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A bacterial culture was diluted to 10000 fold and 0.1 ml of this diluted sample was spread on plate on nutrient agar. In a triplicate run the number of colonies found is 219, 195, 186. The number of colonies in the a signal culture is ..............× 107 cells/ml. [answer in nearest integer]    
    Correct answer is '2'. Can you explain this answer?
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    A bacterial culture was diluted to 10000 fold and 0.1 ml of this dilut...
    Average cell after triplicate run 
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    A bacterial culture was diluted to 10000 fold and 0.1 ml of this dilut...
    Average cell after triplicate run 
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    A bacterial culture was diluted to 10000 fold and 0.1 ml of this dilut...
    To find the number of colonies in the original (undiluted) bacterial culture, we need to calculate the average number of colonies from the triplicate run.

    Average number of colonies = (219 + 195 + 186) / 3 = 200

    Since the sample was diluted 10000 fold, the number of colonies in the original culture can be calculated by multiplying the average number of colonies by the dilution factor:

    Number of colonies in the original culture = Average number of colonies * Dilution factor
    Number of colonies in the original culture = 200 * 10000 = 2,000,000

    Therefore, the number of colonies in the original culture is 2,000,000.
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    A bacterial culture was diluted to 10000 fold and 0.1 ml of this diluted sample was spread on plate on nutrient agar. In a triplicate run the number of colonies found is 219, 195, 186. The number of colonies in the a signal culture is ..............× 107cells/ml. [answer in nearest integer] Correct answer is '2'. Can you explain this answer?
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    A bacterial culture was diluted to 10000 fold and 0.1 ml of this diluted sample was spread on plate on nutrient agar. In a triplicate run the number of colonies found is 219, 195, 186. The number of colonies in the a signal culture is ..............× 107cells/ml. [answer in nearest integer] Correct answer is '2'. 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 A bacterial culture was diluted to 10000 fold and 0.1 ml of this diluted sample was spread on plate on nutrient agar. In a triplicate run the number of colonies found is 219, 195, 186. The number of colonies in the a signal culture is ..............× 107cells/ml. [answer in nearest integer] Correct answer is '2'. 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 A bacterial culture was diluted to 10000 fold and 0.1 ml of this diluted sample was spread on plate on nutrient agar. In a triplicate run the number of colonies found is 219, 195, 186. The number of colonies in the a signal culture is ..............× 107cells/ml. [answer in nearest integer] Correct answer is '2'. Can you explain this answer?.
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