A common inhabitant of human intestines is the bacterium Escherichia coli . A ce

Question

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 59 cells.

(a)  Find the relative growth rate.  (Assume t is measured in hours.)
k =

(b)  Find an expression for the number of cells after t hours.
P(t) =

(c)  Find the number of cells after 8 hours

(d)  Find the rate of growth after 8 hours.  (Round your answer to three decimal places.)

billion cells per hour

Explanation / Answer

a) For each cell present at the beginning of the time period, it will yields 2 after 20min, 4 after 40min, 8 after 60min. So the relative growth rate is k = 8

b) P(t) = c0 * k^t = 60 * 8^t

If you need pessimistic evaluation you should take 60 - 1% ~ 59

For an optimistic evaluation you should take 60 + 1% ~ 61

c) After 5 hours, P(t)=60*8^5=1,966,080

d) After 5 hours, the growth rate is P(6)-P(5) = 13,762,560

e) You reach 20,000 cells when 60*8^t=20000, i.e. t*ln(8) = ln(20000/60)

Hence t=2.7936h, i.e. 2 hours 47 min 37s

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