Biology 112
Population Problem Set
Fall 2005
Due 10/18 or 10/19
This problem set should reinforce and build on your knowledge base from Bio54,
lectures, and Gotelli chapters 1-4. You may consult your notes/book, and during the
population lab, you may discuss problems.
1. What are the assumptions of the exponential growth model?
No I or E
Constant b and d (no variance)
Continuous growth with no time lags
No genetic structure (all are equal)
No age or size structure (all are equal)
How do the assumptions differ between the logistic and exponential growth models?
For the exponential growth model, you assume unlimited resources and therefore
birth and death rates remain constant. However, in the logistic model, resources are
limited, resulting in birth and death rates being dependent on population density.
2. How are r and  related? For what values of r and  will a population increase in size?
Decrease in size? Remain constant?
er =  or r = ln()
 is the geometric growth rate, where Nt+1 = Nt
r is the exponential growth rate (b-d), where dN/dt = ert
er is the increase over one unit of time
increasing population size: r > 0 or  > 1
constant population size: r = 0 or  = 1
decreasing population size: r < 0 or 0 <  < 1
3. Graph population size vs. time for a population with a carrying capacity (K) of 250, for
starting population sizes of N= 500, and N= 50. (Logistic growth)
4. A friend makes you delicious green pepper and cheese omelets before leaving for the
weekend. If the initial Salmonella population size on your friend’s dirty countertop
(unlimited food from all of the egg that was spilled) is 240, what will the population size
be when your friend returns after a weekend away (48 hours later)? Assume the doubling
time for Salmonella bacteria is 40 minutes, a conservative estimate.
[Hint: First calculate r and then use this to calculate Nt] (Exponential growth)
N0 = 240
tdouble = ln(2)/r
tdouble = 40min (60min/ 1 hr)
r = ln(2)/ tdouble = 1.0397
Nt = N0 ert = 240* e (1.0397*48hrs) = 1.244* 1024 Salmonella bacteria
5. A population of salamanders with density-dependent population growth lives in and
around a vernal pool. The current populations size is 27 salamanders and a previous study
determined that r = 0.19 individuals/ individuals*month. (Logistic growth)
a. If the carrying capacity for the population in the vernal pool is 45, what is the
current growth rate for the population?
dN/dt = rN (1- N/K) = 0.19 *27 (1-27/45) = 2.052 salamanders/month
b. What will the salamander population size be in 6 months?
t = 6, K= 45, r = 0.19, No = 27
Nt = K/ (1+((K-No)/No)e-rt) = 45/ (1+((45-27)/27)*e(-0.19*6) = 37.09
~ 37 salamanders
c. When the salamander population is growing at its maximum rate possible, what
is the population size?
Max population growth when N = K/2 = 45/2 = 22.5 salamanders
6. You have the following data on the number of marmots in a threatened population on
Vancouver Island.
Define the terms lx, gx, mx, and ex from the table.
lx = Suvivorship, Proportion surviving at start of age interval = nx/no
gx = age specific survival rate (probability of surviving to the end of the period
mx = qx = age specific mortality rate
ex = age specific life expectancy
Complete the calculations for the following life table so as to calculate age specific life
expectancy. [Hypothetical life table based on estimated population size and life
expectancy in marmots of 3-6 years]
(Note on symbols: remember sx and nx; gx and px are equivalent) (Life Table)
x
sx
dx
lx
gx
Nx
Tx
ex
0
400
52
1
0.87
374
1633
4.0825
1
348
10
0.87
0.971
343
1259
3.61782
2
338
27
0.85
0.92
324.5
605
1.78994
3
311
61
0.78
0.804
280.5
591.5
1.90193
4
250
114
0.63
0.544
193
311
1.244
5
136
86
0.34
0.368
93
118
0.86765
6
50
50
0.13
0
25
25
0.5
7
0
0
Plot the survivorship curve for the marmots. (Ln (lx) vs. Age). Which type of
survivorship curve does the marmot population have?
x
Ln (lx)
lx
0
0
1
1
-0.108
0.8975
2
-0.139
0.87
3
-0.163
0.85
4
-0.621
0.5375
5
-1.079
0.34
6
-1.715
0.18
7
0
The curve is closest to a type I survivorship curve.
Survivorship
0
0
1
2
3
4
Ln(lx)
-0.5
-1
-1.5
-2
Age (X)
5
6
7
Survivorship
1
2
4
6
8
l(x)
0
0.1
Age (X)
7. Given the following transition matrix for a population of trees, what would the
population structure look like at time t+1 if the initial population consists of 500 seeds,
75 saplings, 10 small trees, and 5 adult trees?
Seeds/Saplings/STree/ATree
0.3
0.03
0
0
0
0.4
0.1
0
45
0
0.1
0.2
150
0
0
0.9
Nt+1 = A * Nt
Nt =
500
75
10
5
Nt+1 =
seeds
sapling
small tree
adult tree
1250
45
8.5
6.5
8. Given the following hypothetical population calculate gross reproductive rate (GRR),
net reproductive rate (R0), generation length (G) for the population and the age structure
in 2 years.
x
sx
300
255
238
197
99
0
0
1
2
3
4
5
lx
1
0.9
0.8
0.7
0.3
0
bx
0
0
1
3
2
0
Fx
0
0
238
591
198
0
lxbx
lxbxx
0
0
0.7933
1.97
0.66
0
0
0
1.5867
5.91
2.64
0
gx
0.85
0.93333
0.82773
0.50254
0
N0
5
27
12
5
3
N1
33
4.25
25.2
9.932773
2.51269
0
N2
60.0237
28.05
3.966667
20.85882
4.991597
GRR = bx = 6
R0= lxbx = 3.42
G = lxbxx/lxbx = 2.96
9.Create a Leslie Matrix for the population in the previous problem and project the
population forward in time 2 years given an initial population with an even age
distribution of 25 individuals ages 0-4.
i
1
2
3
4
5
A=
N0
25
25
25
25
25
Pi
Fi
0.85
0.933333
0.827731
0.502538
0
0
0.933333
2.483193
1.005076
0
0
0.85
0
0
0
0.93
0
0.93
0
0
N1
110.5
21.25
23.25
20.75
12.5
N2
98.38
93.93
19.76
19.25
10.375
2.48
0
0
0.83
0
1.01
0
0
0
0.5
0
0
0
0
0