Question

The correct sequence of reactions in Krebs cycle is:

• A. Isocitric acid $ \rightarrow $ $\alpha $ -ketoglutaric acid $ \rightarrow $ oxalosuccinic acid
• B. Isocitric acid $ \rightarrow $ oxalosuccinic acid $ \rightarrow $ $\alpha $ -ketoglutaric acid
• C. Oxalosuccinic acid $ \rightarrow $ Citric acid $ \rightarrow $ $\alpha $ -ketoglutaric acid
• D. $\alpha $ -ketoglutaric acid $ \rightarrow $ isocitric acid $ \rightarrow $ oxalosuccinic acid

Answer 1

Answer: (B)

Acetyl-CoA donates its acetyl group to the four-carbon compound oxaloacetate to form the six-carbon citrate. Citrate is then transformed into isocitrate, also a six-carbon molecule, which is dehydrogenated with loss of CO₂ to yield the five-carbon compound α $-$ ketoglutarate. The latter undergoes loss of CO₂ and ultimately yields the four-carbon compound succinate and a second molecule of CO₂. Succinate is then enzymatically converted in three steps into the four-carbon oxaloacetate, with which the cycle began; thus, oxaloacetate is ready to react with another molecule of acetyl-CoA to start a second turn. In each turn of the cycle, one acetyl group (two carbons) enters as acetyl-CoA and two molecules of CO₂ leave. In each turn, one molecule of oxaloacetate is used to form citrate but-after a series of reactions the oxaloacetate is regenerated. Therefore no net removal of oxaloacetate occurs; one molecule of oxaloacetate can theoretically suffice to bring about oxidation of an infinite number of acetyl groups. Four of the eight steps in this process are oxidations, in which the energy of oxidation is conserved, with high efficiency, in the formation of reduced cofactors (NADH and FADH₂).