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Most Probable Number (MPN)

Most Probable Number (MPN). Begin with Broth to detect desired characteristic Inoculate different dilutions of sample to be tested in each of three tubes. Dilution. - 1 -2 -3 -4 -5 -6. 3 Tubes/Dilution. 1 ml of Each Dilution into Each Tube.

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Most Probable Number (MPN)

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  1. Most Probable Number (MPN) • Begin with Broth to detect desired characteristic • Inoculate different dilutions of sample to be tested in each of three tubes Dilution -1 -2 -3 -4 -5 -6 3 Tubes/Dilution 1 ml of Each Dilution into Each Tube After suitable incubation period, record POSITIVE TUBES (Have GROWTH and desired characteristics)

  2. MPN - Continued • Objective is to “DILUTE OUT” the organism to zero • Following the incubation, the number of tubes showing the desired characteristics are recorded • Example of results for a suspension of 1g/10 ml of soil • Dilutions: 10-1 10-2 10-3 10-4 • Positive tubes: 3 2 1 0 P : Number of positive tubes N : Total quantity in (g) or (ml) of sample in all negative tubes T : Total quantity in (g) or (ml) of sample in all tubes

  3. MPN • Example of results for a suspension of 1g/10 ml of soil • =1g/10mL • =0.1g/mL in the stock tube 1ml 1ml 1ml 1ml • Calculation • Dilutions: 10-1 10-2 10-3 10-4 • Positive tubes: 3 2 1 0 • P = 6 • N = (3 X (0,1 X 10-4g)) + (2 X (0,1 X 10-3g)) + (1 X (0,1X10-2g)) = 1.23X10-3g • T= (3 X (0,1 X 10-1g)) + (3 X (0,1 X 10-2g)) + (3 X (0,1 X 10-3g)) + (3 X (0,1 X 10-4g)) = 3.33X10-2 g 1/10 1/10 1/10 1/10 9ml 9ml 9ml 9ml 0.1g/mL 0.01g =0.1x10-1g 0.001g =0.1x10-2g 0.0001g =0.1x10-3g 0.00001g =0.1x10-4g = 6/0.0064 =9.38X102bacteria/g

  4. Nombre le Plus Probable (NPP) • Commencer avec un bouillon qui permet de déceler les caractéristiques désirées • Inoculer differentes dilutions de l’échantillon à être testé dans chacun de 3 tubes Dilution -1 -2 -3 -4 -5 -6 3 Tubes/Dilution 1 ml de chaque dilution dans chaque tube Après unepérioded’incubationapproprié, enregistrer les TUBES POSITIFS (qui ont de la croissance et les caractéristiquesdésirées)

  5. NPP- Suite • L’objectifest de DILUÉ l’organisme à zero • Après l’incubation, le nombre de tubes qui démontre les caractéristiques désirées sont enregistré • Exemple de résultats pour une suspension de 1g/10 ml de terre • Dilutions: 10-110-2 10-3 10-4 • Tubes positifs: 3 2 1 0 P : Nombre de tubes positifs N : Quantité totale en (g) ou (ml) d’échantillon dans tous les tubes négatifs T : Quantité totale en (g) ou (ml) d’échantillon dans tous les tubes

  6. NPP • Exemple de résultats pour une suspension de 1g/10 ml de terre • =1g/10mL • =0.1g/mLdans la solution mère 1ml 1ml 1ml 1ml • Calcul • Dilutions: 10-1 10-2 10-3 10-4 • Tubes positifs: 3 2 1 0 • P = 6 • N = (3 X (0,1 X 10-4g)) + (2 X (0,1 X 10-3g)) + (1 X (0,1X10-2g)) = 1.23X10-3g • T= (3 X (0,1 X 10-1g)) + (3 X (0,1 X 10-2g)) + (3 X (0,1 X 10-3g)) + (3 X (0,1 X 10-4g)) = 3.33X10-2 g 1/10 1/10 1/10 1/10 9ml 9ml 9ml 9ml 0.1g/mL 0.01g =0.1x10-1g 0.001g =0.1x10-2g 0.0001g =0.1x10-3g 0.00001g =0.1x10-4g = 6/0.0064 =9.38X102bacteries/g

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