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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?

Find the size of the baterial population after 20 minutes.

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Jim S. | Physics (and math) are fun, reallyPhysics (and math) are fun, really
4.7 4.7 (197 lesson ratings) (197)
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.
Steve S. | Tutoring in Precalculus, Trig, and Differential CalculusTutoring in Precalculus, Trig, and Diffe...
5.0 5.0 (3 lesson ratings) (3)
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

Stanton D. | Tutor to Pique Your Sciences InterestTutor to Pique Your Sciences Interest
4.6 4.6 (42 lesson ratings) (42)
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.
Adam S. | Professional and Proficient Math TutorProfessional and Proficient Math Tutor
Q:A bacteria culture initially contains 2500 bacteria and doubles every half hour. What is the size of the baterial population after 20 minutes?
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.