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QveST [7]
2 years ago
14

More ATP per carbon is produced from oxidation of fatty acids as compared to glucose. However, fatty acids are used primarily to

provide fuel for ATP synthesis during long periods of exercise. Why can't fatty acids be used to fuel short, intense periods of exercise?
A. Short, intense periods of exercise preferentially activate ATP synthesis from glucose.
B. Short, intense periods of exercise do not signal release of fatty acids from adipose tissue.
C. Short, intense periods of exercise do not require ATP synthesis.
D. Short, intense periods of exercise do not provide sufficient time for oxidative phosphorylation to occur.
Chemistry
1 answer:
Luda [366]2 years ago
8 0

Answer:

The correct option is D

Explanation:

Although, option A appears correct but the reason for option A is actually option D. What this means is that, during short intense exercise (anaerobic exercise), the body preferentially activates ATP synthesis from glucose (glycolysis) even though the beta oxidation of fatty acid produces more energy (ATP) - this is because <u>glycolysis occurs many times faster (about a 100 times faster) than the beta oxidation of fatty acid/oxidative phosphorylation</u>. Thus, it can be said that short, intense periods of exercise do not provide sufficient time for oxidative phosphorylation to occur.

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Answer:

ν = 5,765x10¹⁴ Hz

λ = 520 nm

230,0 kJ/mole

Explanation:

To convert energy yo frequency you need to use:

E = hν

Where E is energy (3,820x10⁻¹⁹ J)

h is Planck's constant (6,626x10⁻³⁴ Js)

And ν is frequency, replacing ν = 5,765x10¹⁴ s⁻¹ ≡ 5,765x10¹⁴ Hz

To convert frequency to wavelength:

c = λν

Where s is speed of light (2,998x10⁸ ms⁻¹)

ν is frequency (5,765x10¹⁴ s⁻¹)

And λ is wavelength, replacing: λ = 5,200x10⁻⁷ ≡ 520 nm

If 1 photon produce 3,820x10⁻¹⁹ J, in mole of photons produce:

3,820x10⁻¹⁹ J ×\frac{6,022x10^{23}}{1 mole} = 230040 J/mole ≡ 230,0 kJ/mole

I hope it helps!

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