GATE 2019 BT Q.54: solution and answer

GATE 2019 · BTQ.54 · 2 marks · NAT

An industrial fermentor containing 10,000 L of medium needs to be sterilized. The initial spore concentration in the medium is $1 0^{6}$ spores $mL^{- 1}$ . The desired probability of contamination after sterilization is $1 0^{- 3}$ . The death rate of spores at 121 °C is $4 min^{- 1}$ . 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).

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Answer and solution

ANSWER

The holding time is $t_{h o l d} = \frac{ln ( 1 0 ^{13} /1 0 ^{- 3} )}{4 min ^{- 1}} = 9.21 min \approx 9 min$ .

Official answer: 9 to 10Source: Official answer key, GATE 2019 BT (IIT Madras)

Total spores at the start, $N_{0}$ = concentration × volume. Here $10, 000 L = 1 0^{4} L \times 1 0^{3} mL L^{- 1} = 1 0^{7} mL$ , so $N_{0} = 1 0^{6} spores mL^{- 1} \times 1 0^{7} mL = 1 0^{13}$ spores.
A contamination probability of $1 0^{- 3}$ means the expected number of surviving spores in the whole batch after sterilization is $N = 1 0^{- 3}$ . In other words, one batch in 1000 still carries a viable spore.
Thermal death of spores follows first-order kinetics: $\frac{d N}{d t} = - k_{d} N$ , which integrates at constant temperature to $ln \frac{N _{0}}{N} = k_{d} t$ . Here $k_{d} = 4 min^{- 1}$ is the specific death rate constant of the spores at 121 °C.
No spores die during heating or cooling, so the whole required reduction, the design (Del) factor $\nabla = ln \frac{N _{0}}{N}$ , must be delivered by the holding stage: $\nabla = ln \frac{1 0 ^{13}}{1 0 ^{- 3}} = ln 1 0^{16} = 16 \times 2.3026 = 36.84$
Holding time at 121 °C: $t_{h o l d} = \frac{ln ( N _{0} / N )}{k _{d}} = \frac{36.84}{4 min ^{- 1}} = 9.21 min$
Rounded to the nearest integer, $t_{h o l d} \approx 9 min$ .

KEY IDEA

In batch sterilization, $N_{0}$ is the TOTAL spore count in the whole vessel (concentration × volume), and the target $N$ is the contamination probability, read as the expected number of survivors per batch. With first-order death at constant temperature, $ln (N_{0} / N) = k_{d} t$ . Since heating and cooling kill nothing here, the holding stage has to deliver the whole Del factor.

WATCH OUT FOR

Using the spore concentration ( $1 0^{6}$ per mL) as $N_{0}$ instead of the total count in 10,000 L ( $1 0^{13}$ ), which gives $ln 1 0^{9} /4 = 5.2$ min. Two other slips: forgetting that 1 L = 1000 mL, and using $lo g_{10}$ instead of $ln$ (this gives $16/4 = 4$ min).

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GATE 2019 BT Q.54: solution and answer

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