The growth rate of Escherichia coli , a common bacterium found in the human inte

Question

The growth rate of Escherichia coli, a common bacterium found in the human intestine, is proportional to its size. Under ideal laboratory conditions, when this bacterium is grown in a nutrient broth medium, the number of cells in a culture doubles approximately every 20 min.

(a) If the initial population is 50, determine the function Q(t) that expresses the growth of the number of cells of this bacterium as a function of time t (in minutes).

Q(t)=______

(b) How long would it take for a colony of 50 cells to increase to a population of 1 million? (Round your answer to the nearest whole number.)

_____ min

(c) If the initial cell population were 500, what is our model?

Q(t)+_____

Thanks!

Explanation / Answer

Answer:-

Part (a):-

Exponential growth questions are normally of the form

Final Amount = Initial Amount * e^ (rate * time)

so we must know the initial amount, the time, and the final amount for proceeding.

2*Initial = Initial e^ (rate * 20 min)

According to statement , time t is in minute,

solving yield:

2 = e^20r => ln (2) = 20r

=>

r = ln2/20 = .035.

so plugging back into the general equation of growth:

Final Amount = Initial Amount e ^(0.035 t)

Q(t) = 50 e^(.035t)      Answer.

Part (b);-

So now you have the final amount, the initial amount, the rate , and you're looking for time.

1 million = 1000000

and number of cell = 50

then , we can write as

1000000 = 50 e^(.035*t)

after calculation, we get

20000 = e^(.035t)

ln(20000) = .035 t

t = ln(20000) / 0.035

t = 9.9095 / 0.035

t = 282.957 min

t = 283 min      ( round the answer in whole number)   answer.

Part (c):-

Now initial cell population = 500

then our model becomes

Q(t) = 500 e^(0.035t)   answer.

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