During the citric acid cycle, what happens to acetyl-coa? it enters the citric acid cycle and gains carbon dioxide to form citric acid, and gains more carbon dioxide through redox reactions to form a 4-carbon molecule. it enters the citric acid cycle and associates with a 4-carbon molecule, forming citric acid, and then through redox reactions regenerates the 4-carbon molecule. it enters glycolysis and associates with a 5-carbon molecule through redox reactions, forming another acetyl-coa molecule. it enters the citric acid cycle and associates with a 4-carbon molecule, forming a 5-carbon compound, and then through oxidation reactions regenerates the 4-carbon molecule.

The correct answer is: It enters the citric acid cycle and associates with a 4-carbon molecule (OXALOACETATE), forming citric acid, and then through redox reactions regenerates the 4-carbon molecule.

The citric acid cycle is a cyclic process (part of the cellular respiration) where acetyl CoA is modified in the mitochondria to produce high-energy intermediate molecules for the oxidative phosphorilation (production of ATP).

Citric acid cycle is also known as Kreb's cycle. The cycle involve the series of chemical reactions to oxidise the acetyl coenzyme A to yield energy. The acetyl coenzyme A when entering the Kreb's cycle:

Citric acid cycle or tricarboxylic acid cycle takes place in the mitochondria of the cell. The process involves the oxidation of acetyl coenzyme A to form a four carbon molecule, known as oxaloacetate. The cycle is one of the processes occurring during the cellular respiration. The molecules produced at the end of the cycle are utilised for oxidative phosphorylation to generate high-energy molecules, ATP.

The acetyl Co A combines with oxaloacetate in the presence of enzyme citrate synthase. The enzyme facilitates the formation of citrate.

Therefore, the acetyl coenzyme A enters the cycle and associates with the four carbon compound oxaloacetate to form citrate.

To know more about citric acid cycle, refer to the following link:

link

answer; the correct answer is (beta)

i think the answer is g1

hope i

-lavira: )