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marshall27 [118]
3 years ago
6

PLZZZ Over a period of time, a behavioral adaptation occurred that altered the mating behavior of some O. glucosi males. A small

number of male beetles no longer challenged other males for access to the tunnel and females. Instead of dueling, these beetles secretly dug their own tunnels to intersect with and enter those dug by females. In this way, these males were able to mate and produce offspring without combat. Explain how this change in reproductive behavior could influence the evolution of the O. glucosi beetle population
Biology
1 answer:
mestny [16]3 years ago
4 0

This phenomenon is not rare in the animal kingdom. Known as alternative mating strategies, it is employed by a subgroup of males that are usually incapable of facing the larger males in direct confrontations.

These "sneaky males" as they are sometimes called, will employ alternative strategies to mate with females.  Which strategy will be dominant is dependent on the environmental factors and is not permanent. In some conditions where food is not that abundant, there will be less larger males and thus, the smaller males employing this strategy will be more common. At other times the opposite will occur.

However, in most cases and most probably here, the end result will be a coexistence of this traditional and alternative mating strategy, where males will exhibit a greater diversity in phenotype.

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Answer:

Explanation:

From the information given:

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By using Goldman's equation:

E_m = \dfrac{RT}{zF}In \Big (\dfrac{P_K[K^+]_{out}+P_{Na}[Na^+]_{out}+P_{Cl}[Cl^-]_{out} }{P_K[K^+]_{in}+P_{Na}[Na^+]_{in}+ P_{Cl}[Cl^-]_{in}}      \Big )

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and the two species be Na⁺ and Cl⁻

E_m = \dfrac{RT}{zF} In \dfrac{[K^+]_{out}}{[K^+]_{in}}

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100 \times 10^{-3} = \Big (\dfrac{8.314 \times 298}{1\times 96485} \Big) \mathtt{In}  \Big ( \dfrac{4}{[K^+]_{in}}   \Big)

100 \times 10^{-3} = 0.0257 \Big ( \dfrac{4}{[K^+]_{in}}   \Big)

3.981= \mathtt{In} \Big ( \dfrac{4}{[K^+]_{in}}   \Big)

exp ( 3.981) = \dfrac{4}{[K^+]_{in}} \\ \\  53.57 = \dfrac{4}{[K^+]_{in}}

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For [Cl⁻]:

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[Cl^-]_{in} = \dfrac{120}{0.01867}

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100 \times 10^{-3} = 0.0257 \Big ( \dfrac{145}{[Na^+]_{in}}   \Big)

53.57= \Big ( \dfrac{145}{[Na^+]_{in}}   \Big)

[Na^+]_{in}= 2.70

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