Question

If $9ATP$ and $6NADPH$ are utilized for photosynthetic carbon assimilation through the Calvin cycle, what would be the ratio of erythrose$4$- phosphate, xylulose$5$- phosphate, and ribulose $5$- phosphate molecules formed as intermediates in the regeneration phase of the Calvin cycle?
A. $3:1:2$
B. $2:2:1$
C. $2:3:1$
D. $1:2:3$

Answer 1

The photosynthesis in prokaryotes occurs by two general methods- light-dependent reactions and the Calvin cycle, which initiates in the chloroplasts.
You may have seen that the general response for photosynthesis:
$6C{O_2} + 6{H_2}O \to {C_6}{H_{12}}{O_6} + 6{O_2}$
is the converse of the general response for cell respiration:
$6{O_2} + {C_6}{H_{12}}{O_6} \to 6C{O_2} + 6{H_2}O$
Photosynthesis produces oxygen as a result, and respiration produces carbon dioxide as a side-effect.

Complete answer:
The regeneration stage can be separated into steps.

Triosephosphate isomerase changes over the entirety of the $G3P$ reversibly into dihydroxyacetone phosphate (DHAP), additionally a 3-carbon atom.
$Aldolase and fructose - 1,6 - bisphosphatase$ convert a $G3P$ and a DHAP into fructose $6$ -phosphate ($6C$). A phosphate particle is lost into the arrangement.
At that point, the fixation of another $C{O_2}$ creates two more $G3P$.
F6P has two carbons eliminated by transketolase, giving erythrose-$4$ phosphate. The two carbons on transketolase are added to a $G3P$, giving the ketose xylulose- $5$ -phosphate ($Xu5P$).
E4P and a DHAP (framed from one of the $G3P$from the subsequent $C{O_2}$ fixation) are changed over into sedoheptulose$- 1,7 -$bisphosphate ($7$C) by aldolase protein.
Sedoheptulose$- 1,7 -$bisphosphatase (one of just three proteins of the Calvin cycle that are extraordinary to plants) cuts sedoheptulose$- 1,7 -$bisphosphate into sedoheptulose-$7$-phosphate, delivering an inorganic phosphate particle into the arrangement.
Regeneration of a third $C{O_2}$

​

Creates two more $G3P$. The ketose $S7P$ has two carbons eliminated by transketolase, giving ribose-5-phosphate (R5P), and the two carbons staying on transketolase are moved to one of the $G3P$, giving another Xu5P. This leaves one $G3P$ due to an obsession of 3 $C{O_2}$, with the age of three pentoses that can be changed over to Ru$5$P.
R5P is changed over into ribulose-$5$-phosphate (Ru$5$P, RuP) by phosphopeptide isomerase. Xu5P is changed over into RuP by phosphopentose epimerase.
Finally, phosphoribulokinase (another plant-interesting chemical of the pathway) phosphorylates RuP into RuBP, ribulose$- 1,5 -$bisphosphate, finishing the Calvin cycle. This requires the contribution of one ATP.

Hence, the correct answer is option (D).

Note: Utilizing the energy transporters framed in the primary phase of photosynthesis, the Calvin cycle responses fix $C{O_2}$from the climate to assemble starch particles. A compound, RuBisCO, catalyzes the obsession response by joining $C{O_2}$ with RuBP. The subsequent six-carbon mixture is separated into two three-carbon mixes, and the energy in ATP and NADPH is utilized to change over these atoms into $G3P$.

Plants are equipped for both photosynthesis and cell respiration since they contain the two chloroplasts and mitochondria.