Measuring and Modeling Population Changes Chapter 11.2 McGraw-Hill Ryerson 2011

Measuring and Modeling Population Changes An ecosystem has a finite pool of abiotic and biotic resources. Biotic factors tend to vary greatly over time, whereas most abiotic factors vary little over time (some exceptions: temp, water availability, etc...)

Carrying Capacity: the maximum number of individuals an ecosystem can support based on resources available. Carrying capacity is dynamic, always changing, since resource levels are never constant.

POPULATION CHANGE A negative result means population is declining. Population change (%) = [(b + i) – (d+e)] x 100 n b = births, d = deaths, i = immigration, e = emigration, n = initial population size A negative result means population is declining. A positive result means population is growing. In an open population all four factors come into play (i.e. in the wild). In a closed population only births and deaths are a factor (i.e. in a zoo)

POPULATION GROWTH There are 3 main types of population growth. Geometric, Exponential and Logistic.

Geometric Growth Many populations grow rapidly during the breeding season and then decline slowly the rest of the year. Geometric growth is a pattern of population growth where organisms reproduce at fixed intervals at a constant rate. TIME

Geometric Growth λ = N (t + 1) N(t) The growth rate is a constant (λ) and can be determined by comparing the population size in one year to the population size at the same time the previous year. λ = N (t + 1) N(t) λ is the fixed growth rate (one year vs next) N is the population size at a given year (t+1) or (t)

To find the population size at any given year, the formula is: N(t) = N(0)λt N(0) is the initial population size

EXAMPLE The initial Puffin population on Gull Island, Newfoundland is 88 000. Over the course of the year they have 33 000 births and 20 000 deaths. a) What is their growth rate? b) What will the population size be in 10 years at this current growth rate?

Answer to a) ANSWER a) N (0) = 88 000 λ = N(t + 1) = 101 000 = 1.15 Therefore the growth rate is 1.15.

Answer to b) From a) growth rate, or λ = 1.15 = 88 000 (1.15)10 N(10) = N(0)λ10 * BEDMAS = 88 000 (1.15)10 = 356 009 Therefore the population size will be 356 009 in 10 years.

EXPONENTIAL GROWTH A wide variety of species are able to reproduce on a continuous, rather than intermittent basis. (ex. humans, bacteria, cancer cells) Exponential growth is growing continuously at a fixed rate in a fixed time interval J – shaped curve

Calculating Exponential Growth The same formulas can be used as geometric growth, however, since the time interval is not restricted to breeding periods, biologists are able to calculate the instantaneous growth rate. It is expressed in terms of the intrinsic (per capita) growth rate, r, where r = b – d (per capita birth rate – per capita death rate)

Exponential population growth rate is: dN = rN dt Doubling time: to find the time it takes a population that is reproducing exponentially to double, we use the equation: td = 0.69 r Population Size Intrinsic growth rate Instantaneous growth rate

Sample Problem A population of 2500 yeast cells in a culture tube is growing exponentially. If the intrinsic growth rate is 0.030 per hour, calculate: a) the initial instantaneous growth rate of the yeast population. b) the time it will take for the population to double in size. c) the population size after four doubling periods.

Answer A population of 2500 yeast cells in a culture tube is growing exponentially. If the intrinsic growth rate is 0.030 per hour, calculate: N = 2500 yeast, r = 0.030 yeast /hr dN/dt = rN = 0.030 yeast/hr x 2500 yeast = 75 /hr When the population size is 2500 the instantaneous population growth rate is 75 per hour

Answer A population of 2500 yeast cells in a culture tube is growing exponentially. If the intrinsic growth rate is 0.030 per hour, calculate: N = 2500 yeast, r = 0.030 yeast /hr b) td = 0.69/r = 0.69 / 0.03/hr the population size after four doubling periods.

Answer A population of 2500 yeast cells in a culture tube is growing exponentially. If the intrinsic growth rate is 0.030 per hour, calculate: N = 2500 yeast, r = 0.030 yeast /hr c) the population size after four doubling periods.

LOGISTIC GROWTH The previous two models assume an unlimited resource supply, which is never the case in the real world. When a population is just starting out, resources are plentiful and the population grows rapidly (geometrically/exponentially). BUT, as the population grows, resources are being used up and the population nears the ecosystem's carrying capacity.

The growth rate drops and a stable equilibrium exists b/w births and deaths. The population size is now the carrying capacity (K). This is known as a sigmoidal curve.

Logistic Growth Curve Carrying capacity Dynamic Equilibrium: population reaches carrying capacity (b = d) LOG phase: greatest increases, lots of resources LAG phase: Initial growth is slow, due to small numbers

LOGISTIC GROWTH EQUATION Logistic growth represents the effect of carrying capacity on the growth of a population. It is the most common growth pattern in nature. dN = rmaxN K – N dt K Population size at given time Carrying capacity Max intrinsic growth rate Population growth at a given time

continued... If the population size is close to the carrying capacity, there is virtually no growth (K-N = 0), thus the equation takes into account declining resources as the population increases.

Sample Problem A population of humans on a deserted island is growing continuously. The carrying capacity of that island is 1000 individuals and the maximum growth rate is 0.50. Determine the population growth rates over 5 years if the initial population size is 200.

Homework Pg 519 #3, 5, 6, 7, 8, 11, 12