31 



The point has been raised concerning the extent of ionization and the 

 ionic yield. This is of special importance because we have to remember that 

 when we pass an ionizing track through an aqueous solution containing a number 

 of reactants, we have not only the action of the free radicals produced on ioniza- 

 tion of water but also the action of free radicals produced during the oxidation 

 of the reactants. If we accept Michaelis' theory of compulsory univalent oxida- 

 tion, there must be formation of intermediate free radicals whenever we oxidize 

 a bivalent compound. These free radicals will then produce oxidation-reduc- 

 tions, perhaps of different systems than those reacting with the OH and O^H 

 radicals. The high yield of oxidation of glutathione may be explained by a chain 

 of reactions produced by the free radical RS, besides the free radicals OH and 

 O2H: 



(1) 



(2) 



(3) 



RSH + H2O2— ^-RS + H2O + OH (4) 



RSH + OH — *- RS + H^O (5) 



RS + RS — »- RSSR (6) 



There are thus 4 molecules of RSH capable of being oxidized by the 2 

 radicals, which would give 12 molecules per 100 ev. if we assume that 32. 5 ev. 

 produce the 2 oxidizing radicals: 



H^O ^ H + OH 



PLATZMAN: How did you compute the ionic yield? 



BARRON: The ionic yield was computed by measuring the oxidation of 

 ferrous sulfate in acid solutions. 



The great sensitivity of the sulfhydryl groups to ionizing radiations is 

 clearly shown on irradiation of phosphoglyceraldehyde dehydrogenase. With 100 

 r, there was 21 percent inactivation. That this enzyme inhibition was produced 

 entirely by oxidation of -SH groups in the protein molecule was shown by the 

 complete reactivation of the enzyme on addition of glutathione. When the ex- 

 posure was increased to 500 r, there was 94 percent inhibition of enzyme activi- 

 ty and only 10 percent reactivation on addition of glutathione. (Table I) We may 

 assume that the irreversible inhibition was due to action on other groups of the 

 protein, such as the OH groups of tyrosine or serine, the NH^ groups, or to 

 rupture of hydrogen bonds. How much of this second action -- irreversible in- 

 hibition -- is due to the direct collision between the ionizing track and the pro- 

 tein molecule cannot be calculated from these experiments. 



POLLARD: I would answer that none of this is due to the direct effect 

 of the track going through the enzyme. 



PLATZMAN: Are these solutions? 



