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