a) The value P for a cell with the equivalent of 50 monomer plasmids at division. The multimerization results in 40% of plasmid DNA as dimers and 16% as tetramers is
= 0.40 × 50 = 20 monomer equivalent
1 dimer = 2 monomer equivalent
Therefore dimers =10
Tetramer = 0.16×50 = 8 monomer equivalent
1 tetramer = 4 monomer equivalent
Therefore, tetramers = 2
Total independent replication units
Z = M+D+T = 22+10+2 = 34
Thus, plasmid-free cells/division P = 2(1-Z) = 2(1-34) =-10 .
b) In this equation we can calculate how the distribution of plasmids alters P.
Letbe the population-averaged value.
Then, ; & where is the fraction of the population in category i with a plasmid j.
is the probability that cells with plasmid j will form a plasmid-free cell upon division.
For example, category 1 is 4% of the population with an average plasmid of 6 (since the data give a range of plasmid number of 4 to 8).
It should be noted that only the first two terms are significant and the others are very small.
It should also be noted that the average copy number is about 20. If the probability of plasmid loss was calculated based on the population averaged copy number.
= 1.9×10-6 – a difference of 3 orders of magnitude.