Q.28 If the doubling time of a bacterial population is 3 hours, then its average specific
growth rate during this period is _____ h-1.
(Round off to two decimal places)
For a doubling time of 3 hours, the average specific growth rate is 0.23 h⁻¹.
Core Formula and Answer
Bacterial growth follows exponential kinetics:
Nt = N0 eμt
where μ is the specific growth rate (h⁻¹).
Doubling time (td) occurs when:
Nt = 2N0
Solving gives:
td = ln(2) / μ
or
μ = ln(2) / td
For td = 3 hours:
ln(2) ≈ 0.693
μ = 0.693 / 3 = 0.231 h⁻¹ ≈ 0.23 h⁻¹
This calculation follows microbiology standards and is more accurate than approximations such as the Rule of 70.
Explanation of Calculation Steps
- Identify the doubling time: td = 3 h
- Apply the formula: μ = 0.693147 / 3
- Calculate: μ = 0.231049 h⁻¹
- Round to two decimals: 0.23 h⁻¹
Common error: Using base-10 logarithms:
log102 / td gives ~0.10 h⁻¹, which is incorrect for natural exponential growth models.
Common Options and Why They Are Incorrect
| Option | Value (h⁻¹) | Reason Incorrect |
|---|---|---|
| A | 0.10 | Uses common logarithm: log10(2) / 3 ≈ 0.10 |
| B | 0.23 | Correct: ln(2) / 3 |
| C | 0.69 | Equals ln(2) only; ignores division by doubling time |
| D | 2.00 | Incorrect inverse or misapplied Rule of 72 |
Applications in Microbiology
The specific growth rate (μ) quantifies microbial growth during the exponential phase in batch cultures.
It is essential for:
- Bioreactor design
- Antibiotic and stress-response studies
- Modeling microbial inoculants in plant and rhizosphere research
Experimentally, μ can be verified from the slope of a ln(N) vs. time plot.
