1 / 6

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.

zagiri
Télécharger la présentation

Most Probable Number (MPN)

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related