A particular bacterium is found to have a doubling time of 20 minutes. If a laboratory begins with a population of 300 of these bacteria, and there is no change in the growth rate, how many bacteria will be present in 55 minutes?
I expect about 8 times as many, since 55 min is about 3 half-lives.
p(t) = 300*2^(t/20)
so,
p(55) = 300*2^(55/20) = ?
To find out how many bacteria will be present in 55 minutes, we need to calculate the number of doubling cycles that will occur in that time period.
The doubling time of the bacterium is given as 20 minutes, which means that every 20 minutes, the population of bacteria doubles. Let's calculate the number of doubling cycles in 55 minutes:
Number of doubling cycles = (Total time) / (Doubling time)
Number of doubling cycles = 55 minutes / 20 minutes = 2.75 cycles
Since we can't have a fraction of a cycle, we will consider the integer value, which is 2 cycles.
Now, we can find the number of bacteria at the end of 55 minutes using the formula:
Final population = Initial population * (2^Number of doubling cycles)
Considering an initial population of 300 bacteria and 2 doubling cycles:
Final population = 300 * (2^2)
Final population = 300 * 4
Final population = 1200
Therefore, there will be 1200 bacteria present at the end of 55 minutes.
To determine the number of bacteria present after 55 minutes, we need to calculate the number of doubling cycles that occur within that time period and multiply it by the initial population.
The doubling time is given as 20 minutes, which means that every 20 minutes, the population of bacteria doubles.
First, let's calculate the number of doubling cycles that occur within 55 minutes:
55 minutes / 20 minutes per doubling cycle = 2.75 doubling cycles
Since we cannot have a fractional number of doubling cycles, we can round it to the nearest whole number. In this case, it would be 3 doubling cycles.
Now, we can calculate the final population of bacteria after 3 doubling cycles:
Final population = Initial population * (2 ^ Number of doubling cycles)
In our case, the initial population is 300 bacteria, and the number of doubling cycles is 3.
Final population = 300 * (2 ^ 3)
Final population = 300 * 8
Final population = 2400 bacteria
Therefore, after 55 minutes, there will be 2400 bacteria present in the laboratory.