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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
[tex] \frac{dy}{dt}=ky [/tex]
where y is some function, k is a constant and dy/dt is a change in that function
[tex] y(t)=y_0e^k^t[/tex]
where y of t is a function, y of 0 is an initial value, k is a constant and t is time
[tex] y^-^1* \frac{dy}{dt}=k[/tex]
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-
[tex] \frac{1}{63}*\frac{2}{20}[/tex]
[tex] \frac{1}{63}*\frac{1}{10}[/tex]
[tex] \frac{1}{630} [/tex] cells per minute
[tex] \frac{60}{630} [/tex]cells per hour
[tex] .095 [/tex]cells 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 :)