Estimating Abundance Weight Sub-sampleCapture-recapture MethodPeterson Method – Single CensusAssumptions For All Capture-Recapture StudiesExampleMark Recapture – Multiple CensusSlide 7Slide 8Slide 9Slide 10Slide 11Quadrant EstimationEstimating AbundanceWeight Sub-sample•Used to estimate total number in a sample•Method:–Weigh a known number of individuals to get a mean weight–Weigh the entire sample, then divide the total weight by mean weight to get total number of individuals.•Example:–10 individuals weigh 68g, so mean weight = 68 / 10 = 6.8–Total weight = 528, so total number = 528 / 6.8 = 77.6 individualsCapture-recapture Method•Important tool for estimating density, birth rate, and death rate for mobile animals.•Method:–Collect a sample of individuals, mark them, and then release them–After a period, collect more individuals from the wild and count the number that have marks–We assume that a sample, if random, will contain the same proportion of marked individuals as the population does –Estimate population densityMarked animals in second sample (R)Total caught in second sample (C)Marked animals in first sample (M)Total population size (N)=520 N16= N = 64Peterson Method – Single Census= ProportionsAssumptions For All Capture-Recapture Studies•Marked and unmarked animals are captured randomly.•Marked animals are subject to the same mortality rates as unmarked animals. The Peterson method assumes no mortality during the sampling period.•Marked animals are neither lost or overlooked.ExampleA fish biologist goes out and samples (sample 1) a population of trout. A total of 109 (M) trout were marked and released. At this time, a proportion of fish in the total population has a mark and we assume that this proportion remains constant. On a second sampling trip (sample 2), the biologist collected 177 (C) trout and 57 (R) of those were marked from the initial sample. How large is the population (N)? R/C = M/N  N=MC/R  N = (109)(177)/57  N=338Mark Recapture – Multiple Census1 2 3 4 5 6Time of Last Capture 1 10 3 5 2 22 34 18 8 43 33 13 84 30 205 43Total Marked (mi) 0 10 37 56 53 77Total Unmarked 54 136 132 153 167 132Total Caught (Ci) 54 146 169 209 220 209Total Released (Si)54 143 164 202 214 207Time of CaptureMi = Marked population size at time imi = Marked animals actually caught at time iCi = Total number of animals caught at time I Si / Ri = proportion caughtSi = Total animals released at time iZi = Number of individuals marked before time i, not caught in the ith sample but caught in a sample after time iRi = Number of the Si individuals released at time i that are caught in a later sample1 2 3 4 5 6Time of Last Capture 1 10 3 5 2 22 34 18 8 43 33 13 8R34 30 205 43Total Marked (mi) 0 10 37 56 53 77Total Unmarked 54 136 132 153 167 132Total Caught (Ci) 54 146 169 209 220 209Total Released (Si)54 143 164 202 214 207Time of CaptureZ3Mi =Si ZiRi+ mi1 2 3 4 5 6Time of Last Capture 1 10 3 5 2 22 34 18 8 43 33 13 8R34 30 205 43Total Marked (mi) 0 10 37 56 53 77Total Unmarked 54 136 132 153 167 132Total Caught (Ci) 54 146 169 209 220 209Total Released (Si)54 143 164 202 214 207Time of CaptureZ3M3 = {(164)(39)/(54)} + 37 = 155.4Mi =Si ZiRi+ miEstimate Pop. Size at Time 3S3 = 164C3 = 169Z3 = 5+2+2+18+8+4 = 39R3 = 33+13+8 = 54m3 = 37Population Estimation:N3=M3C3/m3 = (155.4)(169)/37 = 710M4 = {(202)(37)/(50)} + 56 = 205.5Mi =Si ZiRi+ miEstimate Pop. Size at Time 4S4 = 202C4 = 209Z4 = 2+2+8+4+13+8 = 37R4 = 30+20 = 50m4 = 56Population Estimation:N4=M4C4/m4 = (205.5)(209)/56 = 7671 2 3 4 5 6Time of Last Capture 1 10 3 5 2 22 34 18 8 43 33 13 84 30 20R45 43Total Marked (mi) 0 10 37 56 53 77Total Unmarked 54 136 132 153 167 132Total Caught (Ci) 54 146 169 209 220 209Total Released (Si)54 143 164 202 214 207Time of CaptureZ4Estimating mortality: We can compare the estimated number of marks in the wild versus a known amount to get mortality rates. For example, in year 3 we estimated that there were 155 marked individuals. We released a total of 132 newly marked individuals, for a total of 287 marked individuals. We estimated the number of marked individuals to be 206 for year 4. 206 is less than 287, so the survival rate is 206/287=0.718. Mortality is then 1-0.718=0.282.Survival3 = Number Marks Estimated3 / Total Marks ReleasedSurvival3 = 206 / 287 = 0.718Mortality3 = 1 – Survival3 = 1 – 0.718 = 0.282Estimating Natality: We estimated that our mortality from year 3  4 was 28%, but our population estimation increased by 57 individuals from 710 to 767. Given an initial population of 710 (N3) and mortality of 28%, we should only have 511 (710-0.28*710) individuals in the population for year 4. However, our estimated population size in year 4 is 767, 256 more individuals than 511. So, we had 256 individuals added to the population!N3 = 710, 28% Mortality3 = (710)(0.28) = 199Expected N4 = 710 – 199 = 511Estimated N4 = 767Therefore, 767 – 511 = 256 new individualsQuadrant Estimation•Individuals evenly spread over a known area•Use a known area quadrant to sample•Determine the mean number per square area•Multiply times total area to get total number of