18. A bacterial cell suspension contains 2×10⁵ cells mL⁻¹. The volume of this suspension required to obtain 1.4×10⁶ cells is __________ mL (rounded off to the nearest integer).
Bacterial Cell Suspension Volume Calculation
Introduction
Determining the number of microorganisms present in a bacterial suspension is one of the most fundamental calculations in microbiology. Whether researchers are preparing bacterial cultures for cloning, transformation, fermentation, antimicrobial susceptibility testing, or laboratory experiments, they frequently need to calculate the volume of a culture that contains a desired number of bacterial cells. These calculations are based on simple relationships between cell concentration, volume, and total number of cells.
Correct Answer
Correct Answer: 7 mL
Detailed Explanation
The bacterial suspension contains:
Cell concentration = 2 × 105 cells mL−1
The required number of bacterial cells is:
Required cells = 1.4 × 106
To calculate the required volume, we use the basic formula:
Volume = Total Number of Cells ÷ Cell Concentration
Substituting the given values:
Volume = (1.4 × 106) ÷ (2 × 105)
Separate the numerical values and powers of ten:
Volume = (1.4 ÷ 2) × (106 ÷ 105)
Volume = 0.7 × 10
Volume = 7 mL
Since the answer is already a whole number, rounding to the nearest integer gives:
Required Volume = 7 mL
Step-by-Step Calculation
| Parameter | Value |
|---|---|
| Cell Concentration | 2 × 105 cells mL−1 |
| Required Cells | 1.4 × 106 cells |
| Formula | Volume = Cells ÷ Concentration |
| Calculation | (1.4 × 106) ÷ (2 × 105) |
| Final Answer | 7 mL |
Formula Used in Cell Suspension Calculations
| Formula | Application |
|---|---|
| Volume = Total Cells ÷ Cell Concentration | Finding required culture volume |
| Total Cells = Concentration × Volume | Calculating total cells present |
| Cell Concentration = Total Cells ÷ Volume | Determining cell density |
Understanding the Calculation
The bacterial suspension contains 2 × 105 cells in every milliliter. Therefore, each milliliter contributes exactly 200,000 bacterial cells. To obtain 1.4 × 106 cells, the required volume is calculated by dividing the desired number of cells by the number of cells present in one milliliter. This direct proportional relationship forms the basis of almost every microbial dilution and inoculum preparation calculation.
Applications of Cell Concentration Calculations
| Field | Application |
|---|---|
| Clinical Microbiology | Preparation of standardized bacterial inoculum |
| Molecular Biology | Bacterial transformation and cloning experiments |
| Industrial Biotechnology | Preparation of fermentation starter cultures |
| Food Microbiology | Determination of microbial contamination |
| Research Laboratories | Cell counting and culture preparation |
Related Numerical Example
If a bacterial suspension contains 5 × 106 cells mL−1 and you require 2 × 107 cells, then:
Volume = (2 × 107) ÷ (5 × 106)
= 4 mL
This demonstrates the same principle used in the present question.
Final Answer
Given:
Cell concentration = 2 × 105 cells mL−1
Required cells = 1.4 × 106
Volume = (1.4 × 106) ÷ (2 × 105) = 7 mL
Correct Answer: 7 mL
