Question

You are studying aging in two species of fish. You bring them into a common laboratory environment free from factors causing extrinsic mortality and find that species A lives twice as long as species B. Based on the evolutionary theory of aging, you would predict that?

1) Species B has lower extrinsic mortality in its natural habitat than species A
2) Species B reproduces later in life than species A
3) The average age of individuals in wild populations of species B is higher than species A
4) Species B has higher extrinsic mortality in its natural habitat than species A
5) A and C are correct

Answer 1

Based on the evolutionary theory of aging, species B has higher extrinsic mortality in its natural habitat than species A (option 4).

Step-by-step explanation:

Based on the evolutionary theory of aging, the correct prediction is that species B has higher extrinsic mortality in its natural habitat than species A. The theory suggests that species with shorter lifespans, like species B, allocate less energy towards maintenance and defense mechanisms, resulting in higher extrinsic mortality rates. In contrast, species with longer lifespans, like species A, allocate more energy towards these mechanisms, leading to lower extrinsic mortality rates.

This difference in mortality rates contributes to the discrepancy in lifespan observed between the two species.The high extrinsic mortality of species B would select for traits that lead to an early age of reproduction, thus compensating for the increased risk of not surviving to reproduce.

This suggests that species B has adapted to a high-risk environment by evolving to reproduce quickly, foregoing long-term survival for immediate reproductive success.