Answer
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Hint: The Calvin cycle is a process in which the first stable product of $CO_2$ fixation is $C_3$ acid that is phosphoglyceric acid. Calvin cycle and Hatch-Slack cycles are two carbon dioxide fixation pathways.
Complete answer:
> 4-C acceptor is oxalo-acetic acid and is the first product of the hatch and slack pathway. The 3-C compound is PGA which is the first stable product of the Calvin cycle. So this option is not correct.
> 4-C compound is again OAA and is the first product of $C_4$ cycle and 6-C compound is not formed in any cycle. So this option is not correct
> 3-C compound is PGA which is the first stable product of $C_3$ cycle and 4-C compound is OAA which is the first stable product of $C_4$ cycle. So this option is correct
> 5-C compound is RUBP but it is not a product that is a carbon acceptor so this option is also wrong.
Hence, The correct answer is option (C).
Additional information:
The Calvin cycle has three main steps namely:
- CARBOXYLATION- It is the. Fixation of carbon dioxide into a stable organic intermediate. It is the most crucial step of the Calvin cycle.
- REDUCTION: This step involves use of two molecules of ATP and two of NADPH for reduction of one $CO_2$
- REGENERATION: This Step requires some ATP for phosphorylation to form RuBP.
Note: C-4 cycle is known as hatch and slack pathway. They have 4 carbon compounds oxalo acetic acid but they use 3 carbon compounds for carbon dioxide fixation That is phosphoenolpyruvate and is present in Mesophyll cells. The enzyme is responsible for the fixation. The $C_4$ acid that is oxalo acetic acid is formed in mesophyll Cells.
Complete answer:
> 4-C acceptor is oxalo-acetic acid and is the first product of the hatch and slack pathway. The 3-C compound is PGA which is the first stable product of the Calvin cycle. So this option is not correct.
> 4-C compound is again OAA and is the first product of $C_4$ cycle and 6-C compound is not formed in any cycle. So this option is not correct
> 3-C compound is PGA which is the first stable product of $C_3$ cycle and 4-C compound is OAA which is the first stable product of $C_4$ cycle. So this option is correct
> 5-C compound is RUBP but it is not a product that is a carbon acceptor so this option is also wrong.
Hence, The correct answer is option (C).
Additional information:
The Calvin cycle has three main steps namely:
- CARBOXYLATION- It is the. Fixation of carbon dioxide into a stable organic intermediate. It is the most crucial step of the Calvin cycle.
- REDUCTION: This step involves use of two molecules of ATP and two of NADPH for reduction of one $CO_2$
- REGENERATION: This Step requires some ATP for phosphorylation to form RuBP.
Note: C-4 cycle is known as hatch and slack pathway. They have 4 carbon compounds oxalo acetic acid but they use 3 carbon compounds for carbon dioxide fixation That is phosphoenolpyruvate and is present in Mesophyll cells. The enzyme is responsible for the fixation. The $C_4$ acid that is oxalo acetic acid is formed in mesophyll Cells.
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