## Really need help with exponential microbial growth problem!!

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JHUT
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Posts: 6
Joined: Mon Sep 24, 2007 4:15 am

### Really need help with exponential microbial growth problem!!

Here is the problem:

Normal growth rate = 3g/h but it decreases by 50% so it would be 1.5g/h. You want to grow up a culture of your strain to harvest the protein, so you inoculate a broth with 1,000 cells; but there is 1 additional cell that is a mutant that is no longer expresses your protein and has a wild-type growth rate. if the culture grows for 10 hours, what proportion of the culture will consist of wild type versus mutant bacteria when you have the culture????

So here is what my partner did:

g=1/k=1/1.5=.6666
k=(log(2No)-logNo)/.301g=1.5
k=n/t=1.
Nt=No*2^n=1000*2^1=2000
so 2,000 cells at 40 min
10hrs=600min 600min/40min=15
15*1000 cells = 15,000 cells of normal e coli
so.. Nt=No*2^n=1*2^1=2
15*1 = 15 at end of time for wildtype
so 15000:15 equal 1000:1 ratio.. but that is what we started with.. What do you think??
I dont know if it's right or not and that is what I am hoping someone can tell me.. I wouldnt expect the ratio to be the same but this stuff is not my cup of tea. Thanks for any help!

mith
Inland Taipan
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Location: Nashville, TN
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if its simply 1.5g/hr, its not exponential...
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Enjoying one moment at a time;
Accepting hardships as the pathway to peace;
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blcr11
Viper
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Joined: Fri Mar 30, 2007 4:23 am
An expression for the future size of an exponentially growing population would be:

N = N0*2^(t/T) where t is the time of growth and T is the doubling time.

I think the point of the calculation is to show how small differences in doubling time can make a big difference in the composition of the final population. It’s not all that uncommon for heterologous plasmids to reduce the growth rate of their host. Nor is it uncommon for plasmids to be spontaneously lost by the host. When this happens in a large fermentor growth from which you’re hoping to harvest mgs to gms to kgs of protein, it can come as a rude—not to mention expensive—dissapointment to find that your stock has been overgrown with some sort of revertant strain and no longer expresses the recombinant protein you were after.

I think your calculations are correct in their intention, but something happened in the arithematic. It looks like you simply multiplied 2 times the number of generations. It is an exponentiation you need to do. I get different numbers for the final populations.

1.5g/h  T=0.667 h, so t/T = 15 (OK, a little more than 15, but 15 is close enough)
N10 = 1000*2^15 = 1000*32768 = 32.8 x 10^6 cells of the slower growing bacteria.

3g/h  T=0.333 h, so t/T = 30 (again, a little more than 30, but twice as fast as the other strain)
N10 = 1*2^30 = 1.07 x 10^9 cells of the faster growing mutant strain.

So even though you start with 1000 times more of the slower growing bacteria, after ten hours you have roughly 32 times more of the faster growing mutant strain. (1.07 x10^9/32.8x10^6 = 32.7 and growing, at least within the limits of the nutrient supply)

I wouldn't rule out mistakes in my arithematic either. You should confirm (or show them to be wrong) for yourself.

JHUT
Garter
Posts: 6
Joined: Mon Sep 24, 2007 4:15 am
ok... so that makes more sense to me.. so the wild type growth rate would not decrease? I did the work and it come out to a little over 32 in terms of a ratio. thanks for your help. I am trying to confirm if it's the right method but it makes a lot more sense to me

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