The effect of phenotypic variation on metapopulation persistence

Abstract

Demographic stochasticity (due to the probabilistic nature of the birth–death process) and demographic heterogeneity (between-individual differences in demographic parameters) have long been seen as factors affecting extinction risk. While demographic stochasticity can be independent of underlying species traits, demographic heterogeneity may strongly depend on phenotypic variation. However, how phenotypic variation can affect extinction risk is largely unknown. Here, I develop a stochastic metapopulation model that takes into account the effects of demographic stochasticity and phenotypic variation in the traits controlling colonization rates to assess what the effect of phenotypic variation may be on the persistence of the metapopulation. Although phenotypic variation can lead to a decrease in metapopulation persistence under some conditions, it also may lead to an increase in persistence whenever phenotypic mismatch—or the distance between the optimal trait value and the population mean—is large. This mismatch can in turn arise from a variety of ecological and evolutionary reasons, including weak selection or a recent history of invasion. Last, the effect of phenotypic variation has a deterministic component on colonization rates, and a stochastic component on persistence through colonization rates, but both are important to understand the overall effect. These results have important implications for the conservation of threatened species and management practices that may historically have overlooked phenotypic variation as unimportant noise around mean values of interest.

Appendices

Appendix 1

In this appendix, I show how Eq. (4) of the main text can be derived from Eq. (3). Assuming that as $t \to \infty$, the system will go to a stationary distribution $P(n,\infty )$, we can set $\frac{dP(n,\infty )}{dt}\,\, = \,\,0$ and solve:

$$C_{n - 1} P(n - 1,\infty ) + E_{n + 1} P(n + 1,\infty ) - P(n,\infty )\left( {C_{n} + E_{n} } \right)\,\, = \,\,0 .$$

(17)

At n = 0, $C_{ - 1} P( - 1,\infty ) + E_{1} P(1,\infty ) - P(0,\infty )\left( {C_{0} + E_{0} } \right)\,\, = \,\,0$. Because $C_{ - 1} \,\, = \,\,0$ and $E_{0} \,\, = \,\,0$ (see main text), we obtain:

$$P(1,\infty )\,\, = \,\,\frac{{C_{0} }}{{E_{1} }}P(0,\infty )\, .$$

(18)

At n = 1, we obtain:

$$P(2,\infty )\,\, = \,\,\frac{{C_{1} C_{0} }}{{E_{2} E_{1} }}P(0,\infty ) .$$

(19)

So, by recurrence, we obtain:

$$P(n,\infty )\,\, = \,\,\frac{{C_{0} \ldots C_{n - 1} }}{{E_{1} \ldots E_{n} }}P(0,\infty ).$$

(20)

Now, $P(0,\infty )$ can be determined from the normalization condition, $\sum\nolimits_{n = 0}^{N} {P(n,\infty ){\mkern 1mu} {\mkern 1mu} = {\mkern 1mu} {\mkern 1mu} 1}$:

$$P(0,\infty ) + \sum\limits_{n = 1}^{N} {P(n,\infty )\,\, = \,\,1} .$$

(21)

Then, we replace with Eq. (20) to obtain:

$$P(0,\infty ) + P(0,\infty )\sum\limits_{n = 1}^{N} {\frac{{C_{0} \ldots C_{n - 1} }}{{E_{1} \ldots E_{n} }}\,\, = \,\,1} ,$$

(22)

which reduces to,

$$P(0,\infty ){\mkern 1mu} {\mkern 1mu} = {\mkern 1mu} {\mkern 1mu} \frac{1}{{1 + \sum\nolimits_{n = 1}^{N} {\frac{{C_{0} \ldots C_{n - 1} }}{{E_{1} \ldots E_{n} }}} }}.$$

(23)

Appendix 2

In this appendix, I show that by replacing Eqs. (1) and (2) from the main text in Eqs. (4) and (5) we can obtain the stationary distribution in Eq. (6) of the main text. Replacing Eqs. (1) and (2) on Eq. (20), and assuming $\phi \,\, = \,\,1$ we obtain:

$$P(n,\infty )\,\, = \,\,\frac{{c\left( {1 - \frac{1}{N}} \right) \times 2\,c\left( {1 - \frac{2}{N}} \right) \times \ldots \times (n - 1)c\,\left( {1 - \frac{n - 1}{N}} \right)}}{e \times 2e \times \ldots \times n\,e}P(0,\infty ).$$

(24)

Which can be rearranged as follows:

$$\begin{aligned} P(n,\infty )\,\, = \,\,\frac{1}{{e^{n} }}\frac{{\frac{c}{N}\left( {N - 1} \right) \times 2\,\frac{c}{N}\left( {N - 2} \right) \times \ldots \times \frac{(n - 1)}{N}c\,\left( {N - n + 1} \right)}}{1 \times 2 \times \ldots \times n}P(0,\infty ) \hfill \\ \Leftrightarrow \,\,P(n,\infty )\,\, = \,\,\frac{1}{{e^{n} }}\left( {\frac{c}{N}} \right)^{n - 1} \frac{{\left( {N - 1} \right) \times \left( {N - 2} \right) \times \ldots \times \,\left( {N - n + 1} \right)}}{n}P(0,\infty ) \hfill \\ \Leftrightarrow \,\,P(n,\infty )\,\, = \,\,\frac{1}{{n\,e^{n} }}\left( {\frac{c}{N}} \right)^{n - 1} \frac{{\left( {N - 1} \right) \times \left( {N - 2} \right) \times \ldots \times 1}}{{\left( {N - n + 2} \right) \times \ldots \times 1}}P(0,\infty ) \hfill \\ \Leftrightarrow \,\,P(n,\infty )\,\, = \,\,\frac{1}{{n\,e^{n} }}\left( {\frac{c}{N}} \right)^{n - 1} \frac{\varGamma (N)}{\varGamma (N - n + 1)}P(0,\infty ). \hfill \\ \end{aligned}$$

(25)

By replacing $P(0,\infty )$ by Eq. (23) we obtain Eq. (6) of the main text.