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A student working with Myxococcus xanthus wants to run an experiment on a popula

ID: 281678 • Letter: A

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)

Explanation / Answer

Given information,

Volume of inoculum = 10 mL

Number of cells = 1500

Lag phase = 2 hrs

Duplication time in log phase = 3 hrs

Stationary phase cell density = 5 X 10^7 cells/mL

After 2+3 hrs, 1500 cells would double = 3000

After 2+3+3 hrs, 3000 cells would double = 6000

After 2+3+3+3 hrs, 6000 cells would double = 12000

After 2+3+3+3+3 hrs, 12000 cells would double = 24000

After 2+3+3+3+3+3 hrs, 24000 cells would double = 48000

After 2+3+3+3+3+3+3 hrs, 48000 cells would double = 96000

After 2+3+3+3+3+3+3+3 hrs, 96000 cells would double = 192000

After 2+3+3+3+3+3+3+3+3 hrs, 192000 cells would double = 384000

After 2+3+3+3+3+3+3+3+3+3 hrs, 384000 cells would double = 768000

After 2+3+3+3+3+3+3+3+3+3+3 hrs, 768000 cells would double = 1536000

After 2+3+3+3+3+3+3+3+3+3+3+3 hrs, 1536000 cells would double = 3072000

After 2+3+3+3+3+3+3+3+3+3+3+3+3 hrs, 3072000 cells would double = 6144000

After 2+3+3+3+3+3+3+3+3+3+3+3+3+3 hrs, 6144000 cells would double = 12288000

After 2+3+3+3+3+3+3+3+3+3+3+3+3+3 hrs, 12288000 cells would double = 24576000

After 2+3+3+3+3+3+3+3+3+3+3+3+3+3+3 hrs = 44 hrs, 24576000 cells would double = 49152000 = 4.9152 X 10^7