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Mark Recapture lecture 1
An example from Sockeye salmon….
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Mark recapture lectures
Petersen method Schnabel method Schumacher-Eschmeyer Jolly Seber Closed population Open population Overview of methods to help your reading of Krebs Chp 2
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Closed populations No individuals enter or leave the population between surveys Survey 1 Survey 2
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Individuals enter or leave the population between surveys
Open populations Individuals enter or leave the population between surveys Survey 1 Survey 2
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What makes a population closed?
Dispersal barriers Philopatry Large surveyed area Slow reproductive/death rate Short time between surveys
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What type of population are snow geese?
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Snow Goose La Pérouse Bay
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Recording which birds are marked, and marking new birds
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LPB Colony size Year
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Petersen method: Closed population
Survey 1: Survey 2: Catch several animals Catch C animals Count recaptures (R) Mark all M animals Return animals to population Return animals to population
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Survey 2: C = 15 R = 4 Survey 1: M = 12
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What is the total population size (N)?
Note that the proportion marked in the population equals the proportion marked in the 2nd sample M = R N C N = M C R M = 12 C = 15 R = 4
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What is the total population size (N)?
Note that the proportion marked in the population equals the proportion marked in the 2nd sample N = 13 * 16 5 -1 N = (M+1) (C+1) (R+1) -1 M = 12 C = 15 R = 4
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When would Petersen give you a bad estimate?
Population not closed Marked animals likely to be re-trapped Marked animals likely to die Marks fall off
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Schnabel method: closed population
Survey 1 Survey 2 Survey 3 Survey 4 Survey 5 Essentially, Petersen estimates on multiple surveys
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Schnabel method: closed population
Catch Ct animals Survey t: Mark Ut unmarked animals Return animals to population Record Rt recaptures
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Schnabel method: closed population
Survey t: Catch Ct animals What’s the relationship between Ct, Rt, and Ut ? Record Rt recaptures Mark Ut unmarked animals Ct = Rt + Ut Return animals to population
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Schnabel method: example
Time (t) Ct Rt Ut 1 20 2 5 3 13 4 10 How many individuals marked by beginning of time 5?
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Schnabel method: example
Time (t) Ct Rt Ut 1 20 2 5 15 3 7 13 4 10 How many individuals marked by beginning of time 5?
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Schnabel method: example
Time (t) Ct Rt Ut 1 20 2 5 15 3 7 13 4 10 Σ = 58
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Schnabel method: example
Time (t) Ct Rt Ut 1 20 2 5 15 3 7 13 4 10 In general: Mt = U1 + U2..Ut-1 Σ = 58
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Schnabel formulas: N = Σ (Ct Mt) Σ Rt N = Σ (Ct Mt) Σ Rt+1
Marked > 10% of population N = Σ (Ct Mt) Σ Rt+1 Marked < 10% of population ( just weighted average of Petersen estimates!)
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Schnabel method: example
Time (t) Ct Rt Ut Mt CtMt 1 20 2 5 15 3 7 13 35 4 10 48
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Schnabel method: example
Time (t) Ct Rt Ut Mt CtMt 1 20 2 5 15 400 3 7 13 35 700 4 10 960 48 Σ = 22 Σ = 2060
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Schnabel method: example
Time (t) Ct Rt Ut Mt CtMt 1 20 2 5 15 400 3 7 13 35 700 4 10 960 N = Σ (Ct Mt) = 2060 = 94 Σ Rt 48 Σ = 22 Σ = 2060
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Schnabel method: example
Time (t) Ct Rt Ut Mt CtMt 1 20 2 5 15 400 3 7 13 35 700 4 10 960 What proportion of total population marked by end? 48 Σ = 22 Σ = 2060
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Scumacher-Eschmeyer method (for Schnabel experiment)
N = C M R R = 1 * M C N y = mx + b
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Scumacher-Eschmeyer method (for Schnabel experiment)
N = C M R R = 1 * M C N R C Slope = ? M
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