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The Lincoln Index. CAPTURE – MARK – RECAPTURE METHOD. The Lincoln Index. Used to estimate the size of an animal population. Capture animals Mark them with a tag Release them and wait a period of time Capture animals again. Assumptions.

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## The Lincoln Index

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**The Lincoln Index**CAPTURE – MARK – RECAPTURE METHOD**The Lincoln Index**• Used to estimate the size of an animal population. • Capture animals • Mark them with a tag • Release them and wait a period of time • Capture 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.**Vole population calculation**n1 = number in the “first sample” and marked. 8+19+10-2 = 35 You 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:**year Captured Marked pond turtles1992: 18 1993: 34 (12) 1994: 30 (18)**Results1992: 18 1993: 34 (12) 1994: 30 (18)**1992-93: N = 18(34) = 51 turtles 12 ____________________________________ The total number marked is 18 + 34 – 12 = 40 1992-94: N = 40(30) = 67 turtles 18 This assumes that none of the paint came off of the turtles over the course of two 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 beans • Count 150 white beans into a cup • Capture 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.**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

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