The document describes how to calculate the multiplicity of infection (MOI) for an experiment using E. coli and bacteriophage. It shows that a 0.25 mL aliquot of E. coli culture containing 2 x 108 cells requires 0.05 mL of a bacteriophage stock with a concentration of 2 x 109 phage/mL to achieve an MOI of 0.5. The calculation involves determining the number of cells, calculating the number of phage needed based on the MOI and cell count, and then calculating the volume of stock solution required to provide that number of phage.

1. Calculation of Multiplicity of
Infection(moi) - ii
Dr. S. Sivasankara Narayani.,M.Sc.,M.Phil.,Ph.D.,MRSB (UK)

2. A 0.25 mL aliquot of an E. coli culture having a
concentration of 8 x 108 cells/mL is placed into a tube. What
volume of a bacteriophage stock having a concentration of 2
x 109 phage/mL should be added to the cell sample to give
an moi of 0.5?
First, calculate the number of cells in the 0.25 mL aliquot:
0.25mL x 8x108 cells = 2 x 108 cells
mL

3. calculate how many bacteriophage particles are required
for an moi of 0.5 when 2 x 108 cells are used
x phage = 0.5 phage
2x108cells cell
Solve for x:
x = 0.5 Phages (2 x 108 cells) = 1 x 108
phages
cell

4. calculate the volume of the bacteriophage stock that will
contain 1 x 108 phage:
(2 x 109 Phages) X ml = 1 x 108 phages.
ml
x = 1 x 108 phage = 0.05mL
2 x 109 phage/mL

5. Therefore, 0.05 mL of the
phage stock added to the
aliquot of cells will give an
moi of 0.5.

6. Reference
 Calculations for molecular biology and Biotechnology – A Guide to mathematics in
the laboratory, Second Edition – Frank. H. Stephenson
21-09-2021
Dr.SS
sivasan91@gmail.com

7. 21-09-2021
Dr.SS