Question

E. coli was inoculated into the Nutrient broth.
It entered the algebraic proliferator within two hours of incubation, and entered a period of suspension from 20 hours of incubation.
The cells in the incubation period of 4 hours and 8 hours were 106 cells/ml and 1012 cells/ml, respectively
What is E. coli's specific growth rate and generation time?
(However, ln 1 = 0, ln 10 = 2.3, ln 20 = 3.0)
- plz, Describe the process in detail

Answer 1

Solution

Nt (Final concentration of bacterial cells at time t) = 1012 cells/ml

N0 (Initial concentration of bacterial cells at time t0) = 106 cells/ml

n = Number of generations

Equation I ( to solve generation number) :

Nt=N0 X 2n

log Nt=log N0 + n log 2

log 1012 = log 106 X n log 2

log 1012 - log 106 = n X .301

(log 9.547)/.301 = n

n = 0.979/0.301

n = 3.25 or 3.

Equation II ( to solve generation time) :

G (Generation time) = (t - t0)/n

G = (8 - 4)/3

G = 4 hr/3 or 240min/3

G = 80 min

Equation III ( to solve specific growth rate ) :

Integrating both sides, we get

	Denotation	Definition
Generation number	n	Number of times bacteria doubled its population at a specific time interval
Generation time	G	The time taken by bacteria to double its number
Specific growth rate		The measure of number of divisions per cell per unit time.