One-sixth of the total PGAL produced is used for the synthesis of
(a) Glucose
(b) RuBP
(c) RuMP
(d) DHAP
Answer
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Hint: To make one six-carbon compound, it takes 6 Calvin cycles, and for each ${CO }_{ 2 }$ molecule it requires 3ATP and 2NADPH molecules. An intermediate three-carbon compound is formed during the Calvin cycle.
Complete answer:
In the Calvin cycle, RuBP is carboxylated by utilizing ${ $CO }_{ 2 }$resulting in the formation of phosphoglyceric acid. It is reduced into phosphoglyceraldehyde by using ATP and NADPH. In the regulation step, this phosphoglycerate is used in the synthesis of glucose. To make one molecule of glucose 6 turns of the cycle is required
Additional information:
Carboxylation:
- ${ 6RUBP }+{ 6CO }_{ 2 }+{ 6H }_{ 2 }O\xrightarrow { RuBisCo } { 12Phosphoglyceric\quad acid }$
Reduction:
- $12Phosphoglyceric\quad acid+{ 12ATP\xrightarrow { Phosphoglycerokinase } { 12Bisphosphoglyceric\quad acid+{ 12ADP } } }$
- ${ 12Bisphosphoglyceric\quad acid+{ 12NADPH\xrightarrow { G-3-P\quad dehydrogenase } { 12Glyceraldehyde-3-phosphate{ { 12NADP }^{ + }+{ 12H }_{ 2 }P{ O }_{ 4 }^{ - } } } } }$
Regeneration:
- ${ 2G-3-P+{ 2DHAP\xrightarrow { Aldolase } }{ 2Fructose-1,6-bisphosphate } }$
-$2Fructose-1,6-bisphosphate\xrightarrow { Fructose-1,6-bisphosphatase } { 2-Fructose-6\quad phosphate+{ 2Pi } }\\ 2Fructose-6phosphate+2G-3-P\xrightarrow { Transketolase } 2Xylulose-5-phosphate(X-5-P)+2Erythrose-4-Phosphate(E-4-P)\\ 2E-4-P+2DHAP\xrightarrow { Aldolase } 2Sedoheptulose-1,7-biphosphate\\ Sedoheptulose-1,7-biphosphate\xrightarrow { Sedoheptulose-1,7-biphosphatase } Sedoheptulose-7-phosphate+2pi\\ Sedoheptulose-7-phosphate+2G-3-P\xrightarrow { Transketolase } 2Xylulose-5-p+2Ribose-5-Phosphate\\ 4Xylulose-5-p\xrightarrow { Ribulose-5-phosphate\quad epimerase } 4Ribulose-5-phosphate\\ 2Ribose-5-P\xrightarrow { Ribose-5-phosphate\quad isomerase } 2Ribulose-5-phosphate\\ 6Ribulose-5-phosphate+6ATP\xrightarrow { Ribulose-5-phospho\quad kinase } 6Ribulose-1,5-bisphosphate+6ADP\\$
Thus, to make one molecule of glucose, six cycles of calvin cycle are required
So, the correct answer is ‘Glucose’.
Note: From $12G-3-P$, the net gain of ATP is two, and to regenerate 6RUBP, $10G-3-P$ are used.
3 molecules of ATP and 2 molecules of NADPH are required for every $CO_2$ molecule that enters the Calvin cycle. To produce one molecule of glucose it takes 18ATP and 12NADPH molecules.
Complete answer:
In the Calvin cycle, RuBP is carboxylated by utilizing ${ $CO }_{ 2 }$resulting in the formation of phosphoglyceric acid. It is reduced into phosphoglyceraldehyde by using ATP and NADPH. In the regulation step, this phosphoglycerate is used in the synthesis of glucose. To make one molecule of glucose 6 turns of the cycle is required
Additional information:
Carboxylation:
- ${ 6RUBP }+{ 6CO }_{ 2 }+{ 6H }_{ 2 }O\xrightarrow { RuBisCo } { 12Phosphoglyceric\quad acid }$
Reduction:
- $12Phosphoglyceric\quad acid+{ 12ATP\xrightarrow { Phosphoglycerokinase } { 12Bisphosphoglyceric\quad acid+{ 12ADP } } }$
- ${ 12Bisphosphoglyceric\quad acid+{ 12NADPH\xrightarrow { G-3-P\quad dehydrogenase } { 12Glyceraldehyde-3-phosphate{ { 12NADP }^{ + }+{ 12H }_{ 2 }P{ O }_{ 4 }^{ - } } } } }$
Regeneration:
- ${ 2G-3-P+{ 2DHAP\xrightarrow { Aldolase } }{ 2Fructose-1,6-bisphosphate } }$
-$2Fructose-1,6-bisphosphate\xrightarrow { Fructose-1,6-bisphosphatase } { 2-Fructose-6\quad phosphate+{ 2Pi } }\\ 2Fructose-6phosphate+2G-3-P\xrightarrow { Transketolase } 2Xylulose-5-phosphate(X-5-P)+2Erythrose-4-Phosphate(E-4-P)\\ 2E-4-P+2DHAP\xrightarrow { Aldolase } 2Sedoheptulose-1,7-biphosphate\\ Sedoheptulose-1,7-biphosphate\xrightarrow { Sedoheptulose-1,7-biphosphatase } Sedoheptulose-7-phosphate+2pi\\ Sedoheptulose-7-phosphate+2G-3-P\xrightarrow { Transketolase } 2Xylulose-5-p+2Ribose-5-Phosphate\\ 4Xylulose-5-p\xrightarrow { Ribulose-5-phosphate\quad epimerase } 4Ribulose-5-phosphate\\ 2Ribose-5-P\xrightarrow { Ribose-5-phosphate\quad isomerase } 2Ribulose-5-phosphate\\ 6Ribulose-5-phosphate+6ATP\xrightarrow { Ribulose-5-phospho\quad kinase } 6Ribulose-1,5-bisphosphate+6ADP\\$
Thus, to make one molecule of glucose, six cycles of calvin cycle are required
So, the correct answer is ‘Glucose’.
Note: From $12G-3-P$, the net gain of ATP is two, and to regenerate 6RUBP, $10G-3-P$ are used.
3 molecules of ATP and 2 molecules of NADPH are required for every $CO_2$ molecule that enters the Calvin cycle. To produce one molecule of glucose it takes 18ATP and 12NADPH molecules.
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