Bacteriology/Food Science 324:
KEY TO THE SAMPLE PLATING PROBLEMS
to go along with Exercise 1 and Appendix F.1
in the 1999 edition of the lab manual


  1. (This is Example No. 2 on page 88.) One ml of a bacterial culture was pipetted into a 9 ml dilution blank. One-tenth ml of this dilution was pipetted into a 9.9 ml dilution blank. From this dilution, one-tenth ml was plated with 25 ml of culture medium. 219 colonies arose after incubation. How many colony-forming units were present per ml of the original culture? By whatever method you work it through, you should result in this answer: 2.19 X 106 CFUs/ml – which is better rounded off to 2.2 X 106. (Does the amount of medium in the plate matter? No.)
    dilutions
    made
     X  amount
    inoculated
     =  plated
    dilution
    1/10 X 1/100  X  1/10  =  1/10,000 or 10–4

    dilution
    factor
     X  no. of
    colonies
     =  no. of
    CFUs/ml
    104  X  219  =  2.19 X 106

  2. Five grams of yogurt were mixed with 45 ml of sterile diluent. Two successive 1/100 dilutions were then made. One-tenth ml was plated from the last (most dilute) dilution onto each of two plates of an all-purpose medium. After incubation, 75 colonies were counted on one plate and 73 were counted on the other. Calculate the number of CFUs per one gram of the yogurt.
    dilutions
    made
     X  amount
    inoculated
     =  plated
    dilution
    1/10 X 1/100 X 1/100  X  1/10  =  1/1,000,000 or 10–6

    dilution
    factor
     X  no. of
    colonies
     =  no. of
    CFUs/g
    106  X  74  =  7.4 X 107

  3. If the first dilution made in the problem above involved one gram of yogurt being added to nine ml of diluent, would there be any change in the answer to the problem? Briefly explain why or why not.

    There should be no change in the expected outcome of the problem. In each case – whether 1 gram of yogurt is added to 9 ml of diluent or 5 grams of yogurt are added to 45 ml of diluent – you have a 1/10 dilution as the initial dilution. A 1/10 dilution can be made in an infinite number of ways, provided that you add 1 part of sample to 9 parts of diluent.


  4. You are given 10 ml of a 10–1 dilution of sauerkraut juice. You wish to do a "total aerobic plate count" such that each plate is inoculated with 0.1 ml, and the following plated dilutions are made: 10–2, 10–3, 10–4 and 10–5. Diagram the procedure you would use to accomplish this, and clearly indicate the amounts of diluents and inocula.

    (The numbers 1, 9 and 0.1 shown below are ml amounts.)



  5. In the setup for the problem above, you obtain 120 colonies on the plate inoculated with 0.1 ml of the 10–2 dilution of juice. What would be the no. of CFUs per ml of the undiluted sauerkraut juice?
    dilutions
    made
     X  amount
    inoculated
     =  plated
    dilution
    10–2  X  1/10  =  1/1000 or 10–3

    dilution
    factor
     X  no. of
    colonies
     =  no. of
    CFUs/ml
    103  X  120  =  1.2 X 105

  6. You took 5 ml of a sample and added it to a 64 ml dilution blank. You would then express the dilution as 5/69. The total amount is (64 + 5 =) 69, of which 5 was the amount being diluted.


  7. One should expect the same number of CFUs in 10 ml of a 10–2 dilution of a hamburger sample as in   0.1   gram of the undiluted sample.


  8. The same dilution can be obtained in each of the following situations:

    1. The addition of 1 ml of a sample to 9 ml of sterile diluent.

    2. The addition of   11   ml of the same sample to 99 ml of diluent.

    3. The addition of 10 ml of the same sample to   90   ml of diluent.


  9. Eleven grams of cheese were mixed with 99 ml of sterile diluent. A 1/100 dilution was then made. One-tenth ml was plated from the last (most dilute) dilution onto each of 3 plates of PCA. An average of 50 colonies per plate was counted. Calculate the number of CFUs per gram of the cheese.
    dilutions
    made
     X  amount
    inoculated
     =  plated
    dilution
    1/10 X 1/100  X  1/10  =  1/10,000 or 10–4

    dilution
    factor
     X  no. of
    colonies
     =  no. of
    CFUs/g
    104  X  50  =  5.0 X 105

  10. One gram of cheese was added to a 99 ml dilution blank. One 1/100 dilution was then made. From the last dilution made, 0.2 ml was plated onto each of two plates of MacConkey Agar. After incubation, one plate had 90 red colonies and 192 white colonies, and the other plate had 94 red colonies and 188 white colonies. Determine the number of gram-negative, lactose-fermenting CFUs per gram of the cheese.

    Remember from your use of this medium that only gram-negative bacteria are expected to grow (form colonies) on MacConkey Agar. The red colonies are the lactose-fermenters. In this problem, the average number of colonies of lactose-fermenting, gram-negative bacteria on the plates is 92. Also, note how the 0.2 ml inoculum is simply "plugged into" the formula.

    dilutions
    made
     X  amount
    inoculated
     =  plated
    dilution
    1/100 X 1/100  X  2/10  =  2/105 (or 2 X 10–5)

    dilution
    factor
     X  no. of
    colonies
     =  no. of
    CFUs/g
    105/2 (or 5 X 104)  X  92  =  (46 X 105 =) 4.6 X 106

  11. One gram of yogurt was added to 99 ml of sterile diluent. Four decimal (1/10) dilutions were then made, and one-tenth ml was plated in duplicate on Plate Count Agar. After incubation, the following colony counts were made:
    dilution made of the yogurt colony count
    initial dilution TNTC, TNTC
    first 1/10 dilution TNTC, TNTC
    second 1/10 dilution 442, 432
    third 1/10 dilution 52, 48
    fourth 1/10 dilution 7, 3

    Using this data, calculate the number of CFUs per gram of the yogurt.

    Note that the "dilution made of the yogurt" in the results table would not factor in the 0.1 ml inoculated into each plate. Also, the "fourth 1/10 dilution" did not yield the countable plates, so it is ignored.

    dilutions
    made
     X  amount
    inoculated
     =  plated
    dilution
    1/100 X 1/10 X 1/10 X 1/10  X  1/10  =  1/1,000,000 (or 1 X 10–6)

    dilution
    factor
     X  no. of
    colonies
     =  no. of
    CFUs/g
    106  X  50  =  (46 X 105 =) 5.0 X 107

  12. One-tenth ml of milk was plated, resulting in 42 colonies. What was the number of CFUs per ml of the milk?

    If the inoculation of 0.1 ml of milk ultimately resulted in 42 colonies, then there would have been 420 colony-forming units per ml of the milk.


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Page last modified on 7/20/00 at 6:00 PM, CDT.
John Lindquist, Department of Bacteriology,
University of Wisconsin – Madison