Why does a fatty acid have a greater energy content than does carbohydrate (per

Question

Why does a fatty acid have a greater energy content than does carbohydrate (per gram or per C atom -- the concept does not differ, only the values of the energy)? The answer has to do with energy released in combustion -- whether that combustion is aerobic metabolism in a living organism or chemical combustion in a bomb calorimeter.

secondly...

1. Write all individual reactions of the pentose phosphate pathway. My feeling is that you have to start with converting 3 (three) glucose-6-phosphate to 3 ribulose-5-phosphate -- because ribulose-5-phosphate is used as reactant in three separate reactions.

You should have exactly as many reactions as there are arrows as in the diagram in the Figure. NOTE: a double arrow is just one reaction. So, you should have 7 reactions, by my count. do your best and see what you get! So the question is (to be answered by your overall or net reaction) is: what are the products of the pathway, starting with 3 glucose-6-phosphate?

2. In approaching this creation of a net reaction, use the same steps as in previous "overall/net reaction" problems to obtain an overall reaction for the scheme shown in Figure 26.1.

3. Finally, from this overall reaction -- just the overall reaction itself -- comment on this. Is the process shown in Figure 26.1 an anabolic or catabolic process (or perhaps both? Remember, the citric acid cycle is called anaplerotic and can be -- can function in -- both anabolism and catabolism. Is this the case for the process in Figure 26.1?)

Explanation / Answer

First of all I would like to say that I can see none of the images that you might have uploaded so I would request you to re-upload them in a separate question if you can.

Then coming to your question, complete oxidation of Fatty acids to CO2 and H2O takes place in 3 stages:
* Oxidation of long-chain FAs to two-carbon fragments, in the form of acetyl-CoA ( oxidation).
* Oxidation of acetyl-CoA to CO2 in the citric acid cycle.
* Transfer of electrons from reduced electron carriers to the mitochondrial respiratory chain.
Now each molecule of acetyl-CoA yields approximately 10 molecules of ATP... which is massive! FAD and NADH are also produced which yields around 1.5 and 2.5 molecules of ATP each respectively.
For a 16-Carbon chain fatty acid you get 8 acetyl-CoAs, 7 FADs and 7 NADHs whchi produce a total of 108 ATPs minus the 2 ATPs that are required in the preparatory stage.

Comparing this to glycolysis, where a glucose mole (6-Carbon) only 2 NADHs are produced while 2 ATPs are required in the preparatory step. So net yield is only 2 ATPs. This energy yield is way lower than beta-oxidation of fatty acids.

And then to answer your question about the pentose phosphate pathway:
Step 1: Glucose-6-phosphate Dehydrogenase catalyzes oxidation of the aldehyde (hemiacetal) at C1 of glucose-6-phosphate, to a carboxylic acid, in ester linkage (lactone-a cyclic ester). NADP+ serves as electron acceptor.

Step 2: 6-Phosphogluconolactonase (lactonase) catalyzes hydrolysis of the ester linkage, resulting in ring opening. The product is 6-phosphogluconate. Although ring opening occurs in the absence of a catalyst, lactonase speeds up the reaction, decreasing the lifetime of highly reactive & potentially toxic, 6-phosphogluconolactone.

Step 3: Phosphogluconate Dehydrogenase catalyzes oxidative decarboxylation of
6-phosphogluconate to yield 5-C ketose ribulose-5-phosphate ("ketose version" of ribose-5-P).The OH at C3 (C2 of product) is oxidized to a ketone. This promotes loss of the carboxyl at C1 as CO2. NADP+ serves as oxidant.

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