Monday, April 27, 2015

B5 calculation

FNSC 4180 2013-2014 2ndterm

B3.
a.         Roasted Turkey is the most suspected food. Pizza is kept at 65-70oC. Therefore,    both Staphylococcus aureus and its toxin won't appear in food. The toxin is          decomposed at 60 oC. Sushi is stored with ice, it is unlikely that Salmonella can    grow. Also, the attack rate for roast turkey is quite high, which is 0.35.

Answer: Sorry. You are wrong. Need to use the difference of attack-rates between ate and did not eat to deduce the answer. 

b.   The microorganism inviolved is C. perfringens, which incubation period is usually at typically 8-12 hours. The calculated incubated period is 8 hours as dinner starts at 7:00pm on sat and the last patient went to hospital at 4:00am on Sunday. The food involved is roasted turkey. The bacteria is found on vegetable and used as turkey stuffing. The surface of the turkey is cooked. However, the things inside the turkey may not be cooked sufficiently with temperature higher than 60 degree. The bacteria can form spore, which is activated by heating 40-50 degree. Therefore, the bacteria present inside the chicken after it is cooked. It usually presents in small number. However, the turkey transported to picnic site at room temperature, which allow them to proliferate to infective dose, which is 106 -107 cell per gram.

Answer: Make the story according to the answer in a. above. Use microbe properties and food process information to make the story.


B5.
a. bacteria: 100logD                  Not logD, just D; need to show calculation: 25min/0.25min=100 (D)
  Bacterial spore: 10logD                                        25min/2.5min=10   (D)
  Fungal spore: 50logD                                           25min/0.5min=50   (D)

b. bacteria: 103 /ml                 10^5/ml/10^100=10^-95/ml  OR power 5-100= -95--     10^-95/ml
  Bacterial spore: 103/ ml                      10^4/ml/10^10=10^-6/ml
  Fungal spore: 10-50 / ml                       10/ml/10^50=10^-49/ml

c. bacteria: No, higher than the acceptable cell density   Incorrect
  Bacterial spore: No, higher than the acceptable cell density  Correct!!!
  Fungal spore: Yes, lower than the acceptable cell density   Correct!!!

d.

D value at 120oC
Z value
Spoilage bacteria
0.025 mins incorrect--0.0025min
7.5 oC
Bacterial spores
0.25 mins Correct
15 oC
Fungal spore
0.05 mins Correct
15 oC

Acceptable cell density: Incorrect
  bacteria: 8logD                     10^-8/ml
  Bacterial spore: 8logD          10^-8/ml
  Fungal spore: 6logD             10^-6/ml

Treatment time required at 120 to achieve accepted cell density foe contaminated lot:
  bacteria: 8 x 0.025 = 0.2 mins      Incorrect  13*0.0025= 0.02 mins
  Bacterial spore: 8 x 0.25 = 2 mins   Incorrect  12*0.25=3 min
  Fungal spore: 6 x 0.05 = 0.3 mins    Incorrect  7*0.05=0.35 min

Only half of the answer provided here. Need to reduce the density of the most heat resistant microbe to the acceptable density. The densities of the other microbes would meet the requirement simultaneously.

Answer: 3 min at 120C

density of the microorganism of the regular lot of ham after canning with new process:

bacteria: 0.001 /ml                   
bacterial spore: 0.1 /ml
fungal spore: 1X10-500 /ml

Answer: 
  1. Calculate new D values at 120C for 3 min as a. above
  2. Use new D values to calculate cell densities as b. above

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