MathBench- Australia In search of ... the exact doubling Dec 2015 page 1

Microbiology:

In search of ... the exact doubling time

URL: http://mathbench.org.au/microbiology/matt-gets-messy-in-search-of-exact-doubling-time/

Learning Outcomes

When you have completed this module, you should be able to:

 Use the growth equation to calculate generation time from a set of data  Know how to use logarithms (logs) to solve equations with exponents

So far, we've been guesstimating the doubling or generation time from a graph of the cell numbers over time. But there is a problem...

Guesstimating a parameter from a graph is not really accurate scientific procedure. It would be hard to imagine publishing a report in an academic journal that started out "we looked at the graph, squinted a little, and decided that the doubling time was 23 ...".

It is possible to look-up the doubling time in a table of data, but only if the table happens to include an exactly doubled population. If the table does NOT contain a doubled population entry, then you're in trouble. In other words, we can figure out the exact doubling time for the “time series with convenient numbers below, but not for the one with any general sort of number:

Time convenient general 10:00 am 10 million 10 million 10:20 am 20 million 15 million

Clearly, the convenient example doubled in 20 minutes. And at first glance, it appears that the more general example got "halfway" to doubling, so doubling time should be 40 minutes. Let's see if that's true -- if the population keeps growing at the same rate, will it double in 40 minutes?

time general 10:00 10 million ... multiply by 1.5 to get 10:20 15 million ... multiply by 1.5 to get 10:40 22.5 million

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Oops, we got to 22.5, not 20 million. Granted, this is a small difference, but it's still not the right answer. As the old joke goes, a million here, a million there, pretty soon you're talking about a real epidemic. What we need is a foolproof way to determine exactly what the doubling time is.

Finding doubling time with general numbers

Recall that we started out with an equation:

n Nt = N0 2

Remember the exponential part of this equation is that we raise 2 to the power of the number of generations, n. Another way of saying this is that, for every generation, we multiply the original population by 2.

Let's say I told you that N0 was 10 million and Nt was 80 million one later. Luckily for you, these are nice neat numbers. You would probably think to yourself, "well, the population doubled three , from 10 to 20 million, from 20 to 40 million, and from 40 to 80 million. Since three doublings took 1 hour, each doubling took 20 minutes."

Another somewhat fancier way of getting the same answer would be to substitute the information that you know (Nt and N0) into the equation above, and solve for the information that you don't know (number of generations, n). Here's how that would look with some conveniently chosen special numbers:

What is the doubling time? (10 million -> 80 million in 1 hour) Hint Explanation Substitute in the known info 80 million = 10 million x 2n Divide both sides of the equation by 10 million 8 = 2n Use trial and error! n must be 3 3 doublings in 60 minutes means... 60min/3doublings = 20min/doubling, so the doubling time (g) is 20 min

Let's try the same basic procedure with more general numbers. Let's say the population increased from 10 to 70 million in one hour.

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What is the doubling time? (10 million -> 70 million in 1 hour) Hint Explanation Substitute in the known info 70 million = 10 million x 2n Divide both sides by the initial population (10 7 = 2n million) Use trial and error...? n must be ... hmmm... a little less than 3...

We have not solved our problem yet, and what's in the way is that 2 is raised to some power in order to get 7 -- but we don't have a way to calculate what that power is. In other words, the exponent is the trouble. We need a way to get the "exponential" out of "exponential growth" so we can deal with the maths. But how?

Logarithms = Exponent-busters

In a word: when you have a problem with exponents, try logs. The great thing about logs is that they help make sense of very large and very small numbers. They also help in solving equations with exponents.

If you're already comfortable with log expressions, just keep going. Otherwise, click on the link here for a quick reminder and an attempt to explain that exponent-busting magic.

Now let’s try out our knowledge of logs

So let's try using these rules where we ran into trouble before. Recall that we got as far as finding the log in our equation -- now we can bust right it...

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What is the doubling time? (10 million -> 70 million in 1 hour) Hint Explanation Substitute in the known info 70 million = 10 million x 2n Divide both sides by 10 million 7 = 2n Bust that exponent log(7) = n x log(2) This looks complicated, but really all you need is 0.84 = n x 0.30 a calculator Now some rearranging... n = 0.84 / 0.30 = 2.8 So there were 2.8 generations in 60 minutes, so.... 60 / 2.8 = 21.4 minutes each!!

Bottom line: Logs plus a calculator are a biologist's best friend....

Let's bust some exponents

So, give that a try on your own.

Let's say the oysters were infected by a single Vibrio at 3 pm, and by the next at 3pm, when they were harvested, there were 10,000 bacteria per oyster. Assuming exponential growth, how fast was the population doubling? n  Start with the basic equation for exponential growth: Nt = N0 x 2 , where n means the number of generations.  Remember that you already know both population sizes, so plug them in  You should get to an equation that has only 1 variable, but that variable is an exponent. Time to bust an exponent!  Once you've got an equation with only one variable and a few logs, use your calculator to get the value of the logs.  Once you know n (how many generations or doublings there were), you should be able to use common sense to figure out how long each generation was (g). Answers: No. of generations (n) = log (10000)/log (2) = 13.3 Generation time (g) = 24 x 60 / 13.3 = 108 minutes

Extended problem #1: Exponential growth ends (with a whimper)

Here's a set of data on bacterial growth rates.

Your instructions are 1) to determine the exponential growth rate of the bacteria, and 2) to determine when growth stops being exponential.

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Time (min) No. of cells 0 4.3 * 106 20 9.7 * 106 40 22 * 106 60 48 * 106 80 97 * 106 100 116 * 106 120 118 * 106 140 67 * 106

Just look at the data first... what can you see with your bare eyes, so to speak?

About doubling time: The population doubles a little more than once in the first 20 minutes, so doubling time is probably between 15 and 20 minutes.

About the general shape of the population trajectory: In general the population raises, and then levels out (around 100 minutes), and then falls (around 120 minutes).

About when growth begins to slow down: Each 20 minutes, the population more than doubles, until 80 minutes, when it just barely doubles. So maybe at 80 minutes it is no longer exponential?

The online version of this module contains an interactive applet, which allows you plot bacterial growth and calculate doubling time. To find this applet go to: http://mathbench.org.au/microbiology/matts- holiday-nightmare/6-looking-at-data/

Extended Problem #2: How long before the oysters are unsafe??

Based on lab tests, each oyster has 400,000 Vibrio at 9am. Assuming it was infected by a single cell at 5pm, then

1. What is the doubling rate of the Vibrio cells? 2. How long till the infective dose of 1 million is reached?

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Doubling rate?

 1*219 = about 524,000, so just under 19 doublings  5pm to 9am = 16 = 960 min  960 minutes / 19 doublings = about 51 minutes/doubling

Answer: about 51 minutes

If the Vibrio keep multiplying at this rate, how long before they reach the infective dose?

 400K --> 800K --> 1.6 million  Less than 2 doublings  2*51 = 102 minutes

Answer: less than 2 hours

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Summary

In order to find the exact doubling time, you first need to understand how to use logs to get rid of exponents. You need to know this, because the number of doublings (n) is the exponent in the exponential growth equation:

n Nt = N0 2

So, to determine the exact doubling time of a population, you need a two-step procedure:

1. Substitute the two population sizes into the equation, then divide both sides of the equation by N0 and finally take the log of both sides. This gives you the number of doublings. 2. Divide the amount of time passed by the number of doublings to find the doubling time.

This same procedure works on any kind of exponential growth, as we saw in examples about world population and about the growth of technology.

Learning Outcomes

Now that you have completed this module, you should be able to:

 Use the growth equation to calculate generation time from a set of data  Know how to use logarithms (logs) to solve equations with exponents