Question

A 0.1 mL aliquot of a bacteriophage stock having a concentration of \(4 \times 10^9\) phages mL\(^{-1}\) is added to 0.5 mL of E. coli culture having \(2 \times 10^8\) cells mL\(^{-1}\). The multiplicity of infection is \(\underline{\hspace{1cm}}\).

Hint:

MOI = number of infecting phages per host cell.

Answer 1

Correct Answer: 4

Total phages added:
\[ 0.1 \times 4 \times 10^9 = 4 \times 10^8 \]
Total cells present:
\[ 0.5 \times 2 \times 10^8 = 1 \times 10^8 \]
Multiplicity of infection (MOI):
\[ \text{MOI} = \frac{\text{phages}}{\text{cells}} = \frac{4 \times 10^8}{1 \times 10^8} = 4 \]