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Question: You inoculate a liquid culture with 100 bacterial cells and allow the bacteria to grow for 6 hours. If the lag phase lasts 1 hour and the generation time is 30 minutes, how many bacterial cells will be present in tube at the end of the 6 hours?– Free Chegg Question Answer

You inoculate a liquid culture with 100 bacterial cells and allow the bacteria to grow for 6 hours. If the lag phase lasts 1 hour and the generation time is 30 minutes, how many bacterial cells will be present in tube at the end of the 6 hours?

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The growth of bacterial populations has four characteristic phases namely, lag phase, log or exponential phase, stationary phase and death phase. The lag phase shows no change in the bacterial population when inoculated into fresh medium, although the cells may grow in volume or mass, synthesizing enzymes, proteins, RNA, etc and increasing in physiological activity. The log or expotential phase involves a steady growth of all the cells as they divide regularly by binary fission and grows by geometrical progression. This phase is followed by stationary phase where the population growth is limited by different factors and the number of new cells created by division equals to the number of cells dying. The last phase is the death phase which is characterized by the death of cells due to accumulation of toxic waste in the culture medium and depletion of nutrients.

We know, the formula of the bacterial population growth is governed by the following relation,

\log N-\log N_{0}=\frac{\mu (t-t_{0})}{2.303}

 —-(i)  and \mu =\frac{0.693}{g} ; where N= population after t time; N0= initial population = 100; g =generation time = 30 min =0.5 hr;  \mu= specific growth rate constant; t= total time for growth = 6 hr; t0= time in lag phase = 1 hr.; Hence, (t-t_{0})=(6-1)=5 hr
 . and the specifc growth rate constant, \mu =\frac{0.693}{g}=\frac{0.693}{0.5}=1.386 hr^{-1}

We have to find out the value of N, therefore putting the above values in the given relation in (i), we get,

\log N-\log N_{0}=\frac{\mu (t-t_{0})}{2.303} or, \log\left ( \frac{N}{100} \right )=\frac{1.386\times 5}{2.303}

   

\therefore N=100\times 10^{3}=10^{5}.

Hence, number of bacterial cells in tube at the end of 6 hrs = 105= 100000.


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