Embden Meyerhof pathway
The initial stage of biological oxidiation includes a series of nine main sequential steps of reactions known as Embden Meyerhof pathway, after the name of the german biochemist who worked out these steps. Enzyme required -for this pathway are should components ofcytosol .Hence this process occur in cytosol and requires no oxygen. Anaerobes some lower animals and plants and some animals tissues produce energy only by the above pathway due to the absence of O2 .This process is then called sugar fermentation .In other organisms using O2
And hence called aerobes also this process occurs as such in cytosol only as preliminary stage of cellurlar respiration In this case it is termed glycolysis ,Presumably therefore this process is more primitive perhaps the earliest method of intracellular energy production adopted by cells during early evolution ,The main difference in sugar fermentation and glycolysis is that of the endproucts lactic acid or ethyl alcohol in the case of sugar fermentation and pyruvic acid in the case of glycolysis .
The nine steps of Embden Meyerhof pathway are as follows
1. In first reaction a glucose molecule is phosphorylated to glucose 6- phosphate by the terminal phosphate bond of an ATP molecule which is consequently hydrolysed to ADP, The enzyme hexokinase activates the glucose molecule and catalyzes this reaction.
C6H12O6+ATP → glucose 6-phosphate +ADP
2. In the second step. Phosphohexose isomerise enzyme catalyzes conversion of glucose 6- phosphate into its structural isomer fructose fructose 6- phosphate.
3. Now the enzyme, phopshofructokinase , catalyzes further phosphorylation of fructose 6- phosphate into fructose 1,6- disphosphate using the terminal bond of another ATP molecule.
4. In fourth step aldolase emzyme catalyzes the splitting of fructose 1,6-diphosphogly molecule into 3 carbon molecules one of glyceraldehydes 3 phosphate or 3 phoshoglyceraldehde (PGAL=C3H5O3 -phosphate) and one of DHAP which j is reconverted to PGAL by triose phosphate isomerase.
5. Now each PGAL molecule is oxidized into 1,3 diphospholyceric acid with the help of 3 phosphoglyceraldehyde dehydrogenase enzyme in presence of inorganic phosphate (pi) and NAD+ ,This reaction yields two electrons and twoprotons (equiavalent to two hydrogen atoms) from each PGAL molecule acting as the hydrogen acceptor NAD+ ( nicotinamide adenine dinucleotide ) accepts two ,electrons and one proton the remaining proton being liberated into the cytosol. NAD + is the thus , reduced to NADH;
2(PGAL) + 2(H3 PO4) +2NNAD+→ 2(1,3,diphosphoglyceric acid ) + 2NADH + 2H+
6. 1,3- diphosphoglyceric acid is a high energy compound the enzyme phosho glyceric acid kinase, now catalyzes the transfer of the high energy phosphate group of this acid to ADP forming a molecule each of 3- phosphoglyceric acid and ATP.
7. Now the enzyme e phosphoglyceromutase , catalyzes the transformation of 3- phosphoglyceric acid to 2- pphosphoglyceric acid
8. Next ,the enzyme enolase, catalyzes dehydration of 2- phosphoglyceric acid, resulting into the formation again of a high energy compound phosphoenolpyruvic acid.
9. The enzyme pyruvic acid kinase, now catalyzes the transfer of the high energy phosphate group of phosphoenolpyruvic acid to ADP .Thus a molecule each of pyrivic acid and ATP are formed .
According to the above account, Embden Meyerhof pathway brings about a gradual degradation of glucose molecule into two molecules of pyruvic acid In anaerobic conditions. Each pyruvic acid molecule is reduced by NADH. H+ to ethyl alcohol (and CO2) in yeast cells , and to lactic acid some bacteria and animals tissues ( muscles and RBCS) certain bacteria convert milk into curd to producing lactic acid in this fashion, NADH. H+ obviously reverts back to its original oxidized form (NAD) to be used again for the same function.
Efficiency of embden Meyerhof pathway : As is clear from the above account the balance sheet of anaerobic oxidation is that two molecules of ATP are spent in it per glucose molecule , and four mols of ATP are formed. Thus there is a net again of two molecules. Thus,or 14600 clories (14.6keal) of energy about 2.0% of the total energy content of a glucose molecule.