population dynamics - bacterioplankton

Question

Bacterioplankton is a very important component of the oceans ecosystem. Bacterioplankton reproduce by dividing at regular time intervals. In a particular coral reef, bacterioplankton serve as a source of food for a group of small fish. The fish feed every morning at 08:00. Immediately following the school’s feeding population of bacterioplankton is 20% of the population immediately before they were eaten.

a)      Assume that the feeding is instantaneous, and that the bacterioplankton are constantly dividing. Determine the average doubling time (d) of the bacterioplankton in order to have a sustainable ecosystem (that is, the population at each feeding time must be equal)

b)      Graph the relationship between the population of bacterioplankton and time for three 24 hour periods (you do not need to determine the equation for multiple days)

Answer 1

(a) Let t=number of times in 24 hours it takes to double the population. After 24 hours the population must grow by a factor of 5 to restore its original numbers, because the fish deplete the population to ⅕ its original population. 

2^t=5, t=log₂(5)=2.3219 approx. So the time period d for doubling is 24/log₂(5)=10.3362 hours, approximately, (about 10hr 20min). 

(b) P=20×5^T, where P is the population percent, and where 0<T≤1, so that when T=0 P=20% and at T=1 P=100%. T=0, 1, 2 represent 8am on the first and successive days, and T=1, 2, 3 represents 8am on the second and successive days. For the second day the formula is P=20×5^T-1 where 1<T≤2 and for the third day P=20×5^T-2 where 2<T≤3. These equations correspond graphically to 3 curves: 1st curve starts at (0,20) representing 8am on the first day, rising to 100% at (1,100); the 2nd curve starts (1,20) representing a time just after 8am on the second day, rising to 100% at (2,100); the 3rd curve starts at (2,20) representing a time just after 8am on the third day, rising to 100% at (3,100). So each curve steps down each day from 100% to 20% as the fish eat the plankton.