Estimating Abundance Weight 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 individuals

Capture-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 density

Peterson Method – Single Census Marked animals in second sample (R) Total caught in second sample (C) Marked animals in first sample (M) Total population size (N) = 5 20 N 16 = N = 64 = Proportions

Assumptions 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.

Example A 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=338

Mark Recapture – Multiple Census

Mi = Si Zi Ri + mi Mi = Marked population size at time i mi = Marked animals actually caught at time i Ci = Total number of animals caught at time I Si / Ri = proportion caught Si = Total animals released at time i Zi = Number of individuals marked before time i, not caught in the ith sample but caught in a sample after time i Ri = Number of the Si individuals released at time i that are caught in a later sample Mi = Si Zi Ri + mi

Estimate Pop. Size at Time 3 Mi = Si Zi Ri + mi S3 = 164 C3 = 169 Z3 = 5+2+2+18+8+4 = 39 R3 = 33+13+8 = 54 m3 = 37 Population Estimation: N3=M3C3/m3 = (155.4)(169)/37 = 710

Estimate Pop. Size at Time 4 Mi = Si Zi Ri + mi S4 = 202 C4 = 209 Z4 = 2+2+8+4+13+8 = 37 R4 = 30+20 = 50 m4 = 56 Population Estimation: N4=M4C4/m4 = (205.5)(209)/56 = 767

Estimating 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 Released Survival3 = 206 / 287 = 0.718 Mortality3 = 1 – Survival3 = 1 – 0.718 = 0.282

Therefore, 767 – 511 = 256 new individuals Estimating 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) = 199 Expected N4 = 710 – 199 = 511 Estimated N4 = 767 Therefore, 767 – 511 = 256 new individuals

Quadrant 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 individuals