Dilution Theory / Supplementary Pages:

Solutions to the General
Problem Set No. 2

John L's Bacteriology Pages > 
Selected General Topics  >  Dilution Theory:
Dilution Theory per se:
• Page 1 – Dilution Plating
• Page 2 – More Dilution Plating
• Page 3 – The MPN Method
Supplementary Pages:
• Five-Tube MPN Table
• Practice Set 1 (Plating)
• Practice Set 2 (Plating&MPN)

This page was clarified on July 22, 2009 in order to eliminate troublesome references to one of our older lab manuals which you probably would not have at hand. Information about enrichment media and procedures for coliforms can be found here.


1. You added 10 ml of a lake water sample to 90 ml of sterile diluent; this is the "first dilution" indicated in the table below. After mixing thoroughly, one ml of this dilution was added to 99 ml of sterile diluent to make the second dilution. To make the third dilution, one ml of the second dilution was added to 99 ml of sterile diluent. From each of these three dilutions, 1 ml or 0.1 ml amounts were inoculated into pour-plates of Plate Count Agar (PCA) and tubes of Lactose Lauryl Tryptose Broth (LLTB) as shown in the table below.

After appropriate incubation, colonies were counted and the tubes were checked for positive results, and the data are summarized below. Subsequent inoculations into BGLB and EC Broth were appropriately made and incubated; their results are also shown below.

dilution of lake water first dilution
=10–1
second dilution
=10–3
third dilution
=10–5
Amount inoculated into each of two plates of
PCA and each of three tubes of LLTB:
1 ml 0.1 ml 1 ml 0.1 ml 1 ml 0.1 ml
Plated dilutions: (dilution of sample multiplied
by the amount inoculated)
10–1 10–2 10–3 10–4 10–5 10–6
Dilution factors: (inverse of the plated dilution) 10 102 103 104 105 106
Colony count on PCA plates: TNTC TNTC 205
215
18
22
0
2
0
0
No. of positive LLTB tubes: 3 2 1 0 0 0
No. of positive BGLB tubes: 3 2 0 0 0 0
No. of positive EC Broth tubes: 1 1 0 0 0 0

a. (2 points) What was the "total aerobic plate count" (in CFUs per ml) of the original, undiluted sample of water?

dilutions made  X  amount inoculated  =  "plated dilution"
1/10 X 1/100  X  1  =  1/103 = 10–3

dilution factor  X  (ave.) no. colonies  =  no. CFUs/ml
103  X  210  =  2.1 X 105

b. (2 points) What was the confirmed, most probable number of fecal coliforms per ml of the original sample of water.

Use the EC Broth results, choosing the results of the 1st, 2nd and 3rd set of MPN tubes, as their results are 1, 1 and 0. (1,1,0 tells us more about how the coliforms are being diluted to extinction than do any other set of results, such as 1,0,0 or 0,0,0.) Note that the EC Broth and the BGLB Broth are each testing the LLTB tube from which they were inoculated to confirm whether or not there was a population of coliforms in that LLTB tube (the EC Broth tests for "fecal" coliforms). Note how the number of positive results decreases with increasing dilution as does the colony count. (What plates you count have no bearing on which MPN tubes you choose or vice-versa.)

According to the table: 1,1,0 means there was an average of 0.073 positive organisms (in our case, fecal coliforms) originally in the inoculum of each of the tubes in the middle set of the 3 sets chosen, i.e., the 2nd set of tubes into which was inoculated 0.1 ml of a 10–1 dilution.

Therefore, the most probable number of fecal coliforms is 0.073 per 0.1 ml of a 10–1 dilution of the sample
= 0.73 per 1 ml of the 10–1 dilution
= 0.73 X 101 or 7.3 per ml of the undiluted sample.

Alternate Method: You can simply multiply 0.073 by the dilution factor of that middle set of tubes (102) and you will get the same answer – i.e., 0.073 X 102 = 7.3 per ml of the undiluted sample.




2. (2 points) A sample of yogurt prepared with the usual organisms (i.e., Streptococcus thermophilus and Lactobacillus bulgaricus) was bacteriologically tested as follows: One tablespoon of yogurt was mixed in a sterile flask with four tablespoons of sterile milk. One ml of this mixture was added to 99 ml of saline. One-tenth ml of this last dilution was plated onto a medium known to support the growth of yogurt organisms.

After appropriate incubation, 250 colonies were counted on the plate. Fifty colonies were then picked at random (a "random sample"), and half of these colonies were found to be composed of cocci in chains. As half of the randomly-tested colonies were determined to be S. thermophilus, we could then say there were (probably) 125 S. thermophilus colonies on the plate.

What was the number of Streptococcus thermophilus CFUs per gram of the yogurt?

dilutions made  X  amount inoculated  =  "plated dilution"
1/5 X 1/100  X  1/10  = 
1/5000 = 1/(5 X 103)
= 2 X 10–4

dilution factor  X  no. colonies  =  no. CFUs/ml
5 X 103  X  125  = 
6.25 X 105
better rounded to 6.3 X 105

During the grading of this problem set, one student suggested this sequential solution, keeping in mind that the so-called "plated dilution" (2 X 10–4 in this problem) really represents the amount of undiluted sample being tested (as we show here):

  • 125 CFUs / 2 X 10–4 g
  • 62.5 CFUs / 1 X 10–4 g
  • 62.5 X 104 CFUs / g
  • 6.25 CFUs X 105 CFUs / g

This solution almost makes me consider throwing away the formulas. I'm still thinking about changing the term "plated dilution" to "virtual dilution" or – better yet – "virtual amount."




3. (2 points) One-tenth ml of drinking water was added to a petri dish to which 19.9 ml of melted Plate Count Agar were added. After incubation, 240 colonies arose on the plate. What was the count of CFUs per one ml of the water?

As 240 colonies arose from the 0.1 ml inoculum, then there had been 2400 or 2.4 X 103 CFUs per ml of the spring water. You have ten times the number of CFUs in ten times the amount of sample. No Appendix C formulas are needed – but if they are used, the "dilution factor" would be 10. Note that there is no dilution in this problem, as the whole 0.1 ml of the sample gets inoculated into the plate. We could have used twice the amount of medium, and the answer would still be the same.




4. (1 point) One should expect the same number of CFUs in 10 ml of an undiluted sample as in  100  ml of a 1/10 dilution of the same sample.




5. (1 point) The same dilution can be obtained in each of the following situations:

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

b. The addition of  3  ml of the same sample to 27 ml of diluent.

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

Be sure to keep the same proportions – like in a recipe!


Page last modified on 7/22/09 at 3:45 PM, CDT.
John Lindquist:  new homepage, complete site outline.
Department of Bacteriology, U.W.-Madison