A 100 ml of contaminated drinking water was given to you by your lab supervisor
ID: 93290 • Letter: A
Question
A 100 ml of contaminated drinking water was given to you by your lab supervisor for analysis. You were instructed by your supervisor to determine the number of bacterial cells present in the water sample. You prepared three test tubes, each tube contained 9 ml of diluent broth. You then pipetted 1 ml of the original sample of water and transferred it to test tube 1. After shaking test tube 1 (to mix broth and sample), you pipetted 1 ml from tube 1 and transferred it to tube 2. After mixing the contents of tube 2, you pipetted 1 ml from tube 2 and transferred it to tube 3. You also plated out the following: 1 ml from tube 1 onto agar plate A: 0.1 ml from tube 2 onto plate B: 1 ml from tube 2 onto plate C: 0.1 ml from tube 3 onto plate D: and 1 ml from tube 3 onto plate E. After incubating the plates for 4 days in the incubator, 80 colonies of bacteria formed on agar plate D. Pls. answer the following questions: What is the sample dilution in Tube 1 Tube 2 Tube 3 Plate A Plate B Plate C Plate D Plate E What is the number of bacteria/ml in the sample What is the total number of bacteria in the 100 ml contaminated drinking water? II. Calculate the generation time if 50 Escherichia coli cells growing for 6 hours produced 100,000 cells.Explanation / Answer
Answer:
I)
Tube 1: 10-1
Tube 2: 10-2
Tube 3: 10-3
Plate A: 10-1
Plate B : 10-3
Plate C : 10-2
Plate D: 10-4
Plate E: 10-3
9)
Total Number of bacteria= (80/0.1 x 10-3 )
= 8 x 105 Cfu/ml
10) 8 x 107 Cfu/100ml
II)
Bo = Initial bacterial at zero= 50 cells
Bn= Population at time t= 100,000 cells
K = log Bn –log B0 /t log 2
K= exponential growth rate constant
if we substitute above values in given equatiion, we get
K= 2.5 per hour
G = 1/K
G= Growth rate
G= 1/ 2.5
= 0.4 hours
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