Question

A medium was inoculated with 5 x 106 cells/ml of E.coli cells. Following a 1-hour lag, the population grew exponentially for 5 hours, after which the population was 5.4 x 109 cells/ml. Calculate g and k for this growth experiment. (10%)

Answer 1

The value of g = 36

The value of k =
`= 8.361 * 10^ -03`

Step-by-step explanation:

`n =3.3 log b/B`

Where,

n = number of generation during the period of exponential growth

b = number bacteria at the end of time

B = number of bacteria at the beginning of time

`n =3.3 log b/B`

=
`3.3 log 5.4 *10 ^(9) / 5 * 10 ^(6)`

=
`3.3 log 1.08 * 10^3`

= 3.3 × 3.033

n = 10

`g = t / n`

Where,

g = generation time

t = duration of exponential growth

n = number of generation

`g = t / n`

= ( 1+5) × 60 /10

= 6 × 60 /10

g = 36

`k = 0.301 / g`

Where,

k = specific growth growth rate

g = generation time

`k = 0.301 / g`

= 0.301 / 36

k
`= 8.361 * 10^ -03`