Fatty acid oxidation

[DeathandTaxes]

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How many ATP would be produced if a 16-carbon fatty acid were completely oxidized to CO2 and H2o?
A. 28
B. 35
C. 106
D. 108

Relevant information from passage:


The equation for the cycle is: Cn- acyl-CoA +FAD + NAD + H2O+ CoA = Cn-2-acyl-CoA +FADH2+NADH + Acetyl CoA + H+

Citric acid cycle yields 3 NADH + 1 FADH2 per acetyl-CoA

The answer says that you would have 7 turns (why not 8?) and this would produce 7 NADH, 7 FADH2, and 8 acetyl CoA (why not 7?). Then you use 1.5 * number of FADH2, and 2.5 * NADH to get 108 ATP, and you subtract 2 from the ATP needed to initiate the reaction.

Can anyone explain the questions I have in paranthesis?

 

[quicheduty]

You have 7 turns because in the last turn, you yield two molecules of acetyl-CoA, making a total of 8. But you only make NADH/FADH2 when you go through the B-ox mechanism, so 7 for those.
Think of it like this: you have a molecule that's x--x--x and you want to break it down just into single x's
How many times will you break a bond, aka have a "turn"? Two, not three. 2 "turns" will yield your 3 x's.
Hope that helps.

 

[Czarcasm]

How many ATP would be produced if a 16-carbon fatty acid were completely oxidized to CO2 and H2o?
A. 28
B. 35
C. 106
D. 108

Relevant information from passage:


The equation for the cycle is: Cn- acyl-CoA +FAD + NAD + H2O+ CoA = Cn-2-acyl-CoA +FADH2+NADH + Acetyl CoA + H+

Citric acid cycle yields 3 NADH + 1 FADH2 per acetyl-CoA

The answer says that you would have 7 turns (why not 8?) and this would produce 7 NADH, 7 FADH2, and 8 acetyl CoA (why not 7?). Then you use 1.5 * number of FADH2, and 2.5 * NADH to get 108 ATP, and you subtract 2 from the ATP needed to initiate the reaction.

Can anyone explain the questions I have in paranthesis?

I admit I struggled with this when I first learned it too. Hopefully this might help a bit. First, a FA is cleaved from a TAG. In order to enter the mitochondria, the FA must be modified. CoA is attached to the FA (which consumes 1 ATP in the process). This forms an Acyl-CoA. The Acyl-CoA, now in the mitochondria, is ready to be oxidized by FAD+ and NAD+.

IN MITOCHONDRIA:

First, FAD+ comes in and steals two hydrogens and two electrons from the c-c bond in red. So consider a chain of carbons: (...)H2C-H2C-H2C-HC=O(CoA. This forms an alkene, which is then hydrated by H2O to form an alcohol. The alcohol then gets oxidized to reduce NAD+ to NADH (and form a ketone in the process). So at this point, we formed 1 FADH2 and 1 NADH2. Then another CoA comes in and displaces the first acetyl-CoA from the chain, leaving a shorted acyl-CoA (by two carbons). And this process continues, until 7 more acetyl-CoA are produced. I guess it might helpful to realize that first acetyl-CoA didn't really produce any energy. It was just there and released once a new ketone was formed. Every acetyl-CoA formed after that resulted in the production of NADH and FADH2. So the fact that there is 7 cycles and not 8 is why we produce 7 NADH and 7 FADH2 during the formation of acetyl-CoA groups.

CITRIC ACID CYCLE:

The second thing to realize is that in total, we have 8 acetyl-CoA, and these molecules will enter the Citric Acid Cycle to produce more energy (3 NADH and 1 FADH2 per).

CALCULATION:

It then becomes a math problem:

1 ATP is consumed to form Acyl-CoA
7 NADH and 7 FADH2 produced during Acetyl-CoA formation
8 Acetyl-CoA formed enter CAC to produce: 8(3 NADH) + 8(1 FADH2) = 24 NADH + 8 FADH2
Don't forget 8 ATP is produced during the CAC.

So in total we have:
31 NADH (x2.5 ATP) = 77.5 ATP
15 FADH2 (x1.5 ATP) = 22.5 ATP
+8 ATP
= 108 ATP Produced (but the net ATP produced is actually 107 ATP, because 1 was consumed to form Acyl-CoA).