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;
[1]
R. Dulbecco,et al.
Genetic Recombinations Leading to Production of Active Bacteriophage from Ultraviolet Inactivated Bacteriophage Particles.
,
1949,
Genetics.
[2]
S. Luria.
Reactivation of Irradiated Bacteriophage by Transfer of Self-Reproducing Units.
,
1947,
Proceedings of the National Academy of Sciences of the United States of America.
[3]
S. Luria,et al.
Ultraviolet Irradiation of Bacteriophage During Intracellular Growth
,
1947,
Journal of bacteriology.
[4]
A. D. Hershey,et al.
Mutation of Bacteriophage with Respect to Type of Plaque.
,
1946,
Genetics.
[5]
Maurice G. Kendall,et al.
The advanced theory of statistics
,
1945
.
[6]
M. Delbrück,et al.
THE GROWTH OF BACTERIOPHAGE AND LYSIS OF THE HOST
,
1940,
The Journal of general physiology.