Answer:
Total number of ATP molecules generated from a 32-carbon fatty acid = 206 ATP molecules
Explanation:
A 32 carbon fatty acid which undergoes complete beta-oxidation assuming that the fatty acid is fully saturated will pass through the beta-oxidation cycle 14 times to produce the following:
15 molecules of acetylCoA, 14 molecules of FADH₂, and 14 molecules of NADH.
Each of the 15 acetylCoA molecules can be further oxidized in the citric acid cycle to yield the following: 15 × 3 NADH; 15 × 1 FADH₂, and 15 ATP molecules from the substrate level phosphorylation occuring at the succinylCoA synthetase catalyzed-reaction.
Total FADH₂ produced = 15 + 14 = 29 molecules of FADH₂
Total NADH produced = 45 + 14 = 59 molecules of NADH
The FADH₂ and NADH will each donate a pair of electrons to the electron transfer flavoprotein and mitochondrial NADH dehydrogenase respectively of the electron transport chain, and about 1.5 and 2.5 molecules of ATP are generated respectively when these electrons are transfered to molecular oxygen.
Thus, number of molecules of ATP generated by 29 molecules of FADH₂ = 1.5 × 29 = 43.5 molecules of ATP.
Number of molecules of ATP generated by 59 molecules of NADH = 2.5 × 59 = 147.5
Sum of ATP generated from FADH₂ and NADH = 43.5 + 147.5 = 191 ATP molecules
Total number of ATP molecules generated = 191 + 15 = 206 ATP molecules
Total number of ATP molecules generated from a 32-carbon fatty acid = 206 ATP molecules
Answer:
Here you go!
Explanation:
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There is a dark energy that is making the Universe expand and accelerate at a larger rate than it did many years ago. ...
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According to the scientists the planets, stars and galaxies include only 4% of what a Universe consists of. 96% of the things in the Universe cannot be seen.
Answer:
0 g.
Explanation:
Hello,
In this case, since the reaction between methane and oxygen is:
If 0.963 g of methane react with 7.5 g of oxygen the first step is to identify the limiting reactant for which we compute the available moles of methane and the moles of methane consumed by the 7.5 g of oxygen:
Thus, since oxygen theoretically consumes more methane than the available, we conclude the methane is the limiting reactant, for which it will be completely consumed, therefore, no remaining methane will be left over.
Regards.