Question

The bacteria you are using grow with a generation time of 18 minutes in the medium...

The bacteria you are using grow with a generation time of 18 minutes in the medium that you are using. You inoculate the flask of medium with 8.00×104 bacteria. The flask is incubated for 180 minutes. Assume there is no lag phase and the culture does not reach stationary phase during the period of incubation.

Within 5%, what is final number of bacteria in the flask?

(Do not use scientific notation to enter your answer. Enter 4.2 x 107 as 42000000.)

I've tried twice, and I only have three tries! How do you solve this?

Someone said: "
The formula is N(f) = N(i) x (2^n)

N(f) = the total number of bacteria
N(i) = the initial number of bacteria
n = the number of generations

(n = growth rate (in gen/hour) x time (in hours))"

But I don't know if these equations are right.
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Answer #1

The generation time is also the doubling time. So, if you have 1 bacterium, after 18 minutes, the number will be 2, and after 36 minutes (2 * 18), the number of bacteria will be 4, and after 54 minutes (3 *18), the bacteria count is 8. That is, if n is the number of generations of bacteria, then 2n is the number of bacteria will be produced after n generations.

If we start our calculation with one bacterium, then it will result to 1 * 2n to get the number of bacterial count. The starting number of bacterial cells is, therefore, multiplied with 2n numbers because every cells doubling time or generation time is countable.

Now, in this question, using the formular, Nf = Ni * 2n, we get Nf = 8.00 * 104 * 210 = 81920000 or 8.192 * 107.

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