Question

For practice in the use of these equations, solve the following problems: 1. From the data given, calculate (a) the number of doublings over the total growth period and (b) the generation time during the exponential growth phase. Cells per ml 2.0 x 106 2.5 x 106 8.8 x 106 7.4 x 107 6.3 x 108 1.2 x 109 1.2 x 109 Time in Hours 0 2 4 10 12 2. You need, for experimental purposes at 8 a.m. Monday morning, an exponentially growing culture containing about 2 x 108 cells/ml of an organism that has a generation time of two hours. You plan to be on campus the preceding Sunday at 8 p.m. Assuming no lag period, how large an inoculum (cells/ml) must you use to start your culture? 3. A culture with an average doubling time of 30 min is inoculated with 3 x 103 cells/ml and incubated for eight hours. What is the final population?

Answer 1

Ans 1 Total growth phase is from 0 to 10 hrs period.

Growth rate overall (k) = (2.303/t) log(Number of final cells/Number of initial cells)

= (2.303/10) log (1.2 x 10⁹/2 x 10⁶)

= (0.230)log (0.6 x 10³)

= 0.638 hr^-1

Doubling time = 0.693/k

= 0.693/0.533 = 1.09 hours or 65.17 minutes

Expontential phase is from 2 to 12 hours

So, Growth rate in exponential phase (k_exp) = (2.303/t) log(Number of final cells/Number of initial cells at t = 2 hrs)

k_exp = (2.303/10) log(1.2 x 10⁹/2.5 x 10⁶) = 0.6174 hr^-1

For exponential growth, No. of cells at 10 hours = No. of cells at 2 hours x 2ⁿ

where n is no. of generations

1.2 x 10⁹ = 2.5 x 10⁶ x 2ⁿ

2ⁿ = 1.2 x 10⁹/2.5 x 10⁶

2ⁿ = 480

On taking log both sides

nlog2 = log 480

0.3010 x n = 2.681

n = 2.681/0.3010

n = 8.90  

No of generations = 8.90

Total time period = 8 hours

Generation time = Total time/No. of generations = 8/8.90 = 0.899 hror 53.93 min

Answer 2

Generation time = 2 hours

Total time = 12 hours

Final cells count = 2 x 10⁸ cells per ml

Number of generations = total time/generation time = 12/2 = 6

So, n = 6

Putting in equation, No. of cells at after 12 hours = No. of cells initially x 2ⁿ

2 x 10⁸ = No. of cells initially x 2⁶

2 x 10⁸ = No. of cells initially x 64

(2 x 10⁸)/64 = No. of cells initially

Initial Inoculum should be 3.125 x 10⁶ cells per ml (Answer 2)

Answer 3

Doubling time = 30 min

Initial cells = 3 x 10³

Time = 8 hours or 480 minutes

Final cells = ??

Now, growth rate = 0.693/doubling time = 0.693/30 = 0.0231 min^-1

Now, Growth rate (k) = (2.303/t) log(Number of final cells/Number of initial cells)

Putting values, 0.0231 = (2.303/480) log(Number of final cells/3 x 10³)

4.814 = log(Number of final cells/3 x 10³)

10^4.814 = Number of final cells/3 x 10³

65162 = Number of final cells/3 x 10³

Number of final cells = 65162 x 3 x 10³ = 1.95 x 10⁸ cells/ml (Answer 3)