118 



PROBLEMS OF PHOTOSYNTHESIS 



where the ketol group is transferred from ribulose-5-phosphate to erythrose- 

 4-phosphate to produce fructose-6-phosphate and phosphoglyceraldehyde : 



CH2OH 



TK 



— > 



H2COPO3H2 H2COPO3H2 



ribuhse- erythrose- 



5-phosphate 4-phosphate 



The transaldolase (TA) transfers the C3 fraction — CHOH . CO . CH2OH 

 and therefore catalyses the formation of tetrose and hexose from heptose and 

 triose (13, 26). The following reaction shows how this fraction of sedoheptu- 

 lose-7-phosphate is transferred to phosphoglyceraldehyde to produce fructose- 

 6-phosphate: 

 CH2OH 



CO 

 HOCH 



CHO 



CHO 



CH2OH 

 CO 



TA 



Transketolase and transaldolase need TPP+ as a cofactor. It has now 

 been found that transketolation and subsequent transaldolation result in an 

 over-all reaction representing the above-mentioned oxidation of one mole- 

 cule pentose phosphate to one molecule triose phosphate and two molecules 

 CO2. Starting from one molecule glucose-6-phosphate as the primary sub- 

 stance, this oxidation reaction produces one molecule triose phosphate and 

 three molecules CO2. Figure 45 shows the reaction scheme as indicated by 

 Racker (25). The ketol donor in the formation of sedoheptulose-7-phosphate 

 is xylulose-5-phosphate and not ribulose-5-phosphate as was formerly as- 

 sumed. An equilibrium state exists between both pentose phosphates; this 

 is maintained by the enzyme epimerase (12, 27). 



If the fate of one molecule glucose-6-phosphate in the presence of one mole- 

 cule ribose-5-phosphate is studied, it is seen that the first three C atoms of glu- 

 cose-6-phosphate are oxidized to CO2, whereas the other three C atoms re- 



