Question

For fixing one molecule of $C{O_2}$ in Calvin cycle, …….. are required.
A. $3ATP + 1NADP{H_2}$
B. $3ATP + 2NADP{H_2}$
C. $2ATP + 3NADP{H_2}$
D. $3ATP + 3NADP{H_2}$

Answer 1

Hint: The Calvin cycle is a reaction that occurs in the stroma during photosynthesis in order to fix the atmospheric carbon dioxide. It consists of three steps and each step requires energy either in the form of $ATP$, or $NADPH$, or both. One molecule of carbon dioxide can be fixed in one cycle.

Complete answer:
For fixing one molecule of $C{O_2}$ in the Calvin cycle, $3ATP$ and $2NADP{H_2}$ are required.

The Calvin cycle comprises three main steps:
> Carbon Fixation
In this step, a $C{O_2}$ molecule binds with a \[Ribulose - 1,5 - bisphosphate\] $(RuBP)$, which is an acceptor molecule. The binding results in the formation of a six-carbon compound that eventually splits to form 2 compounds of 3 carbons each known as \[3 - phosphoglyceric\] acid $(3 - PGA)$.

> Reduction
In this step, one molecule of $3 - PGA$ is converted into \[glyceraldehyde - 3 - phosphate\]
$(G_{3}P)$. This reaction is facilitated by $1ATP$ and $1NADPH$.
So, when 2 molecules of $3 - PGA$are converted into 2 molecules of $G_{3}P$, $2ATP$ and $2NADPH$ are utilized.

> Regeneration
In this reaction, some molecules of $G_{3}P$ are used to make glucose and some are recycled to produce $RuBP$ acceptor. This requires 1 molecule of $ATP$.

So, when we combine all three reactions, we can see that in one cycle, 3 molecules of $ATP$ and 2 molecules of $NADPH$ are required.
Hence, the correct answer is option (B).

Note: 3 turns of the Calvin cycle are required to fix 3 molecules of $C{O_2}$., which results in the production of 6 $G_{3}P$ molecules. Out of these, 1 molecule exits the cycle to form glucose and 5 molecules are recycled to form $RuBP$ acceptors. So, it would take 6 turns of the Calvin cycle to produce glucose (\[6 - carbon \] sugar).