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5 mL of a culture was added to 45 mL of distilled water. This suspension was further diluted to two serial, 1/100 dilutions. 0.1 mL of this final mixture was plated onto an agar plate. Suppose, an average of 137 colonies appeared on the plate after incubation. The CFU/mL of the original sample is _______ ×108.
Correct answer is between '1.35,1.39'. Can you explain this answer?
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5 mL of a culture was added to 45 mL of distilled water. This suspensi...
Total dilutions = 5mL sample in 45 mL diluent = 1/10 = 10-1
It is next diluted by1/100 factor. Hence. 10-3.
Further dilution of 1/100 leads to 10-5.
Hence, Dilution factor = 10-5
Volume plated = 0.1 mL
No. of colonies = 137
CFU = ?
It is next diluted by1/100 factor. Hence. 10-3.
Further dilution of 1/100 leads to 10-5.
Hence, Dilution factor = 10-5
Volume plated = 0.1 mL
No. of colonies = 137
CFU = ?
Most Upvoted Answer
5 mL of a culture was added to 45 mL of distilled water. This suspensi...
To calculate the CFU/mL of the original sample, we need to consider the dilution factor.
The dilution factor for the first dilution (5 mL added to 45 mL) is 1/10 (since 5 mL is 1/10 of 50 mL).
The dilution factor for each of the two serial dilutions (1/100) is 1/100.
To find the total dilution factor, we multiply the dilution factors together:
1/10 x 1/100 x 1/100 = 1/10000
Since 0.1 mL was plated onto the agar plate, we need to divide the number of colonies by the volume plated to get the CFU/mL.
CFU/mL = 137 colonies / 0.1 mL = 1370 CFU/mL
Now we can calculate the CFU/mL of the original sample by multiplying the CFU/mL from the agar plate by the total dilution factor:
CFU/mL of the original sample = 1370 CFU/mL x 1/10000 = 0.137 CFU/mL
The dilution factor for the first dilution (5 mL added to 45 mL) is 1/10 (since 5 mL is 1/10 of 50 mL).
The dilution factor for each of the two serial dilutions (1/100) is 1/100.
To find the total dilution factor, we multiply the dilution factors together:
1/10 x 1/100 x 1/100 = 1/10000
Since 0.1 mL was plated onto the agar plate, we need to divide the number of colonies by the volume plated to get the CFU/mL.
CFU/mL = 137 colonies / 0.1 mL = 1370 CFU/mL
Now we can calculate the CFU/mL of the original sample by multiplying the CFU/mL from the agar plate by the total dilution factor:
CFU/mL of the original sample = 1370 CFU/mL x 1/10000 = 0.137 CFU/mL
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5 mL of a culture was added to 45 mL of distilled water. This suspension was further diluted to two serial, 1/100 dilutions. 0.1 mL of this final mixture was plated onto an agar plate. Suppose, an average of 137 colonies appeared on the plate after incubation. The CFU/mL of the original sample is _______ ×108.Correct answer is between '1.35,1.39'. Can you explain this answer?
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5 mL of a culture was added to 45 mL of distilled water. This suspension was further diluted to two serial, 1/100 dilutions. 0.1 mL of this final mixture was plated onto an agar plate. Suppose, an average of 137 colonies appeared on the plate after incubation. The CFU/mL of the original sample is _______ ×108.Correct answer is between '1.35,1.39'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about 5 mL of a culture was added to 45 mL of distilled water. This suspension was further diluted to two serial, 1/100 dilutions. 0.1 mL of this final mixture was plated onto an agar plate. Suppose, an average of 137 colonies appeared on the plate after incubation. The CFU/mL of the original sample is _______ ×108.Correct answer is between '1.35,1.39'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 5 mL of a culture was added to 45 mL of distilled water. This suspension was further diluted to two serial, 1/100 dilutions. 0.1 mL of this final mixture was plated onto an agar plate. Suppose, an average of 137 colonies appeared on the plate after incubation. The CFU/mL of the original sample is _______ ×108.Correct answer is between '1.35,1.39'. Can you explain this answer?.
5 mL of a culture was added to 45 mL of distilled water. This suspension was further diluted to two serial, 1/100 dilutions. 0.1 mL of this final mixture was plated onto an agar plate. Suppose, an average of 137 colonies appeared on the plate after incubation. The CFU/mL of the original sample is _______ ×108.Correct answer is between '1.35,1.39'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about 5 mL of a culture was added to 45 mL of distilled water. This suspension was further diluted to two serial, 1/100 dilutions. 0.1 mL of this final mixture was plated onto an agar plate. Suppose, an average of 137 colonies appeared on the plate after incubation. The CFU/mL of the original sample is _______ ×108.Correct answer is between '1.35,1.39'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 5 mL of a culture was added to 45 mL of distilled water. This suspension was further diluted to two serial, 1/100 dilutions. 0.1 mL of this final mixture was plated onto an agar plate. Suppose, an average of 137 colonies appeared on the plate after incubation. The CFU/mL of the original sample is _______ ×108.Correct answer is between '1.35,1.39'. Can you explain this answer?.
Solutions for 5 mL of a culture was added to 45 mL of distilled water. This suspension was further diluted to two serial, 1/100 dilutions. 0.1 mL of this final mixture was plated onto an agar plate. Suppose, an average of 137 colonies appeared on the plate after incubation. The CFU/mL of the original sample is _______ ×108.Correct answer is between '1.35,1.39'. Can you explain this answer? in English & in Hindi are available as part of our courses for IIT JAM.
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