Stoichiometry of the Calvin Cycle (With Diagram) | Photosynthesis

In this article, we will discussion about the stoichiometry of the Calvin cycle.

Stoichiometry is that part of chemistry which deals with the determination of combin­ing proportions or chemical equivalents.

Fig. 11.20 shows the simplified version of the Calvin Cycle. It is obvious from it that to produce one molecule of hexose sugar (6C), six molecules of CO₂ are required.

To begin with, 6 mols. of RuBP accept 6 mols. of CO₂ to produce 6 mols. of 6-C addition compound. The latter after hydrolysis yield 12 mols. of PGA (12 x 3C).

12 PGA mols. are converted into 12 mols. of 3-PGAld (12 x 3C) which require 12 mols. of ATP and 12 mols. of NADPH + H⁺.

5 of the 12 mols. of 3PGAld are converted into 5 mols. of DiHOAcP; 3 mols. of 3PGAld condense with 3 of the 5 mols. of DiHOAcP to yield 3 mols. of F-1,6-bisP. The latter are dephosphorylated to yield 3 mols. of F-6-P. Thus, there remain 4 mols. of 3 PGAld and 2 mols. of DiHOAcP.

Out of the 3 mols. of F-6-P, one mol. (1 x 6C) is removed which is hexose gain rep­resenting the fixation of 6 CO₂ mols. Thus, there will be left 2 mols. of F-6-P (2 x 6C) which enter the cycle further.

2 mols. of F-6-P react with 2 of the 4 remaining mols. of 3PGAld to form 2 mols. of Xu-5-P (2 x 5C) and 2 mols. of E-4-P (2 x 4C).

2 mols. of E-4-P condense with remaining 2 mols. of DiHOAcP to yield 2 mols. of Su-1, 7-bis P (2 x 7C). The latter on dephosphorylation yield 2 mols. of Su-7P.

Remaining 2 mols. of 3PGAld condense with 2 mols. of Su-7P to produce 2 mols. of Xu-5-P and 2 mols of R-5-P (i.e., 4 x 5C). The latter get isomerised into 2 mols. of Ru- 5-P.

Thus, there are now 4 mols. of Xu-5-P and 2 mols. of Ru-5-P. 4 mols. of Xu-5-P are epimerised into 4 mols. of Ru-5-P (hence there is a total of 6 x 5C = 30 carbons fixed in pentose sugars.

All 6 mols of Ru-5-P get phosphorylated utilizing 6 ATPs to regenerate 6 mols. of RuBP (6 x 5C) and thus completing the cycle.

Thus, for the fixation of 6 CO₂ mols. into one hexose sugar mol. (F-6-P) through Calvin Cycle, 12(NADPH + H⁺) and 18 ATPs are required:

6CO₂ + 12(NADPH + H⁺) + 18 ATP F-6-P + 12NADPHT + 18 ADP + 17Pi

In other words, to produce the equivalent of one mol. of hexose sugar, 6 CO₂ mols. are needed and the cycle is turned once or the cycle must be turned six times (then 1 CO₂ mol. being fixed in each turn):

(RuBP = Ribulose 1, 5-bisphosphate; 3PGA = 3 phosphoglyceric acid; 1,3-bis PGA = 1, 3- bisphosphoglyeric acid ; 3-PGAld = 3-phosphoglyceraldehyde; DiHOAcP = 1, 3-dihydroxyacetone phosphate; F-l, 6-bisP = Fructose 1, 6-bisphosphate; F-6-P = Fructose-6-phosphate; Xu-5-P = Xylu­lose 5-phosphate; E-4-P = Erythrose-4-phosphate; Su-1, 7-bisP = Sedoheptulose 1-7-bisphosphate ; Su-7-P = Sedoheptulose-7-phosphate; R-5-P = Ribose-5-phosphate ; Ru-5-P = Ribulose-5-phosphate).

(It has been mentioned earlier that during non-cyclic electron transport and non-cyclic photo­phosphorylation, NADPH and ATP molecules are produced in the ratio 1:1. The fixation of CO₂ into hexose sugars however, needs NADPH and ATP in the ratio of 1: 1.5. The extra requirement of ATP is probably fulfilled through cyclic photophosphorylation.