For aerobic metabolism, the main pathway for ATP production in eukaryotes, there are two substrates which compete with each other: glucose and fatty acids. Amino acids can also be used but are not (or should not) be an important source of fuel. For knowing which fuel is more efficient (giving the higher amount of energy per mole, joule per mole) we must review the pathways for glucose and fatty acid catabolism before they converge into the Krebs Cycle.
Glucose oxidation, ie. glycolysis, is the pathway for oxidizing glucose to pyruvate. The preparatory phase of glycolysis requires the "investment" of two molecules of ATP. This is "payed off" in the following steps, producing 4 ATPs. So the net gain of the conversion glucose --> pyruvate is 2 ATPs. This process also produces 2 NADH+. Pyruvate must be converted to acetyl CoA, producing one more NADH+ and CO2. So the total oxidation of one molecule of glucose to acetyl CoA yields:
1 glucose molecule to 2 pyruvates: 2 NADH+
2 pyruvates to 2 Acetyl CoA: 2NADH+, 2CO2
Total = 4 NADH+ plus 2CO2
On the other hand, fatty acid oxidation is relatively simpler. Once triglycerides are hydrolyzed, free fatty acids are taken up by cells. In the cytosol, they are esterified to conezyme A to form a fatty acyl CoA. The fatty acyl group is transferred to carnitine and moved across the inner mitochondrial membrane by an acylcarnitine transporter protein. Once in the matrix side, the fatty acyl group is released from carnitine and reattached to another CoA molecule. The mitochondrial fatty acyl CoA is oxidized in a series of reactions to produce Acetyl CoA. The conversion of a fatty acid to acetyl CoA yields (taking palmitate as an example):
1 palmitate to 1 palmitoyl CoA: 2 ATP
1 palmitoyl CoA to 8 Acetyl CoA: 7 FADH2 + 7 NADH+
Overall, in these preliminary reactions before entering the Krebs Cycle we have that 1 molecule of glucose can yield 2 Acetyl CoA while 1 molecule of palmitate can produce 8 Acetyl CoA. Oxidation of glucose until this step produces 2 molecules of CO2 and 4 NADH+, while oxidation of palmitate produces an equal number of FADH2 and NADH+, 7.
Once acetyl CoA reacts with oxaloacetate, forming citrate, it enters the Krebs Cycle. This pathway produces 3 NADH+, 1FADH2, 1 GTP and 2CO2. So, overall we have:
1 molecule of glucose produces:
2 Acetyl CoA
6 CO2
10 NADH+
2 FADH2
Ratio NADH+:FADH2 = 5:1
ATPs produced from complete oxidation: 30-32 (assuming 2.5 ATP from NADH+ and 1.5 ATP from FADH2)
1 molecule of palmitate produces:
8 Acetyl CoA
16 CO2
31 NADH+
15 FADH2
Ratio NADH+:FADH2 = 2:1 (depending on carbon length)
ATPs produced from complete oxidation: 108 (assuming 2.5 ATP from NADH+ and 1.5 ATP from FADH2)
The final electron acceptors are the coenzymes NADH+ and FADH2 which transfer electrons to the mitochondrial respiratory chain. Understanding this process is of utmost importance for understanding the differences between glucose and fatty acid metabolism. For a basic tutorial in oxidative phosphorylation please read this.
Electrons carried by NADH+ are transferred to Complex I, also called NADH dehydrogenase. Complex II, succinate dehydrogenase, is the same enzyme encountered in the Krebs Cycle, which oxidizes succinate and transfers electrons to FADH2. Studies have shown that Complex I and complex III are the main producers of ROS in the electron transport chain (1, 2, 3, 4)*. So, the ratio of NADH+:FADH2 is important for the potential ROS production in the mitochondria. Because glucose generates 5 times more NADH+ than FADH2, complex I activity is increased. Fatty acid oxidation produces only twice the amount of NADH+ than FADH2, shifting to a more balanced utilization between complex I and complex II. There are other ways to pass electrons into the respiratory chain. In the first step of the beta oxidation pathway, catalyzed by acyl-CoA dehydrogenase, electrons from the substrate are transferred to the FAD of the dehydrogenase, then to electron-transferring flavoprotein (ETF), which in turn passes its electrons to ETF:ubiquinone oxidoreductase. This enzyme reduces ubiquinone directly, without the need for other complexes. Glycerol-3-phosphate derived either from glycerol or from glycolysis is oxidized by glycerol-3-phosphate dehydrogenase and reduces ubiquinone directly.
The overall message is that glucose's complete oxidation involves more complex I activity, ultimately promoting more ROS production by the mitochondria. A more in depth review of the implications of glucose and fatty acid metabolism, and mitochondrial complex utilization in aging and disease can be found in this paper by Mobbs et al. (yes, the same guy from the nephropathy reversal study):
"Therefore shifting away from glucose utilization toward lipid and amino acid utilization would be expected to substantially reduce the production of reactive oxygen species, without necessarily reducing ATP production. As described below, other beneficial effects also occur as a result of this altered pattern of glucose fuel use, including a shift toward producing antioxidizing NADPH and increased protein and lipid turnover, which reduces the accumulation of oxidized protein and lipids."Ok, so we have seen why depending on fat as an energy source seems a better idea than depending on glucose. But what about ketones?
Kurt Harris thinks ketosis is metabolically stressful. Considering that the ketogenesis pathway only demands one molecule of NADH+ and one H2O, I dont see the "stressful" part.
Let's see how ketones alter mitochondrial energy metabolism**. Remember that the respiratory electron transport chain is the process which generates ATP by creating an electrochemical proton gradient. Electrons carried through redox couples are ultimately combined to H+ and O2 to form H2O. The total energy available from the movement of electrons up the respiratory chain is determined by the difference between the variable redox potential of the mitochondrial NAD couple and the O2 couple (this is because the first step in the chain is NADH dehydrogenase and the last step is production of H2O). We know experimentally the reduction potential of both couples (5):
[NAD]/[NADH+] couple: -0.28V
[O2]/[H2O] couple: +1.2V
So we can calculate the total energy yield given by:
∆G' = -nF∆E
Where: n is the number of electrons, F is 96.485 kJ/mol/V (Faraday's constant) and ∆E the difference between potentials. Replacing in the equation with the values we have:
∆G' = -(2)(96.485)(0.92)
∆G' = -177.5324 or -178kJ/2mol
The energy of the proton gradient created by the electron transport chain is given by:
∆G' [H+] cyto/[H+] mito = RTln[H+] cyto/[H+] mito + FEmito/cyto
Where Emito/cyto is the electric potential of the mitochondrial relative to the cytosolic phase.
3 ATPs are generated by the transport of one electron pair through the respiratory chain. Since the maximum energy available from the redox reactions of the chain is -178kJ, the energy available for the synthesis of each ATP cannot be more than -59.2kJ/mol (178/3). Experiments done on different cells have shown that this energy is in the range of -54 to -58kJ/mol (very efficient).
As the electrons liberated by complex I are carried by coenzyme Q to complex III, the difference between the redox potential of the [NAD]/[NADH+] mitochondrial couple and the coenzyme Q couple determines the energy of the proton gradient generated by the mitochondria. This difference determines the energy of hydrolysis of ATP generated by the mitochondrial F1 ATPase.
bOHB and AcAc are in near-equilibrium with the free mitochondrial [NAD]/[NADH+] ratio. Veech and colleages showed that the metabolism of ketone bodies in the heart produced a reduction of the mitochondrial NAD couple and oxidation of the coenzyme Q couple, increasing the redox span between the two couples, making more energy available for ATP synthesis because:
Where Emito/cyto is the electric potential of the mitochondrial relative to the cytosolic phase.
3 ATPs are generated by the transport of one electron pair through the respiratory chain. Since the maximum energy available from the redox reactions of the chain is -178kJ, the energy available for the synthesis of each ATP cannot be more than -59.2kJ/mol (178/3). Experiments done on different cells have shown that this energy is in the range of -54 to -58kJ/mol (very efficient).
As the electrons liberated by complex I are carried by coenzyme Q to complex III, the difference between the redox potential of the [NAD]/[NADH+] mitochondrial couple and the coenzyme Q couple determines the energy of the proton gradient generated by the mitochondria. This difference determines the energy of hydrolysis of ATP generated by the mitochondrial F1 ATPase.
bOHB and AcAc are in near-equilibrium with the free mitochondrial [NAD]/[NADH+] ratio. Veech and colleages showed that the metabolism of ketone bodies in the heart produced a reduction of the mitochondrial NAD couple and oxidation of the coenzyme Q couple, increasing the redox span between the two couples, making more energy available for ATP synthesis because:
∆G = -nF∆EQ/NADH
In this classic study (6), the authors found that***:
"The Eh7 of the mitochondrial NAD couple, Eh7 NAD+/NADH (Eq. 16), was -280 mV and decreased to about - 300 mV on addition of insulin, ketones, and the combination.
The ratio [fumarate2-]/[succinate2-] increased about 1.5-fold on addition of insulin, about 2.5-fold on addition of ketones, and about 4-fold on addition of the combination (Fig. 1). Taking this change to indicate the redox state of the [Q]/[QH2] couple, which is the cofactor for the succinate dehydrogenase (EC 1.3.5.1) reaction, the observed increase in [fumarate2-]/[succinate2-] indicates a 1.5- to 4-fold oxidation of mitochondrial [Q]/[QH2] on addition of insulin, ketones, or the combination. The Eh[Q]/[QH2], calculated from the succinate dehydrogenase reaction (Eqs. 20-22, Fig. 2), increased progressively from -4 mV during perfusion with glucose alone to + 15 mV on the addition of insulin and ketones.
When the oxidation of the mitochondrial Q couple was combined with the reduction of the mitochondrial NAD couple resulting from these additions to the glucose perfused hearts, the estimated energy available in the transfer of 2 e from the mitochondrial NAD to the Q couple (Eq. 23 and Eq. 24) catalyzed by the NADH dehydrogenase multienzyme complex, ∆G QH2/NAD+, increased from -53 kJ/2 mol e during perfusion with glucose alone to -60 kJ/ 2 mol e with addition of insulin and ketones. This increase was paralleled by an increase in the cytosolic free energy of ATP hydrolysis, ∆GATP (Eq. 7 and Eq. 8), determined independently using NMR spectroscopy, which increased from -56 kJ/mol in control hearts to about -59 kJ/mol on addition of insulin or insulin plus ketones.
The potential between mitochondrial and cytosolic phases, E mito/cyto (Eq. 9, Fig. 2, Table 3), was -143 mV in hearts perfused with glucose alone and unchanged on the addition of insulin but decreased to -120 mV on addition of ketones or -130 mV on addition of the combination. "Summarizing (from the study data)****:
For ketones (without insulin):
Eh7 NAD+/NADH = -299mV or -0.299V
Eh7 Q/QH2 = +7.8mV or 0.078V
Cytosolic pH = 7.05
Mitochondrial pH = 7.52
∆G = -nF∆E
∆G = -(2)*(96.485)(0.015-(-0.3))
∆G = -59.2 kJ/2mol
For glucose:
Eh7 NAD+/NADH = -280mV or -0.28V
Eh7 Q/QH2 = -0.4mV or -0.004V
Cytosolic pH = 7.05
Mitochondrial pH = 7.09
∆G = -nF∆E
∆G = -(2)*(96.485)(-0.004-(-0.28))
∆G = -53 kJ/2mol
Recall that the energy of the proton gradient is calculated by:
∆G' [H+] cyto/[H+] mito = RTln[H+] cyto/[H+] mito + FEmito/cyto
Expressing ∆G in terms of pH instead of concentration, we get:
∆G' [H+] cyto/[H+] mito = 2.3RT (∆pH) + FEmito/cyto
where, R is the gas constant (8.315 x 10^-3 kJ/mol.K) and T the absolute temperature (in this study, 311.15K).
For ketones, Emito/cyto = -0.12V, so:
∆G = 2.3 (8.315*10^-3)(311.15) (∆pH) + (96.485)(-0.12)
∆G = (5.9 kJ/mol) (∆pH) + (-11.52)
∆G = (5.9 kJ/mol) (7.05-7.52) + (-11.52)
∆G = -14.2 kJ/mol
For glucose, Emito/cyto = -0.14V, so:
∆G = 2.3 (8.315*10^-3)(311.15) (∆pH) + (96.485)(-0.14)
∆G = (5.9 kJ/mol) (∆pH) + (-13.5)
∆G = (5.9 kJ/mol) (7.05-7.09) + (-13.5)
∆G = -13.7 kJ/mol
The mean calculated ∆G of cytosolic ATP hydrolysis was:
∆G ATP = -57.6 kJ/mol for ketones
∆G ATP = -56.6 kJ/mol for glucose
Concluding from this data we see that ketones cause a decrease in the potential between the mitochondria and cytosol (ie. Emito/cyto) while increasing the ∆G of ATP hydrolisis, paralleled by the increase in ∆G QH2/NAD+ and ∆G[H+]. Increased efficiency.
But why bOHB is such a "superfuel"? This can be illustrated in the following table (7):
The table above shows the heats of combustion for different nutrients metabolized by the Krebs cycle. As shown, there is a wide variation in the inherent energy of different C2 units which enter the cycle, derived from different sources. For instance, as we have seen, one molecule of pyruvate produces 15 ATP. The ∆G' of ATP hydrolisis cannot exceed 59-60kJ/mol and the range of variation in cells is very narrow, less than 10% (-53 to -60kJ/mol). So, by simple maths, for producing 15 ATP, -795 to -900 kJ/15 moles are needed (15*-53; 15*-60). There are only -778kJ available from pyruvate, so the ∆G' of the ATP produced is of low energy (-778/15 = -51kJ/mol). C2 from fatty acids (ie. palmitate) have too much inherent energy to be accomodated in the 15 ATPs, metabolically implausible (-1247/15 = 83). If we look at the table, bOHB has -1021kJ/C2, only 13% excess energy (-1021/15=68). Compared to pyruvate, bOHB is more reduced (ie. more H+ per C).
When interpreting this data, one obvious question arise. If the inherent energy of palmitic acid is greater than bOHB, why does palmitic acid is not a "superfuel"? The answer lies in the nature of beta oxidation. If we remember the sites for entering the respiratory chain, we can see that some reducing equivalents produced by the oxidation of fatty acids enter the chain via the ETF:ubiquinone oxidoreductase, resulting in a loss of approximately 1 of the 6 possible ATP produced, because of its redox potential (8). The remaining reducing equivalents are metabolized in the Krebs cycle normally, but in contrast with metabolizing bOHB, the Q couple is reduced and not oxidized, decreasing ∆E available for ATP synthesis. Last but not least, free fatty acids produce enzymatic/transcriptional changes which account for the difference in efficiency between palmitic acid and bOHB. Elevation of plasma FFA induce the expression of uncoupling proteins (9, 10) and peroxisomal beta-oxidation (11).
Summing up
The response Dr. Harris gave me a couple of months ago included some statements like:
"It [bOHB] is only efficient if you think the steps it took to get it don't count."
""Glycolysis" is anaerobic and only happens by itself when speed is necessary - like weight lifting."
"Ketosis is an adaptive state that beats the alternative of glycogen depletion or muscle wasting that it is designed to counter. This makes it "good". In the same way, muscle wasting to keep your brain supplied with glucose is "good". Good in the sense of beating the alternatives. That does not mean we should seek to live continuously in either state, though does it?"
I hope that after this post (and after comparing the "alternatives"), my position on the subject is clear.
* This is a subject of much debate. Studying ROS production by mitochondria under physiological conditions is not an easy task. Ultimately, every complex is capable of producing ROS.
** This explanation is summarized from the referenced article by Veech. A nice article on the subject can be found here.
*** Eh stands for the redox potential of a half reaction. Eh7 = Eh at pH 7.
****Im omitting some units for the sake of clarity.
Sato K, Kashiwaya Y, Keon CA, Tsuchiya N, King MT, Radda GK, Chance B, Clarke K, & Veech RL (1995). Insulin, ketone bodies, and mitochondrial energy transduction. The FASEB journal : official publication of the Federation of American Societies for Experimental Biology, 9 (8), 651-8 PMID: 7768357
Okay, that was a lot to try and go through--good work.
ReplyDeleteHow does the greater ∆G of cytosolic ATP hydrolysis for ketones relate to uncoupling?
Do your conclusions revolve around the ideas of Nick Lane? How does the mitochondrial mutator mouse fit in?
Hello john,
ReplyDeleteBy making ketones out of excess acetyl CoA, uncoupling produced by high FFA might be counterbalanced so most energy is not lost as heat. I havent read the work of Nick Lane deeply, but agree with most of what i have read.
Mutator mice seem to have lower ATP production and lower membrane potential by dysfunctional mitochondrial complexes, not related to increased ROS. From the data I have read, it seems that increased ROS production might increase mtDNA mutations, some which affect complex assembly and energy production, impairing mitochondrial and cell bioenergetics.
Lucas I see your comment is gone. Anyway, are we sure that a metabolism that "wastes" energy as heat is a bad thing? Many long-lived animals generate high heat per unit (cell, gram of body weight) and have high uncoupling. I was under the impression that this was true on keto--is it more so on "optimal diet," being just out of keto? We would be getting max fat oxidation without high ketones; but, Peter has posted a paper showing ketones raise uncoupling ("Metabolism Nuts and Bolts"), and I think there is enough evidence that points towards them being beneficial, which is in agreement with you anyway.
ReplyDeleteYes, blogger ate my comment. I dont see any problem with energy dissipation as heat, it is required for the body homeostasis. The problem is when excess energy is dissipated as heat, ie. lack of ATP (think about mitochondrial uncouplers). I havent seen any direct evidence for ketones increasing UCP expression, only FFA.
ReplyDelete"I havent seen any direct evidence for ketones increasing UCP expression, only FFA."
ReplyDeleteMy mistake, I found the same when I re-checked.
Is the CoA usually synthesized or donated from pyruvate during the Krebs cycle? If the latter, would it make CHO/FA comparisons more complex?
ReplyDeleteTravis,
ReplyDeleteThe Krebs Cycle "starts" with the condensation of acetyl CoA and oxaloacetate. PDH catalyzes the formation of acetyl CoA from pyruvate. This step produces one more reduced NAD.
Downstream from acetyl CoA, the metabolism of both fatty acids and glucose is nearly the same. The differences are between glycolysis and beta-oxidation.
I mean the CoA itself that is cleaved off of pyruvate. I'm wondering if that CoA that is used to create acyl CoA is usually synthesized or is from glucose metabolism before the Krebs Cycle begins.
ReplyDeleteAcetyl CoA is also formed by beta-oxidation. Acetyl CoA is formed by CoA (Coenzyme A) + an acetyl group. CoA synthesis is independent from glucose metabolism.
ReplyDeleteThe phenomena which make it difficult to build conclusions on this, even after how thoroughly you present your case, is based on the fact that we, as homo sapiens, human, or whatever, have such meager knowledge when it comes to the big picture. Could be in 5-10 years that a new revelation is made and makes it very clear that ketosis is actually extremely stressful, you never know. The fact that the only really healthy population eating a keto diet are the inuit, and given the rather crazy selection of foods they eat I think it's extremely prudent to dip into ketosis on an intermittent basis.
ReplyDeletePeter (www.high-fat-nutrition.blogspot.com) had the following comment:
ReplyDeleteLucas, I have just spent several hours working through Veech's 1995 paper here in the early hours of the morning. I did a quick google to get a clearer idea of what, exactly, Eh7 was and landed on this post of yours as first clickable hit. I will pack up and go home now! Great post.
BTW Liz fwded me this paper by Veech's group, getting a bit beyond physiological ketosis here, but interesting never the less. It's what sent me back to the 1995 paper:
http://www.ncbi.nlm.nih.gov/pubmed/22362892
I have the full text, if you would like the pdf just email me.
Dawn now, time to go feed the chickens!
Peter
Thanks for the link! Veech's group has been working on that for a while and if the KE makes it, there should be no need to eat a very restricted ketogenic diet for achieving the benefits of ketosis. Lets see how it goes.
DeleteLucas your work is fabulous.
ReplyDeleteThank you for this.
ReplyDelete