Question

1. Describe the reaction steps of the citric acid cycle with a focus on their control. In addition discuss the energy provided to a living cell from the metabolism of one acetyl-CoA entering the citric acid cycle.

biochemistry

Answer 1

The Citric Acid Cycle:

step 1: The cycle begins by joining Acetyl CoA with oxaloacetate, and forming Citrate with a release of a CoA molecule. This step happens in the presence of Citrate Synthase.

Step 2: The citrate is then converted into Isocitrate, which is its isomer. This happens in the presence of the enzyme Aconitase.

Step 3: in this step, Oxidation of Isocitrate takes place, releasing a CO₂ molecule and converting into ?-ketoglutarate. Occurring in the presence of Isocitrate dehydrogenase, NAD⁺ is reduced and NADH is formed. This step regulates the speed of the cycle

step 4: Being similar to the previous step,  ?-ketoglutarate is oxidized and converted into an unstable compound called succinyl CoA. the enzyme ?-ketoglutarate dehydrogenase catalyzes this reaction and is also invovled in the regulation of the cycle.

step 5: The CoA from succinyl CoA is removed and replaced by a phosphate which will later be consumed by ADP to convert into ATP. The molecule formed is Succinate.

step 6: Succinate is oxidized to form Fumarate. here electron exchange occurs, converting FAD to FADH₂ .

Step 7: Water is added to Fumarate to form Malate.

step 8: the oxaloacetate is regenerated by the oxidation of Malate.

With only one molecule of Acetyl CoA, 3 molecules of NADH, 1 molecule of FADH2 and 1 molecule of ATP/GTP is formed.

Now energy in the direct form of ATP is less in the Citric acid cycle, but indirect production of energy is high, in the form of NADH and FADH2.