Well, welcome to episode 36 on the binomial theorem. Today we're going to be looking at the expansion of binomials, powers of binomials, to read the large powers, and how we can compute them quickly. Let's go to our list of objectives for this episode. First of all, we're going to be looking at patterns that we find in the expansion of binomials. What I mean by that is if you were expanding like a plus b squared, a plus b cubed, a plus b to the fourth power, how we could find patterns in the coefficients. And then we'll look at something that we call Pascal's triangle. This is named for a French mathematician named Blaise Pascal. Then we'll move on to factorials and what we call binomial coefficients. And then finally, we'll look at a theorem that was first proved by Isaac Newton back in the around 1700, just before 1700, called the binomial theorem. Okay, well, let's begin by looking at patterns in binomials and the expansion of binomials. And let's go to the next graphic and I think you'll see what we mean. Okay, first of all, we have a binomial, a plus b, and we're going to raise it to the zero power. So what is a plus b to the zero power? One is one. A plus b isn't zero. That's exactly right. If a plus b is not zero, because zero to the zero power is undefined. That's a good point, Stephen. So this answer is one otherwise. And whoops, let's see. I must better go to a different marker for that. So we get a one there. Yeah, so in all of these, I'll be assuming a plus b isn't zero. Now, in the next case, a plus b to the first power is a plus b. So I'll write a plus b right here. You notice I'm kind of spreading it out. You'll see why in a moment. What is a plus b squared? A squared. Okay, a squared. Plus two a b. Plus two a b. Plus b squared. Plus b squared, yeah. And now finally, a plus b cubed. That's not necessarily a formula people generally remember.  A cubed. Okay, a cubed. Plus three a squared b. You're right. Three a squared b. Plus three a b squared. Three a b squared. Right, very good. Plus b cubed. Plus b cubed. Okay, now what I wanna do is look at some patterns that we see in these expansions of binomials. First of all, you notice that the number of terms, like here I have three terms, is one more than the exponent. If there's a square there, I get three terms. If there's a cube, I get four terms. It even worked up here. When there's a zero, I get one term, namely, namely one. So we always have one more term than we have for the power. The next thing we might notice is the powers on a go down by one. I have a cube, a squared, a to the first power, and then essentially a to the zero. They're gone, a to the zero is one. At the same time the powers on a are going down, the powers on b are going up, I have b to the zero here, or one, b to the first power, b to the second, b to the third. So if the a's are going down and the b's are going up, what that means is I have the same number of variables in every term. So here I have a three times. Here I have two a's and a b's, that's a total of three. So this is referred to as a term of degree three. This is a term of degree three. I have an a and two b's. And this is a term of degree three. All the terms are of degree three. So if in the future, if you're expanding a binomial to some larger power, one of the things you want to notice is every term should be of that same degree as the initial power, the cube, right here. So what portion of this would be the really awkward part to find then? The coefficient. It would be the coefficients, right. So what we need to do is turn our attention to the coefficients and see how we could find the coefficients, like in this case, the coefficients are one, three, three, and one. Now you know, the way we can find the coefficients, say for this a plus b cubed, is what if I go back up here to a plus b squared and multiply it by a plus b? By a plus b. Now, for example, to get the a cubed, I just multiply a squared times a, and that gives me a cubed. But I think there are several ways I can get terms that say a squared b. Let's see, here's an a squared. If I multiply that times b, so if I multiply a squared times b, I'll come up with one a squared b. But what's another way I can come up with some a square b's? Two a b times a. Two a b times a, exactly right. I can multiply these two terms together, and I'll get a squared b's. In that case, I'll get two of them. And so I get one, let me just put a one on top of that, I get one from this product, and I get two from this product. Now, one plus two, that makes the total of three that we have right here. And you know, since I'm multiplying by one a and one b, the coefficients one and two are the same as the coefficients that I see right here, one a squared and two a b. So I'm thinking there's a pattern in this, and that is that if I add the two coefficients that I see right here, I'll get the coefficient that I need right below it. Let's just try the three right here. If you look at the two coefficients right up above it, there's a two and a one. And if I add the two and the one, I get a three. And you might say, well, Dennis, what's the connection? So just because that's a two and a one, why should this be a three? Well, let's just remove these arrows, and let's see if we can verify this pattern for this other term. Here we have a b squares. Now, how can I get a b squares? Well, I can take the two a b times b, and I'll get a b squared. I'll get two of them, which is exactly what that coefficient is. I get two of these. And I can also get some a b squares right here with this product, and I'll get one of those. And that's exactly the coefficients that's on the b squared. You see, this coefficient won't change because I'm just multiplying by one a or by one b. So that won't affect the coefficient. So if I take the two and the one and add them together, I'll get a three right here. That even works on the very ends because the only way I can get an a cube is to take one a squared times one a, and that gives me a cube, or one b, and one b squared gives me b cube. So the pattern is that if I add the two coefficients directly above, I'll get the coefficient on the term just below it. Now, this leads us to what we call Puskal's Triangle. And let's go to the next graphic and we'll see this. Blaise Puskal was a French mathematician who lived in the early 17th century. First half of the 17th century, obviously in France, I guess, Blaise Puskal, and his triangle begins with a one at the top, and then two ones down below. And then if I continue the ones down the diagonals, ones down the diagonals, the numbers in the middle are merely the sum of the two numbers directly above it. One and one is two. And the same thing here, one and two is three. And two and one is three. And on the next row, we'll see the ones are coming down on the ends. One and three is four. Three and three is six. Three and one is four. And this pattern continues, so actually there's really no end to Puskal's Triangle. What would be the next row of the triangle? What number would go right here? One. One, okay, and then here? Five. Five. Okay, and then? Ten. Ten. Ten. Five. Five. One. And one. Now with those numbers, we can now expand A plus B to the fifth power. These are the coefficients of the terms. And now I know it's the fifth power because if you look in the second diagonal, one, two, three, four, five, that's the power on the exponent. These were the coefficients for A plus B to the first power. Here we have the coefficients for A plus B squared. A plus B cubed. A plus B to the fourth, and now here's the fifth power. You notice there are six numbers written across there, and we said that if you expand A plus B to a power, you get one more term, you get one more term than you get the power. So a fifth power, I should get six terms. Okay, now I'm gonna write those numbers down. A, one, five, ten, ten, five, one, and expand A plus B to the fifth. In fact, maybe there's enough room right below this problem to write this out. So I think you'll probably have to remove the box over on the lower right for us to squeeze this in. But A plus, oops, A plus B to the fifth power is going to be, first of all, one, A to the fifth. The next term is gonna be A to the fourth B. And what's the coefficient? Five. A five, okay. And the next term is gonna be A cubed B squared. You notice the As are going down, B's are going up, the coefficient is? Ten. Ten. And the next term will be ten A squared B cubed. And the next term would be five A, B to the fourth. And then the last term is B to the fifth. So by writing out Pascal's Triangle, I have the coefficients for the expansion of the binomial. And that's really the only subtle part because everything else is pretty straightforward. The As go down, the B's go up. Every term has degree five. Degree five here, degree five here, degree five here, et cetera, all the way across. Okay, looking at Pascal's Triangle, there are some other interesting patterns that we can find in the Triangle. You notice coming down here, we have ones on these two diagonals. What about this diagonal? What did we say? Consecutive integers. They're just consecutive integers and they happen to be the exponents of the A plus B that's being expanded. What about the numbers we see here? One, three, and six. One, three, and six. Have you ever seen numbers like that before? The next number was gonna be ten coming down that way. Also one, three, six, ten coming down this way. These numbers are called triangular numbers. Let me just show you up above. Here's why they're called triangular. One, I could represent with one dot. Three, I could represent with three dots. Six, I could represent with six dots like this. In other words, I could make triangles out of them. Could I represent ten as a triangle? Sure, I'd start with four dots. One, two, three, four. One, two, three, one, two, one. In what sport shall we say do you see an arrangement like that? Pool?  So when you rack the balls, it's one through five and there are 15 balls. But where do you see these 10? Well, this is something I used to do a lot when I was in high school, let's go bowling. So 10 dots, you see the arrangement of the bowling pins that way. Okay, so those are called triangular numbers and along the diagonals we see those triangular numbers. Another pattern is look at the sum along every row. Well, the sum on this row is one. The sum on this row is two. The sum on this row is four. What's the sum on this row? Eight. Eight, okay. And the sum on, let's say I'll have to go to the very end. What's the sum on, I'm gonna call this row four because there's a four there. What's the sum on row four? 16. 16. Let's say we have one, two, four, eight, 16. What pattern do you see there? Powers of two. Those are powers of two, yeah. And I've erased row five, but those would add up to be 32. So in other words, we could hazard a guess that if you expand an expression like A plus B to the nth power, that the sum of the coefficients in A plus B to the nth power is two to the nth. You see this is two to the fourth, and this is two to the third, this is two to the second, and this is two to the first, and this one is two to the zero right here. So imagine there's a zero right there. So if you're expanding A plus B to the nth, then all the coefficients should add up to be two raised to that power. So yet another pattern that we find in Pascal's Triangle, and there are still other patterns in it, but we don't need to get into all that. Okay, well rather than just talk about it, let's actually use it. So let's go to the next graphic and see how we could use Pascal's Triangle to expand a few terms. I'm gonna work these out on the green board, but the first problem says expand A plus B to the fifth. You know, I think we've actually done that one already, so I think I'll skip that, A plus B to the fifth. But then we wanna find X plus three to the sixth, five Y minus one to the fourth, and now here's a complicated one, U cubed minus 10 V squared cubed. Now I say it's complicated, but actually it's no more complicated than any of these others, and they're all fairly easy to work out. So I'm gonna go to the green screen and work these individually, except for that one, because we've already done A plus B to the fifth, and let's see how we would expand every one of these using Pascal's Triangle. Okay, so if we come to the green board, I'm gonna write up here in the corner the numbers that I see in Pascal's Triangle. One, two, one was that row. What was the next row? Could you tell me what that was? One, three, three, one. One, three, three, one. Yeah, and the next row was... One, four, six, four, one. One, four, six, four, one. Yeah, and then I think that's as far as we had printed on the graphic, but then we found one more row a little bit earlier. And of course, this just keeps on going over here. Now, the first problem that we had was to find X plus three to the sixth power. So let's do that right here. X plus three to the sixth power. So, do I have enough rows in my triangle to work this problem? Not quite yet. I don't think so, because I need to get to row number six. That's the one in the... If you look at the second diagonal, I need to go to row six to go with that six. Now, that means I'm calling this one row zero up here, and then row one, two, three, four, five, six. So let's fill in row six. Who wants to tell us what row six will be? Susan, could you tell us what row six is? One, six, 15, 20, 15, six. And one, right. You know, one of the patterns that we have in Pascal's Triangle, I should have mentioned, it's fairly obvious, is that the triangle is perfectly symmetrical. So if you go from the middle out, 15, six, one, and here 15, six, one. So if you go right down the middle, if you could draw a line right down the middle, of course it would split it on the row five, then it's perfectly symmetrical. So if you know the first half, you also know the second half of the triangle. Okay, now when I expand this, I'm gonna get, well, I'm gonna treat X as if it were A, and I'm gonna treat three as if it were B. So this would be, actually maybe I better go down here to make sure I have plenty of room. This would be one, X to the sixth, plus six, X to the fifth, times three to the first. See, that's my A, B term. A to the fifth, B to the first. And then I'll have 15 X to the fourth times what? Oh, nine, okay, I'm gonna write it as three squared just so we see that that's a sixth degree term where I'm thinking of this as my B. And then I'll have 20, 20 X to the third times three cubed. And then I'll have 15 X squared times three to the fourth. And then I'll have six X times three to the fifth. Plus, I'll have to squeeze this one in underneath, plus one times three to the sixth. Now you might say, Dennis, this is terribly long and complicated. Well, what's the alternative to writing it out this way? What's the alternative to finding X plus three to the sixth? Multiply it out. Multiply it out, yeah, you'd have to write this down.  And multiply it out, multiply that out by regular multiplication. Now that's what takes a long time. This is fairly simple child's play. Okay, so if I go back and simplify this, I have X to the sixth plus 18 X to the fifth plus nine times 15. What's 10 times 15? 150. Okay, so what's nine times 15? 135. Yeah, just be 15 less. That's gonna be 135 X to the fourth plus 20 times, what's three cubed? 27. Yeah, 20 times 27. It's gonna end in a zero. What's two times 27? 54. 54, that'll be 540 X cubed. Okay, so we're doing pretty good here. Now three to the fourth is 81, 81 times 15. Oh, 81 times 15, let's see if we can do that without a calculator here. If I wanna find 81 times 15, I'll tell you what, what is 80 times 15? 80, 15. Well, wouldn't that be the same thing as 30 times 40? Yeah, what I did was I doubled that number and I cut this one in half. 30 times 40 is 1200, so this is 1200. Okay, but that's 80, 15s. What's 81, 15s? 1200, 15? 1215, yeah. Isn't that the year the Magna Carta was signed, I think? Okay, 1215. Just had to see if you were paying attention there. Okay, 1215 X squared plus, now let's see, 81, that was three to the fourth. What's three to the fifth? Well, I have to put another three in there. To 243. 243, 243 times six, I think I've reached the point where I'm gonna have to go to regular multiplication. 243 times six is eight, five, two, six. Did I do that right? I think I did. So two, six, five, eight times X, plus, okay, now we come to three to the sixth. Oh my gosh, three to the sixth. What's three times this number? That's three to the fifth. I think it's 729, plus 729 right here. Okay, so this is the expansion of X plus three to the sixth power. Did you see a mistake? Yeah, 243 times six is 1458. Oh yeah, of course, yes. Thank you. You know, I'm better off using my shortcuts than in regular multiplication. Yeah, that's gonna be 18, carry one, 25. This is 12, so not 24. So that's gonna be 1458. What happened in 1428? 1458, can anyone tell us? Good, I can't either. Okay, 1458 and 729. Okay, so I think now we have the coefficients right. You know, what was the hardest part of this problem? It was the multiplication. That's where I made a mistake. The actual expansion is pretty simple, actually. If you know Pascal's Triangle. Now you might say, Dennis, when was the last time I ever had to expand a binomial to the sixth power? You may be asking yourself that. When are you gonna have to expand X plus three to the sixth power or something comparable to it? Well, you just never know, but I think what's important is that you become familiar with the patterns in the triangle because these things come up, well, in other courses now and then, and you'll see our references to Pascal's Triangle. Let's go back to that graphic that had the examples on it and let's pick another problem from those examples to work. Yeah, we've just done part B. Let's go to part C, five Y minus one to the fourth. Now, here are some of the differences that we see in this problem. It's a difference rather than a sum. And the first term is not a simple A or X, but it's a five Y, but we do have a rather simple second term raised to the fourth power. Okay, so here we go. Let's try working out. If you go back to the green board, let's try, oops. You know what, I just erased Pascal's Triangle. Let me just fill it in right here. Somebody help me out. What's the next row? One, two, one. One, two, one. Thanks a lot. What's the next row? One, three, three, one. One, three, three, one. The next row. One, four, six, four, one. Oh, that's cheating. He's reading it all off his paper. I kept mine. Lene, what's the next row going to be? One, five, 10. 10. 10. Five, one. Five, one. Very good. Okay, and we could keep on going as far as necessary, but I think this is enough. We wanted to find five Y minus one to the fourth power. Yeah. Well, let's see, how are we going to handle that negative? I think we could write this as a sum if I write it this way. Five Y plus what? Negative one. Plus a negative one, okay? So what I'm going to do is just treat A as if it were five Y, and I'm going to choose B as if it were negative one, and let's expand it. The very first term is going to be one times five Y raised to the fourth power. Now that's everything there, both the five and the Y are raised to the fourth power. Plus, what's the next coefficient going to be? Four times five Y to the third, right? Times negative one. Exactly, very good. You see, we're using this row because this is a fourth power, the one, four, six, four, one. So that's where we get the four from right here. Five Y cubed, negative one to the first. Then the next term will be six times five Y squared times negative one squared. And then the next term will be what? Anybody, somebody other than Stephen, he answered several of these others. What's the next term going to be? What's the coefficient? Four. Four, yeah. What else, Susan? Five Y. Five Y. Times negative one to the third. Times negative one to the third, and I see I keep crowding myself in on the last term, and the last term is going to be plus one times negative one to the fourth. Okay, so what I do is I write out these terms in factored form, then I go back and reduce them. And let's see, five to the fourth. That would be 25 squared. Does anybody know what 25 squared is? Is it 625? He's asking me, is it 625? It is, it is, 625. You know, the way I do that is I think if I had 25 quarters, I know I'm weird. If I had 25 quarters, how much money would I have? Well, let's see, that'd be $6 and 25 cents. Yeah, that's it. Okay, now the next term, well, there's a negative over here. Let's bring that up to the front. Five cubed, that's 125 times four. That's 500, 500 Y cubed. Well, the next term be positive or negative? Positive. Positive, because the negative one is squared. And this is 25 times six. Well, if you had six quarters. You'd have $1.50. You'd have $1.50, I wish. Okay, so that would be 150 Y squared. The next term will be negative. Okay, are you sitting down? Good. What's five times four? Your bet is 20, 20 times Y. And then plus one. Well, let's see, does anybody see a shortcut that would explain what happens when you have a difference rather than a sum in this answer? Alternating signs. The signs alternate. So rather than putting in a negative one and a negative one and a negative one, what you could do is expand this using five Y and plus one. If you just remember to alternate the signs. You start off with a plus. And in this case, I happen to end up with a plus. But for other powers, I could end up with a negative on the end. Just go plus, minus, plus, minus, plus. And wherever the alternating signs lead you, so be it. So this is the expansion of five Y minus one. Now, you know, those of you at home and those people in the class may say, well, now Dennis, you know, we probably could have multiplied that out by hand faster than it took you to write all this down. But you know, I've been trying to explain it as I go. And there have been several of us contributing information here. I've been adding all the fancy jokes along the way. You can't forget that. So that's why it took me a little longer than usual to work this out. But I think with practice, Pascal's Triangle gives you a real advantage, especially if you have a power bigger than four. A fourth power isn't really that complicated to compute. But what if it had been a fourteenth power or a fourteenth power? You merely have to write Pascal's Triangle out to the appropriate row. And then you have to have a wide enough sheet of paper or enough lines to record all those terms on. But it only takes one line to write it out in factored form and the second line then to write it out in its simplified form. Okay, let's go back to the example screen and I'll try to remember to save Pascal's Triangle here. Now, see in the last example, what makes it more complicated is not so much the power. It's only a third power. But inside I've got a u cubed minus ten v squared. Let's try working that one out here on the green screen. We have u cubed minus ten v squared, and this is raised to the third power. Okay, well the first thing I would say is there are going to be how many terms in this answer? Four. Four terms, yeah, because this is a third power. So let's make room for four terms. Okay, now the next thing is there's a negative in here. So what can you tell me about the signs? They're going to alternate. So let's go back and make a negative here and make a negative over here. Okay, now I'm going to be taking these terms to be a and b. So I'll begin with my first term cubed. By the way, I'm using this row right here. So I'll have one times u cubed cubed.  And then I'll have another three and a u cubed to the first power and a ten v squared squared. Let's move that negative over a little bit. And the last term will be ten v squared cubed. Okay, so I have my terms written out. That's my first stage.  What's u cubed cubed? Is it u to the 27th? U to the ninth. It's u to the ninth, thank you. Yeah, u to the 27th is a common wrong answer. Minus, now here I have a three and a ten. That makes a 30. Rather, u to the sixth and v squared. Plus, what will be the next coefficient? 30. Not 30. 300. Yes, because we have a ten squared times three. Plus 300 u cubed v to the fourth. v squared squared. And then minus 1,000 v to the sixth. So this is the expansion of this binomial. And I think that most people would find this approach quicker, especially for larger powers, that is using Pascal's triangle, than to do straight multiplication for these sorts of problems. Okay, well now we know how to use Pascal's triangle, but there's some more subtleties to it that we haven't investigated yet. Let's go to the next graphic and we'll introduce factorials. Now we'll be using factorials in the rest of this episode and in the next two episodes when we talk about permutations and combinations and probability. But right now I'm just introducing some terms. Now if you take the product of the first n positive integers, that's denoted by n factorial. And we put an exclamation mark after the n. So this does not mean n. No, it means n factorial. So n factorial means 1 times 2 times 3 times 4 up to n. Let me just use this space right here to take an example. Suppose I had 4 factorial. Well then that would mean 1 times 2 times 3 times 4. Which is how much? 24. Yeah, it looks like that's 6 times 4 is 24. Times 24. On the other hand, what if I wanted to change this to 5 factorial? What would I do differently? Times 5. Have to put a 5 in there, yeah. Okay, now how much is this? Well, let's see, here's a 12. And here's a 10. 12 and 10, the one's not going to affect the answer. That's 120. So we've gone from 24 to 120. If you were to go to 6 factorial, that would be 720. And if you go to 8 factorial, you're into the thousands. And so these factorials grow very rapidly. And many of you have a factorial key on your calculator. You may have, if you look at your calculator, you may have a key that looks like this. It may say x factorial with a little exclamation mark by it. And if you enter a 5 and then you push that key, immediately you should see 120. Now if this isn't written on a key, it may be written above a key on your calculator. Or it may be written under a menu, like if you have a TI-82 or a TI-89, you may see x factorial written under a menu that you could call up. And if you press that button, it will immediately multiply, in this case 1 through 5, and give you 120. Now you might say, well, why would my calculator have a factorial key on it? I've never used a factorial before. Well, maybe not so far, but in later courses, factorials are used sometimes, quite a bit. And in fact, in the rest of this course, we're going to be using factorials. So if you have this key available, you might keep it in mind. Okay, now, just to complete this graphic, there's one special case, and that's 0 factorial, or 0 with an exclamation mark. And this is defined to be 1, so we say that 0 factorial is equal to 1. Now, what's wrong with the definition right here? Why couldn't I just write 0 factorial using this definition? Because it doesn't start at 0. It doesn't start at 0. It starts at 1, and it goes up. So if you put a 0 in here, you can't start at 1 and go up to 0. So you see, this technically doesn't make any sense. But we define 0 factorial to be 1 so that it will fit with some of the some of the formulas and definitions that we see later. So we'll just take this to be 1. That's just the definition of it. Okay, now with this introduction to factorials, let's go to the next graphic and look at what are called binomial coefficients. For each pair of integers in an R where R is somewhere between 0 and N, so we're talking about integers. These would have to necessarily be non-negative integers because they're both at least 0 and R is less than or equal to N. Then we will be writing some expressions where I put parentheses and I stack two numbers in here, N and R. So if N is larger, I'll be putting the N on top. And the way we define this expression is we take the upper number factorial and I divide it by R factorial times the difference factorial. Now you might say, why in the world would you want to introduce something like this? Well, it's going to become quite useful here in the rest of this episode. It may not be something you've seen before, but it's something that you will be seeing now. Okay, if I use that definition, let's go to the green screen and calculate a few values that look like this. Suppose I had 4, 2. 4, 2. I'd like to evaluate this expression. This is sometimes referred to as a binomial coefficient, which we'll explain in a moment why it has that name. Now according to the definition, I'm going to put 4 factorial over 2 factorial times the difference factorial. 4 minus 2 factorial. That'll be 4 factorial over 2 factorial times 2 factorial again. Now to evaluate this, what I would do is to cancel off two of the factorials. You see in the numerator I have 1 times 2 times 3 times 4. In the denominator I have 1 times 2 and I have another 1 times 2. So I'm going to cancel off two portions of the factorials, this and this. And so what I'm left with is 3 times 4 over 2, which is 6. So this expression has a value of 6. And I'm using the formula, I probably should have written this again at the beginning, I'm using the formula n factorial over r factorial times n minus r factorial. Okay, I think we should do one more like this. What if we had, oh, let's see, what if we had 10, 7, 10, 7? Well, let's see, this would be 10 factorial over what? 7 factorial times 3 factorial times the difference factorial, 3 factorial. Okay, now on top I have how many factors? How many factors are there in 10 factorial? 10. 10 factors. Why don't I cancel 7 of those factors with 7 factorial? Because there is a 7 factorial within the 10 factorial. And if I cancel off 1 through 7, what would be left on top? 10 times, or 8 times 10. Yeah, or 10 times 9 times 8, I think Jeff was going to say, or 8 times 9 times 10. So I've canceled off the 7 through 1 and I've canceled off down here, 7 down to 1. And I'm left with 3 factorial on the bottom. So, you know, 3 factorial, if I were to write that out, would be 1 times 2 times 3. And the 2, I can cancel with the 8 and leave a 4. And the 3, I can cancel with the 9 and leave a 3. And so this answer is how much? 120. 120. Yeah. Now, you know, just above it here, let me just erase this upper portion and ask you a very similar question. Suppose I wanted to calculate this value, 10, but instead of 7, I'm going to put the difference in there, 3. 10 minus 7 is 3. What if I put in a 3 instead of a 7 there? So I'm replacing 7 with its, shall we say, with its complement. This will be 10 factorial over 3 factorial times what? 7 factorial. Times 7 factorial, because now 7 becomes the difference. Well, if you compare these two, these are exactly the same expressions. And therefore, this should also equal 120. So I think what we're demonstrating is a fact about factorials that I can summarize in this way. And that is that if I have n, r, if I have the binomial coefficient n, r, this is the same thing as n and put the complement of r on the bottom, n minus r. In other words, if I subtract 7 from n, rather r from n, I get n minus r. These numbers are the same thing. Let's write these out and see why. n, r is n factorial over r factorial times n minus r factorial. That was the definition. And the other expression, n, n minus r, is n factorial over n minus r factorial, that's the number that I have on bottom, times the difference. Now the difference is n minus n minus r factorial. So here I take my lower number n minus r factorial, and I take the difference. But you know if I reduce this, that's just going to be an r factorial. n factorial, n minus r factorial, and r factorial. Now if you compare this answer with that answer, those are the same. So we have this fact right here that says that n, r is the same as n, n minus r. We're going to see how this comes up in just a moment. In fact, let's look at it right now. Let's go to the next graphic on the binomial theorem. Okay, now this is a theorem that was first proved by Isaac Newton in a more general form. And it says that if you want to expand a plus b to the nth power, then you're going to get an a to the n, you're going to get an a to the n minus 1b, you're going to get an a to the n minus 2b squared. You notice here my powers on a going down, the powers on a go down by 1, the powers on b are going up. There aren't any b's, b to the first, b squared, all the way up to, finally, b to the n, and the a's are gone. Every term here has degree n. I have n variables there. I have n variables here, n variables here, all the way down. But now my coefficients do not come directly from Pascal's Triangle. These are the binomial coefficients that we just introduced. In my first term, I have zero b's, and so I call this n zero. In my next term, I have 1b, and I call this n1. In the next term, I have 2b's, and the binomial coefficient is n2. And so it goes. When I get to the n minus, when I get to the term that has n minus 1b's, it's n, n minus 1. And when there are nb's, I have the term nn. Okay. So, if I calculate each one of those binomial coefficients, that will tell me the number that came out of Pascal's Triangle. If we come to the green screen, let's just write out something like this using binomial coefficients. Suppose I had a plus b cubed. Now, we already know what the answer to that is. We've seen it come up today. But I'm going to write this as three zero, because n is three a cubed, plus three one a squared b. What would be the next coefficient? Three two. Three two, and what would be the variables? A b squared. A b squared. And finally, three three b cubed. Now, you know, this number and this number should be the same, because what I've done is I've replaced one with its complement two. That's the complement in three. So I've replaced one with the two. Those numbers should be alike. These numbers should be alike, because I've replaced zero with its complement in three. That would be a three. Okay, now, how much are each one of these numbers? I think I only need to calculate that one and that one, and then these two will be the same as the two that I've just calculated. So how much is three zero? Well, that's three factorial over zero factorial times the difference factorial, three minus zero factorial. That's three factorial over zero factorial times three factorial. Well, these two cancel off. And how much did we say zero factorial is defined to be? One. We call it a one. Yeah, just so that at moments like this, we can evaluate it and I get one over one or one. So the first coefficient is a one. Now, what's the coefficient here? What would you guess this should be when you cube A plus B? What should be the second coefficient? I think it should be a three. I think that's what Pascal's triangle told us. So this, if I calculate three one, there will be three factorial over one factorial times the difference factorial. That's two factorial. Now, to reduce this, let's cancel off the two factorial and what will be left on top? Three. There's a three. Yeah, we're canceling off the one and the two and the one and the two. There's a three left on top over one factorial, which is three over one, which is three. So what I can conclude from this is that A plus B to the third power should be one A cubed, so I'll write A cubed, plus three A squared B and then because of symmetry, this coefficient should be the same as that one because I have the complement as my lower number. So that'll be three A B squared and then this binomial coefficient should be the same as the first one because I've replaced zero with its complement. That'll be one B cubed and that's exactly what Pascal's Triangle gives me. We've seen this answer before. Now, you might say, well, Dennis, what is the point of this if we can use Pascal's Triangle to expand binomial? Why would I want to calculate the coefficients in this way? Well, you know, what if I weren't wanting to compute A plus B to the third power? What if I wanted to compute A plus B to the 30th power? Then I would have 31 terms and when I write down Pascal's Triangle, including the zero row, I'd have to write 31 rows all the way down. An alternative would be to calculate individual coefficients in this manner. We won't do A plus B to the 30th power, but this gives me a way of picking out individual coefficients and computing them directly without having to write out all of Pascal's Triangle. Okay. Let me just take another example like this and let's see how computing binomial coefficients can help me evaluate a problem. Suppose we had, let's say, 10C minus 2 to the fifth power. 10C minus 2 to the fifth power. Can anyone tell me how many terms this is going to have in this expansion? Six. It'll have six terms. It has one more term than the fifth power because I have to have a C to the fifth, a C to the fourth, third, squared, first power, and then the zero power. So there are actually six terms to write down. Now the first term will be 5,0 times 10C to the fifth power. And now the signs are going to alternate because of the negatives. I'll put a negative here. 5,1, 10C to the fourth power times 2. And then I'll put a plus. And then 5,2, 10C to the third power times 2 squared, minus. Okay, I'll just bring that minus down to the next row. Minus 5,3 times 10C squared times 2 cubed. You notice the degree is always going to be a total of five, five factors there. Plus 5,4, 10C to the first power times 2 to the fourth. Minus 5,5 times 2 to the fifth because my 10Cs now drop out. Okay, now let's see. These two coefficients will be the same. These two coefficients will be the same. And these two coefficients will be the same. I'm using that symmetry property. These are alike, these are alike. And now I'll do a little crisscross. These are alike. So I really only need to calculate three coefficients, not five. Let's calculate 5,0. I bet you know what that's going to be. 5,0, that's 5 factorial over 0 factorial times 5 factorial. This being the difference right here and I get a 1. Yeah, so that lead term should have a 1 on it. And then 5,1 is 5 factorial over 1 factorial times the difference factorial. And if I cancel off 4 factorial, I have a 5 left on top. 5 over 1 is 5. And then if I go to the coefficient here, 5,2, I'll have 5 factorial over 2 factorial times the difference 3 factorial. What would you cancel off here to make this simpler to compute? What would be better to cancel the 2 factorial or the 3 factorial? The 3 factorial. Yeah, let's cancel as many as we can. Let's cancel 3 factorial and that leaves me with 4 times 5. But I still have the 2 factorial, 1 times 2. That's 20 over 2 is 10. So that tells me the first coefficient here is a 1. The second coefficient is a 5. The next coefficient is a 10. But now what's this coefficient? 10. That's a 10 because, see, I'm using the complement of the 2 that came just before it. That's a 10. This is a 5 because it agrees with that one and this one is a 1. Okay, I have all my coefficients. So, excuse me. So now I'm going to expand this and write down my final answer. Let's see, 10 c to the fifth power. What's 10 to the fifth? 100,000. 100,000. Yeah, it's a 1 with 5 zeros. 100,000 c to the fifth. Minus. Now next I have 10 to the fourth. That's going to be, what's 10 to the fourth? 10,000. 10,000. But I have another 10, a 5 times 2. So that makes 100,000 again c to the fourth. Plus. And now I have 10 to the third. Let's see, 10 to the third, that's 1,000. Times 10 is 10,000. Times 4 is 40,000. That's 40,000 c cubed. Minus. Okay, now the next term. This is going to be 100, because there's a 10 squared there. Times 10 to the thousand, times 8 is 8,000 c squared. Plus. Let's go down to the next line. 10 times 5. I'll tell you what, let's take one of the 2's in that 5. One of these 2's in that 5's makes a 10. There's another 10, that's 100. And I have 3 2's left over. That's 100 times 8, 800 c. Minus. Now all my 10's are gone. 2 to the fifth is 32. So this is 10 c minus 2 to the fifth power. It's this 6 term polynomial. Now you might say, well you know Pascal's triangle might have made this a little bit quicker. But I do have an alternate way now of picking out individual terms and calculating them. And so this gives me now a second way to compute these binomials. Okay, let's go to the last graphic. Here I ask you a few questions about the expansion of a binomial. And I think we've been answering questions sort of like this, but here we see it in a concrete form. How many terms are there in the expansion of a plus b to the 12th power? 13. There would be 13 terms. Yeah, 13 terms. Because it's a 12th power and we get an extra power to include an extra term because there's a zero power involved. What's going to be the very first term of that expansion? Jeff, what would you say? A to the 12th. A to the 12th. Yeah, it would be 1 times A to the 12th. And what's going to be the 13th term? B to the 12th. B to the 12th. Yes, that's going to be done at the far right end. B to the 12th. Now what's the fifth term going to be? Well, let's see now. The fifth term is going to be, we started off with A to the 12th and we're going to reduce it four times so we'll have A to the what? A to the 8th, I think. And the B's, which started on zero, are going to come up to B to the 4th. So it would be A to the 8th, B to the 4th. Now what we need is a coefficient. Let's come to the green screen and try calculating the fifth term. So let's write down that question. We're wondering what is the fifth term of A plus B to the 12th power? Well, let's see. We have a term and we have a term and we have a term and we have a term. Here's the fifth term. That's the one we want. Now what we found out was the very first term is A to the 12th and we found out that the very last term down here is B to the 12th. But the question is what is the fifth term right there? Well, that term is going to have to have an A to the 8th and a B to the 4th. Now the way I figured that is we started on 12 and we got 11, 10, 9, 8. 8 to the 8th. And the B's are going up. B to the 0, 1st, 2nd, 3rd, 4th. B to the 4th. And the coefficient will have what two numbers in the binomial coefficient? Twelve and... Yeah, what number are we going to put there? Four. Twelve and four. Because you see the very first coefficient is 12, 0. So here's 12, 0, 12, 1, 12, 2, 12, 3, 12, 4. So you can't go by the fifth term and put a 5 here. You go by... Well, what I go by is the power on the B. So B, 12, 4. So how much is 12, 4? Well, let's just write it out. That's 12 factorial over 4 factorial times 8 factorial A to the 8th, B to the 4th. Now, if you were going to cancel here, which of the denominator factors would you cancel? Eight factorial. Eight factorial. Get rid of as many as you can. So that's going to leave us with 9, 10, 11, and 12 on top. Because we canceled 1 through 8. On the bottom I have 1, 2, 3, and 4. Well, let's see. 3 and 4 is 12. So that gets rid of that guy. Whether than dividing by 2, let's go ahead and multiply this out and divide by 2 afterwards. I'm thinking this is 99 times 10 is 990 over... Oh, you know what? I left off my variables. Let me just put my variables on the end of that. A to the 8th, B to the 4th. Now, when I multiply that out, I get 990 over 2, A to the 8th, B to the 4th. And now let's divide by 2, so I don't have to do any further multiplication. 4, 95, A to the 8th, B to the 4th. That is the 5th term. That's what goes right here. That's going to be 4, 95, A to the 8th, B to the 4th. I have to squeeze that in that position. Okay, one more example. And I'll do this one on the green screen. I'm wondering how much is 11 raised to the 6th power? 11 raised to the 6th power. Now, I'll tell you what. Let me make that a little bit smaller. That may be a little bit too aggressive there. Let's go to the 4th power, and then we'll do the 6th power after this. Now, in this case, I'm going to go back and use Pascal's Triangle one more time to illustrate, just to remind people how this goes. And I'm going to raise this number to the 4th power using Pascal's Triangle. The way I'll do it is I'll think of this as being 10 plus 1 to the 4th power. So this is my A plus B. So when I expand this, I get 10 to the 4th plus 4 times 10 to the 3rd times 1 plus 6, oops, better put a multiplication dot in there, plus 6 times 10 squared times 1 squared. I won't bother writing that one out. 1 squared is just 1. Plus 4 times 10 plus 1. Now, you notice this is an abbreviated form. I didn't show my B factors because those are all 1s. And so this is going to give me 10,000 plus 4,000 plus 600 plus 40 plus 1. Now, if you add that all up, that's going to be 10,000, 4,641. Where have you seen these numbers before? In Pascal's Triangle. That is Pascal's Triangle. You see that row of the triangle represents 11 raised to that power. What would you guess would be 11 cubed? 1,331. 1,331. Here's 11 squared. Look, here's 11. And here's 11 to the 0 power. Well, it was nice seeing you today, and I'll see you next time for permutations and combinations in episode 37.


Welcome to the crypto minute. The Delity is growing their digital asset unit. Interpol is getting into the metaverse. Not see that one coming. Monetary 30 of Singapore proposes a ban on cryptocurrency credit. Can't throw mashed potatoes on JPEGs. Red NFT trading hit an all-time high and there's almost 3 million wallets. Looks like Bitcoin got the memo that it's spooky season. Fain becomes the third largest crypto ATM hub. Happy Wood said that she bought 100K Bitcoin years ago at $250 and is never sold. Do you believe that is the definition of diamond hands? Makeford always had to break up into smaller, more decentralized pieces called meta-dubs. Robin Hood is adding support for Ave and Tezos. And Apple's tightening rules on cryptos and NFTs in the App Store. How much to tell them that fighting innovation never works out? 75% of people who own Bitcoin are long-term holders. I reckon there's a new sheriff in Twitter town. There's now an inflation peg coin that's tied to the cost of living. The UK Prime Minister once commissioned a royal mint NFT. So cringey. Apparently Sam's going to refund some people that fell for an FTX phishing scam. Winners reportedly building a crypto wallet. Google searches for Bitcoin is dead reached an all-time high. And FT's versus digital collectibles. Keep stacking and I'll see you next week.


The speaker is Andrea Waise and she's going to talk about mechanistic models of bacterial growth. Thanks very much. Do you hear me well? Alright, so I will talk about yet another approach of modeling bacterial growth. Perhaps most similar to what Sanjay presented yesterday. Right, so just before I start, I wanted to mention that in my group, what we do is we work to use our skills and do our humble bid to try and tackle the AMR crisis, the resistance crisis. So we do this not only at the level of bacterial physiology, we also work at the patient level where we work with clinicians to understand how transmission of the worst forms of antibiotic resistance works in hospitals. So this is pretty much data driven whereas most of what we do at the cellular level is a mechanistic approach. So you've seen that in Holly's talk on Tuesday on the RNA repair and trying to understand how that influences tolerance. There's also a poster outside by Elena where we're trying to better understand antibiotic responses and the dose responses. So another project that we're working on which is not mechanistic but rather data driven is one by Emily where we're looking at the early adaptive response of MRSA, so methicillin resistant staphylococcus aureus and the role of different RNAs in dynamically adapting to a host environment. So mechanistic projects are mainly based around this approach that I will present to you now which is about before I start with that. So if you find that things I present here or things that you hear from Elena and Holly are interesting to you, please talk to us. We do have PhD openings and also spread the word if you find that there might be candidates who might be interested. So basically the models that we work with, we see them sort of as a framework to predict growth rate in response to different kinds of inputs or perturbations. So those inputs could be different kinds of nutrient environments or drugs that they are exposed to as well as genetic modifications. The framework that we want to look at should ideally be dynamic and what I will talk about now will be split in two parts. In the first part I will talk about a framework that we developed of modeling bacterial growth responses deterministically, sort of the average responses of cells in a population. And the second part then will focus on single cell responses and how stochasticity drives heterogeneous responses in growth. Before I run out of time I would also like to thank my coworkers on those projects. So those are mainly spread between Edinburgh and Imperial College which were the two places I have been flip-flopping between with the exception of Vincent who deserted us to France. So we've seen a lot of different approaches and obviously we know that there's a lot of activity going on in the cell and it's highly complex and we're trying to coarse-grain this in some way. And there are various approaches that we've already seen here at the workshop. And for us now the choice was which kind of avenue to go for with the goal that we had of predicting emergent responses. So as I said already with the inclusion of different kinds of perturbations or inputs, actually a mechanistic approach lends itself very well to this. And so although we've actually taken a lot of input from the kinds of approaches that Terry has presented and has developed, so we decided to go for a mechanistic approach which kind of on this slide, the only truly mechanistic model that you see here is the whole cell model where there's an attempt of modeling everything in the cell which was also not the avenue we wanted to go for. The last approach down in the bottom is one that in terms of the coarse-grainness was actually most appealing to us, but it was still based on optimization which was something that we wanted to shy away from. So the approach that we took was one that had a very similar level of coarse-grainness as the third approach I've shown you there. But instead of relying on optimization we wanted to fully remain mechanistic. So the model that we developed here is one that largely builds on the central dogma, the central dogma based around a coarse-grain genome much inspired by the works of Matt Scott and Terrewah. So this basically describing the central dogma on this coarse-grain genome dynamically and then combining it with some sort of nutrient and energy influx. So the import of a nutrient is conversion to energy that can then be used to fuel the biosynthesis of proteins. And we focused our modeling approach on three fundamental constraints that all living cells will face. So one is that they will only have a finite amount of energetic resources that we coarse-grain with this one species A here. They all rely on a finite pool of ribosomes for their protein biosynthesis. And then their proteome is finite and it needs to include all the proteins that the cell needs to perform the different tasks. Now I will show you now, I will go through how we implemented those different constraints. So in the first one about the finite pool of energy, so basically we build this model as an ordinary differential equation model where we describe the temporal evolution of the different species. So here it's this energy species and this basically comes in at a rate that depends on the enzyme concentration of this green guy here as well as the concentration of the internal nutrient. And then to model different kinds of nutrient environments, we have this parameter that basically tells us how much energy do we get out from that type of nutrient. And then we have all the processes that are feeding from this energy resource and here we're making a huge simplification. We do assume that it is only a translation that dominates energy consumption. And this is based on some estimates that stated that energy consumption by translation is around 70% of the energy budget. And then finally all our cellular species are being diluted through growth of the cell. We have a mechanistic derivation as well of the translation rate of the different species that we're considering. So for that derivation we're considering a mechanism where basically a ribosome bound to an mRNA can reversibly bind to the energetic resource and then once it's bound one elongation step can happen. So obviously this has to happen in times within the number of amino acids in that protein. And then at the end the peptide is released and the complex falls apart. And so we do not model all of this explicitly, we only use that to derive the net rate of translation, which then turns out to be this kind of Michaelis Menten type kinetic where we have a maximal translational elongation rate and sort of an energy threshold at which elongation is half maximal. And further it depends on the levels of bound ribosomes of that type of mRNA as well as the length of that protein. Right, so in the next trade-off we look at the finite number of ribosomes and for that first we need to model how mRNAs are being transcribed and then how they competitively bind free ribosomes. So here we have the equation for the free mRNAs being produced through transcription and through the same logic as I showed you in the previous slide about translation we can argue that each elongation step in transcription consumes a finite amount of energy even though we neglect it we can assume this kind of same shape of the transcription function. And then I want to emphasize that we have those parameters theta here that principally can apply to all different genes but in our framework we infer them for ribosomal and non-ribosomal genes separately. And only for the kind of housekeeping proteins that are supposed to have a more or less stable concentration throughout different growth conditions we assume this kind of negative auto-regulation that keeps them more or less stable. And then the way we implement the constraint in ribosomes is in the equation of the free ribosomes where basically apart from the production and dilution obviously we have to sum over all the processes that either take away free ribosomes or feedback into the pool of free ribosomes. And that is basically summing over all possible genes, all possible mRNA types that ribosomes can interact with and we have the binding of mRNAs and free ribosomes as well as their unbinding and then the free up at the end of translation. Right and finally we have the constraint on the finite mass of the cell. Again here we're assuming that this mass is only made up of protein and we can sum up basically our total mass of the cell. So this is given basically by looking at all the different proteins that we have and multiplying them by their length. So we're considering the mass as a total of amino acids incorporated into the proteome. And then obviously we also have to consider the ribosomes that are being found by mRNAs. So when we write this we can then also write the ODE for the mass and that obviously is determined by how much protein is being produced and then how much of the mass is being diluted through growth. Now so here we still have this kind of undetermined growth rate and it is quite obvious that when we set M to zero so we are at kind of steady state growth. Then this mass is not the total mass. It's being diluted so it's like a concentration. It's different from... I mean we're not considering volume, we're kind of considering that... That mass is proportional to volume but this M I should think of it as a fraction or a concentration. There's a mass fraction or a concentration. Yeah. Thanks. So obviously when we set this to zero then we get that lambda should be proportional should be equal to the total translation rate divided by the current mass. And to define this rate dynamically we set the dynamic growth rate to the total translation rate divided by a mass that we take as the typical mass of an exponentially growing cell. And then what we guarantee is that at steady state so at balanced growth we will reach that typical mass. Right so this is just to show you how the full model then looks. We have a set of coupled oddities here. We have except for this auto inhibition term in the housekeeping transcription. We have no sort of switching functions or anything. And overall we then have 14 oddities so basically 3 oddities per gene that we're considering in terms of the free mRNA, the bound mRNA to the ribosome and then the protein plus the internal nutrient as well as the energy. Now obviously this model has a lot of parameters but the advantage of this mechanistic approach is that we could actually harvest quite a lot of those parameters from the literature. And the parameters that we were left to infer were actually those associated with the expression of those coarse grained genome sectors. Right so we used the data from Terry's lab to fit our model and what you see here is those lines are simulations based on samples from our posterior distribution of the parameters and overall we see that a lot of different parameter values seem to produce this behavior very robustly. We could also look at the actual values of those parameters and we saw that there was a huge range of variation in the actual parameters. So this was a very robust behavior which then led us to actually derive those growth laws analytically from the model and we could which I guess is not so surprising anymore. We've also seen that from Sanjay's model yesterday which has kind of the same structure although we fleshed it out a bit more with mechanisms. And then what this allows us is to kind of derive the relationships where we can determine the parameters in those growth relationships. We can link them back to parameters in our model. So for example here the slope related to the time it takes to translate the efficiency of the ribosome. This is something that Terry also mentioned already. And then the second growth law when inhibiting translation we could relate it back to the efficiency of the enzymes. The amount of housekeeping load that we need to produce. And then finally also related to the previous talk now the mono growth where we found that the maximal growth rate depends heavily on the housekeeping load as well as the enzymatic and the ribosomal efficiency. Right, so we also use this model now to study different systems. So as an example of one system where we kind of use the model to contextualize, provide a cellular context to systems that are often studied in isolation and that may behave very differently when we actually consider them within a dynamic environment of a cell. So one example that we saw on Tuesday was Holly's RNA repair system which we're trying to currently integrate into this growth framework to kind of see how her system dynamically interacts with the growth apparatus and kind of invokes those different kinds of tolerance states. Obviously as I told you in the very beginning we're very interested in antibiotic responses so this framework really allows us to model how the different antibiotics tackle different sports targets inside the cell and the framework has also been taken up quite a lot by the synthetic biology community or I have to say now engineering biology community it's what we have to say in the UK now. So where basically people use it to quantify the interaction of the synthetic circuits that they build with the host cell and kind of quantify the burden that they impose on the cell as well as the limits how the host physiology constrains the performance of the synthetic circuits. And I'm aware this, I'll mention it I know this is kind of makes me a target here but we've also used it to model competition experiments so I believe this is really a very flexible framework that allows us to modularly put different aspects that one might be interested in together and investigate them. So if there are no questions for this part I'll move on to the second part. We know that the growth of single cells can vary quite a lot so what you see here is the growth trajectories of individual cells and you can see that those differ quite a lot like how fast they grow, how big they grow, et cetera but also if we look at the growth rate of individual cells then those can vary considerably over time and this variation of growth rates can be influenced as well by the environments that the cells are exposed to. Now why is this interesting? This is interesting to us because it can have quite an impact in real environments when cells are exposed to antibiotics and when they might display tolerance to antibiotics which then leaves the treatment ineffective. So the framework, the reason traditionally people believe is how heterogeneity among isogenic cells arises is through gene expression because it is the process where we have mRNAs at very low copy numbers and this then typically results in a variation of levels at the protein level and then this can lead to different phenotypic responses. This has been studied extensively but typically at the level of a single gene or maybe a combination of a few genes. Now we're looking at this messy system and to study how variation in some of these components lead to variation in growth rate we again go back to our framework but now we consider it in a stochastic setting. So I apologize it's a different figure but it's exactly the same reaction, it's exactly the same species and now we consider every reaction as a stochastic reaction. So this really allows us to not impose stochasticity phenomenologically but kind of predict its emergent effects from first principles. Now I think at the single cell level it's not enough to only consider stochastic reaction kinetics but a major stochastic event in the cell is also cell division so what we did for that is we coupled the stochastic growth model with an established cell cycle model which is the Donicky model so basically we coupled cell growth to DNA replication where we used the Donicky model which basically says that a new round of replication is initiated whenever the initiation mass is passed the way we interpreted it was that the concentration of DNA origins goes below a certain threshold. One comment, so there's a lot of single cell data that challenges this initiation model probably it's irrelevant for what your purpose is but just to mention. So after initiation then there's a fixed period of time that takes for the cell to replicate. The chromosome, this is called the C period and after termination of replication so basically when this replication reaches this yellow point here there's the D phase, the division phase where the cell prepares for division most likely and those we consider fixed. There's also some contention about that. Sorry? Are your models into models that resemble what you actually measure in single cells which are sort of equivalently complex? So this now allows us to simulate the growth and division of a cell lineage over time. Just to clarify how we do the division so there are different ways of doing that and basically we assume that we have the cell as a bag of different types of molecules and what we do is then that we say that the cell decides on a septum position where the cell will divide and that septum position will on average be one-half but there is some variation around that and we found that this little bit of variation was actually important to reproduce some of the data that we looked at and then once this proportion of the daughter cell is determined then we do a binomial partitioning of all the different chemical species separately with that proportion. Right, so now we can stochastically grow our cell. We can divide it and we can simulate it over time. So here you see one such simulation where we can see how fluctuations and some mRNA species might have like a slow fluctuation over time and we can see how the slow fluctuation kind of propagates to a slow fluctuation in growth rate so we see that molecular fluctuations move on to more physiological fluctuations and we can look at different kinds of macro variables in our cell. Right, now this is hugely inefficient and we have to, in each of those cell cycles we have to simulate the production of millions of molecules so this is computationally infeasible and that's why we also looked at some approximation method for that. Again, I'll just quickly go over this. So here we have our variables which are the molecule numbers of different types each of those molecules is associated with a certain mass and I'll remind you that in the current setting we're only considering proteins that have mass and then we can sum up the masses of all the proteins to arrive to the total mass of the cell. And then assuming that mass density is fixed and mass is proportional to volume we can look at the concentrations of different species and also write down the cellular growth rate as the rate at which mass is being produced. Now the first approximation that we do is that we assume that those concentrations although they are discrete of course they are continuous and if we ignore division and growth for a moment then this is just a standard set of chemical launch event equations but of course we have division and growth and so we have some extra terms in our equation which account for the dilution of species through growth the biomass production as well as the partitioning of the cellular contents at growth. We also do have an equation for the mass and that overall gives us a couple of set of launch event equations which we still cannot solve. So we did another approximation here which is basically our small noise approximation where we say that fluctuations are small as compared to our system size so basically our M is typically very large. That's what we assume and then we can derive some closed form estimates that allow us really to use this model effectively. So by being able to solve this model quickly we can integrate different kinds of data and estimate parameters we can analyze the model and we can test different kinds of hypothesis as I will show you now. Right, so the data that we integrated is a mix of population, both data so this is the data that we've seen over and over again so the data from Matt Scott's paper on mean growth rates and ribosomal contents as well as single cell growth data and the fluctuations in terms of the coefficient of variation of the growth rate in different growth conditions. So we were able to fit the data that we used and then we used the inferred model to do some sort of sanity check on how we're able to reproduce independent data so one thing that we looked at is basically the unit size of the cell that's the size of a cell per number of origins DNA origins, replication origins and this has been observed to be fairly constant over different conditions which is what we reproduced this is not a surprise because it's a consequence of the Donnecke model that we assumed for the cell cervical and DNA replication but when we looked at our estimates of the unit size which were in the units of amino acids incorporated into the proteome compared to volume then actually we could see that we're reproducing very nicely the protein density that's been estimated in the literature and we also looked at cell mass which we're producing very nicely so we typically assumed here the black ones oh I forgot to say sorry so the different things that you see here is the squares are always the data that we used the dots are stochastic simulations and the line is the small noise approximation that we use which agrees very well with the stochastic simulations just as a sanity check so for the cell mass in most of our simulations we used a C plus D period of 60 minutes and there was another set of data where they had observed a longer C plus D period so when we inserted that we could reproduce also that set of data quite well and then I was very glad to see Ludovico's talk the other day which made me understand why our predictions of total mRNA in different growth rates have this linear type of growth across different conditions so I haven't updated here the data to the newer estimates so this is from the Bremer and Dennis data which were indirect measurements right so this makes us confident that the model does something right and so we went on to actually analyze where the noise comes from that we're observing so maybe as a first observation one prediction of the model is that the noise for maximal growth rate goes to zero so a question that I always get is so does that mean that there are no fluctuations anymore? I have a question that's not this question so your model should give a prediction for the effective noise of the growth rate along the lineage of single cells so is it additive or mixed or additive, multiplicative? Did you look at that? No we didn't so the other question I get is that noise goes down to zero which doesn't mean there are no fluctuations in the cell anymore it basically means that we're now in saturated levels of the growth rate where fluctuations don't lead to changes in the growth rate anymore so we do not observe noise in the growth rate anymore and then our noise, the decomposition so basically the noise that is predicted by the model we can decompose into the contributions of different reactions that cause that noise and basically one prediction that was made by the model is that throughout basically all of the growth conditions we have half of the noise coming from basically the decay of mRNAs and that is either through mRNA degradation or through their partitioning at division and the other half is coming through their production now this is just any type of mRNA but we can also further pin that down and one thing that we found when asking which mRNAs causing the growth noise was that this is quiet condition dependent so basically when we assumed that we have one rate limiting step in the import of the nutrient so this is the slowest reaction here then basically the model predicted that all of the noise pretty much comes from the mRNA that produces this transporter here and then conversely if the metabolic enzyme that converts the internal nutrient to energy is the rate limiting step then all of the noise came from the mRNA for this enzyme now since ribosomes are so essential to growth we also asked if ribosomes were maybe able to produce the observed levels of noise and in principle that was sort of possible but it meant that the mRNAs coding for ribosomal proteins had to be at extremely low levels and that the blue ones here didn't agree with values that had been reported in the literature whereas values that we estimated in those two conditions actually had a fairly good agreement so we could kind of discard this hypothesis here right so we can look at the sources of noise here but we can also look at how initial fluctuations those sources kind of trickle through the growth apparatus of the cell I actually didn't want to talk about this because it always takes a long time to explain so basically I will just sketch it and if you're interested in how we did that precisely then talk to me or check out our paper so basically this is looking at the stochastic simulations and then computing the cross correlation between species and then based on the lag time between different species we build those kind of minimal delay graphs so basically species that are next to each other in this graph have a minimal delay between in their correlation and we interpret that as it's the species that first feels fluctuations in the other species and now the reason I wanted to mention this was because Dan gave this very beautiful talk yesterday about how growth rate should influence the switching rate between different phenotypes and perhaps like at a mechanistic level one thing that we found was that growth rate was never at the bottom of the propagation so it was not a pure receiver of noise in this propagation rather it was usually quite high up and it would transmit noise to a lot of downstream processes which also means that if there are some switching gene systems that I don't know switch between persistence and susceptibility those could be affected by the noise that is transmitted through growth rate which would be higher at lower growth rates and finally I wanted to talk about antibiotic responses so the stochastic model can reproduce as well the inhibition data through chloramphenicol this is no surprise because the deterministic model did that already and here we are looking at the average responses of the stochastic model we also looked at this in more detail so how different like the whole space of different nutrient qualities and chloramphenicol doses how that affected average growth rate we didn't see any surprises here so basically with higher nutrient quality growth rate goes up and with increasing antibiotic dose growth rate goes down what was surprising to us was how complex the noise landscape was so basically the heterogeneity in growth rate which might indicate that cells that exhibit different growth rates might also exhibit tolerance when they are exposed to antibiotics so here we saw that this was a very non-monotonic landscape and basically if we're looking at any fixed nutrient condition in this range here and we're kind of in this context thinking as a nutrient and condition as a location in the body where we have an infection or so so those might be very different in terms of nutrients if it's in the bladder so if we look at any different any fixed nutrient condition then increasing the antibiotic dose doesn't have the expected effect of monotonically increasing noise because this is what we expect from what we've seen before that at lower growth rates we see higher noise but rather we see that actually we might be initially going up a noise and then the fluctuations take a dip so meaning that actually for many conditions there's this non-zero antibiotic dose that minimizes the growth heterogeneity which might be something desirable when we're trying to devise a treatment strategy that minimizes the possibility of invoking tolerance so with that I'm coming to the end and basically the key messages that I'd like you to take home is that by looking at those cellular mechanisms key cellular mechanisms and the kinds of constraints that we considered we were able to capture the fundamental growth relations that we've been talking all week about and this mechanistic approach gives us a lot of flexibility to modulally study different systems of interest and their emerging growth responses and also that this kind of stochastic framework that we developed I think is very powerful to pinpoint sources of noise integrate different kinds of data and then analyze how this noise goes to the cell and what kind of potential consequences this might have and with that I'd like to finish and thank you for your attention We have time for questions Thank you for your talk and thank you for putting in the slide on the noise transmitter which is the growth rate I was wondering you have this coarse grained model and that also per gene I guess quite high compared to what it would normally be per gene do you think this would really change the model if you have mRNAs that are most of the time zero can you still do the small noise approximation for example Yes we can I mean so I would assume that if we inserted now some sort of circuit of interest then and we would re-estimate the different like expression parameters of the different sectors I would assume that by fitting the actual system that we're investigating then we will get much higher expression levels on the other sectors What about the noise you have when you add antibiotics the very sharp transition increasing noise at a very high concentration of antibiotics is that increased due to that you have a growth rate equals to zero or is there another reason Well we think actually in that regime we're getting into what we investigated before this ribosome limiting regime when we get noise and ribosome availability as well as getting very close to zero Yes so in your model you can let's say the deterministic one you can measure the fraction of ribosomes that are translating a particular gene or set of genes and ask whether it's the ratio of messengers of that particular class of genes or genes to the total is it and do you have regimes where it's different things I'm not sure I understand What is the fraction of ribosomes that are translating a certain gene or class of genes in your model and how it's related to messenger concentrations Did you look into that? Well I mean we just have mass action binding and unbinding so it is completely related to the relative abundance of the mRNA But because you have the complete level of mass action you could have different regimes probably I was wondering if you get to different regimes or not In terms of what Ludovico for example I was very glad to hear his explanation because it made me understand much better why we're seeing this and that actually what he presented made a lot of sense to me because we did actually because we're modeling the binding explicitly we do account for this complex formation regime and then as far as I understood the linearity was only explained at the same time the elongation rate increases and this is also something that we had included already in our energy dependence from your distribution it sounds like you're probably not aware of our recent work on the total mRNA measurements but then from that study we were able