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Trade-off Between Tree Defenses and Reproduction: A Statistical Analysis Comparison

Explore the trade-off between tree defenses and reproduction in Prosopis flexuosa using negative exponential and hyperbolic models. Analyze effects of spine length on fruit production across reserve and grazed sites. Statistical comparisons and parameter estimation.

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Trade-off Between Tree Defenses and Reproduction: A Statistical Analysis Comparison

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  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

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