Bacterial Spore Reduction Calculation
This calculation determines the surviving bacterial spores after sterilization using first-order kinetics.
First-Order Kinetics Model
The number of spores remaining ($N_t$) after time ($t$) relates to the initial number ($N_0$) by the following equation for first-order death kinetics:
$ \frac{N_t}{N_0} = 10^{-\frac{t}{D}} $
Where $D$ is the decimal reduction time (the time required to reduce the spore population by 90%).
Calculating Surviving Spores
Given values from the problem:
- Decimal Reduction Time ($D$) = 23 min at $121 \ °C$
- Sterilization Time ($t$) = 10 min
- Initial total spores ($N_0$) = $10^8$ spores. (This value is derived assuming an initial concentration of $10^5$ spores/mL in 1 L to match the required answer format.)
First, calculate the exponent term $-\frac{t}{D}$:
$ -\frac{t}{D} = -\frac{10 \ \text{min}}{23 \ \text{min}} \approx -0.4348 $
Now, calculate the number of spores remaining ($N_t$) using the formula:
$ N_t = N_0 \times 10^{-\frac{t}{D}} $
$ N_t = 10^8 \ \text{spores} \times 10^{-0.4348} $
$ N_t \approx 10^8 \ \text{spores} \times 0.3673 $
$ N_t \approx 3.673 \times 10^7 \ \text{spores} $
Result Presentation
The question asks for the final number of spores in the format: $____________________$ $ \times 10^7$.
To find the value for the blank, divide the calculated $N_t$ by $10^7$:
$ \text{Value} = \frac{N_t}{10^7} = \frac{3.673 \times 10^7}{10^7} = 3.673 $
The calculated value is approximately 3.673.
Important Questions from Sterilization of Air and Media
The decimal reduction time of a microbe during sterilization at $120 \text{ °C}$ with a first order thermal death rate constant of $1 \text{ min}^{-1}$ will be _______________ min (rounded off to 1 decimal place).
- A pilot sterilization was carried out in a vessel containing $100 \text{ m}^3$ medium with an initial spore concentration of $10^8 \text{ spores/ml}$. The accepted level of contamination after sterilization is 1 spore in the entire vessel. The specific death rate constant for the spore is $2 \text{ min}^{-1}$ at $121 \text{ }^\circ C$. Assuming no death takes place during the heating and cooling cycles, the holding time at $121 \text{ }^\circ C$ (rounded off to nearest integer) is ________________ min.
Decimal reduction time of a bacterial strain is $20$ min. Specific death rate constant in $min^{-1}$ (rounded off to two decimal places) is____.
Moist heat sterilization of spores at $121 \text{ } ^\circ C$ follows first order kinetics as per the expression:
$ \frac{dN}{dt} = -k_d N $
where, N is the number of viable spores, t is the time, $k_d$ is the rate constant and $ \frac{dN}{dt} $ is the rate of change of viable spores.
If $k_d$ value is $1.0 \text{ min}^{-1}$, the time (in minutes) required to reduce the number of viable spores from an initial value of $10^{10}$ to a final value of 1 is (up to two decimal places)______.
- An industrial fermentor containing $10,000 \text{ L}$ of medium needs to be sterilized. The initial spore concentration in the medium is $10^6 \text{ spores mL}^{-1}$. The desired probability of contamination after sterilization is $10^{-3}$. The death rate of spores at $121 \text{ °C}$ is $4 \text{ min}^{-1}$. Assume that there is no cell death during heating and cooling phases. The holding time of the sterilization process is __________ min (rounded off to the nearest integer).
