4) Hint for MPN calculation: The following gives an estimate of the MPN. First,

Question

4) Hint for MPN calculation: The following gives an estimate of the MPN. First, select 5 highest the lowest dilution that doesn't have all positive tubes. Second, select the dilution with at least one positive tube. Finally, select all the dilutions between them. Use only the selected dilutions in the following formula of Thomas (1942): . MPN/100 ml: ( gj) x 100 / ( tim, (ti-gum) (s) where the summation is over the selected dilutions and 8; denotes the nurmber of positive tubes in the selected dilutions, 2 tm, denotes the mL of sample in all tubes in the selected dilutions (qg,)m, denotes the mL of sample in all negative tubes in the selected dilutions. The following examples will illustrate the application of Thomas's formula. We assume that the dilutions are 1.0, 0.1, 0.01, 0.001, and 0.0001 mL. 2 Example (1). For outcome (5/5, 10/10, 4/10, 2/10, 0/5) use only (-,4/10, 2/10,-); so tm = 10*0.01 + 10*0.001 = 0.11. Where * means multiplication. There are 6 negative tubes at 0.01 and 8 negative tubes at 0.001, so (t-g)m, 6*0.01 8*0.001 0.068. There are 6 positive tubes, so MPN/100 mL = 6x100/(0.068 * 0.11)(xlz 600/0.0862 7000MPN/100mL .

Explanation / Answer

5) Sample B: The lowest dilution that does not have any positive tubes in sample B is 0.0001 (0/10). The highest dilution that has atleast one positive tube is 1 mL ( 5/5 or 10/10). The dilutions between them are 0.1 mL (7/10), 0.01 mL (4/10) and 0.001 mL (1/10)

Summation TjMj= ml of samples in all selected dilution of 0.1, 0.01 and 0.001 m = 10*0.1 + 10*0.01+10*0.001= 1+0.1+ 0.01=1.11

Summation gj= Number of selected tubes in the selected dilutions= 7+4+1= 12

0.1 mL dilution has 10-7= 3 negative tubes

0.01 mL has 10-4= 6 negative tubes

0.001 mL has 10-1=9 negative tubes

Summation (tj-gj)mj= mL of samples in all negative tubes in selected dilution= (3*0.1 mL) + (6*0.01) + (9*0.001) = 0.3+ 0.06+0.009=0.369

MPN/100 mL= Summation gj X 100/ [Summation TjMj X Summation (tj-gj)mj] 1/2=12 *100/[ 1.11*0.369]1/2= 1200/(0.41)0.5=1200/0.64= 1875 MPN/100 ml

Sample A: MPN/100 ml for sample A is already calculated in the example shown in 4. It is 7000 MPN/ 100 ml

The lowest dilution that does not have any positive tubes in sample B is 0.0001 (0/10). The highest dilution that has atleast one positive tube is 0.1 mL (10/10). The dilutions between them are 0.01 mL (4/10), and 0.001 mL (2/10).

Summation gj= 4+2=6

0.01 mL has 10-4= 6 negative tubes

0.001 mL has 10-2=8 negative tubes

Summation (tj-gj)mj= (6*0.01)+ (8*0.001)= 0.06+0.008=0.068

Summation TjMj= Summation TjMj=(10*0.01)+(10*0.001)= 0.1+0.01=0.11

MPN/100 mL= Summation gj X 100/ [Summation TjMj X Summation (tj-gj)mj] ½

= 6*100/(0.11*0.068)1/2=600/(0.00748)1/2 = 600/0.086= 6976.6= 7000 MPN/100 mL

Sample A is more coliforms than sample B.

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