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A bacteria culture initially contains 2500 bacteria and doubles every half hour. What is the size of the baterial population after 20 minutes?

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4 Answers

Hey Sam,
         First thing we have to do is find the growth rate k….the exponential model for this situation is B(t)=B0 ekt We are given that B0=2500 and the population doubles every 30 min (half hour)
so we know that at t=30 min that B(30) = 2*2500. Putting all this in the model equation gives
 
(2*2500/2500)=e30k solving for k gives k=Ln(2)/30=.02310 min-1 now we can calculate the size of the population at 20 min from
 
B(20)=2500e.02310*20 =1.587*2500=3968.
 
Jim
A = P(2)^(kt), t in minutes
 
2P = P(2)^(30t)
 
2^1 = (2)^(30t)
 
30t = 1
 
t = 1/30
 
A = P(2)^(t/30)
 
A(20) = 2500(2)^(20/30)
 
A(20) = 2500(2)^(2/3) ≈ 3968.5026299205075

 
Of course, the standard answer assumes that the replication of the bacteria is randomly phased within the culture. However, you could (conceivably) have stopped their reproduction at a particular point in the cell cycle (synchronously), and then turned them loose all at once. In that case, the bacterial population would be a step-function of time (double, abruptly, every 30 minutes). Granted, that would be the rare exception -- but other facets of bacterial behavior (such as production of toxins) are subject to that type of "on/off" regulation, via quorum sensing, so it's not theoretically impossible. And the reproductive cycle can be halted at various phases by various chemical treatments.
Q:A bacteria culture initially contains 2500 bacteria and doubles every half hour. What is the size of the baterial population after 20 minutes?
 
A: 
 
Known: Ao = initial bacterial quantity of 2500; and the bacteria quantity is doubling every half hour.
 
Unknown: We need to find the amount of bacteria after 20 minutes of growth.
 
To do this we must first find a function of the bacteria quantity for a time: A(t). 
 
The first step is to express the relationship between quantity and rate of change measured in bacteria per hour as a differential equation:
 
dA/dt = kA. where A is the quantity of bacteria at a given time and k is the relative growth rate.
 
The formula for finding the relative growth rate is:
 
(ln(y2)-ln(y1))/(x2-x1)->(ln(2Ao)-ln(Ao))/(1/2)=2ln(2)= 1.386 
 
This is a seperable equation so: dA/A=kdt. Integrating both sides yields -> ln (A) = kt + C where C is a constant.
 
Taking the exponential of both sides gives us: A = Ce^kt.
 
We have a function for the quantity of bacteria but we must find the constant. We can find the constant by using the initial conditions at time zero.
 
2500 = (C*e^0) -> C=2500=Ao.
 
A = Ao*e^kt where Ao = 2500 bacteria and k = 1.386
 
 
So plugging in 20 minutes or 1/3 hour into the function gives us:
 
A=2500*e^(1.386*1/3)= 3968 bacteria.

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