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Question 1: Binary Fission, theory and reality When in optimalcondition, E. coli divides every 20 minutes. You just started aliquid culture with 1 bacterium= No.1-H...

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Question 1: Binary Fission, theory and reality When in optimalcondition, E. coli divides every 20 minutes. You just started aliquid culture with 1 bacterium= No.1-How many bacteria Nn will be produced by binary fission after24 hour-culture, in other words after 72 cycles (n) of 20 minutes?Use the formula Nn = No*2n -2-If on bacterium weighs 1 picogram (that’s 10-12 Kg), what isthe mass of bacteria generated after 24 hours? -3- The estimated mass of earth is 6. 1024 Kg, compare the massof bacte

Question 1: Binary Fission, theory and reality When in optimal condition, E. coli divides every 20 minutes. You just started a liquid culture with 1 bacterium= No. 1-How many bacteria Nn will be produced by binary fission after 24 hour-culture, in other words after 72 cycles (n) of 20 minutes? Use the formula Nn = No*2n - 2-If on bacterium weighs 1 picogram (that’s 10-12 Kg), what is the mass of bacteria generated after 24 hours? - 3- The estimated mass of earth is 6. 1024 Kg, compare the mass of bacteria generated after 24 hours with the mass of earth. - 4-Explain why after 24 hours of culture, such a huge amount of bacteria is not seen.



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A single cell of bacteria reproduces through a process called binary fission. Escherichia coli cells divide into two every 20 minutes. Suppose the same rate of division is maintained for 12 hours after the original cell enters the body. How many $E .$ coli bacteria cells would be in the body 12 hours later? Suppose there is an infinite nutrient source so that the E. coli bacteria cells maintain the same rate of division for 48 hours after the original cell enters the body. How many $E .$ coli bacteria cells would be in the body 48 hours later?

The number of cells at a specific time is defined by the equation in equals in zero times to to the t divided by G, where n zero is equal to the number of cells at time zero and G is equal to the doubling time. And so in in this equation is equal to the number of bacterial cells that are needed to equate to the mass of the earth. And we don't have this value. So before we can solve this equation, we need to find this value and to solve for in all you need to do is divide the mass of the earth which is given in the problem mass of the earth, which is six times 10 to the 24th kilograms, divided by the mass of the bacterial cell which I have converted 2 kg here, so it would be 10 to the negative 15 kg. And so this gives you the number of bacterial cells needed to equate to this value, which is the mass of the earth. And so we get six times 10 to the 39 cells. This is equal toe end. So then we can solve this equation because we want to find teeth. That's what we're solving for in this equation. T is the time, the amount of time it takes for one cell dividing at 20 minutes. It's doubling times 20 minutes to reach this number of cells. So when we saw for tea, we get 2642 0.8 minutes and you can convert that toe 1.84 days. So, realistically, the mass of the bacterial cells on Earth does not equate to the actual mass of Earth because food is scarce and the exponential growth of cells Onley occurs for a short amount of time, Um, Onley. When food is available. Eventually, food runs out. Individuals of the same species start to compete for this limited food supply. Also, natural selection will favor those cells with viable mutants mutations so cells will start to die off. So what this problem is showing is that exponential growth occurs on Lee during a limited time, and there are factors that play that determine how many cells can actually survive

Given problem. We're looking at the growth of bacteria. So we have the growth rate of bacterium, E. Coli, which is found in the human intestine. So we know that it doubles every 20 minutes. So Q. Of T. The natural population is going to be given to us as 100. Then we're gonna multiply that times Since it's doubling every 20 minutes. That's gonna be times two To the t. over 20 for Tias and minutes. We want to know how long it will take for a colony of 100 cells to each population of one million. Well, we can determine this by looking at why of or y equals to one million. Or we could also let Qfc equal a million. And solve either way, we end up seeing that it only takes 266 days approximately, or 266 minutes. That is to reach a colony of a million. Which is extremely fast.

So in the given caution we have that bacteria growth is uh is doubled inside in every 40 minutes. So there are three parts of the question. We have to, In the first part we have to find a common formula for the geometric segments. The second, but we have to find the value of N. For which A. And the income is greater than then look right. And in the third part we have to find how much time it will take for the number of bacteria to exceed one million. Right? So coming to the question, if you're upset here, the first time of the cities is a initial number. Right? The second after 40 minutes it will get doubled like again after 40 minutes. It will get to times of A. Two which is equal to two, multiplied by to a one. That is who is square everyone. Similarly, after 40 minutes it will become two times of a three. That is equals two to kill both A one. Right? So if you observe the pattern here, you can see that it is forming a geometric sequence whose common issue is two and the first time this year one. So the end time of the geometric sequence is given by he went into a to the power and minus one. The first time you see one here, the common ratio is too, so do power than minus one. This is a formula further and term of the geometric sequence. Now, coming to the second part, we have to find the in term of the cities for which A. N. Is greater than. Yeah, yeah, yeah. And it's greater than this value. Right? If the first time of the series is 100 right? So we know that A. N. Is everyone into two power and minus one. The first time is 100. So 100 multiplied by to power and minus one. This is the end of time. This term has to be greater than the given value. Late. So we can from this equation we have to find the value of an so when we simplify this we get to power than minus one. Yeah, 2000 minus one. Greater than to power. Land minus fun should be uh Mhm. Yeah. Right. So when we simplify this, we get the absolute approximate value of venice 15. Like now coming to the sea bad we have to find after. How much time does it take to for the back to a number of bacteria to exceed one million late. So we know that. And as equals to 15 for a n gather than uh dan leg. Right? So when we substitute any close to 15, 2 40 in two and minus one. We'll find out the time. Right? Since we know that. Mhm. Here is the number after 14 20 minus one minutes. And and it's better than one million. Right? For value of fairness 15. Therefore, it will take no time. That is 14 20 minus one minutes, where the value of an is 15. So when you simplify this, we get 5 60 minutes. So there's much time it will take for the bacteria to get uh to exceed one million.

Okay, so here we are, given that and sub j is the number after After while 40 times j minus one minutes. Right. So and sub j, um is the number after 40 times J minus one minutes. Okay. Eso in symbols. We can say that and sub j well is equal to 40 times. Um, 40 times J minus one up so equal to 40 times J minus one. And this is for all j greater than equal toe one. So for J greater than you, No one. Okay, so we can write on DSO sub j plus one in terms of and sub J for J grades and ego, one as follows. So we have that n sub j plus one. Okay, well, that's equal to, um 40 times. Well, j plus one is just, um j plus one. So 40 times J plus one minus one. Okay. And then we just distribute here, um, where we clean up. So well, J plus one minus one is just j. So never distribute. This is just equal to just 40 times, um, 40 times J, which is 40 times J minus one plus one. So this is equal to again 40 just times J, which is equal to 40 times J minus one plus one, which is equal to 40 um, times. Well, times J minus one plus one. Right. Which, um which e commerce on twice Right to say. I just, um you might notice that I copied down the same thing twice, but this is just equal to, um, equal to 40 times J minus one plus 40 which is equal to and sub J plus 40. So again, writing it writing and sub J plus one In terms of ends up Jay, we get that this is equal to end sub J plus 40. There is and subjects one in terms off and subject. Okay, um, so now let's see. Were given that and someone is the initial number of bacteria cells. So we have that, and someone is the initial number. Bacterial cells and sub two is the number of bacteria's number of bacteria after 40 minutes. So here we have initial. And then after 40 minutes, andan ends up three is after, um, and so on. Right. And with that end, someone we were given that end, someone is, um, 230 So, you know, the ends of one is 230. Okay, so we know that two hours is 120 minutes, so we can break up two hours into three parts as follows. So we have that, too. Hours is equal to well, 40 minutes, Um, plus 40 minutes. So that's what, 80 minutes, right? Plus another 40 minutes. Right. Um, which is, well, 120 minutes, which is two hours. Okay, so after two hours, the number of bacteria will be on did sub three. So from the, um, equation, we have that end sub j plus one is equal to again and sub J plus 40 right from party. Um, and then we confined and sub two and and sub three so and sub two is just gonna be equal to ends of one plus 40. So that's 2 30 plus 40 which is to 70 and then for end sub three or ends up three is just equal to end sub to write plus 40. So that's gonna be equal Thio to 70 plus 40 which is equal to 310. So therefore, after two hours, the number of bacteria And sub three here will be 310 after again after two hours. Right. Um and then, um Well, we could make a chart here for the next part here, so we can evaluate ends of one ends up to ends up three up two ends up seven recursive lee by basically a table so we could have our our j and then our end Sub j Okay. And we have here well, and J is one we know that ends of one is, um 2. 30. So we have that and someone is 2. 30 right? Our initial value here is 2. 30 and then we just use this recursive Lee, we just basically add 40 for every j. So we have when Jay is too. We have ends up to is just equal to 2. 30 plus 40. So that's 2. 70 and so on. We just add 40 so we can get an, uh, wins. Jay's three. We have that ends up three is just equal to 2. 70 plus 40. That's 3 10 and then you go keep going here. So when j is four, we have that n sub four is equal to just 3. 10 plus 40. So that's 3. 50 and you can continue here and for this out. When it is five, we have the three ends. A five is 3 50 plus 40. So that is what? That's 3 90 and what? Two more here. So when n when Jay is six, we have the end. Sub six is equal to 3. 90 plus 40. So that's 4. 30. And when j. S seven we have the ends of seven is equal to 4. 30 plus 40. That's 4 70. Okay. Eso they would evaluate and step one through ends up seven recursive lee in our table here. Um and well, then, if we graph this sequence so and sub j for J equals one up to seven again by using the window from zero 10. The closed interval here from 0 to 10 on Ben, um, 0 to 15,000. Um, well, would have our table. That would look are our graph here, which would look something like this. So we have one through seven. So we have, um, j values one to three for 56 seven. Okay. On ben, we go from on r R y axis. Here we go from here is 1000 and then, um, because 15,000. So let's say 1000. That's maybe up here today. 5000, 10,000 and 15,000. Uh, oops. 15,000. I had ah, hopes. I don't know. Strata wrists. There we Oh. Um Well, okay, so the the the growth of bacteria will increase as as Jay goes up. Okay, But on this table, um, well, so when jay was one, right, we were where we were, like, 2. 30. Right on. We're going up. We're gonna very I mean, we don't get very high here on a graph at all. Um, we're gonna look something. Something like this. Eso from the from the table and the diagram given we can see that the growth of the bacteria will increase right as Jay goes up. So when there are unlimited nutrients, the growth off the bacteria population will grow. We'll go faster


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