Citrate,the first stable product of the Krebs | vedclass

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(A) In the Krebs cycle (Tricarboxylic Acid Cycle),$1$ molecule of citrate $(6C)$ is converted back to oxaloacetate $(4C)$.During this process,the foll

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Four and two

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(A) In the Krebs cycle (Tricarboxylic Acid Cycle),$1$ molecule of citrate $(6C)$ is converted back to oxaloacetate $(4C)$.During this process,the following steps occur:$1$. Decarboxylation: Two molecules of $CO_2$ are released. One during the conversion of isocitrate to $\alpha$-ketoglutarate,and another during the conversion of $\alpha$-ketoglutarate to succinyl-CoA.$2$. Oxidation: Four oxidation steps occur where $NAD^+$ or $FAD$ are reduced to $NADH$ or $FADH_2$. These are: $(i)$ Isocitrate to $\alpha$-ketoglutarate,(ii) $\alpha$-ketoglutarate to succinyl-CoA,(iii) Succinate to fumarate,and (iv) Malate to oxaloacetate.Therefore,there are $4$ oxidation steps and $2$ decarboxylation steps.

Citrate,the first stable product of the Krebs cycle,undergoes how many oxidation and decarboxylation steps respectively to regenerate oxaloacetate?

Similar Questions

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