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Connie T Connie T
wrote...
6 years ago
A sample of raw ground beef initially contained 57 E. Coli bacteria. After six hours, there are 125,456 cells in the sample.



What is the growth rate of the E. coli population?



Express your answer using scientific notation in the form of x.yz x 10d bacteria/hr. 
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Answer rejected by topic starter
wrote...
Educator
6 years ago
It's not mentioned in the question, but bacteria divide by binary fission, essentially doubling each time.

(A)

Use the formula:

A=A0btc

Let's fill this in:

123 456=57(2)6c

Solve for 'c', which represents the time it takes for the growth factor b to occur.

ln12345657=ln(2)6c

ln12345657=6cln(2)

ln(12345657)ln(2)=6c

c=6ln(12345657)ln(2)

c=6ln(12345657)ln(2)=0.5414

(B) c=0.5414  5.41×101 bacteria / hour.
dri
wrote...
6 years ago
That's wrong Upwards Arrow
wrote...
Educator
6 years ago
Well, nothing here mentions continuous or exponential growth, so I don't know.

Could be you subtract the smaller number by the smaller, then divide by 6. That'd suggest linear growth though, which makes little sense
dri
wrote...
6 years ago
try this

123 456 = 57 b ^ 6

123 456 / 57 = b ^ 6

log ( 123 456 / 57 ) = log ( b ^ 6 )

log ( 123 456 / 57 ) = 6 log b

log ( 123 456 / 57 ) / 6 = log b

10 ^ ( log ( 123 456 / 57 ) / 6 ) = b

b = 3.60 bacteria / hour

Each of the original 57 bacteria will divide 3.60 * 10 ^ 0 times per hour.
wrote...
2 years ago Edited: 2 years ago, Nyko.21
This is what I did:  

Growth rate formula: change in pop size divided by change in time


125,456 (final) - 57 (initial) / by 6 hrs

= 20899.83333 bacteria/hr

=2.08x10^4 bacteria/hr
Anonymous
wrote...
2 years ago
Check out this topic: https://biology-forums.com/index.php?topic=2039663.0
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