Q.75. A 𝟐𝟓𝟎𝝁𝐥 of bacteriophage stock containing 𝟖 × 𝟏𝟎𝟖 phages /𝐦𝐥 is added to 𝟓𝟎𝟎𝝁𝐥 of E. coli
culture containing 𝟒 × 𝟏𝟎𝟖
cells /𝐦𝐥. The multiplicity of infection is ____ .
Bacteriophage MOI Calculation
Multiplicity of infection (MOI) in this bacteriophage experiment is calculated as the ratio of total phages added to total bacterial cells present.
The correct MOI value is 1.
Step-by-Step MOI Calculation
MOI equals total infectious phages divided by total target cells.
1. Calculate Total Phages Added
Phage concentration = 8 × 108 phages/ml
Volume added = 250 μl = 0.25 ml
Total phages = 8 × 108 × 0.25 = 2 × 108 phages
2. Calculate Total E. coli Cells
Cell concentration = 4 × 108 cells/ml
Volume used = 500 μl = 0.5 ml
Total cells = 4 × 108 × 0.5 = 2 × 108 cells
3. Calculate MOI
MOI = Total phages / Total cells
MOI = (2 × 108) / (2 × 108) = 1
This assumes all phages and cells mix uniformly in the final volume without loss.
Why MOI Equals 1
Standard virology defines MOI as the number of infectious agents added per target cell, not per milliliter of mixture.
Since the total number of phages exactly equals the total number of bacterial cells, the MOI is 1:1.
Common mistake: Using concentrations or final volumes instead of total particle counts leads to incorrect values such as 0.4.
Common MCQ Options Explained
| Option | Value | Explanation |
|---|---|---|
| A | 0.2 | Incorrect. Divides concentrations incorrectly: (8 × 0.25) / (4 × 0.5) ≈ 0.4, then halved. Ignores total particle counts. |
| B | 0.4 | Incorrect. Uses volume-adjusted concentration: (8 × 108 × 0.25 / 0.75) / (4 × 108) = 0.4. Mixes density and volume incorrectly. |
| C | 1 | Correct. Total phages = 2 × 108, total cells = 2 × 108. MOI = 1. |
| D | 2 | Incorrect. Assumes double phage volume or reverses the ratio. |
