The Number of Particles of Bacteriophage T2 That Can Participate in Intracellular Growth.

Let us consider a population of bacterial cells simultaneously infected with particles of the two phages TZr+ and TZr. Supposing that for each phage the number of particles adsorbed by every bacterium is the same, for example, 20 particles for TZr+ and one for TZr, the yield from these cells will depend on whether all particles or-only a limited number participate in the process of multiplication. In the case of unlimited participation, we shall find both r and r+ particles in the yield of each bacterium. In the case of limited participation, r particles will fail to be produced in a fraction of bacteria that can be calculated easily by assuming that there is nq advantage for either type of virus in the course of growth. Let us assume that a bacterial cell adsorbs m particles of phage A and n of phage B, and that only K particles can participate in growth. The number of possible groups of K particles is given by ("2). The number of groups containing no A particles is ( z ) . The probability P (A=O) of obtaining yields without phage A i;