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

Answer:

The value of g = 36

The value of k = $= 8.361 \times 10^ -03$

                       

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

$n =3.3 log b/B$

   = $3.3 log 5.4 \times10 ^{9} / 5 \times 10 ^{6}$

   = $3.3 log 1.08 \times 10^3$

   = 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  $= 8.361 \times 10^ -03$