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.

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

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.

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)

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

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

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

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

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

Multiplication Rate Dependent on Population Size dN dt = 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 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

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

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

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 Age Based F = fecundity 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. 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

Matrix Model P1 F2 F3 F4 F5 F6 F7 G1 P2 G2 P3 G3 P4 G4 P5 G5 P6 G6 P7 G2 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

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

X = P1 F2 F3 F4 F5 F6 F7 G1 P2 G2 P3 G3 P4 G4 P5 G5 P6 G6 P7 N1 N2 N3 G2 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)

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.