Six turns of the Calvin cycle are required to generate one mole of glucose. Explain.

(N/A) The Calvin cycle is the primary pathway for carbon fixation in $C_3$ plants,occurring in three main stages: Carboxylation,Reduction,and Regeneration.
$1$. Carboxylation: This is the fixation of $CO_2$ into a stable organic intermediate. $CO_2$ is utilized for the carboxylation of $RuBP$ (Ribulose$-1,5-$bisphosphate) by the enzyme $RuBisCO$,resulting in the formation of two molecules of $3-PGA$ ($3$-phosphoglycerate).
$2$. Reduction: This stage involves a series of reactions leading to the formation of glucose. For every molecule of $CO_2$ fixed,$2$ molecules of $ATP$ are used for phosphorylation and $2$ molecules of $NADPH$ are used for reduction. This converts $2$ molecules of $PGA$ into $2$ molecules of Triose phosphate $(PGAL)$.
$3$. Regeneration: To keep the cycle running,the $CO_2$ acceptor molecule $RuBP$ must be regenerated. This step requires $1$ $ATP$ molecule per $CO_2$ fixed.
Calculation for one glucose molecule:
- Each turn of the Calvin cycle fixes one molecule of $CO_2$.
- Glucose is a $6$-carbon sugar $(C_6H_{12}O_6)$.
- Therefore,to synthesize one molecule of glucose,$6$ molecules of $CO_2$ must be fixed.
- Since one turn of the cycle fixes one $CO_2$,$6$ turns of the cycle are required to produce one molecule of glucose.