
Radioactive ${C^{14}}$ is given to $C{O_2}$ and released to the atmosphere. This $C{O_2}$ is taken by ribulose diphosphate in a C3 plant. First radioactive ${C^{14}}$ is seen in which compound?
A) PGAL
B) PEP
C) RMP
D) PGA
Answer
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Hint: C3 carbon fixation is the most usual of three metabolic pathways for carbon fixation in photosynthesis, down with C4 and CAM. This process transfers carbon dioxide and ribulose bisphosphate (RuBP, a 5-carbon sugar) into two molecules of 3-phosphoglycerate via the following reaction. C3 plants cannot grow in very searing areas because RuBisCO incorporates more oxygen into RuBP as temperatures boost.
Complete answer:
${C^{14}}$ an isotope of carbon has been utilized to depict the path of carbon dioxide fixation in the Kelvin cycle. The principal function of the Calvin cycle is to transform carbon dioxide into usable energy known as glucose.
This labeled ${C^{14}}$ becomes involved with ${}^{14}C{O_2}$ which is admitted by RuBP and a 3-C compound phosphoglyceric acid (3 PGA) is created, PGA is the first stable compound. It is a three- carbon monosaccharide that is an intermediate in the Calvin cycle of photosynthesis and also in glycolysis.
PEP is an important intermediate in biochemistry and is involved in glycolysis and gluconeogenesis. PGAL (phosphoglyceraldehyde) is a triose phosphate or glyceraldehyde phosphate. A three-carbon molecule is generated from fructose diphosphate within glycolysis. The ribulose monophosphate (RMP) is an extremely proficient route for the incorporation of reduced one-carbon compounds.
Hence, the correct answer is option (D).
Note: 3-Phosphoglyceric acid is a fragment that is biochemically major metabolic intermediate in both glycolysis and the Calvin cycle. This anion is often known as PGA when referring to the Calvin cycle. In the Calvin cycle, 3-phosphoglycerate is the creation of the spontaneous scission of an unstable 6-carbon intermediate formed upon carbon dioxide fixation. So, two equivalents of 3-phosphoglycerate are generated for every molecule of carbon dioxide that is fixed.
Complete answer:
${C^{14}}$ an isotope of carbon has been utilized to depict the path of carbon dioxide fixation in the Kelvin cycle. The principal function of the Calvin cycle is to transform carbon dioxide into usable energy known as glucose.
This labeled ${C^{14}}$ becomes involved with ${}^{14}C{O_2}$ which is admitted by RuBP and a 3-C compound phosphoglyceric acid (3 PGA) is created, PGA is the first stable compound. It is a three- carbon monosaccharide that is an intermediate in the Calvin cycle of photosynthesis and also in glycolysis.
PEP is an important intermediate in biochemistry and is involved in glycolysis and gluconeogenesis. PGAL (phosphoglyceraldehyde) is a triose phosphate or glyceraldehyde phosphate. A three-carbon molecule is generated from fructose diphosphate within glycolysis. The ribulose monophosphate (RMP) is an extremely proficient route for the incorporation of reduced one-carbon compounds.
Hence, the correct answer is option (D).
Note: 3-Phosphoglyceric acid is a fragment that is biochemically major metabolic intermediate in both glycolysis and the Calvin cycle. This anion is often known as PGA when referring to the Calvin cycle. In the Calvin cycle, 3-phosphoglycerate is the creation of the spontaneous scission of an unstable 6-carbon intermediate formed upon carbon dioxide fixation. So, two equivalents of 3-phosphoglycerate are generated for every molecule of carbon dioxide that is fixed.
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