D. E. Lea and C. A. Coulson 267 



that one of the mutants shall divide in time dt. Similarly, (r-1) dnjn is the probability 

 that one of the mutants shall divide in time dt in a culture containing (r-1) mutants. 



Let p r (a function of n) be the probability that a culture of n bacteria grown from 

 a single bacterium at t = shall have r mutants (i.e. of a large number of cultures of 

 n bacteria, a proportion p r of cultures will have r mutants). 



Consider the proportion Pr + ^dn of cultures which, at time t + dt, when the culture 



size is n + dn, have r mutants. These will be derived from: 



(a) cultures which, at time t, had r-1 mutants and in which a mutation occurred in the 



interval dt; 

 (6) cultures which, at time t, had r — 1 mutants and in which a mutant divided in the 



interval dt; 

 (c) cultures which, at time t, had r mutants and in which neither mutation nor division 

 of a mutant occurred in the interval dt ; 

 providing that the interval dt is small enough. For (<x/]8) dn and rdn/n to be much less than 

 unity, the possibility of more than one of the rare events mutation and division of a 

 mutant in the interval dt can be neglected. We see, therefore, that 



dp r a r /a r — 1\ 



sothat sV + » Pr= M£ + ~r 



Making, from (3), the substitution m = (a//3) n, we have 



dm rT m \ m J 



Multiplying by the integrating factor e m we have 



^ r +-?r = ?r-i( 1 + r ). where q r = e m p r . 



dm m \ m J 



Now m is the mean number of mutations which have occurred in a culture by the time it con- 

 tains n bacteria. Therefore e~ m , the first term of the Poisson distribution, is the probability 

 p that no mutation shall have occurred. Thus q Q = e m p = 1 for all values of ra. Evidently 

 initially, when m ■= 0, q r = for all r > 0. 



Starting from q = l we can calculate q lt q 2 , q 3 , etc. in succession from the differential 

 equation (6). Thus: 



-^ + —0, = !, whence o, = Am ; 



dtn m 



(5) 



(6) 



dm m 



dm 



H — <7 2 = \m 1 1 + — ) , whence q 2 = hn + hn 2 ; 

 m \ m) 



3 / 2\ 



+ — q z = (\m + \m 2 ) 11+—), whence 5 , 3 = Y^w+Y^m 2 4-^ 



Evidently q r is a polynomial in m, with powers ranging from 1 to r. 



Writing ?r =2C,, r ^ = C 1)r m + C 2ir | [ %...+C r , r ^, (7) 



