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. 2013 Apr 15;8(4):e61813.
doi: 10.1371/journal.pone.0061813. Print 2013.

Life span evolution in eusocial workers--a theoretical approach to understanding the effects of extrinsic mortality in a hierarchical system

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Life span evolution in eusocial workers--a theoretical approach to understanding the effects of extrinsic mortality in a hierarchical system

Boris H Kramer et al. PLoS One. .

Abstract

While the extraordinary life span of queens and division of labor in eusocial societies have been well studied, it is less clear which selective forces act on the short life span of workers. The disparity of life span between the queen and the workers is linked to a basic issue in sociobiology: How are the resources in a colony allocated between colony maintenance and reproduction? Resources for somatic maintenance of the colony can either be invested into quality or quantity of workers. Here, we present a theoretical optimization model that uses a hierarchical trade-off within insect colonies and extrinsic mortality to explain how different aging phenotypes could have evolved to keep resources secure in the colony. The model points to the significance of two factors. First, any investment that would generate a longer intrinsic life span for workers is lost if the individual dies from external causes while foraging. As a consequence, risky environments favor the evolution of workers with a shorter life span. Second, shorter-lived workers require less investment than long-lived ones, allowing the colony to allocate these resources to sexual reproduction or colony growth.

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Conflict of interest statement

Competing Interests: The authors have declared that no competing interests exist.

Figures

Figure 1
Figure 1. Hierarchical trade-off model for eusocial species.
Simplified hierarchical trade-off with a focus on workers for eusocial species, including two trade-offs at the colony and the individual levels. Arrows indicate resource flows. Resources are obtained by workers that do not reproduce and are allocated toward worker maintenance (formula image) and/or the colony (1−formula image). At the colony level, resources that are not consumed by workers can be allocated to sexual reproduction (formula image) and/or maintenance (1−formula image), such as the production of new workers or different levels of worker quality.
Figure 2
Figure 2. Model results under different levels of extrinsic mortality.
The horizontal axis represents the values of extrinsic mortality used to run the model(formula image;formula image). A) and B) show the optimized parameters (formula image, formula image,) from our model. In C)–F) the solid lines indicate the results when the optimal values of formula image and formula image, The dashed lines indicate results if the colony did not change the maintenance investments (formula image) or the switching time (formula image) with increasing extrinsic mortality (formula image = 0.38 formula image = 84). A) Optimal investment into workers (formula image) decreases with increasing extrinsic risk. B) Denotes the switching time (formula image), where the colony switches to the production of sexuals. C) The number of sexuals alive at the end of the season (maximized by finding optimal values for switching time (formula image) and maintenance investments into workers (formula image)) decreases with increasing extrinsic mortality. D) Worker mortality combines intrinsic and extrinsic mortality (formula image). The dashed line denotes the increase of extrinsic mortality. The difference between the dashed and solid lines shows the effect of the changing investment in worker maintenance. E) Worker life expectancy at the beginning of the season with different levels of extrinsic mortality. F) Worker life expectancy at switching time. At switching time, the worker population reaches its maximum. The difference in worker life span between e) and f) is due to the reduction of foraged resources caused by density dependency. Used parameters: formula image = 0.15, formula image = 0.005, formula image = 0.0024, formula image = formula image = 0–0.06.

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