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Alina [70]
4 years ago
4

Which pathway would be used during a one minute, steep climb?

Biology
2 answers:
Dafna1 [17]4 years ago
6 0

Answer;

Glucose metabolism (anaerobic)

The pathway that would be used during a one minute, steep climb is anaerobic glucose metabolism.

Explanation;

-Anaerobic glycolysis does not require oxygen and uses the energy contained in glucose for the formation of ATP. This pathway occurs within the cytoplasm and breaks glucose down into a simpler component called pyruvate.

Murrr4er [49]4 years ago
5 0

Answer:

The best answer to this question: Which pathway would be used during a one minute, steep climb, would be: glycolysis, or anaerobic pathway.

Explanation:

The body has different ways in which it can extract the necessary energy from such molecules, as glucose, to power different activities. While for the most part the body requires oxygen in order to be able to exact the most energy possible from the chemical manipulation of glucose, this aerobic pathway is not the only resource the body has.

When a circumstance, or situation, requires us to move, and our muscles and other organs to receive the necessary energy for such a thing, the body counts with a specific pathway that comes right before the aerobic metabolism, and which does not depend on the presence of oxygen. This is known as glycolysis, or the anaerobic pathway. Although short-lived, and used only at the beginning to power initial movement of muscles, glycolysis is vital to initiate all metabolic processes in the body, and provides enough energy to the body when an action, or circumstance, requires immediate response from it, like is suddenly going up a steep incline for one minute.

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]To calculate the amount of heat entering or leaving a system, the equation  

Q

=

m

c

Δ

T

is used.

Explanation:

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ΔT = change in temperature (°C)

Here, we will use the specific heat capacity for liquid water which is 4.19 J/g°C.

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As for the change in temperature, I will assume that it start off at room temperature, 25°C.

25

°

C

−

0

°

C

=

25

°

C

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=

m

c

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=

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g

r

a

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