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:

• 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

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

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.

Y = mX + b

Y = b – m(X)

Intercept

Net Reproductive rate (R0) as a function of population density:

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

• 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)

Nt+1 = [1.0 – B(z)]Nt

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.

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

11 100.00

12 100.000

20

40

60

80

100

120

1 2 3 4 5 6 7 8 9 10 11 12

Generation

Po

pu

lati

on

Siz

e

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

020406080100120

1 3 5 7 9 11 13 15 17 19

Generation

Po

pu

lati

on

Siz

e

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

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

14 122.50

15 53.59

16 115.77

17 70.12

18 122.50

19 53.59

20 115.77

0

50

100

150

1 3 5 7 9 11 13 15 17 19

Generation

Po

pu

lati

on

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

0

50

100

150

1 3 5 7 9 11 13 15 17 19

Generation

Po

pu

lati

on

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

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.

• 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 – 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

Nt

N0= 2 = ert

We can determine how long it will take for a population to double:

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

dNdt = rN

K - N

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 0.5

Three female offspring 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

Total population in next generation:

14 16 12 10

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 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

F = fecundity

P = probability of surviving and staying in same stage

G = probability of moving to next stage

Age Based

Stage Based

Stage # Class Size

Approx. Age

Annual survivorship

Fecundity (eggs/yr)

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-based life table and fecundity table for the loggerhead sea turtle. #’s assume a 3% population decline / year.

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

Matrix Model

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

1 2 3 4 5 6 7

0 0 0 0 127 4 80

0.6747 0.7370 0 0 0 0 0

0 0.0487 0.6610 0 0 0 0

0 0 0.0147 0.6907 0 0 0

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 #

Approx. Age

Annual survivorship

Fecundity (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

0.0487

0.7857

= P2

= G2

= P2 + G2

= Stage #

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

N1

N2

N3

N4

N5

N6

N7

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)

X =

• 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.