Food Microbiology Laboratory question, **********Question********** (PLease help
ID: 208458 • Letter: F
Question
Food Microbiology Laboratory question,
**********Question********** (PLease help me to figure out the answers after reading/ using following text, thx a lot)
1. Deeply analyse and explain the result of this lab (results is provide in the following text)
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Laboratory: Microbiological analysis of food products (Standard plate count)
Used media: Brain-heart Infusion Agar
Brain Heart Infusion (BHI) Agar is a general-purpose medium suitable for the cultivation of a wide variety of organism types, including bacteria, yeasts and molds. BHI Agar derives its nutrients from the brain heart infusion, peptone and dextrose components. The peptones and infusion are sources of organic nitrogen, carbon, sulfur, vitamins and trace substances. Dextrose is a carbohydrate source that microorganisms utilize by fermentative action. The medium is buffered with the effect of disodium phosphate.
Ingredients: Beef brain and heart, peptone, dextrose, disodium phosphate
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Food products used: 2 Char sui samples (Store at 4°C and room temperature for 20 hours)
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Procedure:
1.1 Label 5 sterile petri dishes 10-2, 10-3, 10-4, 10-5, 10-6 (1 set); prepare 2 sets for each food sample (A total of 20 plates for 2 food samples).
1.2 Place 20 g of the food sample in the stomacher plastic bag
1.3 Use the dilution scheme shown in Figure 3, add 180 ml of sterile water to the bag/blender. This is a 1:10 dilution (10-1)
1.4 Place the bag in the stomacher, mix at medium speed for 30 seconds.
1.5 Pour the liquid of the mixture into a 50 ml Falcon tube.
1.6 Pipette 1 ml of this 10-1 to a 9 ml sterile water. This is a 1:100 dilution (10-2).
1.7 Mix well the 10-2 dilution by vortex, pipette 1 ml of 10-2 to a new 9 ml sterile water. This is a 1:1,000 dilution (10-3).
1.8 Before pipetting, mix well each dilution by vortex.
1.9 Repeat step 1.6 to obtain 1:10,000 dilution (10-4) to 1:1,000,000 dilution (10-6).
1.10 Pipette 0.1 ml of the suspension from the 50 ml Falcon tube (containing the blended food sample) to the bottom of the plate marked 10-2.
1.11 Pipette 0.1 ml of 10-2 dilution to the bottom of the plate marked 10-3.
1.12 Repeat step 1.11 for 10-3 to 10-6 dilution according to Figure. 3
1.13 Add 15 ml of cooled molten BHI agar to all plates.
1.14 Swirl the plates gently 20 times on the bench-top. The goal is to disperse the bacterial cells throughout the agar so you will get isolated colonies when incubated.
1.15 Allow agar in plates to solidify. Then invert and incubate the plates at 37oC.
1.16 After 24 hours of incubation, count the colonies in the BHI agar plates.
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Results:
10-2
10-3
10-4
10-5
10-6
Plate 1
RT
TNTC
37
3
0
0
4 ºC
77
11
0
0
0
Plate 2
RT
TNTC
31
2
1
0
4 ºC
111
5
0
3
0
SPC calculation as follows:
For the Char Siu sample stored at room temperature,
N = (37+3+31+2+1) / [(1 x 2) + (0.1 x 2) x 10-3]
= 74 / 0.0022
=3.36 x 104
SPC Round off to two significant = 30000
For the Char Siu sample stored at 4 ºC,
N = (77+11+111+5+3) / [(1 x 2) + (0.1 x 2) x 10-2]
= 207 / 0.022
=9.41 x 103
SPC Round off to two significant = 9000
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10-2
10-3
10-4
10-5
10-6
Plate 1
RT
TNTC
37
3
0
0
4 ºC
77
11
0
0
0
Plate 2
RT
TNTC
31
2
1
0
4 ºC
111
5
0
3
0
1ml 1ml 20 g food sample180 ml sterile water (10) 1ml 1ml 1ml -6 9 (10) 9 (10) 9 (10) 9(10 2 9 (10) 0.1ml 0.1ml 0.1ml 0.1ml 0.1ml 10 10 10 10 10 Fie 3. Dilution scheme for SPCsExplanation / Answer
In the experiment, the microbial load of char sui stored under different conditions ( room temperature and low temperature i.e. 4 degree C). For the purpose, the samples are homogenized using stomacher and appropriately diluted. The diluted samples are plated on BHI medium and incubated at 37 degree C. The dilution is very significant because it allows the microbial concentration to reach a level where the individual cells are separated from each other and form isolated and countable colonies. The medium provides appropriate conditions for growth of the microbe so that they can form visible and separated colonies. (It is assumed that one colony arises from one microbial cell).
After incubation the number of colonies are counted. However, for calculating the SPC (standard plate count), the plates with more than 250 are ignored.
In calculation of SPC, the values are computed as per the formula provided in the question.
For eg., for the sample stored in room temperature, the total number of colony is calculated. As per the observation table, plate no. 1 of room temperature (rt) sample at dilution 10-2 contains tntc ( this means the number is greater than 250 and hence is ignored). Similarly, the plate no. 2 of the same sample at 10-2 also has tntc. So, both these values are ignored. The plate no. 1 of this sample at dilution 10-3 contains 37 colonies and the plate no. 2 contains 31 colonies. So, at 10-3 dilution 2 plates are considered. In case of dilution 10 -4, plate no.1 contains 3 colonies and plate no.2 contains 2 colonies. So 2 plates are considered at dilution 10-3. In case of dilution 10 -4, only plate no. 2 contains 1 colony. There are no colonies in other plates (0).
So the total no. of colonies= 37+3+ 31+2+1
The denominator has [(1x n1)+ (0.1 x n2) x d]
n1 is the no. of plates in first dilution that was considered, so in this case 2 plates of 10-3 were considered. So, 1x n1 becomes (1x 2). n2 refers to the no. of plates in the second dilution that was considered. In present case, 2 plates of 10-4 dilution were considered. So, (0.1 x n2) becomes (0.1x 2). d denotes the first dilution from which the the first counts were obtained. In other words, it refers to the lowest dilution from which the colony counts were obtained. So, in the present case the d is 10-3 (since we ignored the values of 10-2 dilution as they were tntc).
So the spc (N)= (37+3+31+2+1)/ [(1x2)+ (0.1 x 2) x 10-3)]
on solving, we get 3.36 x 10 4 or 3.36 x 10000.
Rounding off two last digits we get 3x 10000= 30000.
Similarly in case of sample stored at 4 degree C, the total number of colony can be calculated. In this case however, at dilution 10-2, the plate no.1 has 77 colonies and plate no.2 has 111 colonies. So, these counts are considered. And 10-2 becomes the least dilution which is considered. At 10-3, plate no. 1 has 11 and plate no. 2 has 7 colonies. So at n2 ( 10-3) also there are 2 plates. Apart from this, only 10-5 plate no. 2 has 3 colonies. Rest all plates do not show any growth.
So N becomes= 77+ 11+ 111+5+3/[ (1x2)+ (0.1 x2) x 10-2]
on solving we get 9.41 x 10 3 or 9.41 x 1000
On rounding off, 9 x 1000= 9000
In this way, it can be observed that if the sample is stored at 4 degree C, there is less microbial load (as indicated by lower spc of 9000 as compared to 30000 at room temperature). Hence 4 degree C is better for the storage of the sample.