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
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