Calculate Bacteria Growth Rate: Doubling Time

[stegz]

Homework Statement

A beaker contained 2000 bacteira. one hour later the beaker contained 2500 bacteria. What is the doubling time of the bacteria?

Homework Equations

rate = (distance)/(time)
Time to double = .693/((ln(1+r))^t)

The Attempt at a Solution

rate = 2500/2000
My biggest problem is trying to find the rate, I used this at first, but think it is giving me the wrong answer. I know how to finish the problem, i just need help finding the rate. Thanks!

 

[Borek]

You should think in terms of N=N₀e^kt.

 

[Blueshift5]

I would approach this problem by first identifying the key information and variables. In this case, the initial number of bacteria (N0) is 2000 and the final number of bacteria (Nt) is 2500. The time interval between these two measurements is one hour (t = 1 hour).

To calculate the growth rate, we can use the formula: rate = (Nt - N0)/N0 x 100%. Plugging in the numbers, we get: rate = (2500 - 2000)/2000 x 100% = 25%. This means that the bacteria population increased by 25% in one hour.

To find the doubling time, we can use the formula: time to double = 0.693/(ln(1+r)), where r is the growth rate in decimal form. In this case, r = 0.25 (25% expressed as a decimal). Plugging in the numbers, we get: time to double = 0.693/(ln(1+0.25)) = 2.77 hours. This means that it took approximately 2.77 hours for the bacteria population to double.

In summary, the doubling time of the bacteria in this experiment is 2.77 hours. It is important to note that this calculation assumes that the growth rate remains constant over time and does not take into account any external factors that may affect the growth of the bacteria.