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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
tripalmitoylglycerol.

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.

Solution Preview

The oxidation of palmitoyl-CoA involves the following generalized reaction:

Palmitoyl-CoA + 23 O2 + 131 Pi + 131 ADP ----> CoA + 131 ATP + 16 CO2 + 146 H2O

We can't use all of the 146 moles of water formed because most of ...

Solution Summary

The solution examines tripalmitoylglycerol oxidation.

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