Growth and Form 103 



have occurred. Therefore, starting with one cell and after n generations 

 of growth, the number of cells would be 2". If we started with 1000 cells 

 instead of 1, the series would be 1000, 2000, 4000, . . . i.e., 1000 X 2^, 

 1000 X 2S 1000 X 2-, . . . and after n generations, No2", where No is 

 the starting number. The general statement of the number N of organisms 

 would be: 



N = iVo2" 2 



Let us assume that these cells reproduce once every two hours (a con- 

 stant generation time). Then the population would have undergone one 

 generation of growth in 2 hours, two in 4 hours, three in 6 hours, and so 

 on. In short, the number of generations would be proportional to time, 

 and this is stated algebraically as n =: k't where k' is a constant. Equation 

 2 can be restated as : 



N = No2''-'' or ^ = 2fc'f 3 



Taking logarithms of both sides, we get: 



I0ge^=(l0ge2)fc't 



and if we put fc'(loge 2) equal to a new constant, k, the equation becomes: 



log,^ = /cf 4 



which is identical with equation 1. Remember that the only two assump- 

 tions used to derive this relationship are that 1 organism gives rise to 2 

 and that the generation time is constant. Obviously, the shorter the gen- 

 eration time, the faster would the population grow and the higher would 

 be the value of k. Suppose the cells do not divide by binary fission but, 

 instead, each cell gives rise to 3. Then the growth would be 1, 3, 9, 27, 

 81, ... a series in powers of 3, and the equation would become: 



log,^=(Iog,3)fe'i 



Only the value of k in equation 4 would change, but the basic equation 

 would remain the same. In other words, whether 1 organism gives rise 

 to 2 or 3 or more, or whether 2 organisms must cooperate to yield 3 or 4 

 or more does not change the equation but only the value of the constant. 

 But equation 4 as written does not describe the entire growth se- 

 quence. Note that as time passes {t gets larger), N/No would get larger 



