SBI4U Population Dynamics Homework Questions
Review Questions *BRING A CALCULATOR. You will get a Formula Sheet
1. Define: population size, population density, biotic potential or fecundity, limiting factor, ecological niche
2. Compare and contrast the following sets of terms. Give examples for each.
a. Crude density and ecological density
b. Open and closed population
3. Use a simple sketch to draw the 3 patterns of population dispersion and give an example organism for each. Which
pattern do most populations follow?
4. How do biologists determine the population size for something like a wolf? A trillium? Compare and contrast the
two methods.
5. How would the pattern of growth differ between rabbits and a bacteria colony? What type of graph represents their
patterns of growth?
6. Most populations cannot continue to grow indefinitely. What is the upper limit defined as? Why does it exist? What
type of graph demonstrates this pattern? What are the different sections of the graph called and what do they refer
to?
Sample Population Dynamics Problems (Answers on last page)
1. A backyard measuring 3.0 m by 4.0 m contains 215 dandelions. Determine the population density of the plants. A formal solution is
required.
2. A small field having an area of 1.5 ha contains a pond with a surface area of 0.3 ha and is home to 300 field mice. Determine the
ecological density of the mouse population, showing all work, with statements.
3. A lawn-care company wishes to estimate the density of a weed population in a park before spraying. The park measures 200 m  100
m. The technician randomly places four 2.0 m  1.0 m quadrats in the park, and finds weed counts of 22, 46, 32 and 40 plants in the 4
quadrats. Estimate the density of the population. A formal solution is required.
4. Each May, harp seals give birth off the coast of Newfoundland and Labrador. In a hypothetical situation, an initial population of 900
seals gives birth to 390 pups, and during the next 12 months, 60 seals die.
Assuming that the population is growing geometrically, what will the harp seal population be in 12 years? A formal solution is
required.
5. A population of squirrels in a local park exhibits geometric growth. The initial population size is 43 squirrels. During the year being
studied, 77 young are born, and 21 squirrels are lost to predators and accidents.
a. Calculate the geometric growth rate (). A formal solution is required.
b. Is it reasonable to assume that the population will continue to increase at this rate? Explain why you made this conclusion.
6. A population of 1750 yeast cells in a culture tube is growing exponentially. If the intrinsic growth rate (r) is 0.023 per hour, calculate
the initial instantaneous growth rate, dN/dt. A formal solution is required.
7. A population is growing continuously. The carrying capacity of the environment is 800 individuals and the maximum growth rate is
0.35.
a. Determine the population growth rate if the population size is 20 individuals. A formal solution is required.
b.
How would this rate compare to the rate if the population was very near the carrying capacity? A calculation is not required.
Additional Practice
Answer Section: Sample Population Dynamics Problems
1. ANS:
D
= N/S
=LW
S
= 215 dandelions/12 m2
= 3.0 m  4.0 m
= 18 dandelions/m2
= 12 m2
The population density of the dandelions is 18 plants/m2.
2. ANS:
D = N/S
= 300 mice/1.2 ha
S
= total area – usable area
= 1.5 ha – 0.3 ha
= 250 mice/ha
= 1.2 ha
The ecological density of the population is 250 mice/ha.
3. ANS
The density of the weed population is 17.5 weeds/m2
4. ANS:
In year 1, the population change
= births – deaths
= 390 seals 60 seals
= 330 seals
Initial population, N(0) = 900 seals
Population at end of year 1, N(1)
= 900 seals 330 seals
= 1230 seals
Geometric Growth Rate
N(t+1) = 1230 = 1.37
N(1) 900
N(0) = 900 seals, Geometric Growth Rate = 1.37, t = 12 a
N(12) = N(0)  (Geometric Growth Rate)12
= 900  1.3712
= 900  43.7
= 39 330
With the same growth rate, the seal population will reach 39 330 after 12 years.
5. ANS:
a.
During the year the population change
= births – deaths
= (77 21) squirrels
= 56 squirrels
Initial population, N(0)
= 43 squirrels
Population at end of the year
= (43 + 56) squirrels
= 99 squirrels
= N(t+1) = 99 = 2.3
Geometric Growth Rate ()
N(t) 43
The geometric growth rate of the squirrel population is 2.3
6. ANS:
r = 0.023/h and N = 1750
7. ANS:
rmax = 0.35
b. It is not reasonable to assume a constant
growth rate. The park has a carrying
capacity determined by food, shelter and
space requirements of the population. Once
this is reached, the growth rate will decline.
dN = rN
dt
= 0.023 /h  1750 yeast
= 40.25/h
= 40/h
The initial instantaneous
growth rate is 40/h.
N = 20
K = 800
dN/dt = rmaxN[(K - N)/K]
= 0.35  20  [(800 20)/800]
= 0.35  20  0.975
= 6.8
a. The population growth rate is 6.8 when the population has 20 individuals.
b. Nearing the carrying capacity, the rate would be declining /approaching zero.
7.
ANS:
a.
rmax = 0.35
N = 20
K = 800
dN/dt = rmaxN[(K N)/K]
20)/800]
= 6.8
The population growth rate is 6.8 when the population has 20 individuals.
b.
Nearing the carrying capacity, the rate would be declining/approaching zero.