CAPTURE – MARK – RECAPTURE METHOD

The Lincoln Index

The Lincoln Index

Used to estimate the size of an animal population.

Capture animalsMark them with a tagRelease them and wait a period of timeCapture animals again

Assumptions

The population of organisms must be closed, no immigration or emigration.

The time between samples must be small compared to the life span of the organism

The marked organisms must mix completely with the rest of the population during the time between sampling.

Assumption:

The proportion of animals caught to the whole population is the same as the marked animals in the second sample.

First Capture = Marked animals, second capture

Whole population second capture

Equation

n1 = m2

N n2

n1 = number of animals first marked and released.

n2 = number of animals captured in the second sample

m2 = number of marked animals in the second sample

Solve for N = n1 . n2 to find the total population

m2

Limitations

Marking or capturing the animal might change its behavior to avoid recapture

The mark might be toxic or cause predators to see it

Marks might cause the animal to be rejected by its own population

Trapping might cause the animal to avoid the traps in the future or to choose to be trapped to get food!

Assumptions:

No animals are born or die during the period tested

No animals immigrate or emigrate in the period tested.

Animals do not die because of the markings.

The tags do not fall off.

What is the estimated snail population in a woodland?

In a woodland, the undergrowth was sampled for snails and 430 were found and marked. They were then released and the population similarly sampled after a two-week period. This second sampling produced 410 snails, 100 of which were marked.

Answer

n1 = 430

n2 = 410

m2 = 100

N = 430 x 410 = 1763 snails 100

Woodlice

Woodlice are terrestrial crustaceans that live under logs and stones in damp soil. To assess the population of woodlice in an area, students collected as many of the animals as they could find, and marked each with a drop of fluorescent paint. A total of 303 were marked. 24 hours later, woodlice were collected again in the same place. This time 297 were found, of which 99 were marked from the first time. What is the estimated population of woodlice in this area?

Population estimate

N = 303 x 297 = 909 99

Voles…

While studying field voles, an ecologist caught 500 and ringed a foot of each before releasing them. Every day for the next two weeks he examined the waste material found in the nests of their predators. He collected a total of 300 field vole skulls and 15 rings.

How many field voles were probably in the area examined?

Population estimate

N = 300 x 500 = 10 000 15

A harder problem from the 2009 IB exam

A population of meadow voles in an area of woodland was sampled using small mammal traps

over a period of four days. On each occasion any unmarked individuals were marked and all

the voles from the capture were then released. The following table represents the data obtained

on each of the four days.

Estimate the population size

Vole population calculation

n1 = number in the “first sample” and marked.

8+19+10-2 = 35You need to know how many animals were marked!

n2 = number in the “second sample” 23

m2 = number in the second sample with a mark 8

N = 35x23 = 100.6 The size of the population is 101 as 8 100.6 voles is impossible!

Estimate the size of a turtle population:

yearyear CapturedCapturedMarkedMarked

pond turtlespond turtles 1992: 1992: 18181993: 1993: 34 34

(12)(12)1994: 1994: 30 30

(18)(18)

ResultsResults1992: 18 1993: 34 (12) 1994: 30 (18)1992: 18 1993: 34 (12) 1994: 30 (18)

1992-93: N = 1992-93: N = 18(34)18(34) = 51 turtles = 51 turtles

1212____________________________________

The total number marked is 18 + 34 – 12 = The total number marked is 18 + 34 – 12 = 4040

1992-94: N = 1992-94: N = 40(30)40(30) = 67 turtles = 67 turtles

1818

This assumes that none of the paint came This assumes that none of the paint came off of the turtles over the course of two off of the turtles over the course of two years.years.

Mini experiment:

Does the size of the first capture, n1, while the size of the second capture is held constant, affect the accuracy of the population estimate, N?

Does the size of the second capture, n2, while the size of the first capture is held constant, affect the accuracy of the population estimate, N?

Hypothesis?

Do you think that the Lincoln Index is more reliable if many animals are captured and marked in the first capture?

Do you think that the Lincoln Index is more reliable if many animals are captured in the second capture?

Mini experiment with beans

You will use beans instead of animals to study the accuracy of the Lincoln index.

Independent variable:Vary either the size of the first capture OR the second capture while holding the other variable constant

Dependent variable:The value for m2 will change, so the value for

N will change…. maybe

Minilab: Investigation with beans:Separate black and white beansCount 150 white beans into a cupCapture some white beans, n1

Mark the captured beans.Release the marked beans into the cup and mix.Capture a second sample, n2

Count the number of marked beans in the second sample, m2

Repeat for a second and third trial, using the same size captures. ( Three trials of the same experiment should improve your accuracy.)

Change the size of the first or second capture while keeping the same value for the other capture. Repeat

Change the number of beans in one of the captures.. Five times total.

Data table: Numbers of beans + 1

Trial n1 n2 m2

1

2

3

1

2

3

1…

Data Processing: Calculate NCalculate the population estimate for N

Mean value Estimated population N

Percent error

Graph the results

Do the population estimates improve with increasing first or second captures?

The independent variable is on the x axis

The dependent variable is on the y axis

Rubric for grading