Question 9 from 2012-13 examination paper

Question 9              2012-13 

a. F value?

At pH=4.0
Spoilage bacteria : D=0.001min----10/0.001=10000D
Bacterial spores:    D=0.1 min----10/0.1=100D
Fungal spores:        D=0.002 min ----10/0.002=5000D

Actual set F value= D values above

Sterilization:         10 min at 105C

Alternative answer: F=required D.

Bacteria: 10D
Bacterial spores: 8D
Fungal spores: 8D

b. Actual D?

Because actual pH is 4.5
At pH4.5 10 min at 105C

Spoilage bacteria : D=0.01min----10/0.01=1000D
Bacterial spores:    D=2 min----10/2=5D  (----failed)
Fungal spores:        D=0.02 min ----10/0.02=500D

c. Cell densities?

Per ml
Spoilage bacteria : 1000/ml*10^-1000=10^-997
Bacterial spores:    100/ml*10^-5=10^-3
Fungal spores:        100/ml*10^-500=10^-498

d. One more time, 10 min at 105C?

Per ml
Spoilage bacteria : 10^-997/ml*10^-1000=10^-1997
Bacterial spores:    10^-3/ml*10^-5=10^-8
Fungal spores:        10^-498/ml*10^-500=10^-998

OK.  Bacterial spores:    10^-8/ml < 10^-6/ml requirement

e.  The technician was correct.

f.  New process?

1. First calculate time required for 105C sterilization to achieve target reduction of bacterial spores density to 10^-6/ml.

Number of D required for bacterial spores = 8

Bacterial spores D value at 105C = 2 min

Therefore, time required at 105C = 16 min

2. Calculate the cost:   (16min/10min)*$20=$32

3. Calculate time required for 120C sterilization to achieve target reduction of all microbes to acceptable densities.  Use Z values to get to new D values, then calculate as above. 

Time required at 120C = 1.6min

4. Calculate the cost when changed to 120C: (1.6min/1min)*$5=$8

5. Compare the two costs $8<$32

Better to use 120C for 1.6min