the triggering or initiation of a fork of replication. Does that make sense? I mean, not the bottom plot. So he used the bottom plot to infer the upper plot. And then he measured this time this age. Yes. How do you mean? Wait, sorry. Here? No, this is the age. So there might be some transient here. But you set 0 to be the point where you had the first division. So absolute time now falls out of the equation. You're talking about here? Yeah. Yeah? I think we're doing the same thing. Yeah. I'm looking at the first one. Yeah, these are the same points. Oh, yeah. But you're only, yeah, so the mothers are unsynchronized. So, OK, so you have this situation for the number of mothers. Is that all right? Yeah? This might solidify the discussion. Is that OK? Yeah? And then you throw radioactivity on them. And then you allow time to advance by one minute. And so everybody that's at this age plus delta A takes in their amount of radioactivity. Everyone that's at this age for one minute takes in radioactivity. And so it's an unbiased sample of the population of what the person at that age or the mother at that age is doing in terms of their DNA replication. Yeah, OK, yeah, yeah. So this minute is enough for you to get hundreds of bases of DNA. And so you lose any, that's important too. The thymidine is not biased either. Any other questions? Is that OK for you? It's OK? So it's, I mean, there are like five different things happening here. But the thing is that they, so for the one minute that he lets them be exposed to the radioactivity, they have some consumption rate. And so you integrate that rate along that one minute time. And that tells you how much thymidine comes in. So there's an integral going on. And that integral then integrates across this amount of space. And that tells you how much thymidine comes in. But then you put on top of it this, which is telling you, so these guys consume at some low rate. These guys consume at some very high rate. And these at some moderate rate.  That's the idea. Yeah, yeah, yeah, yeah, yeah, right? Exactly, exactly. Let's go, can you just hold it in your mind for one second? And then let me show you what the plot of this looks like. OK, and then, so let me show you the plot. And then we'll break. And you can think about what the plot means. And so now if you look at the doubling time, so this is doubling time. And this is the relative age. So now this is going to be the age of initiation divided by the doubling time. So you count this, it's like five minutes first outgrowing at 30 minutes or something like this. And so this will be between 0 and 1. And this is going to have real units. This is something that you're using your watch to count. And so I'm going to start. So these E. coli, you can't really grow them any faster than say 20 minutes. So 20 minutes, let's say 30, 40, 50, 60, 70. I'll give you one more, 80. All right. And so for very slow-growing cells, this is what he's observed in 1967, this age of initiation goes like that. It increases. And it increases for the reasons that was suggested earlier. There's a gap. And then it starts replicating. But then there's suddenly a discontinuity. And you have it jumping up to 1. And then falling down again. And then jumping up to 1 and falling down again. And if I've timed this correctly, there we go. So my plot is idealized. But this is the data. And he's left off the data from 1967. But I put it in here because I want to be able to possibly solve the puzzle. So does everybody see what I'm saying? He's got these weird shapes, which will come back to in a second. Like after the break. But more importantly, he's got these discontinuities in this age of initiation. And they look exactly like this. So boom, boom, boom. Does that make sense? So he doesn't see a spike here at all, really. But then I'd say 58 minutes. He sees a spike just before division. And then that age sort of creeps closer and closer. And then a spike just before division at about for 28 minute growing sense. All right, does the data make sense? I mean, not the interpretation, but does the actual plot itself, do the axes make sense to what he's measuring? And again, it doesn't really matter that the data itself was creeping up or anything like that because he's measuring jumps in radioactivity. And so that you can get pretty accurately, irrespective of some baseline shift, which is one of those great things. All right, let me pause. Yeah? Oh, get the plot. So what he's doing is he runs this type of experiment at different growth rates. He knows where the cells are dividing because he knows how the number count is going. And then he measures this gap. He says, all right, going backward in time, he actually plotted them flipped around, which is a double mix up. He measured this time, which is an absolute unit. So it's, say, 15 minutes. So he writes that down. Then he does it at a different growth rate. He gets 10 minutes. He writes that down. And then he plots the age of initiation divided by the growth rate. And he gets some number between 0 and 1. And he does that for all these different growth rates. So this he can change by changing the nutrient that the cells are growing in. This he measures from this radioactivity jump. Is that OK? Just where the data is coming from, not necessarily why it's jumping around like crazy. But is the data, where the data is coming from, OK? Pardon me? OK, no, think about it, though. You don't worry. So this is a real puzzle. All right, any questions about what the data means? All right, we'll simultaneously answer four questions right after the break. So your original question, why does a DNA increase with growth rate, has been just lurking in the background for days? So we'll come to that. And your question of why would it ever overshoot is going to happen. We'll talk about that in like five minutes too. All right, a new pause. Oh, and what we've never talked about is why does it take one hour, why is one hour doubling special for that mass per cell plot? That's also going to be cleared up. All right, any questions? All right, OK, I'll see you in like five minutes, 10 minutes. It's how they got this data or how they rationalize this data. But I think it's easier for us in retrospect to figure this out. So what we now know is that to go from here to here to here so the fully replicated DNA takes about 40 minutes. Now to separate. Oh, gosh. Sorry. OK, apologies to the camera guys. They go from, oh heavens. Double apologies. So all right. OK, I think. So to go from initiation of DNA replication to fully replicated chromosome, Molly was right. I mean, it's mostly growth rate independent. And it takes about 40 minutes. And then the cell has to do, I mean, it's physically there's nothing else it can do. It needs to separate that DNA into its daughters before they divide. And so to go from, so these are now just completed, replicated chromosomes to separated and divided, that takes about 20 minutes. OK, so we have these two processes. And so Helmsted and Cooper called these two different names. So they called this the C time and the D time, which will be, it's just a nice shorthand. But the C plus D is going to be the cycle time. Does everybody understand what I've drawn here in terms of cartoons? No matter what you do, these forks will take 40 minutes to get from the beginning to the end. No matter what you do, it's going to take 20 minutes for these DNA strands to separate into their respective daughters before cell division. All right, so to move from initiation of replication to termination of replication takes 40 minutes. To go from termination to cell division takes 20 minutes. And there's nothing you can do about it. Now here's the question, why can a cell double every 20 minutes? It seems impossible. So what's the resolution? Exactly. So you parallelize this. You initiate multiple forks of replication. So how do we divide? Do they divide faster, 60 minutes? And the answer is multiple forks. And so we saw this. I mean, we didn't make a big deal about it when we were talking about protein synthesis. But ribosomes, which are driving protein synthesis, work for the most part at full blast as fast as they can. So then in order to make protein faster, the cell can't turn the rate up. It just adds more ribosomes. It parallelizes the process. It makes a production pipeline. The same thing goes here. We can't make these forks move any faster. All right, add more forks. It's incredible. I mean, it's a really clever engineering strategy. And so what you end up with then is something that looks like this. You have one set of forks here. And there's an origin of replication like that. And then you would have another set of forks here and another set of forks here. Oops, what have I done? What am I doing? Good heavens. Oh, yeah, no, that's good. Forks here, forks here. That's horrible. Let me drop a nice picture. OK, so here's a fork. Here's a fork. Here's an origin. Here's an origin.  Here's a fork. Here's a fork. Here's a fork. Here's a fork. And there's another replication. So here we have simultaneously 1 plus 2 rounds of DNA replication. And so when you have something like this, you'll have one round here for 1, 2, 3. So this would be the synthesis rate. We'll go from 1. Then you initiate the second rate. You'll get 1, 2, 3, 3 times more synthesis rate. And then you'll drop down to 2. So this is with 2 forks. This is with 2 plus 2 to the 2 forks. And now this is just 2 to the 2 forks, if that makes sense. So now you've got 2 forks coming down at the beginning. And then halfway through, well, whatever, a third of the way through the generation, you initiate these new replication sites. So you have these old forks plus these new forks. A blink. And then you keep going. And this fork, these two forks meet at the bottom and terminate. A blink. But you don't fall back to your old rate because you still have these two simultaneous forks proceeding. And so what Helm said it was measuring was this first age. And so each time we get a discontinuity, that's initiation of a new or that's pushing initiation of DNA replication back another generation. So I'm going to unpack this a little bit more and then we'll break lunch. But what's going on here is that the bacteria themselves are self-initiating. So I wait, I wait, I initiate replication, and then I divide. And it's all me. But here, your mother is initiating replication for you or better not anthropomorphize it.  And so the daughter is born with a half replicated chromosome. And that's how they're able to double much faster. And here, the grandmother is initiating the chromosome. The mother keeps growing the chromosome and impasses it onto the daughter almost fully replicated. It's extraordinary. Does that make sense? I mean, it doesn't need to make sense, but just in general. And then we'll look at it quantitatively and we'll sketch it quantitatively today, but then we'll go into it in more detail tomorrow. But now, so that's why the DNA is increasing. It's not that it's a different species. It's that you're carrying redundant information for your daughter later on. But this leads then to a point that notice that the DNA that's close to the origin is effectively amplified by a replication. And the cell can take advantage of that. We'll talk more about that maybe tomorrow. But does everybody understand this parallelization? It's extraordinary. I tell you there's a speed limit? No, there's not. You add more cars to the road, basically. Yeah, yeah, exactly. And so if you look back, so I think it's probably gonna be easiest to sketch it the way Helmstetter did, which is graphically. That's the way that I gravitate to, but then that sometimes doesn't work. And then we'll do it analytically with pencil and paper. But you're exactly right. So as you go past these discontinuities, you're moving from self-initiation, mother initiates, grandmother initiates. And if you could grow faster, your great-grandmother would initiate and so on. And you'd have these hugely branched DNA loops. Any other questions? And so what we have then, I mean, this isn't how Helmstetter described it, although it is graphically how he derived his rules, is a tiling of a continuum of division times. So these are dividing whatever they might be dividing, say, for every 45 minutes. And that's what I do by fixing the chemical composition of their test tubes. And I can change that any way I like. But lurking underneath that is a discrete tiling of that continuum in chunks of 40 and 20 minutes. Tom, tom, tom, tom, tom, tom, tom. And the cell, of course, has to figure out how to do that and it does it, but that's regulation. What we're talking about here is phenomenology. Given that the regulation is there, what can we predict, for example, about the age of initiation? Okay, and so Helmstetter went the other way. He looked at this data and inferred this. Well, he and Cooper, which is an extraordinary act of mental agility, I think. But I'll show you how he did it tomorrow, I guess. I'll show you tomorrow how he did it. And if you feel like it, you can just read the end of this lecture. And as I say, he just literally drew tiles. So one tile that was the division time. All right, I'll sketch it. It'll just take one minute, then we'll break for lunch. So Helmstetter's is tile the generations with strips of 40 minutes or C minutes and D minutes. And so, for example, he would have, let me see what his picture would look like. And again, I'll just start this. So he used squiggly line for tau equals, say, 35 minutes. And this works much better if you've got a piece of graph paper to do this on. So I'm gonna freehand it, and it's not gonna be the scale, but the next line would be the 40 minutes. Oh, it's not bad, but. And then a dashed line that was half as long again, which is D equals 20. And then he moved this over and repeated the tiling. So he would have this and then this and then this. And then again, so, and this is again, this will be the last thing I draw, and then we'll break for lunch. And did I do that okay? Yeah, and you would keep going until, depending on how fast these cells are growing. Does everybody see what I'm doing? I've got a strip, and then I just keep displacing it. Ideally, of course, you would have these squiggly lines as you're in the background, and then you would lay transparent pieces of C and D on top of them, and then just look at the darkness or something of the transparency. But here he's separating them. And then if you look from here to here, you have a complete record of what's going on inside the cell during one division period. And so what you have is DNA replication up to here. So now if you look at the forks herself, you would have, say, one fork, or one pair of forks. Then here you would get, so this is now two forks, let's say. This would jump, this would be two plus two squared forks, because each one of these initiates a new round. Oh, this one terminates, but these guys are still around. So you get two, two forks. Till the end here. What did I do? What have I done? Sorry, this ending should be here. There we go. So from this point to this point is one cell division. These solid lines tell you the history of the DNA replication, two forks. Then each of these forks initiates a fork. This one terminates, but leaves its progeny forks. And so in this way he was able to rationalize those very strange DNA replication graphs. I mean this is graphical, and to try and extract an analytic expression for these ages of initiation, doesn't look like it's gonna be easy, but it will be pretty straightforward. But I think it's best if we let it stew and then we'll talk about it tomorrow. That's okay. But let me pause though, just if there are any clarification questions, and then we can really resolve this tomorrow. Any questions? Yeah. Yeah. And so her question is, what about the media? How does the growth medium affect this? And it seems to not affect it at all for a given growth rate. So irrespective of the composition of your growth medium, if you're doubling it one, doubling every 35 minutes, it will look exactly the same, no matter what the recipe is. And so that's one of the great findings of Mola, and it continues here. So, but we'll talk about that tomorrow. Any other questions? Oh no, there's randomness. Yeah, yeah, yeah. So the question is, is this true at the single cell level? Of course, expected distribution at the single cell level for all of these parameters. All right, and that's stuff that people are measuring these days, but are impossible at this time. Any other questions? All right, great. So let's all have lunch and I'll see you guys tomorrow.


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