Coliform refers to a group of Gram-negative non-spore-forming bacilli that can ferment lactose, produce acid and gas, and are aerobic and partially anaerobic. The number of coliforms in food is expressed as the most probable number (the MPN) of coliforms per 100 g (or ml) of sample. In this experiment, the final determination of coliforms in food was based on the lactose fermentation gas production of coliforms combined with the MPN search form.
Operation method
biostatistical method
Principle
Coliform refers to a group of Gram-negative non-spore-forming bacilli that can ferment lactose, produce acid and gas, and are aerobic and partially anaerobic. This bacterium mainly comes from human and animal feces, so it has wide hygienic significance as an indicator of fecal contamination to evaluate the hygienic quality of food. It reflects whether the food is contaminated by feces, and at the same time indirectly indicates whether the food has the possibility of contamination by enteric pathogenic bacteria. The number of coliforms in food is expressed as the most probable number (the MPN) of coliforms per 100 g (or ml) of sample.
Materials and Instruments
Dairy Meat Eggs Beverages Pastries Fermented Condiments Escherichia coli Aerobacter aerogenes Move I. Sampling and dilution For more product details, please visit Aladdin Scientific website.
Single lactose bile salt fermentation tubes Double lactose bile salt fermentation tubes Lactose bile salt fermentation tubes Erythromycin agar Gram's stain Protein aging water Indocyanine reagent MacConkey
Warming oven Water bath Balance Microscope Homogenizer Milk bowl Thermometer Petri dish Test tubes Fermentation tubes Straws Slides Inoculation needles
1. Aseptically place 25 g (or 25 mL) of the sample in a sterilized glass bottle containing 225 mL of sterilized saline or other diluent (with an appropriate number of glass beads pre-positioned in the bottle) or a sterilized mantle, and then make a 1:10 homogeneous diluent by sufficient shaking or grinding. It is best to use a sterile homogenizer to process the solid sample at 800 r/min-1000 r/min for 1 min to make a 1:10 dilution.
2. Use a 1 mL sterilized pipette to draw up 1 mL of 1:10 dilution solution, inject it into a test tube containing 9 mL of sterilized saline or other diluent, shake and mix well to make 1:100 dilution solution, replace it with a lmL sterilized pipette, and make 10-fold incremental dilutions in accordance with the above operation in turn.
3. Inoculate 3 tubes according to food hygiene requirements or estimation of the contamination of the test sample. Samples may also be inoculated directly.
II. Lactose primary fermentation test
This is commonly known as the presumptive test. Its purpose is to check for the presence of lactose-fermenting gas-producing bacteria in the sample.
The samples to be schooled will be connected to the seed lactose bile salt fermentation tube, the inoculum volume of 1 mL or more, with double lactose bile salt fermentation tube 1 mL and 1 mL or less, with single lactose fermentation tube. Inoculate 3 tubes for each dilution, place them in an incubator at (36 earth 1) ℃ and incubate them for (24 earth 2) h. If all the lactose bile salt fermentation tubes do not produce any gas, then they can be reported as coliform-negative, and if they do not produce any gas, then follow the procedure below. If there is any production, the following procedure will be followed.
III. Isolation culture
The fermentation tubes of gas production were transferred to the erythromelane agar plate or MacConkey agar plate, placed in (36 soil 1) ℃ temperature box, incubated for 18 h a 24 h,: then removed, observed the morphology of the colonies and Gram staining microscopy and re-fermentation test.
IV. Lactose re-fermentation test
This is commonly referred to as the confirmation test, and its purpose is to demonstrate that Gram-negative Bacillus sphaericus-free bacilli isolated from tubes that test positive for lactose priming are indeed capable of fermenting lactose to produce gas.
On the above selective medium, 1-2 suspected coliforms were picked for Gram staining, and lactose fermentation tubes were inoculated at the same time, and incubated in a warm box at (36shl) ℃ for (24 soil 2) h, and the gas production was observed. 
Where lactose fermentation tubes produce gas. Gram stain is negative for no bacilli, it is reported as E. coli positive Sugar fermentation tubes do not produce gas or Gram stain is positive, it is reported as E. coli negative.
V. Reporting
Based on the number of tubes confirmed to be coliform-positive, check the MPN Retrieval Table (see Table 5-4) and report the most probable number of coliforms per 100 mL of food.
Calculation of values in the Most Probable Number (MPN) of Coliforms retrieval table
The values in the Coliform Most Probable Number retrieval table are calculated by chance. Since our coliform test uses three different lo-fold decreasing inoculums for the test samples, each inoculated with three lactose bile salt fermentation tubes. As a result, lactose-positive tubes may occur in only one tube group, in two tube groups, or within three tube groups. For this reason, an example of each of these three different scenarios for the occurrence of lactose-positive tubes is given to illustrate the most probable number of scenarios found in the retrieval table.
1. When coliforms are present in only one set of lactose fermentation tubes. Calculate using equation (1):
Nλ = 2.303 x lgA/B (1)
N - the amount of sample added to the lactose fermentation tubes in which coliforms were detected;
λ - the most probable number of coliforms, individuals/mL or individuals/g;
A - the total amount of samples added to all lactose fermentation tubes of each tube group;
B - total amount of samples in tubes without detectable coliforms for all tube groups.
Example 1; Tube group ① ② ③
Number of tubes 3 3 3
Sample volume per tube 10mI 10mL 10mL
Number of positive tubes 3 0 0
Therefore, N = 10 A = 33.3 B = 3.3
Substituting into equation (1) gives
10λ = 2.303 x lg33.3/3.3
=2.303×lg10.091
= 2.3
λ = 0.23 per mI
So the most likely number of coliforms per 100mL of sample is 23. According to Table 5-4 the sample inoculum is 10 times lower than Example 1, so its coliform effect is 230.
2. two groups of lactose fermentation tubes in the presence of coliform bacteria, using formula (2) to calculate the
N2γ - tube group ② with N2mL(g) sample volume per tube with γ coliform-positive tubes;
γ - possible number of coliforms (persons/mL or persons/g);
A - the total amount of sample added to all the lactose fermentation tubes of each tube group;
B - total amount of samples in tubes without detectable coliforms for all tube groups.
Application of the formula (2), must first assume that the value of λ, with a test algorithm into the formula to calculate the value of k, when the assumed value into the formula and the calculated value of λ in line with (or the closest), then the assumed value of λ, that is, the most likely number of coliforms.
Example 2: tube number ① ② ⑧
Number of tubes 3 3 3
Sample volume per tube 10mL 10mL 0.1mL
Number of positive tubes 0 3 2
Therefore, N1=1 N1P=3 N2=0.1 N2γ=0.2 A=33.3 B=30.1
Setting λ = 0.16 per mL and substituting into the above equation gives:
lgK = a 0.4343 x 0.1 x 0.16
= 1.9930512
Therefore, K = 0.98412712
Substituting the above values into Eq. (2) yields:
= 2.303 x 0.0679963
= 0.1565955
This calculated λ-value is consistent with the assumed λ-value, so the most probable number of coliforms is 0.16/mL, and the number of coliforms should be 160 by the sample inoculum in Table 5-4, which is 10 times lower than that in Example 3.
3. When coliforms are present in all three groups of lactose fermentation tubes, use equation (3) to calculate:
Where A - the total amount of samples added to the lactose fermentation tubes of all groups of tubes;
B - the total amount of samples in the tubes in which no coliforms were detected in all the tubes in each tube group.
N1P - tube group ① N1 mL (g) of sample added to each tube with P coliform-positive tubes;
N2γ - tube group ② with γ coliform-positive tubes by adding N2 mL(g) of sample per tube;
N3t - tube group ③ with t coliform-positive tubes by adding N3 mL(g) samples per tube;
λ - the most probable number of coliforms, individuals/mL or individuals/g.
Application of the formula (3), with the application of the same formula (2), must also be assumed first λ value, calculated by substituting into the formula in order to calculate the value of λ, when the assumed value of the contemporary into the formula and the calculated value of λ in line with the value of (or the closest), then the assumed value of λ, that is, for the most probable number of coliforms.
Example 3: tube number ① ② ⑧
Number of tubes 3 3 3
Sample volume per tube 10mL 10mL 10mL
Number of positive tubes 2 3 1
Therefore, N1=10 N1P=20 N2=1 N2γ=3 N3t=0.1 A=33.3 B=10.2
Set λ = 0.36 / mL, substituting into ① ② ③ formula can be obtained;
lgK1 = -0.4343×l×0.36
= a 0.156348
Therefore, K1 = 0.69767312
lgK2 = 0.4343 x 0.1 x 0.36
= -0.0156348
K2 = 0.96464
lgK2 = 0.4343 × (1 + 0.1) × 0.36
= -0.1719828
Therefore, K3 = 0.6730033
Substituting the above values into Eq. (3) yields:
10λ = 2.303 x
= 2.303 x lg37.705191
=3.6304517
So, λ = 0.363045l7
From the calculation results of the λ value and the assumed λ value is consistent, so the most likely number of coliforms is 0.36 / mL (g) so the most likely number of coliforms in the 100mL (g) sample is 36, according to the sample in Table 5-4 inoculation amount of the sample is 3010 times more examples, so the most likely number of coliforms is 360.