Micro Lab Report 1-Serial Dilution PDF

Title Micro Lab Report 1-Serial Dilution
Author Mary Abraham
Course Microbiology
Institution Northern Illinois University
Pages 4
File Size 204.4 KB
File Type PDF
Total Downloads 120
Total Views 153

Summary

Lab Report 1...


Description

Serial Dilution Mary Abraham BIOS 313 L004 09/24/2020

Purpose: In a serial dilution, where a measured amount of a sample is transferred from one vessel to the next in succession. This begins with a culture of microorganisms, and each measured transfer reduces the density of cells contain therein. The dilutions are repeated until a sample can be used in a counting procedure, usually containing between thirty and three hundred microbial colonies. This counting process is commonly called a standard plate count.

Procedure: Seven tubes containing 09. mL of water each were individually labeled 10 -1 through 10-7. One part (0.1 mL) of the E. coli broth sample was added to nine parts dilution liquid (0.9 mL water) in each test tube in order for the new sample to be ten times more diluted than the original sample. The first sample is called the 10 -1 dilution. This dilution is then repeated, taking care to use a new pipette for each transfer. The new pipette is used to transfer 0.1 mL of the 10 -1 dilution to the 0.9 mL of water in the second test tube in order to create another sample, called 10-2. This newest sample is now 10 times more diluted than the previously diluted sample, or 100 times more diluted than the original sample. These steps were repeated until a total of seven dilutions were made. Four nutrient agar plates were also labeled for use. The plates were needed to culture the corresponding dilutions of 10 -4, 10-5, 10-6, and 10-7 respectively, using the spread plate technique. Once inoculated, the plates were allowed to incubate, undisturbed, at a temperature of 37o C for 24 hours.

Results:

# of CFU’s: TNTC

# of CFU’s: 187 Colonies

# of CFU’s: 70 Colonies

# of CFU’s: TFTC

ml volume of sample taken(¿) Calculating CFU/mL: ¿ CFU per plate x dilution factor −1 CFU mL = ¿

NA Plate 1

Labele d 10-4 -5

CFU's Counted

Dilution Factor

CFU/mL

TNTC

1:10,000 or 104

TNTC 5

2

10

187 colonies

1:100,000 or 10

1.87 x 108

3

10-6

70 colonies

1:1,000,000 or 106

7 x 108

4

10-7

TFTC

1:10,000,000 or 107

TFTC

5

NA Plate 2:

187 x 10 8 =1.87 x 10 or 187,000,000 CFU/mL 0.1

NA Plate 3:

7 0 x 10 6 8 =7 x 1 0 or 700,000,000 CFU/mL 0.1

Conclusions:

This experiment demonstrated the benefits of using serial dilution in order to determine the number of CFU’s in a sample. While the first several NA plates had samples that either would not grow on the plates due to crowding and lack of sufficient resources, the final four plates had populations where the cultures would be visible and, on two of the four plates, could also be counted. For the first plate with the 1:10,000 dilution factor, the population was too high to obtain an accurate count. With such a significant amount of growth in the culture it becomes difficult to differentiate between separate CFU’s. The plate containing the 1:100,000 dilution factor had a significant growth as well but the population was separated enough to accurately count and a total of 187 colonies were found. The third plate containing the dilution factor of 1:1,000,000 had the final number that could be accurately counted. It contained 70 colonies. The final plate containing the most diluted sample, with a dilution factor of 1:10,000,000 appeared to have about 12 colonies present but the margin of error on such a small population is too high to consider this count accurate. Therefore, the plate with the highest dilution was considered too few to count. During the calculations I had trouble understanding the concept of CFU/mL where in the PowerPoint listed the calculation to mean that the final answer would result in the bacterial load present in the original sample. When I followed the calculation in the presentation and listed above for plates 2 and 3, my final answers differed between the two plates, even though they both contained dilutions of the original sample. This is likely due to an error in calculation but further understanding of the formula and application is necessary to determine what steps are needed to correctly compute the data and achieve the necessary results....


Similar Free PDFs