Question:

The first stable product of $C_3$ pathway is

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$C_3$ plants fix $\text{CO}_2$ using Rubisco to form 3-phosphoglycerate (3-C).
$C_4$ plants fix $\text{CO}_2$ using PEP carboxylase to form oxaloacetate (4-C).
  • Phosphoglycerate
  • Oxaloacetate
  • Pyruvic acid
  • Malate dehydrogenase
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The Correct Option is A

Solution and Explanation

Step 1: Understanding the Concept:
The $C_3$ pathway, also known as the Calvin cycle, is the chemical cycle of carbon fixation in photosynthetic plants.
The pathway is named "$C_3$" because the first stable organic intermediate formed contains three carbon atoms.

Step 2: Detailed Explanation:

In the first phase of the Calvin cycle (carboxylation), carbon dioxide ($\text{CO}_2$) is fixed onto a 5-carbon acceptor molecule, ribulose-1,5-bisphosphate (RuBP).
This reaction is catalyzed by the enzyme ribulose-1,5-bisphosphate carboxylase/oxygenase (Rubisco).
The resulting unstable 6-carbon intermediate immediately splits into two molecules of 3-phosphoglyceric acid (3-PGA or 3-phosphoglycerate).
Because 3-PGA is a 3-carbon compound and is the first stable intermediate of the process, the pathway is termed the $C_3$ pathway.
In $C_4$ plants, the first stable product is oxaloacetate (OAA), which contains 4 carbon atoms.
Pyruvic acid is a key intermediate in glycolysis, and malate dehydrogenase is an enzyme that regulates the interconversion of malate and oxaloacetate.

Step 3: Final Answer:

The first stable product of the $C_3$ pathway is Phosphoglycerate, which corresponds to Option (A).
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