SNIPES BIOLOGY 115
POPULATION DYNAMICS PROBLEM ANSWERS

23. The population of pocket gophers in Hagerman NWR was 10,000 in 1980.  In 1990 the population was 12,750.  What was the intrinsic rate of growth (biotic potential) for the population between 1980 and 1990 (carrying capacity unknown)?

[exponential growth equation]    

24. The pocket gopher carrying capacity for Hagerman NWR was calculated to be 14,250 in 1990.  Assuming the same biotic potential as in previous question, how much time will elapse until the gopher population is at carrying capacity (no change in biotic potential)?

The change in N = K-N = dN; because the top of both sides of the equation are the same (dN and K-N), you can cancel them both out.  You then flip the simplified equation over to solve for dt.

log. growth eqation

If you use the 1980 population (10,000 gophers) as N, then get 51.8 years.  If you use the 1990 population (12,750 gophers), then 40.65 years.

25. The doubling time of a pathogenic virus is 4 hrs.  An initial infection of 500,000 viruses invades a host.  The available cells for it to infect number 5x109.  How long before all the cells are infected by the virus (assume no replacement of infected cells by host)?  Treatment for the infection increases the viral doubling time by 2 hrs. by reducing the viral birth rate by 50%.  How much longer does the treatment extend the life of the host's cells?
[r calculation based on doubling time]

[r calculation based on doubling time]  

[r calculation based on doubling time]

[new log. growth calculation]

*The increased life expectancy is 9.5 - 6.7 = 2.8 years.

New Problem:  A virus has a doubling time of 45min.  A person is an initial infection of 1500 viruses.  The viral carrying capacity is 4.5 million.  How many viruses will be inside the person 3 days after exposure?  Assume no immune response by person.  **Try to solve before looking below for answer!



ANSWER:
r calc from 2x time

log. growth equation

Add the new viruses to initial population of 1500 = 101,262 viruses in person 3 days later.  

You may also use the other anti-derivative equation for finding population size at a given time:
answer using anti-derivative equation

The difference in the two answers is because the second has no carrying capacity effect (no slowing of growth as population gets closer to K).