If at noon today someone has 1 E. coli cell, in their gut/large intestine, and E. Coli does binary fission, in unison each thirty minutes, how many will one have in their gut at 11:55 A.M. tomorrow? (five minutes short of a whole day)
I'm confused on how to begin setting this up. I saw an example of a problem similar to this one, and it said N=the initial count of how many E. Coli were in the stomach the first time, multiplied by the number of hours it was in the stomach, which in this case would be 23.5hrs. So, would I set it up like N=1(2^23.5)? If not could you help me set it up correctly? I don't believe this is the right way because my answer comes out to 552.25, which seems like less than I would believe to be in the stomach after all of that time, but maybe not? Please help asap! Thank You!
To calculate the number of E. coli cells in the gut at 11:55 A.M. the next day, you need to determine the number of divisions that occur in the given time period.
Let's break down the problem step by step:
1. First, consider the time period between noon today and 11:55 A.M. tomorrow. This is approximately 23 hours and 55 minutes, or 23.92 hours.
2. Next, determine the number of divisions that occur within the given time frame. Since E. coli cells divide every 30 minutes, we need to calculate the number of 30-minute intervals within the total time period.
Number of intervals = Total time period / Division time
Number of intervals = 23.92 hours / 0.5 hours (30 minutes) = 47.84 intervals
3. Now, since each E. coli cell divides into two cells during each division cycle, you can calculate the total number of cells by raising 2 to the power of the number of intervals.
Total number of cells = 2^number of intervals
= 2^47.84
Using a calculator, the approximate value of 2^47.84 is approximately 1.816 x 10^14.
Therefore, at 11:55 A.M. tomorrow, the estimated number of E. coli cells in the gut will be around 1.816 x 10^14.