Describe the Calvin cycle.

(N/A) Melvin Calvin and his co-workers worked out the whole pathway and showed that the pathway operated in a cyclic manner,where the $RuBP$ was regenerated.
The Calvin pathway occurs in all photosynthetic plants,regardless of whether they have $C_3$ or $C_4$ (or any other) pathways.
The Calvin cycle can be described under three stages: $(1)$ Carboxylation,$(2)$ Reduction,and $(3)$ Regeneration.
$(1)$ Carboxylation: Carboxylation is the fixation of $CO_2$ into a stable organic intermediate. It is the most crucial step of the Calvin cycle where $CO_2$ is utilised for the carboxylation of $RuBP$. This reaction is catalysed by the enzyme $RuBP$ carboxylase,which results in the formation of two molecules of $3-PGA$. Since this enzyme also has an oxygenation activity,it is more correctly called $RuBP$ carboxylase-oxygenase or $RuBisCO$.
$(2)$ Reduction: These are a series of reactions that lead to the formation of glucose. These steps involve the utilisation of $2$ molecules of $ATP$ for phosphorylation and $2$ molecules of $NADPH$ for reduction per $CO_2$ molecule fixed.
$3-PGA + 2 ATP + 2 NADPH \rightarrow 2$ Triose phosphate $(PGAL)$
The fixation of six molecules of $CO_2$ and $6$ turns of the cycle are required for the removal of one molecule of glucose from the pathway.
$(3)$ Regeneration: Regeneration of the $CO_2$ acceptor molecule $RuBP$ is crucial if the cycle is to continue uninterrupted. The regeneration steps require one $ATP$ for phosphorylation to form $RuBP$.
$PGAL \xrightarrow{ATP} RuBP$
Hence,for every $CO_2$ molecule entering the Calvin cycle,$3$ molecules of $ATP$ and $2$ molecules of $NADPH$ are required. It is probably to meet this difference in the number of $ATP$ and $NADPH$ used in the dark reaction that cyclic photophosphorylation takes place. To make one molecule of glucose,$6$ turns of the cycle are required.