Subtopic: Fundamentals of Cell structure and signaling
Question Statement:
A 0.1 mL aliquot of a bacteriophage stock having a concentration of
4x10°
phages mE' is added to 0.5 mL of E. coli culture having a
concentration of 2x10% cells mL"'. The multiplicity of infection is _
Official Answer Key:
4
Step-by-Step Solution & Explanation
### **Step-by-Step Solution & Mathematical Derivation (GATE BT 2021 Q.50)**
**1. Governing Concept & Definition**:
- **Topic**: Virology & Multiplicity of Infection (MOI)
- Multiplicity of infection (MOI) is defined as the ratio of infectious virus particles (phages) to host bacterial target cells:
$\text{MOI} = \frac{\text{Total number of phages}}{\text{Total number of host cells}}$
**2. Detailed Step-by-Step Calculation**:
- **Phage population**:
$\text{Volume} = 0.1\text{ mL}$
$\text{Titer} = 4 \times 10^9\text{ phages mL}^{-1}$
$\text{Total phages} = 0.1\text{ mL} \times 4 \times 10^9\text{ mL}^{-1} = 4 \times 10^8\text{ phages}$
- **Bacterial host cell population**:
$\text{Volume} = 0.5\text{ mL}$
$\text{Titer} = 2 \times 10^8\text{ cells mL}^{-1}$
$\text{Total host cells} = 0.5\text{ mL} \times 2 \times 10^8\text{ mL}^{-1} = 1 \times 10^8\text{ cells}$
- **MOI Calculation**:
$\text{MOI} = \frac{4 \times 10^8}{1 \times 10^8} = 4$
**3. Final Answer**:
- **Calculated Value**: **4**
- **Official GATE Accepted Range**: **4 to 4**
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