Question (B-15): Flavoproteins in Krebs Cycle


 

In this representation of Krebs Cycle, each reaction, is marked with a number.

Select the two reactions that require the participation of Flavoproteins.

 

a)     1 and 3

 

b)     2 and 4

 

c)      3 and 5

 

d)     4 and 6

 

e)     5 and 7

 

f)       6 and 8

Bioenergetics Question B-14


 

 

This compound dissipates the electrochemical gradient between the intermembrane space and the mitochondrial matrix, consequently the electron transportation takes place but not the synthesis of ATP

 

a)     Antimycin B

b)     Carbon Monoxide

c)      Oligomycin

d)     Rotenone

e)     UCP

Q: About the production of NADH in the Krebs Cycle


 

Cycle_de_krebs numerado

In this representation of Krebs Cycle, each reaction is marked with a number.  The reactions where NAD+ is reduced to NADH.H+ are those marked with the numbers:

 

a)     1,3 and 4

 

b)     3,4 and 5

 

c)      3,4 and 6

 

d)     3,4 and 8

 

e)     5,6 and 7

 

f)       4,5 and 6

Oxidation of a fatty acid with 17 atoms of carbon


(This post analise the energetic balance considering that the Propionyl CoA follows an anaplerotic fate)

 

Apply the equations described in the previous post:

 

N= Number of Carbons

 

(N/2) -1.5 = Number of rounds in Beta-oxidation

 

(N/2) -1.5 = Number of acetyl CoA produced in Beta-oxidation

 

So, in terms of production and consumption of ATP of ATPs, the oxidation of a 17-carbons fatty acid will show the following energetic balance:

 

Activation of a fatty acid to Acyl CoA = -2 ATP

 

Number of rounds in Beta-Oxidation:

 

(17/2) – 1.5 = 8.5 -1.5 = 7

7 rounds x 5 ATP/round = 35 ATP

 

Number of produced Acetyl CoA: 7 Acetyl CoA

 

7 Acetyl CoA x 12 ATP/Acetyl CoA = 84 ATP

 

Additionally, the Beta-oxidation has produced 1 Propionyl CoA. The conversion of Propionyl CoA to Succinyl CoA, as described in a former post, will consume 1 ATP (Consider -1 ATP).

 

As described in a previous post:

 

We can consider the conversion of Propionyl CoA to Succinyl CoA as an anaplerotic pathway, in which case, the molecule of Succinyl CoA continue in the Krebs Cycle generating Oxalacetate in the following sequence of reactions:

 

Succinyl CoA + GDP + (P) —> Succinate + CoA + GTP (equivalent to 1 ATP)

 

Succinate   +  FAD ———-> Fumarate +  FADH2 (Generates 2 ATPs in the Respiratory Chain)

 

Fumarate +     H2O————– > Malate

 

Malate +   NAD+ —————->  Oxalacetate + NADH.H+ (Generates 3 ATPs in the Respiratory Chain) “

 

Total = (-1+1+2+3) = 5 ATPs

 

Total of ATPs produced (considering the anaplerotic fate of the Propionyl CoA turned into Succinyl CoA) = -2+35+84-1+6 = 122 ATP                        

 

BUT…

 

We may also consider a total oxidation that would include the carbon atoms of the Propionyl CoA!

 

In our next post, we will consider what happens if, instead of the Succinyl CoA following an anaplerotic pathway, it follows a path that allow the carbon atoms of Propionyl CoA end up being oxidized to CO2. It would allow a real  total oxidation of all the original carbons of the heptadecanoic acid or any other fatty acid with an odd number of carbons.

 

 

Inhibitors of the Electron Transport Chain


 

Answer to Question B-09

 

As described in a former post, the inhibitors of the Electron Transport Chain are substances that bind to some of the components of the ETC blocking its ability to change in a reversible form from an oxidized state to a reduced state.

 

This inhibition results in the accumulation of reduced forms before the inhibitor point, and oxidized forms of the components of the ETC downstream (ahead) the inhibition point.

 

 

Since energy is not released, the synthesis of ATP also stops. The most important known inhibitors of the ETC are Amytal, Rotenone, Antimycin A, CO, Sodium Azide, and Cyanides.

 

Amytal, a barbiturate, and Rotenone, a plant product used as insecticide and pesticide, block the ETC between NADH dehydrogenase (Complex I) and CoQ.

 

Consequently, they prevent the utilization of NADH as a substrate. On the contrary, electron flow resulting from the oxidation of Complex II is not affected, because these electrons enter through QH2, beyond the block.

 

The effect of Amytal has been observed in vitro, since the intoxication with amytal and other barbiturates in vivo affect mainly the CNS by acting on GABA-sensitive ion channels, an effect not related to the action of Amytal on Complex I.

 

Rotenone intoxications are very rare. In facts, some human tribes used to catch fishes by spreading plant extracts containing rotenone in the water, and this substance was easily absorbed by the fishes through the gills. These fishes were eaten later without notable side effects in humans, since rotenone is absorbed very difficult by the gastrointestinal tract. Usually, when taken in a concentrated form, irritating action in mucoses causes vomits.

 

It is interesting to note that Rotenone and MPTP (a neurotoxin), when administered in vein,  cause at the same time interference with the functioning of Complex I and a Parkinson-like disease. These substances affect primary neurons in substancia nigra; apparently the sequence is: impairment of Complex I, impairment of mitochondria metabolism, accumulation of free radicals, cell death, release of toxic compounds and destruction of other cells.

 

Antimycin A is an antibiotic produced by Streptomyces griseous that has been used as a piscicide for the control of some fish species. Antymicine A interferes with electron flow from cytochrome bH in Complex III (Q-cytochrome c oxidoreductase).  In the presence of this substance, cytochrome bH can be reduced but not oxidized, consequently,  in the presence of antimycin A cytochrome c remains oxidized, as do the cytochromes a and a3 that are ahead.

 

Carbon monoxide (CO) is responsible for more than 50 % of death by poisoning worldwide. It is colorless and odorless; high levels can result from incomplete combustion of fuels: engine and furnace exhausts are important sources. Tobacco smoking increases CarboxyHb levels.

 

Carbon monoxide intoxication causes impaired oxygen delivery and utilization at the cellular level. The affinity of Hb for CO is almost 300 times higher than for Oxygen. An environment in which there is 100 ppm of CO is enough to form 16 % carboxyhemoglobin. The situation is worsen since the binding of CO to one of the Hem groups of Hemoglobin increases the affinity of the other three Hem groups for Oxygen, so the delivery of Oxygen to  tissues is very affected. The brain and the heart, that has a high Oxygen consumption, are the most affected. Myoglobin has even a greater affinity for CO than Hemoglobin. As a consequence of the binding of CO to these molecules, the heart functioning is very impaired and the patient presents sever hypotension. As described above, this intoxication is an important cause of death worldwide.

 

The affinity of respiratory chain components for CO is lower than for Oxygen,

but since the clinical status does not correlate very well with the carboxyhemoglobin levels, it is considered that the inhibition of Cytochrome Oxidase by CO also plays a role in CO intoxication. CO binds to the reduced form of iron in Hem groups (Fe++) in cytochrome Oxidase

 

On the contrary, in cyanide intoxication the inhibition of the respiratory chain has a primary role. Intoxication by cyanide can be seen relatively frequent in patients with smoke inhalation from residential or industrial fires. Also in persons related professionally with cyanide or derivatives in certain industries.  Intentional poisoning can be seen in suicidal persons with access to cyanide compounds. Cyanide affects practically all metalloenzymes, but its principal toxicity derives from the binding to the Fe+++ in the Hem groups in cytochrome Oxidase, inhibiting the functioning of the Electron Transport Chain. As a consequence, redox reactions in the respiratory chain will stop, energy will not be released, proton pumps will not function, so they will not return through Complex V, and the production of ATP will cease (Related question here).

 

Azides have an action on the respiratory chain very similar to cyanide, inhibiting the Hem groups of cytochromes in Cytochrome Oxidase (Complex IV). Azides are used as propellants in airbags, in detonant (explosive) industry and as preservative of sera an reagents. Some cases of azide intoxication in humans have been reported.

 

You can find more information about these inhibitors of the Electron Transport Chain in these links:

 

Antimycin A: toxicity, ecological toxicity and regulatory information

 

 

Risk assessment for Piscicidal Formulations of Antimycin

 

Leybell, I: Toxicity: Cyanide

 

Cyanide poisoning

 

 

Azide Toxicity

 

Sodium Azide Toxicity effects

 

Shochat, G.N.: Toxicity, Carbon Monoxide

 

 

How to calculate QUICKLY the energetic balance of the total oxidation of a fatty acid


 

The basis for the calculation and how to do it have been explained in detail in other post.

 

Anyway, since time is an issue in exams, some readers are interested in knowing a method of doing the calculation in a real fast way.

 

For obtaining a quick answer when calculating the energetic balance of the total oxidation of a fatty acid (and also for verifying your answers if you use some other method of calculation) you can apply these formulas:

 

a) If your course use for the reduced cofactors these energetic yields

NADH.H+ = 2.5 ATP

FADH2 = 1.5 ATP

 

Use this formula:

 

[(n/2) -1] x (14) +8 = Number of ATP formed

 

Where n= Number of carbons

 

Example:

 

Balance of the total oxidation of palmitic acid (16 carbons):

(16/2)-1] x (14) +8 = 106 ATP

 

b) If the equivalence used for the energetic yields in your course are:

NADH.H+ = 3 ATP

FADH2 = 2 ATP

 

Then apply this formula:

 

[(n/2) -1] x 17 +10 = Number of ATP formed

 

For the total oxidation of palmitic acid, following the former energetic yielding criteria:

 

[(16/2)-1] x 17 + 10 = 129 ATP

 

Thanks for the questions and comments!