Bacterial Growth and Disinfectant Log Reduction
A microbiology lab is testing a mouthrinse against a planktonic bacterial culture. The culture starts at 4.0 × 10^4 CFU/mL in 50 mL of broth and grows exponentially with a doubling time of 40 minutes at 37 °C. After exactly 2 hours of growth, the technician removes a 1.0 mL sample, performs a serial dilution of 1:100 followed by 1:10, and plates 0.1 mL of the final dilution on an agar plate. In a separate disinfection test, a suspension containing 2.0 × 10^6 CFU/mL is exposed to the rinse. The rinse produces a 4.0-log10 reduction within 5 minutes. The lab's acceptance criterion is that a 1.0 mL sample of the treated suspension, plated after a 1:100 dilution, must yield no more than 5 colonies per plate. The lab manager wants to know whether the rinse passes and how long the untreated culture would take to reach 1.0 × 10^8 CFU/mL.
1. Predict the number of colonies on the plate after 2 hours of growth. Show each dilution step.
Read the worked answer
First find the concentration after 2 hours. The number of doublings is:
120 min ÷ 40 min per doubling = 3 doublings.
Final concentration = 4.0 × 10^4 × 2^3 = 4.0 × 10^4 × 8 = 3.2 × 10^5 CFU/mL.
A 1.0 mL sample therefore contains 3.2 × 10^5 CFU. The serial dilution is 1:100 followed by 1:10, for a total dilution of 1:1,000:
3.2 × 10^5 CFU/mL × (1/1,000) = 320 CFU/mL in the final dilution.
Plating 0.1 mL gives:
320 CFU/mL × 0.1 mL = 32 colonies.
The expected plate count is about 32 colonies, which is in a countable range.
2. Does the rinse meet the acceptance criterion? Support your answer with a calculation.
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A 4.0-log10 reduction means the surviving concentration is reduced by a factor of 10^4:
2.0 × 10^6 CFU/mL ÷ 10^4 = 2.0 × 10^2 = 200 CFU/mL.
The test then plates 1.0 mL of a 1:100 dilution. The diluted concentration is:
200 CFU/mL ÷ 100 = 2 CFU/mL.
Plating 1.0 mL therefore yields an expected 2 colonies, which is not more than the limit of 5 colonies. The rinse passes the stated criterion, with a margin of 3 colonies. If the lab instead plated only 0.1 mL, the expected count would be 0.2 colonies and would be less informative because of Poisson variability.
3. How long from inoculation would the untreated culture take to reach 1.0 × 10^8 CFU/mL?
Read the worked answer
Use the exponential growth relationship:
N(t) = N0 × 2^(t/40)
1.0 × 10^8 = 4.0 × 10^4 × 2^(t/40)
Divide both sides by 4.0 × 10^4:
2^(t/40) = 1.0 × 10^8 ÷ 4.0 × 10^4 = 2,500.
Take log base 2:
t/40 = log2(2,500) = ln(2,500) ÷ ln(2) ≈ 7.824 ÷ 0.693 = 11.29.
t ≈ 11.29 × 40 min = 451.6 min.
451.6 min ÷ 60 ≈ 7.53 hours, or about 7 hours 32 minutes after inoculation. This assumes the culture remains in exponential phase with a constant 40-minute doubling time.