THE CHEMISTRY OF PHOTOSYNTHESIS 



153 



The symmetric distribution of radioactivity in fructose- 1,6-diphosphate is 

 due to the condensation of two similar trioses. Investigations by Kandler and 

 Gibbs (27, 34) show a distinct asymmetrical distribution as can be seen from 

 the following reaction scheme : 



H2COPO3H2 H2COPO:iH2 H2COPO3H2 



COH 



|l 

 COH 



HCOH 



H2COPO3H2 



ribi/lose-7,5- 

 diphosphate 



COH 



+ *C02 ^ COH 



HOO*C— COH 



light hama- light 



*■ melonic — ^ 



TPNH ? acid 



H2COPO3H2 



labile 



intermediate 



compound 



CO 



-HOCH + O2 



1 

 H*COH 



I 

 HCOH 



H2COPO3H2 



fructose-7,6- 

 diphosphate 



The C4 atom of fructose-l,6-diphosphate shows the highest radioactivity. 

 The labile intermediate compound — a carboxylation product of ribulose-1,5- 

 diphosphate (cf. Calvin's intermediate ketoacid) — is reduced in the light, 

 and a polyhydroxyacid, probably hamamelonic acid, is produced as an inter- 

 mediate (31, 32). It may be that, instead of the carboxyl group of the la- 

 bile compound, a peroxide group is formed being able to split off O2 exo- 

 thermically. This splitting would be associated with its reduction to alde- 

 hyde. 



C 



/ 



O 



c 



/ 



o 



+ o, 



^O— OH 



^H 



It is to be noted that according to these views Oo is produced from COo. 

 Kandler suggested that Warburg's labile CO2 would thus more or less find 

 its place in this reaction scheme, a somewhat bold assumption indeed (see 

 § 59). 



Furthermore, Kandler assumes that the labile intermediate compound 

 may partially produce 3-phosphoglyceric acid providing the cells with other 

 important constituents such as amino-acids. This is represented in Figure 

 58. By means of transaldolation and transketolation, a part of the fructose- 

 1,6-diphosphate produced is converted to ribulose-l,5-diphosphate again, so 

 that Kandler's cycle may be written as follows 



light 

 6 ribulose-l,5-diphosphate + 6 CO2 *■ 6-fructose-l,6-diphosphate + 6 O2 



dark 



6 fructose-l,6-diphosphate + ATP 



glucose + 6 ribulose-l,5-diphosphate + ADP + ph 



6 CO2 + ATP -»► glucose + ADP + ph + 6 O2 



According to the over-all reaction, the minimal requirement of ATP is 

 only 1 mole per 6 mole CO2. The mole ATP used in the cycle compensates 



