in glycolysis, why G.6.P is converted into F.6.P?
348749
Glucose 6 phosphate is only immobilised by the attachment of the phophate group so that it cannot escape from the cell. Cells do not have the trasport machinery to export phosphorylated sugars. Further more, gluocse-6-phosphate formation does not guarantee glycolysis will continue, neither does the formation of F-6-P for that matter. This is because these intermediates can join pathways other than glycolysis. G-6-P for instance can join the Pentose Phosphate Pathway. So The G-6-P has to be oxidized to F-6-P and the F-6-P has to be reduced to F-1,6-bisphosphate. The F-1,6-BP formation is the committing step in glycolysis and hence naturally one of the regulation steps. It definitely will go into glycolysis. So oxidation of the intermediate G-6-P to F-6-P is needed and this has to be further oxidized to F-1,6-BP.
215604
C6H12O6(s) ---H2SO4--> 6C(s) + 6H2O(in H2SO4 solu.) + Heat
825956
ATP production per pyruvate that enters the Krebs Cycle:\n         3 NADH =   9 ATP\n         1 FADH2 = 2 ATP\n         1 GTP = 1 ATP\n                      ------------------\n                        12 ATP per pyruvate or 24 ATP per glucose\n\n**note**\n\n1 NADH = 3 ATP generated\n1 FADH2 = 2 ATP generated\n\nthese NADH and FADH2 are converted to ATP when they enter the oxidative phosphorylation.\n\nIf you are confused where the 38 ATPs come from, here it is:\n(1) Glycolysis - 1 mole glucose is converted to 2 moles pyruvate.\n         \n                2 ATP\n                2 NADH = 6 ATP generated\n              ------------------------------------------\nGlycolysis generates 8 ATP\n\n(2) 2 moles of  Pyruvate generated from glycolysis cannot enter the Krebs Cycle unless it will be converted to Acetyl-CoA. In this process, it produces 2 moles of pyruvate generate 2 moles of NADH.\n          2 NADH = 6 ATP\n\nin this process, 6 ATP is generated.\n\n(3) Krebs Cycle: 2 acetyl-CoA then enters the Krebs cycle.\n(see explanation above) generating 24 ATPs.\n\nso for the 3 steps: \n           8 ATP (glycolysis) \n           6 ATP (intermediate step 2)\n          24 ATP (Krebs cycle)\n-------------------------------------------\n          38 ATP
845828
Here you go...\n\nhttp://home.planet.nl/~skok/techniques/laboratory/pka_pkb.html
858478
glyceraldehyde 3-phosphate which has three carbons. two of them can make a glucose molecule after calvin benson cycle.
89857
C6H12O6 represents Glucose and Fructose. Go to http://images.google.com/images?q=Glucose%20structure
317821
(m)\n\nmetabolic pathway analysis, it should be considered that many enzymes operate with low specificity (e.g. nucleoside diphosphokinase, uridine kinase, transketolase, aldolase), so that various substrates and products can be converted. Here, we analyze the effect of enzymes with low substrate specificity on the elementary flux modes (pathways). We also study the benefits of two different approaches to describing multifunctional enzymes. The usual description is in terms of (overall) enzymatic reactions. At a more detailed level, the reaction steps (half-reactions, hemi-reactions) of the formation and conversion of enzyme-substrate complexes are considered. Multifunctional enzymes operate according to various mechanisms. This is illustrated here by the reaction schemes for the different enzyme mechanisms of bifunctional enzymes. For enzymes with two or more functions, it is important to consider only linearly independent functions, because otherwise cyclic elementary modes would occur which do not perform any net transformation. However, the choice of linearly independent functions is not a priori unique. We give a method for making this choice unique by considering the extreme pathways of the hemi-reactions system. A formal application of the algorithm for computing elementary flux modes (pathways) yields the result that the number of such modes sometimes depend on the level of description if some reactions are reversible and the products of the multifunctional enzymes are external metabolites or some multifunctional enzymes partly share the same metabolites. However, this problem can be solved by appropriate interpretation of the definition of elementary modes and the correct choice of independent functions of multifunctional enzymes. The analysis is illustrated by a biochemical example taken from nucleotide metabolism, comparing the two ways of description for nucleoside diphosphokinase and adenylate kinase, and by several smaller examples.
911009
C6H6O + 7O2 yields 3H2O + 6CO2
747676
2C12H22O11 +24O2 ---> 24CO2 +22H2O\na)F\nb)F\nc)T\nd)F\ne)T\nf)T
698351
Phosphoglycerate mutase catalyzes the migration of a phosphoryl group from the C-3 position of 3-phosphoglycerate to the C-2 position. This glycolytic enzyme is a tetramer with four identical subunits.♥ \n\nCheck out this link for more:\nhttp://www.chem.uwec.edu/Modeling/GGProteinStructures/Pages/prs3b11.htm
767270
1. glycolysis\n2. lactic acid\n3. I'm not sure what the question is asking. 4 ATP molecules are produced by glycolysis but 2 are used up so there is only a net of 2 ATP.\n4.Fermentation\n5. NADH and FADH2 (2 is subscript)
926847
Substrate-level phosphorylation is the direct transfer of a phosphate group from some compound onto ADP to make ATP.  \n\nIn the glycolytic pathway, there are two specific steps where substrate-level phosphorylation occurs.  If you're counting ATPs produced from a single glucose molecule, then 4 ATPs are produced by the substrate level phosphorylation reactions during glycolysis. \n\nIn the citric acid cycle, there is another point where substrate-level phosphorylation occurs.  Again, if you're counting ATPs from a single glucose molecule, this accounts for 2 more ATPs.\n\nSo, adding those up, you have 6 ATPs produced from the metabolism of one glucose molecule by substrate-level phosphorylation reactions...
914156
6 CO2 + 12 NADPH + 12 H+ + 18 ATP → C6H12O6 + 6 H2O + 12 NADP+ + 18 ADP + 18 Pi
796693
4PH3 (g)  ---> P4(g) + 6H2(g)\n\n4 moles of PH3 are consumed then 1 mole of P4 is produced and 6 moles of H2 are produced.\n\nIf you are consuming 0.0048 mol/s of PH3, then by the relation above you will be producing:\n\n0.0048/4 = 0.0012 mol/s P4\n0.0048*6/4 = 0.072 mol/s H2.
294234
* The process begins with the oxidation of pyruvate, producing one CO2, and one acetyl-CoA.\n    * Acetyl-CoA reacts with the four carbon carboxylic acid, oxaloacetate--to form the six carbon carboxylic acid, citrate.\n    * Through a series of reactions citrate is converted back to oxaloacetate. This cycle produces 2 CO2 and consumes 3 NAD+, producing 3NADH and 3H+.\n    * It consumes 3 H2O and consumes one FAD, producing one FADH+.\n    * 1st turn end= 1 ATP, 3 NADH, 1 FADH2\n    * Since there are two molecules of Pyruvic acid to deal with, the cycle turns once more.\n    * The complete end result= 2 ATP, 6 NADH, 2 FADH2
747126
6co2+6h2o <---> c6h12o6+6o2\n\nLeft side:\n\n6 CO2  =  6 carbon + 12 oxygen\n6 H2O =  12 hydrogen + 6 oxygen\n\nRight side:\n\nC6H12O6 = 6 carbon + 12 hydrogen + 6 oxygen\n6O2 = 12 oxygen\n\nNicely balanced equation for glucose metabolism!
930311
Two per molecule of glucose