A0.1mLaliquotofabacteriophagestockhavingaconcentration
of4X10
9
phage/mLisaddedto0.5mLofE.colicellshavinga
concentrationof2X10
8
cells/mL.Whatisthemoi?
First, calculate the total number of bacteriophage and the total number of bacteria.
Total number of bacteriophage:
0.1mL X 4x10
9
phage/ml = 4 X10
8
bacteriophage
Total number of cells:
0.5mL X 2x10
8
cells/ml =1 x 10
8
cells
The moi is then calculated as bacteriophage per cell:
moi = 4 x 10
8
phage / 1x 10
8
cells
= 4 phage/cell
Therefore, the moi is 4 phage/cell.
Reference
Calculations for molecular biology and Biotechnology –A Guide to mathematics in
the laboratory, Second Edition –Frank. H. Stephenson
11-09-2021Dr.SS [email protected]