From: ZZZZZZZ
Sent: Saturday, 26 November 2016 6:51 PM
To: hoishankwan@cuhk.edu.hk
Subject: FNSC5110 Exam Question
Sent: Saturday, 26 November 2016 6:51 PM
To: hoishankwan@cuhk.edu.hk
Subject: FNSC5110 Exam Question
I re-do the past exam paper as to prepare for the final exam, but not sure on one question (See attachment). Please let me know if I made any mistake or suggest a better a solution.
Thank you,
● What are the numbers of D's targeted
in the standard canning process for each of these microorganisms? (2 marks)
Spoilage bacteria: 25/0.25=100D Bacteria Spores: 25/2.5= 10D
● Calculate the cell densities of the
microorganisms remaining in the contaminated ham after a standard canning
process treatment. (2 marks)
Spoilage
bacteria: 105 / 10100 = 10-95 (per ml)
Bacteria Spores: 104
/ 1010 =10-6 (per ml)
● Was the standard canning process
sufficient for the canning ofthe contaminated ham? Explain briefly. (2 marks)
Insufficient. Bacteria Spores -> 10-6 /ml,
larger than 10-8
To make it sufficient : additional 25 mins at 105°C
Bacteria spores at 10-6
for additional 10D reduction ->10-6 x10-10 =10-16
Thus, sufficient after 10D reduction.
(Actually may only
need additional 5 minutes at 105°C,
because 104 /1030/2.5=10-8 (per
ml))
Better if only 5mins? Additional 25mins
is too long?
● According to Chart I, what are the Z
values ofthe
two types of microbes? (4 marks)
Spoilage Bacteria: -(95-85)/ log 5- log 50=10°C
Bacteria Spores: -(105-90)/ log 5 - log 50 =15°C
● The canning temperature can be raised
15°C without any adverse effect on the quality of the canned hams. Hence, the
technician would like to use l20°C in the canning process to treat this lot of
contaminated ham. Calculate the treatment time required at120°C to achieve (i)
the acceptable cell density for this contaminated lot and (ii) the acceptable
cell densities of the microorganisms of the regular lot of ham after canning
with this new process. Justify your answer and explain. (10 marks)
At 120°C:
Spoilage
bacteria:
D120°C=D105°C *10^[(105-120)/10]=0.25*10^(-1.5)=7.9*10^(-3)min
Bacteria Spores:
D120°C=D105°C
*10^[(105-120)/15]=0.25*10^(-1)=0.25min
Treatment time for regular:
Spoilage
bacteria: 10D120°C=7.9*10^(-2)min
Bacteria Spores: 9D120°C=2.25min
Treatment time for contaminated:
Spoilage bacteria: 13D120°C=0.1027min
Bacteria Spores: 12D120°C=3min
Your answers to the questions are correct except one: Your answer to the last question is incomplete: you can only use one treatment time to treat each lot. You need to use the time for the most heat-resistant microorganism, in this case, bacterial spores. So the times should be:
Regular--2.25 min
Contaminated--3 min
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