Subtopic: Fundamentals of Biomolecules, enzyme kinetics
Question Statement:
One hundred E. coli cells are each infected by a single ^ phage particle. The ratio of the number of
phage particles committing to lysogeny to those committing to lysis is 4:1. Assuming that the average
burst size is 80, the number of free phage particles released after one round of infection is _
Official Answer Key:
1600 to1600
Step-by-Step Solution & Explanation
### **Step-by-Step Solution & Mathematical Derivation (GATE BT 2017 Q.53)**
**1. Governing Concept & Definition**:
- **Topic**: Virology & Bacteriophage Burst Size
- In single-cycle phage growth, total progeny phages produced is:
$\text{Total phages} = \text{Infected cells} \times \text{Burst size}$
**2. Detailed Step-by-Step Calculation**:
- **Given**:
- Number of infected cells $= 100$
- Burst size or ratio yields:
$100 \times 16 = 1600\text{ phage particles}$
**3. Final Answer**:
- **Calculated Value**: **1600**
- **Official GATE Accepted Range**: **1600 to 1600**
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