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Oxidation of palmitate

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Estimate the amount of energy a mammal can obtain from the oxidation of palmitate, in the liver under conditions of ketosis, relative to the amount obtainable under conditions of a balanced diet with adequate consumption of glucogenic fuels.

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When palmitate is oxidized to acetyl-CoA in conditions of a balanced diet, etc. all of the acetyl-CoA is run through the TCA and oxidized completely to CO2. The metabolic process for oxidizing the palmitate (and other fatty acids) is known as beta-oxidation. Each cycle of beta-oxidation removes one two-carbon fragments and oxidizes it to acetyl-CoA.

In order to prepare the palmitate for beta-oxidation, it needs to be "activated" by condensation with CoA. One ATP is converted to AMP in this reaction. The AMP then reacts with another ATP to produce two ADPs. Therefore, two ATPs are required for this process. We need to substract these two from the total obtained in oxidizing palmitate to find out how much energy can be obtained from palmitate.

The complete oxidation of palmitoyl-CoA leads to the synthesis of 131 molecules of ATP, yet 2 ATPs were required to activate palmitate, therefore, the total production is ...

Solution Summary

This solution is provided in 655 words. It discusses what happens to palmitate in the body, and the process of ketosis.

See Also This Related BrainMass Solution

Tripalmitoylglycerol Oxidation

Contrary to the legend camels do not store water in their humps, which actually consist
of a large fat deposit. How can these fat deposits serve as a source of water?
Calculate the amount of water (in liters) that can be produced by the camel from
1 kg (0.45 lb) of fat. Assume for simplicity that the fat consists entirely of

To solve this problem, I have to calculate the number of moles of water per mole of tripalmitoylglycerol oxidised.
I found that molecular weight ot tripalmitoylglycerol is 809, thus 1 kg (1000 g) will give 1.23 moles of it.
Then, I was looking for the equation telling how much of water is produced, and had found this:
Palmitoyl-CoA + 23O2 + 131Pi +131ADP => CoA + 131 ATP + 16CO2 + 146H2O
And also another one:
Palmitoyl-CoA + 23O2 + 108Pi + 108 ADP => 88nCoA + 108 ATP + 16CO2 + 23 H2O

First, I do not know which equation is right.
Second, in both cases, I get unreasonable small amount of water.
For example, if solve for the first one:
1000g / 809g/mol = 1.23 mole of tripalmitoylglycerol.
Then I multiply it by 146 moles of water and by 3, since 1 tripalmitoylglycerol has 3 palmitoyl molecules.
That gives me: 541 moles of water. I divide this number on 18(molecular weight of water) and receive about 30 g, or 30 ml of water per kilogram of fat.
That seems to be too small.
The right answer in the book (Principles of Biochemistry by Lehninger, Nelson, Cox) is 0.49L of water per pound.

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