
Three carbons of phosphoglyceric acid formed during carbon fixation are derived from
A. \[{\rm{PEP + C}}{{\rm{O}}_{\rm{2}}}\]
B. \[{\rm{RuBP}}\]
C. \[{\rm{C}}{{\rm{O}}_2}\]
D. \[{\rm{RuBP + C}}{{\rm{O}}_2}\]
Answer
568.8k+ views
Hint:The fixation of carbon is a procedure involving the installation of a carbon. The RuBisCO enzyme catalyzes carbon dioxide in a two-step reaction to carboxylation of Ribulose-1,5-bisphosphate. Ribulose-1,5-bisphosphate is the first thing a cell wants.
Complete answer:
Carbon fixation is the mechanism of taking and tackling inorganic carbon to an organic molecule (usually carbohydrate). The carbon fixation process is in fact the first step in the Calvin cycle and also named as the $C_3$ cycle or the carbon fixation cycle.
A molecule known as Ribulose-1,5-bisphosphate or RuBP is the first thing a cell wants. Next is a special enzyme that requires the cell. Enzymes are proteins that accelerate the reaction to chemical substances. Ribulose-1,5-bisphosphate carboxylase has the enzyme that the cell requires to repair it.
Ribulose-1,5-bisphosphate Carboxylase is very mouthful, so it is also shortened up to RuBisCO. The RuBisCO enzyme is a carbon fixation enzyme. 3PG is made of the 3-phosphate (Ga3-P) glyceraldehyde, which is utilized in the processing or cycling of sugar or starch by the plant A \[{\rm{C}}{{\rm{O}}_{\rm{2}}}\] molecule incorporates RuBP with a molecule of 5-carbon acceptor. This step consists of a compound of 3-carbon, 3-phosphoglyceric acid, which is divided in two molecules.
Thus, the correct answer is option ‘D’. i.e., \[{\rm{RuBP + C}}{{\rm{O}}_2}\]
Note:The RuBisCO enzyme catalyzes carbon dioxide in a two-step reaction to carboxylation of Ribulose-1,5-bisphosphate, a 5-carbon compound. The first step result is a complex of enediol-enzymes trapping carbon dioxide or oxygen. The true carboxylase or oxygenase is thus the enediol-enzyme complex. The\[{\rm{C}}{{\rm{O}}_{\rm{2}}}\], which is captured by enediol in the second stage, initially results in a six-carbon intermediate which splits into half immediately forming two 3-phosphoglycerate (3-PGA) molecules, a 3-carbon compound.
Complete answer:
Carbon fixation is the mechanism of taking and tackling inorganic carbon to an organic molecule (usually carbohydrate). The carbon fixation process is in fact the first step in the Calvin cycle and also named as the $C_3$ cycle or the carbon fixation cycle.
A molecule known as Ribulose-1,5-bisphosphate or RuBP is the first thing a cell wants. Next is a special enzyme that requires the cell. Enzymes are proteins that accelerate the reaction to chemical substances. Ribulose-1,5-bisphosphate carboxylase has the enzyme that the cell requires to repair it.
Ribulose-1,5-bisphosphate Carboxylase is very mouthful, so it is also shortened up to RuBisCO. The RuBisCO enzyme is a carbon fixation enzyme. 3PG is made of the 3-phosphate (Ga3-P) glyceraldehyde, which is utilized in the processing or cycling of sugar or starch by the plant A \[{\rm{C}}{{\rm{O}}_{\rm{2}}}\] molecule incorporates RuBP with a molecule of 5-carbon acceptor. This step consists of a compound of 3-carbon, 3-phosphoglyceric acid, which is divided in two molecules.
Thus, the correct answer is option ‘D’. i.e., \[{\rm{RuBP + C}}{{\rm{O}}_2}\]
Note:The RuBisCO enzyme catalyzes carbon dioxide in a two-step reaction to carboxylation of Ribulose-1,5-bisphosphate, a 5-carbon compound. The first step result is a complex of enediol-enzymes trapping carbon dioxide or oxygen. The true carboxylase or oxygenase is thus the enediol-enzyme complex. The\[{\rm{C}}{{\rm{O}}_{\rm{2}}}\], which is captured by enediol in the second stage, initially results in a six-carbon intermediate which splits into half immediately forming two 3-phosphoglycerate (3-PGA) molecules, a 3-carbon compound.
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