Question

In a bacterial growth experiment, the concentration of cells increased from 10,000 cells/ml to 30,000 cells/ml in 3 h during the exponential growth phase. The doubling time (h) of the bacteria is

Answer 1

Answer:

two hours

Explanation:

10,000 cells in 3 hours -> 30,000

this means 1 cell in 3 hours -> 3

therefore 1 cell in 1 hour -> 1 cell

and we need to double the present bacterias

thus 1 cell in 2 hour -> 2 cells

so, 30,000 in 2 hours -> 60,000 cells

Answer 2

Answer:

G = t/n, t= total time, n= no.of generations, G= time required for bacteria to double.

Here, t= 3hr

From the provided data n has to be found out first.

To find n;

b = B x 2n

(B = number of bacteria at the beginning of a time interval

b = number of bacteria at the end of the time interval)

Solve for n:

logb = logB + nlog2

n = logb - logB / log2

n = logb - logB / .301

n = 3.3 logb/B

Now,

G= T/3.3 logb/B

G= 3/3.3 (log 30,000/10000)

G=3/3.3 log3

G= 3/1.57

G= 1.9hr