Overview of Citric Acid Cycle

so now that we discuss the details of the eight steps of the citric acid cycle let's actually summarize our results so we have the citric acid cycle also known as the tricarboxylic acid cycle TCA cycle and also known as the Krebs cycle so we have three different names for the single process that takes place entirely in the matrix of the mitochondria of our cell now before this process can actually begin we have to break down glucose molecules into pyruvate molecules in the process we call glycolysis and that takes place in the cytoplasm of the cell so once we form the pyruvate molecules if we have plenty of oxygen present in the cell then the pyruvate molecules will move into the matrix of the mitochondria via special type of protein found in the membrane of the mitochondria known as pyruvate translocates and once the pyruvate moves into the matrix of the mitochondria before it begins the citric acid cycle we have to transform that pyruvate into acetyl coenzyme a so ultimately a 2 carbon component from the pyruvate is transferred onto a carrier molecule known as coenzyme a or simply co a and so once we form this cetyl coenzyme a molecule in the pyruvate decarboxylation process only then can the citric acid cycle actually begin and so let's focus on step one in step one we have a water molecule this acetyl coenzyme a as well as an oxaloacetate react to ultimately produce a six carbon molecule known as citrate so notice we begin with the oxaloacetate of four carbon molecules 2 1 2 3 4 and we begin with this acetyl coenzyme a that contains this 2 carbon component and so ultimately what happens in step 1 is the enzyme known as citrate synthase synthase catalyzes the transfer of this acetyl group from the cetyl coenzyme a and onto this oxalá acetate and that forms this citric molecule so citrate is actually a conjugate base of citric acid and that's why we call this the citric acid cycle now this is also an example of a tri carboxylic acid and that's why this is sometimes known that as the TCA cycle tricarboxylic acid cycle now this step is an exergonic step and under physiological cell conditions it releases about negative 31.4 kilojoules per mole of energy and this step as we discussed previously actually consists of two different steps the first step is an aldol condensation the second step is a hydration reaction but ultimately we form this citrate molecule from the oxaloacetate and this R and this acetyl coenzyme a now once we form the citrate molecule it must be transformed into its isomer molecule known as Isis citrate why well because only the isocitrate can actually undergo the decarboxylation step that takes place in step 3 so in step 2 we have an isomerization reaction that is catalyzed by a connotates and what acontece does is it basically transfers this hydroxyl group shown in orange from this carbon onto this carbon here so that's the only difference between this citrate molecule and the isomer isocitrate but now this molecule is actually activated and it's ready to undergo the decarboxylation step that takes place in step 3 now by the way I forgot to mention if we go back to this situate molecule here notice that we color-coded this molecule and that's because this violet region basically comes from this section here and this second oxygen comes from this water molecule shown in blue now I color-coded this orange because this is the molecule that is ultimately being transferred onto this carbon here so the only difference between situated isocitrate is the position of this hydroxyl group it is from this location on to this location here now once we form the isocitrate which by the way is actually an endergonic reaction we essentially use up about six point three kilojoules per mole of energy now once we actually form this isocitrate now it is ready to actually undergo the first oxidative decarboxylation step that takes place in the citric acid cycle so what we mean by an oxidative decarboxylation step is we actually have two reactions taking place we have an oxidation reduction reaction and we have a decarboxylation step and this step three is catalyzed by the enzyme known as isocitrate because that's the substrate molecule to the dehydrogenase enzyme so isocitrate dehydrogenase and remember whenever you hear the word dehydrogenase what that means is we're going to have an oxidation reduction reaction which electrons are going to be transferred onto a carrier molecule in this particular case the nicotine ad are the nicotinamide adenine dinucleotide nad plus so in step three we basically reduce the energy plus into NADH and the H ion and the two electrons essentially come from this molecule here so this is the hydride ion that is transferred onto the nad plus to form the NADH in the process we essentially oxidize the isocitrate so this is reduced then this is oxidized and we form the alpha ketoglutarate and also the process this entire component this carbon the oxide region is basically released and this H atom or the H ion attached to this oxygen is basically released as well and so we form the carbonyl group between this carbon shown here and so the alpha keto gluta rate is now basically ready to undergo the next step so in the next step what happens is once again we owe and by the way the amount of energy that is released in step three is equal to negative eight point four kilojoules per mole of energy and actually the formation of the alpha ketoglutarate is the rate determining step and we'll see why we'll talk about that in much more detail in a future lecture now once we form the alpha ketoglutarate if now undergo is a second oxidative decarboxylation step remember in the citric acid cycle we have two oxidative decarboxylation steps and this is the second one so in step four we take the alpha keto gluta rate we essentially reacted with coenzyme a the same coenzyme a that we released in this particular case and we also have the nad plus the carrier for the electron so in this step what we want to do is we want to essentially kick off this carbon dioxide shown in blue and replace that with a coenzyme a which is what we have in this particular case in the process we also abstract our electrons a hydride ion and we place it onto the NADH or the nad to form the NADH so the nad is reduced and the alpha Q to glue to rate is oxidized we release the carbon dioxide and an H+ ion and notice we attach the coenzyme a so we form this very high-energy bond known as the thio ester bond so this is essentially the same bond that we have here so we have that shown here and the key here is because we form this thio ester bond it will be very unstable high in energy and when we cleave this bond it will release a certain amount of free energy that will allow us to actually carry out step 5 now let's let's go back to step 4 in a moment for a moment because we didn't mention the enzyme that catalyzes this step so we have the substrate molecule is alpha ketoglutarate and the enzyme is basically a dehydration a dehydrogenase complex so alpha keto gluta raid hydrogenase complex and it's a complex because we actually have three different enzymes involved in this process and this complex is very similar to the complex that we spoke about now discussion on pyruvate decarboxylation so once we form this molecule this molecule that contains the high-energy thioester bond is known as succinylcholine a and what happens next is we essentially cleave this high-energy bond that releases a certain amount of energy and the energy that is released is used to basically drive this reaction here the addition of the ortho phosphate onto the gdp to form the GTP in the process we release the coenzyme a now by the way step four releases negative thirty point one kilojoules per mole of energy while step five release is negative three point three kilojoules per mole of energy and we also produce the GTP so step five is the only step of the citric acid cycle that actually generates this high energy purine nucleoside triphosphate molecule the GTP now the GTP can either be transformed into ATP or it can actually be used by for instance a g-protein to carry out some type of specific process in the cell for instance a signal transduction pathway now the product molecule of Step five is succinate and notice that even though we had all these color-coded atoms in these steps here we don't have the color-coded atoms in this step and that's because now this molecule is symmetric no look we have both of these ends contain the C double o negatively charged group and then we have the methylene group in between and so this is a completely symmetrical molecule and that's why we no longer actually use these color-coded atoms now once we form the succinate the next reaction step six or actually as should general once we form the succinate molecule notice that we lost the carbon dioxide's two of them and so we went from a six carbon molecule to a four carbon molecule and now in step six seven and eight entire point it's truth to transform this four carbons succinate into a four-carbon oxaloacetate so that the citric sub the citric acid cycle can basically begin all over again so that's the goal in step six seven and eight so step six is actually a hydration area or an oxidation reaction oxidation reduction reaction this is a hydration reaction this is another oxidation reduction reaction and ultimately we transform this methylene group into this carbonyl group and going through it and going from this molecule to this molecule in these in these three steps so let's focus on step six step six is catalyzed by succinate dehydrogenase and what this does is it ultimately abstracts two H atoms and those two H atoms are then carried by the fa d so we form the fadh2 so we essentially reduce the fa d into this and we oxidize this molecule into fumarate which has a double bond so ultimately one h atom and one h atom here so these two H atoms are basically abstracted and they are placed onto the fa d and so then we formed a double bond between these two carbons to form that fumarate and this process actually is at equilibrium it has a Gibbs free energy value of zero kilojoules per mole under the conditions that we find in ourselves now in step seven this is a hydration step and it's calor and it's catalyzed by fumer rates and what fumarate does is it basically attaches a hydroxyl group from water onto this side and the H ion is attached on to this side and so we form the l isomer of malate and once malate is four oh and this reaction releases negative 3.8 kilojoules per mole of energy and once we form the Mawlid now the final enzyme malate dehydrogenase is able to actually reduce the nad plus into NADH releasing an H+ ion in the process we oxidize the malate into oxaloacetate and now we regenerate this same molecule that we begin with and we can use this same oxaloacetate to basically carry out that same process all over again so if we sum up all these steps oh and by the way this final step in oxidation reduction reaction is a very endergonic reaction it requires about 29.7 kilojoules per mole of energy now it fits so and organic why does it actually take place well because it produces the nadh molecule that then goes on into the electron transport chain and when this is oxidized by the proteins of electron transport chain that process is exergonic and that process helps drive this process here in addition once we form the oxaloacetate it goes on to carry out step 1 and step 1 is a very exergonic process so these two steps the oxidation of NADH into nad plus along the electron transport chain and step one of the citric acid cycle helps drive this final step that is a very endergonic step so if we sum up all these reactions this will be the net equation and reaction of the citric acid cycle so we input an acetyl coenzyme a and step one we have three and AZ plus molecules one here one here and one here we have a single fvd molecule here we have a GDP na and an inorganic orthophosphate and two water molecules one here and one here and so ultimately we produce the coenzyme a basically here we have the three nadh is one in step three one in step four and one in step eight we have the fadh2 one in step six we have the gtp that is produced in step five and we have the two carbon dioxides and the two H+ ions these two carbon dioxide's are produced one is produced in step three and the other one is produced in step four

Channel: AK LECTURES

Views: 241,298

Rating: 4.9381442 out of 5

Keywords: citric acid cycle, Krebs cycle, tricarboxylic acid cycle, TCA cycle, overview of citric acid cycle, summary of citric acid cycle, steps of citric acid cycle, biochemsitry, net result of citric acid cycle

Id: fOY03Wl8VG4

Length: 15min 56sec (956 seconds)

Published: Tue May 26 2015