Why only 6 water molecules are formed in the aerobic degradation of glucose?

Why only 6 water molecules are formed in the aerobic degradation of glucose?

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I am studying the aerobic degradation of glucose and it seems that for every glucose molecule we should obtain $ce{10H2O}$ molecules. However, it is known that we only obtain 6.

$ce{C6H12O6 + 6O2 -> 6CO2 + 6H2O}$

(I am not going to focus in all products and reactants, but just in the important ones for the formation of water molecules)

First, in the glycolysis, for each molecule of glucose we obtain two water molecules, $ce{2NADH+}$ and 2 pyruvate molecules. By the oxidation of two pyruvate molecules we obtain $ce{2NADH+}$ and 2 acetyl Co-A molecules. So we are going to pass twice through the Krebs cycle, obtaining $ce{6NADH+}$ and $ce{2FADH2}$, and requiring 4 water molecules.

So, when we arrive to the electron transport chain, we have a negative balance of 2 water molecules, and we have $ce{10NADH+}$ and $ce{2FADH2}$. We have been told that for every of these molecules 2 electrons go to the electron transport chain, that means that a total of 24 electrons go to the system. The problem comes here:

$ce{4e- + 4H+ + O2 = 2H2O}$

So, bearing in mind that we have $24e^-$, 12 water molecules should be formed, so at the end, we have gained 10 water molecules, but we know that the number of water molecules formed should be 6. So, clearly there is something wrong in my explanation. I would be very pleased if you could tell me what is wrong.

Thanks in advance.

Your confusion comes entirely from this equation:

$ce{C6H12O6 + 6O2 -> 6CO2 + 6H2O}$

This reaction is the combustion of glucose. This is not how glucose is oxidized in cells! Why so many biology texts and courses present this equation when introducing metabolism is beyond me.

Indeed, your tracking of water molecules is correct: starting with one glucose molecule, 2 waters are produced in glycolysis, 4 are consumed in the tricarboxylic acid cycle and 12 are produced during oxidation of NADH/QH2 (ie FADH2). This gives a net total of 10 produced.

Why is this different from the combustion of glucose? The answer lies in the oxygens introduced by inorganic phosphate during substrate level phosphorylation. Consider the balanced net reaction for the biological oxidation of glucose (simplified by ignoring ATP produced by oxidative phosphorylation and substituting ADP/ATP for GDP/GTP):

$ce{C6H12O6 + 6O2 + 4ADP + 4P_i + 4H+ -> 6CO2 + 4ATP + 10H2O}$

In particular, consider the formation of ATP from ADP and Pi (HPO42-). In both substrate level phosphorylation reactions (catalyzed by GADPH/PGK in glycolysis and succinate-CoA ligase in the tricarboxylic acid cycle), inorganic phosphate nucleophilically attacks the activated carbonyl (thioester) of the substrate and is then transferred to ADP (to form ATP):

The oxygens of the original inorganic phosphate are coloured red. The key point is that an oxygen atom from HPO42- is transferred to the substrate. This oxygen is later removed in the form of carbon dioxide, via oxidative decarboxylation, during the conversion of pyruvate to acetyl-CoA and in the tricarboxylic acid cycle. This occurs four times for each glucose molecule entering glycolysis and is accompanied by the reduction of NAD+ to NADH. Given that NADH is used to reduce molecular oxygen during the electron transport chain:

$ce{NADH + H+ + 1/2O2 -> NAD+ + H2O}$

… this explains where the four, apparently extra, water molecules come from when comparing the combustion of glucose with its biological oxidation.

I think you are misconceiving the intent assumed within the statement "6 H20 generated from glycolysis". The number 6 is simply in relation to the number of carbons oxidized to CO2 within the TCA… which generates an electron at each oxidative occurance, in which the 6 electrons are then shuttled to OXPHOS requiring 3 O2 to form 6 H20… In short: Whenever they say 6 H20 are produced "aerobically" they are specifically referencing the simplified OXPHOS component (or simply: C6H12O6 + 6 O2 → 6 CO2 + 6 H2O).

Of note:

  1. Stoichometry is relatively useless in real research. Don't overthink it. Metabolism is incredibly dynamic.

  2. Real life H20 involvement within complete glucose oxidation is much more complex: Two turns of the TCA cycle generate: 4 CO2 from 2 acetyl CoA which requires 4+ H20 in total, but only has a net H20 loss of 4. Citrate synthase and fumerase both consume 2 H20 apiece to allow production of 4 CO2. The remaining H20 is used at aconitase, but no loss nor gain of H20 occurs in this reaction. Pyruvate dehydrogenase produces the other 2 CO2's making 6 CO2 in total, but does not require H20 at this step… OXPHOS activity: (following 2 turns of TCA) results in 48 total H20 produced between ATP synthase, cytochrome oxidase, and enolase - and being that 4H20 are consumed in the TCA, the net H20 production per glucose is actually 44 H20.

Two waters come in at each turn of the TCA cycle, one at citrate synthase and one at fumarase. Additionally, I don't think it's right to say that you get a net water from glycolysis (in fact if you look at what you wrote, the oxygens don't balance). I think what you're missing is you also get an ATP, and the water you do get at the 2PG->PEP step is really part of the net reactionADP + phosphate -> ATP + water.

To really balance the waters you have to consider all the substrates and byproducts of glucose oxidation, including the ATP/GTP.

Aerobic Respiration

Aerobic respiration is the process by which organisms use oxygen to turn fuel, such as fats and sugars, into chemical energy. In contrast, anaerobic respiration does not use oxygen.

Respiration is used by all cells to turn fuel into energy that can be used to power cellular processes. The product of respiration is a molecule called adenosine triphosphate (ATP), which uses the energy stored in its phosphate bonds to power chemical reactions. It is often referred to as the “currency” of the cell.

Aerobic respiration is much more efficient, and produces ATP much more quickly, than anaerobic respiration. This is because oxygen is an excellent electron acceptor for the chemical reactions involved in generating ATP.

What is Aerobic Glycolysis? (with pictures)

Aerobic glycolysis is the first of three stages that make up aerobic cellular respiration. Cellular respiration is the process that takes place within all cells to release energy stored in glucose molecules. There are two forms of cellular respiration, aerobic and anaerobic, meaning requires oxygen and doesn’t require oxygen.

All living organisms need energy to survive. That energy is received through food, which for plants also includes energy captured from the sun. Whatever the form of food that is taken in by the organism, it is converted to carbohydrates, glucose in particular. During cellular respiration, glucose is converted to carbon dioxide and water with energy being released into the cell. Breaking down the glucose molecules is an oxidation reaction, so oxygen is required for the process to go ahead.

The three stages of aerobic respiration are aerobic glycolysis, the Krebs cycle and the electron transport system. During each stage, a number of chemical reactions take place which form the cellular respiration overall process. The outcome of aerobic glycolysis is that the glucose molecule is broken down into two pyruvate, or pyruvic acid, molecules, which are broken down further in the Krebs cycle, and two water molecules.

The energy that is released by cellular respiration does not happen all at once. In fact, some energy is released through each of the three main stages. When the energy is released from the glucose molecule, it is not released as free energy. The energy is stored in adenosine triphosphate (ATP) molecules, which are short term energy storage molecules that are easily transported within and between cells.

The energy production begins during aerobic glycolysis. During this process, two of the 36 total ATP molecules are created. All the stages of cellular respiration are made up of a number of complex chemical reactions. Aerobic glycolysis is actually made up of a number of different stages that the glucose molecule moves through. The energy necessary to produce the eight ATP molecules is released at different stages of the process.

During aerobic glycolysis, two ATP molecules are initially used to make the glucose molecule sufficiently reactive. The glucose molecule is phosphorylated, meaning that phosphate molecules are added to the glucose molecule from the ATP molecules. After the glucose has been phosphorylated, it splits from a six carbon sugar molecule into two three carbon sugar molecules. Hydrogen atoms are removed from the resulting three carbon sugars and two phosphates are lost from each, forming four new ATP molecules. After the glucose has gone through all these steps, the final outcome is two three carbon pyruvate molecules, two water molecules and two ATP molecules.

Outcomes of Glycolysis

Glycolysis starts with glucose and ends with two pyruvate molecules, a total of four ATP molecules and two molecules of NADH. Two ATP molecules were used in the first half of the pathway to prepare the six-carbon ring for cleavage, so the cell has a net gain of two ATP molecules and 2 NADH molecules for its use. If the cell cannot catabolize the pyruvate molecules further, it will harvest only two ATP molecules from one molecule of glucose. Mature mammalian red blood cells are not capable of aerobic respiration—the process in which organisms convert energy in the presence of oxygen—and glycolysis is their sole source of ATP. If glycolysis is interrupted, these cells lose their ability to maintain their sodium-potassium pumps, and eventually, they die.

The last step in glycolysis will not occur if pyruvate kinase, the enzyme that catalyzes the formation of pyruvate, is not available in sufficient quantities. In this situation, the entire glycolysis pathway will proceed, but only two ATP molecules will be made in the second half. Thus, pyruvate kinase is a rate-limiting enzyme for glycolysis.

ATP Yield from Oxidation of Glucose in Aerobic Respiration

The net ATP-yield in eukaryotes from glycolysis, TCA cycle, and electron transport and oxidative phosphorylation can be readily calculated.

Before the general acceptance of the chemiosmotic hypothesis for oxidative phosphorylation, this calculation was based on phosphate/oxygen ratio (P/O ratio). Most experiments yielded P/O (ATP to ½ O2) ratio of more than two when NADH was the electron donor, and more than one when succinate was the electron donor.

Given the assumption that P/O ratio should have an integral value most experimenters agreed that the P/O ratios must be 3 for NADH and 2 for succinate (FADH)2.

On the basis of these P/O ratios (the number of ATPs formed per oxygen atom and reduced by 2 electrons in electron transport chain), the total ATP yield from oxidation of one glucose molecule in aerobic respiration was calculated to be a maximum of 36 ATPs. The number goes to 38 when malate-aspartate suttle rather than the glycerol 3- phosphate suttle is used.

With the general acceptance of the chemiosmotic hypothesis for coupling ATP synthesis to oxidative phosphorylation, there was no theoretical requirement for P/O ratio to be integral.

The relevant question now became how many protons (H + ) are pumped outward by electron transport chain from one NADH to oxygen, and how many protons (H + ) must flow inward through the F1/F0 ATPase complex to drive the synthesis of one ATP? The best current estimates for protons pumped out per pair of electrons arc 10 for NADH and 6 for succinate (FADH2).

The most widely accepted experimental value for number of protons required to drive the synthesis of an ATP molecule is 4, of which one is used in transporting Pi (inorganic phosphate), ATP, and ADP across the mitochondrial membrane. If 10 protons are pumped out per NADH and 4 must flow in to produce one ATP, the proton-based P/O ratio is 2.5 (10/4) for NADH and 1.5 (6/4) for succinate (FADH2).

Hence, as given in Table 24.3, 30 molecules of ATP are synthesized when glucose in completely oxidised to CO2. This number goes to 32 when malate-aspartate suttle rather than the glycerol 3-phosphate suttle is used.

ATP-yields in bacteria in aerobic conditions can be less because the bacterial electron transport systems often possess lower P/O ratios than the eukaryotic system. For instance, Escherichia coli with its branched electron transport chains has a P/O ratio around 1.3 when respiring at high oxygen levels and only a ratio of about 0.67 when respiring at low oxygen levels.

In this case ATP synthesis varies with environmental conditions. Perhaps because E. coli normally grows in habitats rich in nutrients it does not have to be particularly efficient in ATP synthesis. Presumably the electron transport chain functions when E. coli is in an aerobic fresh water environment between hosts.

What is Aerobic Fermentation

As mentioned above, aerobic respiration is the more precise and scientific term for aerobic fermentation. Aerobic respiration refers to the set of chemical reactions involved in the production of energy by completely oxidizing food. It releases carbon dioxide and water as by-products. Aerobic respiration mainly occurs in higher animals and plants. It is the most efficient process among various processes of energy production. The three steps of aerobic respiration are glycolysis, Krebs cycle, and electron transport chain.


Glycolysis is the first step of aerobic respiration, which occurs in the cytoplasm. This process breaks down glucose into two pyruvate molecules. The pyruvate molecules undergo oxidative decarboxylation to form acetyl-CoA. 2 ATP and 2 NADH are the yield of this process.

Krebs Cycle

Krebs cycle occurs inside the mitochondrial matrix. A complete breakdown of acetyl-CoA into carbon dioxide occurs in the Krebs cycle, regenerating the starting compound, oxaloacetate. During Krebs cycle, releasing the energy from acetyl-CoA produces 2 GTPs, 6 NADH, and 2 FADH2.

Electron Transport Chain

The production of ATP during the oxidative phosphorylation uses the reducing power of NADH and FADH2. It occurs in the inner membrane of mitochondria. The below figure shows the overall chemical reaction of aerobic respiration.

Figure 1: Aerobic Respiration – Steps

Several Parameters Affect the ΔG of a Reaction

The change in free energy of a reaction (ΔG) is influenced by temperature, pressure, and the initial concentrations of reactants and products. Most biological reactions — like others that take place in aqueous solutions —𠁚lso are affected by the pH of the solution.

The standard free-energy change of a reaction ΔG°′ is the value of the change in free energy under the conditions of 298 K (25 ଌ), 1 atm pressure, pH 7.0 (as in pure water), and initial concentrations of 1 M for all reactants and products except protons, which are kept at pH 7.0. Table 2-4 gives values of ΔG°′ for some typical biochemical reactions. The sign of ΔG°′ depends on the direction in which the reaction is written. If the reaction A →𠁛 has a ΔG°′ of −x kcal/mol, then the reverse reaction B →𠁚 will have a ΔG°′ value of +x kcal/mol.

Table 2-4

Values of ΔG°′, the Standard Free-Energy Change, for Some Important Biochemical Reactions.

Most biological reactions differ from standard conditions, particularly in the concentrations of reactants. However, we can estimate free-energy changes for different temperatures and initial concentrations, using the equation

where R is the gas constant of 1.987 cal/(degree · mol), T is the temperature (in degrees Kelvin), and Q is the initial ratio of products to reactants, which is expressed as in Equation 2-1 defining the equilibrium constant. Again using as our example the interconversion of glyceraldehyde 3-phosphate (G3P) and dihydroxyacetone phosphate (DHAP)

In a reaction A +𠁛 ⇌𠁜, in which two molecules combine to form a third, the equation for ΔG becomes

The direction of the reaction will shift more toward the right (toward formation of C) if either [A] or [B] is increased.


  1. Aerobic respiration is the aerobic catabolism of nutrients to carbon dioxide, water, and energy, and involves an electron transport system in which molecular oxygen is the final electron acceptor.
  2. The overall reaction is: C6H12O6 + 6O2 yields 6CO2 + 6H2O + energy (as ATP). Glucose (C6H12O6 ) is oxidized to produce carbon dioxide (CO2) and oxygen (O2) is reduced to produce water (H2O).
  3. This type of ATP production is seen in aerobes and facultative anaerobes.
  4. Aerobic respiration involves four stages: glycolysis, a transition reaction that forms acetyl coenzyme A, the citric acid (Krebs) cycle, and an electron transport chain and chemiosmosis.

Biochemistry. 5th edition.

Glycogen is a readily mobilized storage form of glucose. It is a very large, branched polymer of glucose residues (Figure 21.1) that can be broken down to yield glucose molecules when energy is needed. Most of the glucose residues in glycogen are linked by α-1,4-glycosidic bonds. Branches at about every tenth residue are created by α-1,6-glycosidic bonds. Recall that α-glycosidic linkages form open helical polymers, whereas β linkages produce nearly straight strands that form structural fibrils, as in cellulose (Section 11.2.3).

Figure 21.1

Glycogen Structure. In this structure of two outer branches of a glycogen molecule, the residues at the nonreducing ends are shown in red and residue that starts a branch is shown in green. The rest of the glycogen molecule is represented by R.

Glycogen is not as reduced as fatty acids are and consequently not as energy rich. Why do animals store any energy as glycogen? Why not convert all excess fuel into fatty acids? Glycogen is an important fuel reserve for several reasons. The controlled breakdown of glycogen and release of glucose increase the amount of glucose that is available between meals. Hence, glycogen serves as a buffer to maintain blood-glucose levels. Glycogen's role in maintaining blood-glucose levels is especially important because glucose is virtually the only fuel used by the brain, except during prolonged starvation. Moreover, the glucose from glycogen is readily mobilized and is therefore a good source of energy for sudden, strenuous activity. Unlike fatty acids, the released glucose can provide energy in the absence of oxygen and can thus supply energy for anaerobic activity.

The two major sites of glycogen storage are the liver and skeletal muscle. The concentration of glycogen is higher in the liver than in muscle (10% versus 2% by weight), but more glycogen is stored in skeletal muscle overall because of its much greater mass. Glycogen is present in the cytosol in the form of granules ranging in diameter from 10 to 40 nm (Figure 21.2). In the liver, glycogen synthesis and degradation are regulated to maintain blood-glucose levels as required to meet the needs of the organism as a whole. In contrast, in muscle, these processes are regulated to meet the energy needs of the muscle itself.

Figure 21.2

Electron Micrograph of a Liver Cell. The dense particles in the cytoplasm are glycogen granules. [Courtesy of Dr. George Palade.]

21.0.1. An Overview of Glycogen Metabolism:

Glycogen degradation and synthesis are relatively simple biochemical processes. Glycogen degradation consists of three steps: (1) the release of glucose 1-phosphate from glycogen, (2) the remodeling of the glycogen substrate to permit further degradation, and (3) the conversion of glucose 1-phosphate into glucose 6-phosphate for further metabolism. The glucose 6-phosphate derived from the breakdown of glycogen has three fates (Figure 21.3): (1) It is the initial substrate for glycolysis, (2) it can be processed by the pentose phosphate pathway to yield NADPH and ribose derivatives and (3) it can be converted into free glucose for release into the bloodstream. This conversion takes place mainly in the liver and to a lesser extent in the intestines and kidneys.

Figure 21.3

Fates of Glucose 6-Phosphate. Glucose 6-phosphate derived from glycogen can (1) be used as a fuel for anaerobic or aerobic metabolism as in, for instance, muscle (2) be converted into free glucose in the liver and subsequently released into the blood (more. )

Glycogen synthesis requires an activated form of glucose, uridine diphosphate glucose (UDP-glucose), which is formed by the reaction of UTP and glucose 1-phosphate. UDP-glucose is added to the nonreducing end of glycogen molecules. As is the case for glycogen degradation, the glycogen molecule must be remodeled for continued synthesis.

The regulation of these processes is quite complex. Several enzymes taking part in glycogen metabolism allosterically respond to metabolites that signal the energy needs of the cell. These allosteric responses allow the adjustment of enzyme activity to meet the needs of the cell in which the enzymes are expressed. Glycogen metabolism is also regulated by hormonally stimulated cascades that lead to the reversible phosphorylation of enzymes, which alters their kinetic properties. Regulation by hormones allows glygogen metabolism to adjust to the needs of the entire organism. By both these mechanisms, glycogen degradation is integrated with glycogen synthesis. We will first examine the metabolism, followed by enzyme regulation and then the elaborate integration of control mechanisms.


Signal cascades lead to the mobilization of glycogen to produce glucose, an energy source for runners. [(Left) Mike Powell/Allsport.]

  • 21.1. Glycogen Breakdown Requires the Interplay of Several Enzymes
  • 21.2. Phosphorylase Is Regulated by Allosteric Interactions and Reversible Phosphorylation
  • 21.3. Epinephrine and Glucagon Signal the Need for Glycogen Breakdown
  • 21.4. Glycogen Is Synthesized and Degraded by Different Pathways
  • 21.5. Glycogen Breakdown and Synthesis Are Reciprocally Regulated
  • Summary
  • Problems
  • Selected Readings

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Multiple Choice

Which is the location of electron transports systems in prokaryotes?

A. the outer mitochondrial membrane
B. the cytoplasm
C. the inner mitochondrial membrane
D. the cytoplasmic membrane

Which is the source of the energy used to make ATP by oxidative phosphorylation?

A. oxygen
B. high-energy phosphate bonds
C. the proton motive force
D. Pi

A cell might perform anaerobic respiration for which of the following reasons?

A. It lacks glucose for degradation.
B. It lacks the transition reaction to convert pyruvate to acetyl-CoA.
C. It lacks Krebs cycle enzymes for processing acetyl-CoA to CO2.
D. It lacks a cytochrome oxidase for passing electrons to oxygen.

In prokaryotes, which of the following is true?

A. As electrons are transferred through an ETS, H + is pumped out of the cell.
B. As electrons are transferred through an ETS, H + is pumped into the cell.
C. As protons are transferred through an ETS, electrons are pumped out of the cell.
D. As protons are transferred through an ETS, electrons are pumped into the cell.

Which of the following is not an electron carrier within an electron transport system?

A. flavoprotein
B. ATP synthase
C. ubiquinone
D. cytochrome oxidase

Pentose Phosphate Pathway (PPP) | Respiration

Major pathway for aerobic respiration of glucose is through glycolysis and Krebs cycle, however, an alternate pathway exists in many organisms. This pathway, which requires the presence of oxygen, is called pentose phosphate pathway (PPP) or hexose monophosphate shunt (HMS).

As shown in figure that reduced NADP is formed in the reactions forming 6-phosphogluconic acid and ribulose-5-R If the equivalent of a molecule of glucose is oxidised to CO2 and Hp via this cyclic pathway (six turns of cycle), then 12 molecules of reduced NADP would be formed. In the presence of enzyme transhydrogenase the hydrogens of NADPH can be transferred to NAD to form NADH.

Hence the formation of 12 molecules of reduced NADP via hexose monophosphate shunt ultimately can lead to synthesis of 36 molecules of ATP. Thus, capture of energy released in oxidation of glucose via this pathway (hexose monophosphate shunt) is as effective as that of glycolytic-Krebs cycle pathway.

This pathway (PPP) is also known as Warburg-Limpam-Dlckens cycle and phosphogluconate shunt. This was first studied by Warburg (1935) and Dickens (1938). This pathway occurs within the cytosol, where all enzymes of pentose phosphate pathway (PPP) are present.

Reactions of Pentose Phosphate Pathway (PPP):

Starting from 6 molecules of glucose 6-phosphate, the various reactions of PPP are as follows:

a. 6 molecules of glucose-6-phosphate in the presence of coenzyme NADP are oxidised into 6 molecules of 6-phosphogluconolactone by the enzyme glucose-6-phosphate dehydrogenase. 6 molecules of NADP are reduced in the reaction which is reversible.

b. 6-phosphogluconolactone is hydrolysed by the enzyme lactonase to produce 6 molecules of 6-phosphogluconic acid.

c. Phosphogluconic acid is oxidatively decarboxylated by the enzyme 6-phosphogluconic acid dehydrogenase. 6 molecule of NADP are reduced, 6 molecule of CO2 are released, and 6 molecule of ribulose-5-phosphate are produced.

d. 6 molecule ob ribulose-5-phosphate isomerise into 4 molecule of xylulose-5-phosphate and 2 molecule of ribose-5-phosphate in the presence of enzymes ribulose phosphate-3-epimerase and pentose phosphate isomeras, respectively.

e. 2 molecule of xylulose-5-phosphate and 2 molecule of ribose-5-phosphate combine In the presence of the enzyme transketolase to form 2 molecule of sedoheptulose-7-phosphate and 2 molecule of phosphoglyceraldehyde.

f. 2 molecule of sedoheptulose-7-phosphate and 2 molecule of 3-phosphoglyceraldehyde combine in the presence of enzyme transketolase to form 2 molecules of fructose-6-phosphate and 2 molecule of erythrose-4-phoshate.

g. 2 molecules of erythrose-4-phoshate reacts with remaining 2 molecules of xylulose-5-phosphate ( reaction 4 and 5 ) In the presence of transketolase to form 2 molecules of fructose-6-phosphate and 2 molecule of 3-phosphoglyceraldehyde.

h. One molecule of phosphoglyceraldehyde isomerise into dihydroxyacetone phosphate, in the presence of enzyme phosphotriose isomerase.

i. Remaining one molecule of 3-phosphoglyceraldehyde unites with dihydroxyacetone phosphate In of enzyme aldolase to form one molecule of fructose 1, 6-diphosphate, which in the presence of phosphatase forms one molecule of fructose-6-phosphate.

j. 5molecule of fructose-6-phosphate produced in reaction 6, 7 and 9, isomerise into molecules of glucose-6-phosphate in presence of enzyme phosphohexose isomerase.

To summarise, 6 molecules of glucose-6-phosphate which enter into this pathway, produce 6 molecules of CO2, after oxidation, and 12 molecules of reduced coenzyme NADPH2, while 5 molecules of glucose-6-phosphate are regenerated.

6 glucose-6-phosphate + 12 NADP + → 5-glucose-6-phosphate + 12 NADPH2 + 6 CO2

Complete oxidation of a molecule of glucose produces 12 molecules of NADPH2, which is equal to 36 ATP molecules. This capture of energy released in oxidation of glucose via this pathway (PPP) is as effective as that of glycolytic-Krebs cycle pathway, where 38 ATP molecules per glucose molecule are produced.

Significance of Pentose Phosphate Pathway (PPP):

a. This pathway provides alternative route for carbohydrate degradation.

b. This pathway (PPP) generates NADPH 2 molecules which are used as reductants in biosynthetic processes under conditions when NADPH molecules are not generated by Fructose-6-p photosynthesis. It is, therefore, important in non-photosynthetic tissues, such as differentiating tissues, germinating seeds and during period of darkness.

Production of NADPH is not linked to ATP generation in pentose phosphate pathway (PPP).

c. It produces ribose sugars for synthesis of nucleic acids.

d. It plays an important role in fixation of CO2 in photosynthesis through ribulose-5- phosphate ribulose 1, 5-phosphate derived from ribulose-5-phosphate is the primary acceptor of CO2 in photosynthesis.

e. It provides erythrose-4-phosphate, which is required for synthesis of shikimic acid. The latter is precursor of aromatic ring compounds.

f. It produces a number of tetroses and pentoses for synthesis of nucleosides, nucleotides, nucleic acids, anthocyanins and other compounds.

Watch the video: 10 λεπτο πρόγραμμα για καύση λίπους! - (July 2022).


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