MULTITARGET ANALYSIS 113 



experiments gives us the actual volume, area, and thickness of the target. 

 From a knowledge of the structure of the organism, it may be possible 

 to use this knowledge of shape to identify the part of the organism in- 

 volved in the effect being studied. 



Still another way in which these radiations have been employed in- 

 volves the use of electrons so low in energy that they can penetrate only 

 small thicknesses of materials. By increasing the energy of these elec- 

 trons, an energy can be found at which the electrons begin to produce a 

 particular effect. From the known penetration thickness of such elec- 

 trons, we then deduce how far under the surface of the organism lies the 

 portion associated with the effect being studied. 



MULTITARGET ANALYSIS 



For those who can follow a little algebra, it is possible to show how 

 the number of targets per organism can be discovered. Such a situation 

 could obtain, for example, in polynucleate cells. If it is necessary to in- 

 activate all the nuclei to kill the organism, several targets must be hit 

 in each cell. Below are given the few mathematical steps needed to make 

 this analysis for a cell with n nuclei : 



Let p be the probability that a nucleus survives a given dose D of 



radiation. 



(1 — p) is the probability that it does not survive this dose D. 



(1 — p) n is the probability that all n nuclei do not survive this dose 



D. 



1 — (1 — p) n is then the probability that at least one of the n nuclei 



survives the dose D. 



This last expression then represents the fraction of cells which survive 

 the dose D because each of such cells has at least one intact nucleus. We 

 write this as 



£-1 -(!-„)", 



where N is the number of cells surviving a dose D if there were A T cells 

 at zero dose. 



If there were only one nucleus (n = 1), this would reduce to p, as 

 expected. When very large doses have been given, the survival prob- 

 ability p becomes very small. In this case, we can expand the term in 

 parentheses by the binomial theorem to give 



(1 - p ) n = l - np + n(n - l)p 2 /2 



If p is very small, then the right-hand side of this expression is ad- 



