Problem
In lab-report-14-bacteriophage-specificity, a bacteriophage is tested on two bacterial hosts using a plaque assay. A volume of 0.10 mL is plated from a \(10^{-6}\) dilution onto each host lawn.
- Host A: 124 plaques
- Host B: 5 plaques
Determine the phage titer (PFU/mL) on each host, compute the efficiency of plating (EOP) of Host B relative to Host A, and interpret what the results imply about bacteriophage specificity.
Core concepts
A plaque is assumed to originate from one infectious phage particle, so plaque counts estimate PFU (plaque-forming units). Host specificity (host range) reflects whether the phage can adsorb to host receptors, inject its genome, replicate, and lyse cells. The EOP compares how efficiently a phage forms plaques on a test host versus a reference host.
Step-by-step solution
Step 1: Use the plaque-count titer formula
The standard titer estimate is:
Step 2: Compute the titer on Host A
For Host A: plaques \(=124\), volume \(=0.10\ \text{mL}\), dilution \(=10^{-6}\).
Step 3: Compute the titer on Host B
For Host B: plaques \(=5\), volume \(=0.10\ \text{mL}\), dilution \(=10^{-6}\).
Step 4: Compute efficiency of plating (EOP) on Host B relative to Host A
A common definition is:
Step 5: Interpret bacteriophage specificity
- Host A is the preferred (more permissive) host because the plaque-forming titer is much higher on Host A.
- Host B is partially susceptible but inefficient: an EOP of about 0.040 means Host B yields about 4% as many infectious plaques as Host A under the same conditions.
- This pattern is consistent with host specificity driven by receptor compatibility and/or intracellular restrictions (e.g., restriction-modification systems, CRISPR defense, or replication constraints).
Results summary
| Quantity | Host A | Host B |
|---|---|---|
| Plaques counted | 124 | 5 |
| Volume plated (mL) | 0.10 | 0.10 |
| Dilution plated | \(10^{-6}\) | \(10^{-6}\) |
| Estimated titer (PFU/mL) | \(1.24\times 10^{9}\) | \(5.0\times 10^{7}\) |
| EOP of Host B relative to Host A | \(\text{EOP}_{B/A}\approx 0.040\) | |
Visualization
Final answers
- Titer on Host A: \(1.24\times 10^{9}\ \text{PFU/mL}\)
- Titer on Host B: \(5.0\times 10^{7}\ \text{PFU/mL}\)
- Efficiency of plating (Host B relative to Host A): \(\text{EOP}_{B/A}\approx 0.040\)
- Interpretation: strong bacteriophage specificity for Host A, with inefficient plaque formation on Host B (narrow host range or partial susceptibility).