Unit 5 Populations [The end is getting nearer!]

A. CHARACTERISTICS OF POPULATIONS 1. Terms a. population – a group of individuals of the same species living in the same geographical area b. population size – the number of individuals of the same species occupying a given area/volume at a given time

c. population density – the number of individuals of the same species that occur per unit area or volume population density = population number space occupied D = N. S

d. crude density vs ecological density crude density = the population number divided by the total area ecological density = the pop number divided by the usable area eg. for fish density in a park, we only want their density in the area of lakes or rivers, not the total area of the park

e. population dispersion the general pattern of individuals within an area clumpeduniformrandom eg. bat colonieseg. nesting penguinseg. maple trees in a forest

2.Calculating population characteristics eg A. A population of 450 porcupines live in an area 12.1 km by 15.3 km in which there are three lakes [7.1 km 2, 15.2 km 2 and 2.7 km 2 ]. What is the crude density of this population? What is the ecological density? area = 12.1 km x 15.3 km useful area = km 2 – [7.1 km km km 2 ] = km 2 – 25.0 km 2 = km 2 = km 2 area of lakes

D = N S D = 450 porc/ km 2 D = 450 porc/ km 2 D = 2.81 porc/km 2 D = 2.43 porc/km 2 crude densityecological density The crude density for this population was 2.43 porcupines per km 2. The ecological density for this population was 2.81 porcupines per km 2

eg. BIf the wolf population density in an area is wolves /km 2 and the study area is 47.3 km by 60.4 km, how many wolves would there be in the area? How many wolves would be in the area if there is a lake of km 2 in the area? S = 47.3 km x 60.4 kmS = km km2 S = km 2 S = km 2 no lakeswith lakesN = D x S N = wolf/km 2 x km 2 N = wolf/km 2 x km 2 N = wolvesN = wolves There are 266 wolves in the area if there are no lakes. There are 232 wolves in the area with the lakes present.

3. Mark – Recapture Method of Counting a. some individuals in a population are captured, tagged or marked in some way Small mammals are usually trapped in cages. Fish are usually netted as hooks will harm them.

b. the marked individuals are then released The tag on this released fish is clearly visible.

c. the experimenters then wait for some time, return and capture individuals in the same area d. they note whether the captured animals are marked [recaptured] or unmarked [new] e. they determine the total population in the area with the formula total# marked [M]=# of recaptures [m] total population [N]size of second sample [n] M = m N n

eg. In a population of unknown size – researchers capture and mark 72 mink. After release, they capture 65 individuals on a second trapping, of which 8 are marked. What is the total population of mink? There were 585 mink in the original population. M = m N n 72 = 8 N 65 8N = 72 x 65 N = 585

4. Quadrat sampling a. a quadrat is a sampling frame used to estimate population size b. the quadrat may be 1.0m x 1.0m for small species or 100m x 100m for large species Quadrats can be used on land or underwater with equal success.

c. the quadrat only works for sessile species [ones that don’t move] eg trees, barnacles d. usually more than one quadrat is counted & then average values are used Having completed one quadrat, it is then moved and a second sample is taken.

eg using a 10m by 10m quadrat, young pines trees were counted in an area with the following results , 17, 19, 25. What is the population of pine trees in an area 2.3 km by 1.7 km? average density = ( ) = 84 pines 4 4 N = D x S N = 0.21 pines/m 2 x (2300m x 1700m) N = 821,100 pine trees There are 821,100 young pine trees in the study area. = 21 pine tree / 100m 2 = 0.21 pines/m 2

HW pg 659 # 3, 5, 6, 4, 2 Do them in the order shown They are to be marked