How Do You Calculate the Relative Growth Rate of E. coli in a Nutrient Broth?

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Homework Statement

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A common inhabitant of human intestines is the bacterium Escherichia coli. A cell of this bacterium in a nutrient-broth medium divides into two cells every 20 minutes. The initial population of a culture is 63 cells.
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(a) Find the relative growth rate.

Homework Equations

Well this has to do with exponential growth and decay. So equations that could apply are

\frac{dy}{dt}=ky
where y is some function, k is a constant and dy/dt is a change in that function

y(t)=y_0e^k^t

where y of t is a function, y of 0 is an initial value, k is a constant and t is time

y^-^1* \frac{dy}{dt}=k

The Attempt at a Solution

(a) Find the relative growth rate in cells per hour.

well I thought since I know the change in the change of the number of bacteria per minute and the starting number of bacteria, I thought i could do this-

\frac{1}{63}*\frac{2}{20}

\frac{1}{63}*\frac{1}{10}

\frac{1}{630} cells per minute

\frac{60}{630}cells per hour

.095cells per hour

that answer is wrong. I don't know how to go about this as you can clearly tell. can someone give me a kick start?

thanks :)

 

[futurebird]

Plug in the values to:

y(t) = y_0e^{kt}

After 20 min you have twice as many...

126 = 63 e^{k(20)}

Now solve for k. Then you will have the equation for how many there are after t min.

 

[arildno]

Start with:
y(t)=y_{0}e^{kt}
Since we have:
63=y(0)=y_{0}, we ave determined ONE of the two constants, y_0.

The relative growt rate is, indeed, k.
We know that after t=20 minutes, the population has doubled.
Thus, we have:
2y_{0}=y(20)=y_{0}e^{20k},
which means:
e^{20k}=2\to{k}=\frac{\ln(2)}{20}
Thus, we get:
y(t)=63e^{\frac{t\ln(2)}{20}}=63*2^{\frac{t}{20}}
if you want to make the doubling time explicit, t being understood to be measured in minutes.

 

[hover]

Thanks guys :). I knew I kind of overlooked this question. I should have known better. k=2.07 cells per hour. I got part a right and once I got that, all the other parts of the question (a-e) fell right out. I thank all of you.

-Hover