1. Population Growth – Chapter 11

2. 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:
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
Nt+1 = R0Nt

3. 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 increases:
Geometric Growth

4. 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.

5. Net Reproductive rate (R0) as a function of population density:
Y = mX + b
Y = b – m(X)
Intercept
N = 100, then R0 = 1.0 population stable
N > 100, then R0 < 1.0 population decreases
N < 100, then R0 > 1.0 population increases
Remember, at R0 = 1.0 birth rates = death rates

6. R0 = 1.0 – B(N – Neq) ( When N = Neq then R0 = 1.0)
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)
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.

7. How much the population will change (R0)
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)

8. Start with an initial population (Nt) of 10, a slope (B) = 0
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
The population reaches stabilization with a smooth approach. 1
10.00 2
18.10 3
31.44 4
50.84 5
73.34 6
90.93 7
98.35 8
99.81 9
99.98 10
100.00 11 12

9. 1
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

10. 1
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 15
53.59 16
115.77 17
70.12 18 19 20
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

11. 1
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

12. B
Population
0.009
Gradually approaches equilibrium
0.018
Convergent oscillation
0.025
Stable limit cycles
0.029
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.

13. 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

14. 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.

15. Multiplication Rate Constant
In a given population, suppose the probability of reproducing (b) is equal to the probability of dying (d).
r = b – d
Then rN = (b – d)N 
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 N0
= ert
Nt+1 = R0Nt

16. We can determine how long it will take for a population to double:
= 2 = ert
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

17. Multiplication Rate Dependent on Population SizedN dt
= rN
K - N K
Where:
N = population size
t = time
r = intrinsic capacity for increase
K = maximal value of N (‘carrying capacity’)

18. K r Pop. Size (K-N/K) Growth Rate 1 99/100 0.99 25 25/100 6.25 50
50/100 75
18.75 95
5/100
4.75 99
1/100
100
0/100 K

19. Logistic population growth has been demonstrated in the lab.

20. Year-to-year environmental fluctuations are one reason that population growth can not be described by the simple logistic equation.

21. 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

22. 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.

23. 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
0.5
Three female offspring

24. 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 2 3 4 5 6
Total population in next generation: 14 16 12 10

25. 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.

26. Population Projection Matrices
Used to calculate population changes from age-specific (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
Age Based
F = fecundity
P = probability of surviving and staying in same stage
G = probability of moving to next stage
Stage Based

27. Stage-based life table and fecundity table for the loggerhead sea turtle. #’s assume a 3% population decline / year.
Stage #
Class
Size
Approx. Age
Annual survivorship
Fecundity (eggs/yr) 1
Eggs, hatchlings
<10
<1
0.6747 2
Small Juv.
10.1 – 58.0
1-7
0.7857 3
Large Juv.
58.1 – 80.0
8-15
0.6758 4
Subadults
80.1 – 87.0
16-21
0.7425 5
Novice Breeders
>87.0 22
0.8091
127 6
1st year remigrants 23 7
Mature breeder
24-54 80

28. Matrix Model P1 F2 F3 F4 F5 F6 F7 G1 P2 G2 P3 G3 P4 G4 P5 G5 P6 G6 P7G2 P3 G3 P4 G4 P5 G5 P6 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

29. Stage #
Approx. Age
Annual survivorship
Fecundity (eggs/yr) 1
<1
0.6747 2
1-7
0.7857 3
8-15
0.6758 4
16-21
0.7425 5 22
0.8091
127 6 23 7
24-54 80
0.7370
0.0487
0.7857
= P2
= G2
= P2 + G2
= Stage # 1 2 3 4 5 6 7
127 80
0.6747
0.7370
0.0487
0.6610
0.0147
0.6907
0.0518
0.8091
0.8089

30. X = P1 F2 F3 F4 F5 F6 F7 G1 P2 G2 P3 G3 P4 G4 P5 G5 P6 G6 P7 N1 N2 N3G2 P3 G3 P4 G4 P5 G5 P6 G6 P7 N1 N2 N3 N4 N5 N6 N7 X = 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)

31. 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.

32. 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.