Transcript Document

Population Growth
– Chapter 11
Growth With Discrete Generations
• Species with a single annual breeding season
and a life span of one year (ex. annual plants).
• Population growth can then be described by the
following equation:
Nt+1 = R0Nt
• Where
– Nt = population size of females at generation t
– Nt+1 = population size of females at generation t + 1
– R0 = net reproductive rate, or number of female offspring
produced per female generation
• Population growth is very dependent on R0
Multiplication Rate (R0) Constant
• If R0 > 1, the population increases geometrically
without limit. If R0 < 1 then the population
decreases to extinction.
• The greater R0 is the faster the population
Geometric Growth
increases:
Multiplication Rate (R0) Dependent
on Population Size
• Carrying Capacity – the maximum population size
that a particular environment is able to maintain
for a given period.
– At population sizes greater than the carrying capacity,
the population decreases
– At population sizes less than the carrying capacity, the
population increases
– At population sizes = the carrying capacity, the
population is stable
• Equilibrium Point – the population density that =
the carrying capacity.
Net Reproductive rate (R0) as a
function of population density:
Y = mX + b
Y = b – m(X)
N = 100, then R0 = 1.0
population stable
N > 100, then R0 < 1.0
population decreases
Intercept
N < 100, then R0 > 1.0
population increases
Remember, at R0 = 1.0
birth rates = death
rates
• We can measure population size in terms of deviation
from the equilibrium density:
z = N – Neq
Where:
z = deviation from equilibrium density
N = observed population size
Neq = equilibrium population size (R0 = 1.0)
• R0 = 1.0 – B(N – Neq) ( When N = Neq then R0 = 1.0)
Where:
R0 = net reproductive rate
y-intercept (b) will always = 1.0; population is
stable
(-)B = slope of line (m; the B comes from a
regression coefficient.
With these equations:
z = N – Neq
R0 = 1.0 – B(N – Neq)
We can substitute R0 in Nt+1 = R0Nt to get:
Nt+1 = [1.0 – B(zt)]Nt
How much the
population will
change (R0)
Start with an initial population (Nt) of 10, a slope (B) =
0.009, and Neq = 100, and the population gradually
reaches 100 and stays there.
Nt+1 = [1.0 – B(z)]Nt
10.00
2
18.10
3
31.44
4
50.84
120
5
73.34
100
6
90.93
7
98.35
8
99.81
9
99.98
10
100.00
11
100.00
12
100.00
Population Size
1
The population reaches
stabilization with a smooth
approach.
80
60
40
20
0
1
2
3
4
5
6
7
Generation
8
9
10
11
12
10.00
2
26.20
3
61.00
4
103.82
5
96.68
6
102.46
7
97.92
8
101.58
9
98.69
10
101.02
11
99.17
12
100.65
13
99.47
14
100.42
15
99.66
16
100.27
17
99.78
18
100.17
19
99.86
20
100.11
Start with an initial population (Nt) of 10, a slope
(B) = 0.018, and Neq = 100, and the population
oscillates a little bit but eventually (64
generations) stabilizes at 100 and stays there.
This is called convergent oscillation.
Nt+1 = [1.0 – B(z)]Nt
Population Size
1
120
100
80
60
40
20
0
1
3
5
7
9
11
13
Generation
15
17
19
10.00
2
32.50
3
87.34
4
114.98
5
71.92
6
122.41
7
53.84
8
115.97
9
69.67
10
122.50
11
53.60
12
115.78
13
70.11
14
122.50
15
53.59
16
115.77
17
70.12
18
122.50
19
53.59
20
115.77
Start with an initial population (Nt) of 10, a slope
(B) = 0.025, and Neq = 100, and the population
oscillates with a stable limit cycle that continues
indefinitely.
Nt+1 = [1.0 – B(z)]Nt
150
Population
1
100
50
0
1
3
5
7
9
11
13
Generation
15
17
19
10.00
2
36.10
3
103.00
4
94.05
5
110.29
6
77.39
7
128.13
8
23.59
9
75.87
10
128.96
11
20.64
12
68.14
13
131.10
14
12.87
15
45.40
16
117.28
17
58.51
18
128.91
19
20.84
20
68.68
Start with an initial population (Nt) of 10, a slope
(B) = 0.029, and Neq = 100, and the population
fluctuates chaotically.
Nt+1 = [1.0 – B(z)]Nt
150
Population
1
100
50
0
1
3
5
7
9
11
13
Generation
15
17
19
B
0.009
0.018
Population
Gradually approaches equilibrium
Convergent oscillation
0.025
0.029
Stable limit cycles
Chaotic fluctuation
As the slope increases, the population fluctuates
more. A high B causes an ‘overshoot’ towards
stabilization. Remember: B is the slope of the line
and represents how much Y changes for each
change in X.
• Define L as B(Neq): The response of the
population at equilibrium
– L between 0 and 1
 Population approaches equilibrium without
oscillations
– L between 1and 2
 Population undergoes convergent
oscillations
– L between 2 and 2.57
 Population exhibits stable limit cycles
– L above 2.57
 Population fluctuates chaotically
Growth With Overlapping Generations
• Previous examples were for species that
live for a year, reproduce then die.
• For populations that have a continuous
breeding season, or prolonged
reproductive period, we can describe
population growth more easily with
differential equations.
Multiplication Rate Constant
• In a given population, suppose the
probability of reproducing (b) is equal to
the probability of dying (d).
– r=b–d
Nt
rt
– Then rN = (b – d)N 
=
e
N0
– Where:
 Nt = population at time t
 t = time
 r = per-capita rate of population growth
 b = instantaneous birth rate
 d = instantaneous death rate
– Population grows geometrically
Nt+1 = R0Nt
We can determine how long it will take for a population
to double:
Nt
rt
=
2
=
e
N0
Loge(2) = rt
Loge(2) / r = t; r = realized rate of population growth per capita
For example:
r
t
0.01
69.3
0.02
34.7
0.03
23.1
0.04
17.3
0.05
13.9
0.06
11.6
Multiplication Rate Dependent on
Population Size
dN
K-N
= rN
dt
K
Where:
N = population size
t = time
r = intrinsic capacity for increase
K = maximal value of N (‘carrying capacity’)
K
r
Pop. Size
(K-N/K)
Growth Rate
1
1
99/100
0.99
1
25
25/100
6.25
1
50
50/100
25
1
75
25/100
18.75
1
95
5/100
4.75
1
99
1/100
0.99
1
100
0/100
0
Logistic population growth has
been demonstrated in the lab.
Year-to-year environmental fluctuations are one
reason that population growth can not be
described by the simple logistic equation.
Time-Lag Models
• Animals and plants do not respond immediately
to environmental conditions.
• Change our assumptions so that a population
responds to t-1 population size, not the t
population size.
L=Bneq
If 0<L<0.25, then stable
equilibrium with no
oscillation
If 0.25<L<1.0, then
convergent oscillation
If L > 1.0, then stable
limit cycles or divergent
oscillation to extinction
Ex. Daphnia
Stochastic Models
• Models discussed so far are deterministic:
given certain conditions, each model
predicts one exact condition.
• However, biological systems are
probabilistic:
– what is the probability that a female will have a
litter in the next unit of time?
– What is the probability that a female will have a
litter of three instead of four?
• Natural population trends are the joint
outcome of many individual probabilities
• These probabilistic models are called
stochastic models.
Basic Nature of Stochastic Models
• Nt+1 = R0Nt
• If R0 = 2, then a population size of 6 will yield a
population of 12 in one generation according to a
deterministic model: Nt+1 = 2(6) = 12
• Suppose our stochastic model says that a female
has an equal probability of having 1 or 3 offspring
(average = 2; so R0 = 2):
Probability
One female offspring
Three female offspring
0.5
0.5
• Since the number of offspring is random, we can
flip a coin and heads = 1 offspring, tails = 3
offspring to determine the total number of
offspring produced:
Outcome
Parent
Trial 1
Trial 2
Trial 3
Trial 4
1
(h)1
(t)3
(h)1
(t)3
2
(t)3
(h)1
(t)3
(h)1
3
(h)1
(t)3
(h)1
(h)1
4
(t)3
(t)3
(t)3
(t)3
5
(t)3
(t)3
(t)3
(h)1
6
(t)3
(t)3
(h)1
(h)1
14
16
12
10
Total population in next
generation:
Frequency Distribution After Several Trials
• Although the most common population size is
twelve as expected, the population could be any
size from 6 to 18.
Population Projection Matrices
• Used to calculate population changes from agespecific (or stage specific) birth and survival
rates.
– Can estimate how population growth will respond to
changes in only one specific age class.
F = fecundity
P = probability of surviving
and moving to next age
class
F = fecundity
Age Based
P = probability of surviving
and staying in same stage
G = probability of moving
to next stage
Stage Based
Stage-based life table and fecundity table for the loggerhead
sea turtle. #’s assume a 3% population decline / year.
Class
Size
Approx.
Age
1
Eggs,
hatchlings
<10
<1
0.6747
0
2
Small Juv.
10.1 – 58.0
1-7
0.7857
0
3
Large Juv.
58.1 – 80.0
8-15
0.6758
0
4
Subadults
80.1 – 87.0
16-21
0.7425
0
5
Novice
Breeders
>87.0
22
0.8091
127
6
1st year
remigrants
>87.0
23
0.8091
4
7
Mature
breeder
>87.0
24-54
0.8091
80
Stage #
Annual
survivorship
Fecundity
(eggs/yr)
Matrix Model
P1
F2
F3
F4
F5
F6
F7
G1
P2
0
0
0
0
0
0
G2
P3
0
0
0
0
0
0
G3
P4
0
0
0
0
0
0
G4
P5
0
0
0
0
0
0
G5
P6
0
0
0
0
0
0
G6
P7
Pi = proportion of that stage that remains in that stage
Gi = proportion of that stage that moves to the next stage
Fi = specific fecundity for that stage
Stage #
Approx.
Annual
Fecundity
Age
survivorship (eggs/yr)
1
<1
0.6747
0
2
1-7
0.7857
0
3
8-15
0.6758
0
4
16-21
0.7425
0
5
22
0.8091
127
6
23
0.8091
4
7
24-54
0.8091
80
0.7370
= P2
0.0487 = G2
0.7857 = P2 + G2
1
2
3
4
5
6
7
0
0
0
0
127
4
80
0
0
0
0
0
0
0
0
0
0
0
0
0.6747 0.7370
0
0.0487 0.6610
0
0
0.0147 0.6907
0
0
0
0.0518
0
0
0
0
0
0
0
0.8091
0
0
0
0
0
0
0
0.8091 0.8089
= Stage #
P1
F2
F3
F4
F5
F6
F7
N1
G1
P2
0
0
0
0
0
N2
0
G2
P3
0
0
0
0
N3
0
0
G3
P4
0
0
0
0
0
0
G4
P5
0
0
N5
0
0
0
0
G5
P6
0
N6
0
0
0
0
0
G6
P7
N7
X
N4
=
N1
= (P1*N1) + (F2*N2) + (F3*N3) + (F4*N4) + (F5*N5) + (F6*N6) + (F7*N7)
N2
= (G1*N1) + (P2*N2) + (0*N3) + (0*N4) + (0*N5) + (0*N6) + (0*N7)
N3
= (0*N1) + (G2*N2) + (P3*N3) + (0*N4) + (0*N5) + (0*N6) + (0*N7)
N4
= (0*N1) + (0*N2) + (G3*N3) + (P4*N4) + (0*N5) + (0*N6) + (0*N7)
N5
= (0*N1) + (0*N2) + (0*N3) + (G4*N4) + (P5*N5) + (0*N6) + (0*N7)
N6
= (0*N1) + (0*N2) + (0*N3) + (0*N4) + (G5*N5) + (P6*N6) + (0*N7)
N7
= (0*N1) + (0*N2) + (0*N3) + (0*N4) + (0*N5) + (G6*N6) + (P7*N7)
• With matrix models, we can simulate an
increase or decrease in survival or
fecundity and then determine what effect
that will have on population growth.
• So what? Well, we can determine what
age class or stage is most important to
population growth for an endangered
species.
By either increasing fecundity by 50% or survival to
100%, we can see that large juvenile survival is most
important to population growth, so put your
management efforts towards protecting large juveniles.