Q.72 The transformation efficiency of competent cells prepared in a laboratory is 𝟏𝟎4 𝐂𝐅𝐔/𝝁𝐠 of plasmid
DNA. If 𝟎. 𝟎𝟏𝝁 𝐠 of this plasmid is used to transform these competent cells, the number of transformed
bacteria in CFU after plating will be ____
Understanding Transformation Efficiency in Bacterial Transformation
Transformation efficiency measures how effectively competent cells take up plasmid DNA, expressed as colony-forming units (CFU) per microgram (μg) of DNA. For the given scenario, competent cells have an efficiency of 𝟏𝟎4 CFU/μg, and 0.01 μg of plasmid DNA is used. The number of transformed bacteria is directly calculated as efficiency multiplied by DNA amount: 𝟏𝟎4 CFU/μg × 0.01 μg = 100 CFU.
Step-by-Step Calculation
Multiply the transformation efficiency by the mass of plasmid DNA used. Here, 10,000 CFU per μg times 0.01 μg equals 100 CFU, representing the total transformants expected after plating, assuming standard conditions without dilution factors specified. This value indicates successful DNA uptake and expression of selectable markers like antibiotic resistance.
Correct Answer and Options Analysis
The correct answer is 100 CFU. Though no explicit options are listed, typical multiple-choice alternatives in such questions might include:
| Option | Value (CFU) | Explanation |
|---|---|---|
| A | 10 | Underestimates by factor of 10; ignores full efficiency scaling . |
| B | 100 | Correct: Direct product of 𝟏𝟎4 × 0.01 . |
| C | 1,000 | Overestimates by factor of 10; common error from misplacing decimals . |
| D | 10,000 | Assumes 1 μg DNA instead of 0.01 μg . |
This table clarifies why only 100 CFU matches the parameters.
Key Factors Affecting Results
Supercoiled plasmids yield higher efficiency than linear forms, and lab-prepared cells often achieve 𝟏𝟎4 CFU/μg, lower than commercial 𝟏𝟎8–𝟏𝟎10. Plating volume and recovery time influence countable colonies, but the query specifies post-plating CFU. Optimizing CaCl2 or electroporation boosts yields for molecular biology applications like cloning.
