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Old 03-03-2013
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Bacteria Bacterial Growth Time Calculation

A laboratory technologist needs to set up a Kirby-Bauer test to evaluate the antibiotic sensitivity of a bacterial isolate from a debrided perirectal abscess. He plates 1 x 10^4 bacteria/mL onto a petri dish containing 40 mL of agar. The lag phase of these bacteria is 20 minutes. If the generation time of the bacteria is 30 minutes, how long after the lag phase will he need to incubate the plate for the plate to have the minimum density of 1 x 10^6 cells/mL necessary to conduct the test?


A) 1.5 hrs
B) 2 hrs
C) 3.5 hrs
D) 5 hrs
E) 6.5 hrs


Please provide the explanations on why the answer is the answer. Thank you.
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Old 03-03-2013
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D) 5 hrs.

it takes 10 doubling time for the sample to go from 10,000 to 1,000,000. So that works out to 5 hrs after the lag phase.
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Old 03-03-2013
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what is the source of this q ?
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Old 03-06-2013
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Thanks slowpoke! But, that's not the correct answer in the key. =/

Please help.


Quote:
Originally Posted by slowpoke View Post
D) 5 hrs.

it takes 10 doubling time for the sample to go from 10,000 to 1,000,000. So that works out to 5 hrs after the lag phase.
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Old 03-06-2013
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sorry my mistake it's 3.5 hrs.

It only takes 7 doubling times to get a minimum of 1,000,000. I don't know what I did to get 10 doubling times. But after 7 doubling times you'll get 1,280,000 bacteria/ml. and that works out to 3.5 hrs.
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Old 03-06-2013
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* thumbs up * slow poke!

The right answer is 3.5 hrs.

But, I'm not getting the math. How do you calculate it to result 7 doubling times?

I do understand that if it is 7 doubling times, then 7 of 30 minutes s 3.5 hr.

Can you please work out the math on how you get 7 doubling time?

Thanks again!
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Old 03-06-2013
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well there's a formula for calculating doubling times and half-times.

For doubling time:

x = y (2)^t/d

where x = final quantity, y = initial quantity, t = total time, d = doubling time

so in this case;

1,000,000 = 10,000 (2)^ t/0.5 {d = 30 mins = 0.5 hrs}
100 = (2)^t/0.5
ln 100 = t/0.5 ln (2)
ln 100/ln 2 = t/0.5
t = 3.3, so the closest answer is 3.5 hrs

If you don't like formulas, you could simply keep doubling 10,000 on your calculator till you reach at least 1,000,000 and that gives u the number of times the bacterial sample must double itself. Then u simply multiply that number by 0.5 to get the number of hrs.
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The above post was thanked by:
excellence (06-28-2014), luschka (03-07-2013), samren17 (03-06-2013)



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