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Question
If 9 ATP and 6 NADPH are utilized for photosynthetic carbon assimilation through Calvin cycle, what would be the ratio of erythrose 4- phosphate, xylulose 5- phosphate and ribulose 5- phosphate molecules formed as intermediates in regeneration phase of Calvin cycle?
If 9 ATP and 6 NADPH are utilized for photosynthetic carbon assimilation through Calvin cycle, what would be the ratio of erythrose 4- phosphate, xylulose 5- phosphate and ribulose 5- phosphate molecules formed as intermediates in regeneration phase of Calvin cycle?
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Solution
The correct option is D 1:2:3
The regeneration stage can be broken down into steps.
- Triose phosphate isomerase converts all of the G3P reversibly into dihydroxyacetone phosphate (DHAP), also a 3-carbon molecule.
- Aldolase and fructose-1,6-bisphosphatase convert a G3P and a DHAP into fructose 6-phosphate (6C). A phosphate ion is lost into solution.
- Then fixation of another CO2 generates two more G3P.
- F6P has two carbons removed 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 (formed from one of the G3P from the second CO2 fixation) are converted into sedoheptulose-1,7-bisphosphate (7C) by aldolase enzyme.
- Sedoheptulose-1,7-bisphosphatase (one of only three enzymes of the Calvin cycle that are unique to plants) cleaves sedoheptulose-1,7-bisphosphate into sedoheptulose-7-phosphate, releasing an inorganic phosphate ion into solution.
- Fixation of a third CO2 generates two more G3P. The ketose S7P has two carbons removed by transketolase, giving ribose-5-phosphate (R5P), and the two carbons remaining on transketolase are transferred to one of the G3P, giving another Xu5P. This leaves one G3P as the product of fixation of 3 CO2, with generation of three pentoses that can be converted to Ru5P.
- R5P is converted into ribulose-5-phosphate (Ru5P, RuP) by phosphopentose isomerase. Xu5P is converted into RuP by phosphopentose epimerase.
Finally, phosphoribulokinase (another plant-unique enzyme of the pathway) phosphorylates RuP into RuBP, ribulose-1,5-bisphosphate, completing the Calvin cycle. This requires the input of one ATP.
So, the answer is 1 : 2 : 3.
The regeneration stage can be broken down into steps.
- Triose phosphate isomerase converts all of the G3P reversibly into dihydroxyacetone phosphate (DHAP), also a 3-carbon molecule.
- Aldolase and fructose-1,6-bisphosphatase convert a G3P and a DHAP into fructose 6-phosphate (6C). A phosphate ion is lost into solution.
- Then fixation of another CO2 generates two more G3P.
- F6P has two carbons removed 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 (formed from one of the G3P from the second CO2 fixation) are converted into sedoheptulose-1,7-bisphosphate (7C) by aldolase enzyme.
- Sedoheptulose-1,7-bisphosphatase (one of only three enzymes of the Calvin cycle that are unique to plants) cleaves sedoheptulose-1,7-bisphosphate into sedoheptulose-7-phosphate, releasing an inorganic phosphate ion into solution.
- Fixation of a third CO2 generates two more G3P. The ketose S7P has two carbons removed by transketolase, giving ribose-5-phosphate (R5P), and the two carbons remaining on transketolase are transferred to one of the G3P, giving another Xu5P. This leaves one G3P as the product of fixation of 3 CO2, with generation of three pentoses that can be converted to Ru5P.
- R5P is converted into ribulose-5-phosphate (Ru5P, RuP) by phosphopentose isomerase. Xu5P is converted into RuP by phosphopentose epimerase.
So, the answer is 1 : 2 : 3.
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