Trade-off between tree defenses and reproduction

1. Valeria Aschero Natacha Chacoff Silvina Velez Valdemar Delhey Trade-off between tree defenses and reproduction

2. Biological background Herbivores Induced responses (spines) Reproduction (fruits)

3. Design: • Prosopis flexuosa • Reserve and cattle grazed sites • Response variable=Fruit production/ind • Explanatory variable= Spine length

4. Looking at our data

5. Imagine the deterministic model

6. Eyeball estimates of the Negative Exponential parameters (estimando a ojo) Deterministic model: #fruits = a e -bspine_length Stochastic model: Negative Binomial (counts, variance higher than mean)

7. Fitting the Negative Exponential Model NUMERICAL OPTIMIZATION USING mle2() fEnBn=function(a,b,k){ media=a*exp(-b*espi$spine) -sum(dnbinom(espi$fruits,mu=media,size=k,log=T)) } m1 = mle2(fEnBn, list(a=750,b=0.1,k=1), data=espi, method="Nelder-Mead") Both approaches yielded equal estimates GENERALIZED LINEAR MODEL (negative binomial) Linearized function (log link): Log(# fruits) = Log(a) –b spine_length glmnb=glm.nb(fruits~spine, data=espi)

8. Negative Exponential Fit

9. Fitting the Hyperbolic model with mle2() Hyperbolic model: Fruits= a / (b + spine length) fHyBn=function(a,b,k){ media=a/(b+espi$spine) -sum(dnbinom(espi$fruits,mu=media,size=k,log=T)) } mHy= mle2(fHyBn, list(a=3000,b=10,k=1), data=espi, method="Nelder-Mead")

10. Plot hyperbolic model Which one do you vote?...doodle.com/espina$#@!%$#@!

11. Estimating CI for the hyperbolic model # Generar valores aleatorios de parámetros usando matriz de varianza y covarianza coefazHy=rmvnorm(1000,coef(mHy),vcov(mHy)) sec.esp=seq(0.05,40,leng=100) curvasHy=NULL for(i in 1:length(sec.esp)){ temp2=coefazHy[i,1]/(coefazHy[i,2]+sec.esp) curvasHy=cbind(curvasHy,temp2)} cinfHy=apply(curvasHy,1,quantile, prob=0.025 ) csupHy=apply(curvasHy,1,quantile, prob=0.975 )

12. Confidence intervals of both models (NegExp vs Hyperbolic)

13. But… Comparing models The Hyperbolic is marginally better: Δ AIC= 0.8

14. Using the Neg. Exp. model: Are parameters different in the reserve and cattle grazed sites? “a” PARAMETERS DIFFERENT : glmnb.a<-glm.nb(fruits~spine+situation,data=espi) fEnBntA=function(aC,aR,b,k){ a=c(aC,aR)[espi$situation]; media=a*exp(-b*espi$spine) -sum(dnbinom(espi$fruits,mu=media,size=k,log=T)) } mtA = mle2(fEnBntA, list(aC=375,aR=375,b=0.1,k=1), data=espi, method="Nelder-Mead") “b” PARAMETERS DIFFERENT: glmnb.b<glm.nb(fruits~spine+spine:situation,data=espi) fEnBntB=function(a,bC,bR,k){ b=c(bC,bR)[espi$situation]; media=a*exp(-b*espi$spine) -sum(dnbinom(espi$fruits,mu=media,size=k,log=T)) } mtB = mle2(fEnBntB, list(a=362,bC=-0.09,bR=0.09,k=1), data=espi, method="Nelder-Mead")

15. Is it worthy to use different parameters for trees in the reserve and in cattle grazed sites? NO! LRTest Results: “a” equals vs “a” different df=1, LRstat= 0.18, p=0.66 “b” equals vs “b” different df=1, LRstat= 0.04 , p=0.83

16. To take home: • Fruit production per individual decreases with spine length • Negative exponential and hyperbolic model both could be used to describe the response • We don't have enough evidence to say that the relationship between fruits and spine length differ between protected and cattle grazed sites