Question

A student working with Myxococcus xanthus wants to run an experiment on a population that is as dense as possible without actually being in stationary phase. In CTTYE (Casitone/Tris/Yeast Extract) broth, commonly used to grow M. xanthus, cells begin entering stationary phase at a cell density of approximately 5 × 10⁷ cells/mL. She inoculates a 10 mL culture with 1500 cells and sets it up in a shaking incubator at 30°C, knowing that after about two hours of lag time M. xanthus should double every 3 hours (this is a particularly quickly growing strain of M. xanthus). How long can she leave to go play Monster Hunter before she has to be back in lab for her experiment? (show work)

Answer 1

Given data

Density at stationary phase = 5 × 10⁷ cells/mL

Inoculum = 10 ml with 1500 cells or 150 cells/mL

Incubator temperature = 30^oC

Lag phase time = 2 hr

Inoculum doubling time = 3 hr

We know that

Generation time – N = No2ⁿ

N₀ = original number of cells

N = number of generations

G = generation time

G = t/n

N = No2ⁿ

log N = log No + nlog2

log N – log No = nlog2

n = [log N – log No] / log2

n = log N – log No]/0.301        ( log 2 = 0.301)

n = [log 5*10⁷ – log 150]/0.301

= [7 log 5 – 2 log 1.5]/0.3001

= [7*0.69 – 2*0.17]/0.301

= 14.9

We know that

G = t/n

3 = t/15

t = 45 hr

Lag phase time = 2 hrs

Total time = 45 + 2 = 47 hrs

Therefore, the girl can leave to go play Monster hunt for 47hr