The doubling period of a baterial population is 10 minutes. At time t=80 minutes, the baterial population was 50000
1What was the initial population at time t=0
2Find the size of the baterial population after 4 hours
Note that 80 minutes is 8 doubling periods, for a growth of 2^8 = 256 times.
So, initially there were 50000/256 bacteria.
I think you can take it from there, no?
To answer these questions, we need to understand the concept of population doubling and how it relates to time.
1. What was the initial population at time t=0?
To find the initial population, we need to trace back to the starting point when the population was zero. Given that the doubling period is 10 minutes, we can calculate the number of doubling periods that occurred from t=0 to t=80 minutes.
Number of doubling periods = (80 minutes) / (10 minutes) = 8 doubling periods
Since the population doubles with each period, we can calculate the initial population by dividing the final population by 2 raised to the power of the number of doubling periods:
Initial population = (Final population) / (2^(Number of doubling periods))
= 50000 / (2^8)
= 50000 / 256
= 195.3125
So, the initial population at time t=0 was approximately 195.3125 (rounded to the nearest whole number).
2. Find the size of the bacterial population after 4 hours.
First, we need to convert 4 hours to minutes since the doubling period is given in minutes.
1 hour = 60 minutes
4 hours = 4 * 60 minutes = 240 minutes
To calculate the size of the bacterial population after 4 hours, we need to determine the number of doubling periods that occurred during this time:
Number of doubling periods = (240 minutes) / (10 minutes) = 24 doubling periods
Using the same formula as above, we can calculate the final population:
Final population = (Initial population) * (2^(Number of doubling periods))
= 195.3125 * (2^24)
≈ 195.3125 * 16,777,216
≈ 3,276,800,000
So, the size of the bacterial population after 4 hours is approximately 3,276,800,000.