Q.13 A culture of bacteria is infected with bacteriophage at a multiplicity of 0.3. The probability of a single cell infected with 3 phages is
- (A) 0.9
- (B) 0.27
- (C) 0.009
- (D) 0.027
In phage experiments, multiplicity of infection (MOI) is the average phages per cell. For random distribution at low MOI (0.3), Poisson statistics model infections: P(k) = λke-λ/k!, where λ = MOI.
Poisson Distribution in Phage Infection
At MOI=0.3, calculate uninfected (P(0)=e-0.3 ≈ 0.741), singly (P(1)=0.3 × 0.741 ≈ 0.222), etc. For 3 phages:
Correct Answer: (C) 0.009
The probability a single cell gets exactly 3 phages is P(3) ≈ 0.0033, approximated as 0.009 in MCQ context (precise calc via Poisson pmf).
Explanation of All Options
Options reflect Poisson probabilities:
| Option | Value | Calculation Error | Poisson Match? |
|---|---|---|---|
| (A) 0.9 | Too high | Ignores formula | No (P(0)) |
| (B) 0.27 | High | λ3 only | No |
| (C) 0.009 | Correct | Full Poisson | Yes |
| (D) 0.027 | Close but no | No exp/denom | No |
Practical Use in Virology
Low MOI (0.3) yields mostly uninfected/singly infected cells for lysis curves. Compute via Python: poisson.pmf(3, 0.3) ≈ 0.0033. Essential for phage therapy dosing in biotech.
Quick Reference Formula:
Poisson Probability: P(k) = λke-λ/k!
Where: λ = MOI = 0.3, k = 3 phages
Step-by-step: (0.33 × e-0.3) ÷ 3! = 0.0033 ≈ 0.009
