1. Deterministic elasticity patterns of age-structured animals: A Review Henry F. Mollet Moss Landing Marine Labs and Monterey Bay Aquarium

2. 7/9/20162 Deterministic elasticity patterns of age-structured animals Deterministic: Vital rates are assumed to be constant. An E–pattern has to be done correctly deterministically before we can do any stochastic calculations. Elasticities are proportional sensitivities. Changing a vital rate by say 1%, will tell us by how how many % the population growth ( ) will change. Age-structured animals have simple rather than complex life histories. Complex life histories are usually found in some plants. Elasticity patterns apply to the stable age distribution (as does population growth ).

3. 7/9/20163 Sharks Down Under, Sidney 1991 VanDykhuisen and Mollet (1992) Growth, Age Estimation, and Feeding of Captive Sevengill Sharks, Notorynchus cepedinanus, at the Monterey Bay Aquarium Cailliet, Mollet, Pittenger, Bedford, and Natanson (1992) Growth and Demography of the Pacific Angel Shark (Squatina californica), Based upon Tag Returns off California

4. 7/9/20164 Pacific Angel Shark (Cailliet et al. 1992) Basic life history table with vital rates age-at-first maturity (  ) = 10 yr; longevity (  ) = 35 yr; survival rate S = 0.8187 (M = 0.2 yr -1 ) assumed to be age independent ; fertility m = 6/2 = 3 female pups every year; Results were = 1. 0596 (r = 0.0579 yr -1 ) Ro = 2.227, T = 13.84 yr,  1 = 14.37 yr (Abar = 13.37 yr, not reported)

5. 7/9/20165 Pacific Angel Shark (Cailliet et al. 1992) We could have provided an E-pattern using empirical calculation. Results would have been: E(m) = E 1 = 0.0748; E(S j ) = E 2 = 0.7481 (E 2/ /E 1 =10); E(S a ) = E 3 = 0.2519 (E 3/ /E 1 =3.37); Sum is 1+ E 1 = 1.0748! (not 1.0)

6. 7/9/20166 Pacific Angel Shark (Cailliet et al. 1992) Today we would calculate E-pattern more quickly using the following formulas: E(m) = E 1 = 1/Abar = 0.0748; E(S j ) = E 2 =  /Abar = 0.7481 (E 2/ /E 1 =10 =  ) E(S a ) = E 3 = (Abar -  )/Abar = 0.2519 (E 3/ /E 1 =3.37 = Abar -  ) Sum is 1+ E 1 = 1.0748! (not 1.0) For graphing in elasticity triangle we need normalized elasticities: E n (m) = E n,1 = 1/(Abar + 1) = 0.0696 E n (S j ) = E n,2 =  /(Abar + 1) = 0.6960 E n (S a ) = E n,3 = (Abar -  )/(Abar + 1) = 0.2343 (Sum is now 1.0)

7. 7/9/20167 Elasticity Triangle for PAS Showing E1=E(m) = ~7%, E2=E(Sjuv) =~ 70%, E3=E(Sadu) = ~23%

8. 7/9/20168 Elasticity Triangle for PAS adding contours  = 1 yr (&10 yr) and Abar/  = 2

9. 7/9/20169 Elasticity triangle for PAS Showing results for  = 8,10, and 13 year (95% confidence band for  = 10 yr) )

10. 7/9/201610 Elasticity triangle for PAS Showing results for increasing fishing of adults (F = 0.0-0.55; 1 = 1.06 - 0.996)

11. 7/9/201611 E-patterns for 66 Elasmobranchs E 2 =E(Sjuv) > E 3 =E(Sadu); Abar/  <2

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13. 7/9/201613 Durban 2001 LHT (or Leslie matrix) vs. Stage-based matrix models Mollet and Cailliet (2002) Comparative population demography of elasmobranchs using life history tables, Leslie matrices and stage-based models Miller, Frisk and Fogarty (2003) Comment on Mollet and Cailliet (2002): confronting models with data Mollet and Caillliet (2003) Reply to comments by Miller et al. (2003) on Mollet and Cailliet (2002) confronting models with data Best to avoid stage-based models for age-structured animals (e.g. elasmobranchs)

14. 7/9/201614 Stage-based models produce incorrect E-patterns: Example Pacific Angel Sharks (used different species in M&C 2002; Abar was not included) LHT or Leslie matrix 1 = 1.060; Ro = 2.23  1 = 14.4 yr; T = 13.8 yr ( Abar = 13.37 yr) E 1 =E(m) = 0.0748 E 2 = 0.748 (10.0) E 3 = 0.252 (3.37) 1/Abar = 0.0748, OK 3x3 stage-based model 1 = 1.060; Ro = 1.80  1 = 11.1 yr; T = 10.1yr (Abar = 9.234) E 1 = E(m) = 0.108 E 2 = 0.534 (4.84) E 3 = 0.476 (4.39) 1/Abar = 0.108, OK but Abar must be wrong as it is smaller than  ! Thus E-pattern is wrong.

15. 7/9/201615 Looking at ‘ generation ’ times  1,T, Abar (  1 & Abar are ages) It was known that stage-based models are not suitable for transient dynamics and convergence (e.g. how long will it take after a catastrophic event to get back to stable age distribution) The E–pattern is determined by Abar and  alone. As per Mollet and Cailliet (2003), stage- based models produce incorrect Abar (and  ) and therefore cannot be used to calculate the E– pattern.

16. 7/9/201616 Why is Abar of stage-based model smaller than Abar from LHT? Stage-based models are speeding juveniles through the juvenile stage. A 3x3 stage-based model will produce pups in 4 years if we start with pups only. (In the LHT it will take 10 yr.) This will lower  and the mean age of the adult females (Abar) at the stable age distribution. This can also be understood by looking at the characteristic function.

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18. 7/9/201618 Returning to LHT or corresponding Leslie matrix Stage-based matrix models are not suitable for the calculation of E-patterns because they provide incorrect Abar. Now we are returning to age-structured models using LHT or corresponding Leslie matrix. The use of the corresponding Leslie matrix presents another problem which again can produce incorrect E- patterns. Fertilities in the LHT versus discounted fertilities in the Leslie matrix are the crux of the problem. Note that the use of our formulas circumvent the problem of incorrect additions of elements in the E-matrix.

19. 7/9/201619 Brazil 2003-Cairns 2010: Current Theory is Wrong Heppell, Caswell, & Crowder (2000);Caswell (2001, p.233) E-patterns using LHT and Leslie matrix should be the same Leslie matrix using post- or pre-breeding should be the same Not acceptable to mix lower- and upper-level parameters Should be able to deal with  < 1 species when using pre-breeding Leslie matrix Solution that eliminates all the above problems: Survival rates in the discounted Fertilities (F ’ s) have to be included when calculating the E-pattern. Therefore, sum of elasticities is 1 + E 1 as per LHT; not 1.0

20. 7/9/201620 Elasticities of vital rates (x) from the Characteristic Function

21. 7/9/201621 E-pattern is determined by  and Abar alone (Mollet and Cailliet 2003 Appendix, slightly modified) E(m) = E 1 = 1/Abar E(S j ) = E 2 =  /Abar [E 2 /E 1 =  ] E(S a ) = E 3 = (Abar -  )/Abar [E 3 /E 2 = (Abar/  ) -1] Normalization needed for graph using E-triangle: E n (m) = E n,1 = 1/(Abar + 1) E n (S j ) = E n,2 =  /(Abar + 1) E n (S a ) = E n,3 = (Abar -  )/(Abar + 1)

22. 7/9/201622 Abar, why not used in Caswell (2001) Abar was not used in the formulation of E– patterns of age-structured animals. Why? It is simple: Abar and  1 are not well-defined for plants with complex reproductive cycles that require stage-based matrix population models.

23. 7/9/201623 Elasticity triangle for hypothetical species with  = 1,2, 5, & 15 yr Demonstration of errors in E-patterns if survival rates in the discounted fertilities are not included We are going to look at Elasticity triangle using hypothetical species with alpha = 1, 2, 5, and 15 year Vital rates used were:  = 1 yr,  = 5 yr, S1 = 0.50, Sj = 0.50, Sa = 0.60 m = 2  = 2 yr,  = 10 yr, S1 = 0.70, Sj = 0.70, Sa = 0.95, m = 2  = 5 yr,  = 25 yr, S1 = 0.87, Sj = 0.87, Sa = 0.95, m = 2  = 15 yr,  = 75 yr, S1 = 0.50, Sj = 0.90, Sa = 0.95, m = 2

24. 7/9/201624 LHT for hypothetical species with  = 1 and  = 5 yr

25. 7/9/201625 Leslie matrices for hypothetical species with  = 1 and  = 5 yr and corresponding E-matrices

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27. 7/9/201627 Acknowledgements Mybar helped to sort out Abar