Bacteriology 102: Solutions to
the Third Batch of Sample
Problems in the Manual

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Page 171: Nos. 12 and 13 (related to Exps. 15 and 13.1, respectively)

12.  From a water sample from beautiful Lake Splammo, you inoculated 10 ml into 90 ml of sterile diluent; this is the "first dilution" indicated in the table below. After mixing thoroughly, 1 ml of this dilution was added to 99 ml of sterile diluent, and a third dilution was made the same way. From each of these dilutions, plates of Plate Count Agar (PCA) and tubes of Lactose Lauryl Tryptose Broth (LLTB) were inoculated with various amounts 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 1st dilution(= 10–1) 2nd dilution(= 10–3) 3rd dilution(= 10–5)
amount inoculated into each of two plates
of PCA and each of three tubes of LLTB
5 ml 0.5 ml 5 ml 0.5 ml 5 ml 0.5 ml
plated dilution (dilutions made
X amount inoculated)
5X10–1 5X10–2 5X10–3 5X10–4 5X10–5 5X10–6
colony count on PCA plates TNTC TNTC 388, 394 42, 46 6, 4 0, 1
no. of positive LLTB tubes 333320
no. of positive BGLB tubes 333310
no. of positive EC Broth tubes 000000

a.  What was the total number of CFUs per ml of the lake water?

On countable plates, average no. of colonies = 44
Plated dilution of plates = 5 X 10–4
Dilution factor of plates (inverse of p.d.) = 2 X 103
No. of CFUs per ml of lake water = 44 X (2X103) = 8.8 X 104

b.  What was the presumptive, most probable number of coliforms per ml of the original sample of water?

Use LLTB results:  3 2 0
From table (p. 97): 0.93 in 5ml of a 10–5 dilution
= 0.19 in 1ml of a 10–5 dilution
= 0.19 X 105/ml of the undiluted sample
= 1.9 X 104/ml of the undiluted sample

c.  What was the confirmed, most probable number of coliforms per ml of the original sample of water?

Use BGLB results:  3 1 0
From table (p. 97): 0.43 in 5ml of a 10–5 dilution
= 0.09 in 1ml of a 10–5 dilution
= 0.09 X 105/ml of the undiluted sample
= 9 X 103/ml of the undiluted sample

d.  Why can't you say there are zero fecal coliforms in the water sample?

For one thing, a 0 0 0 result doesn't match up with a zero value on the MPN table.
(Nor would a 3 3 3 result match up with an infinite value!)
There is still a possibility of fecal coliforms, and one could have made inoculations of greater amounts of the lake water (e.g., inoculating a set of tubes each with 5 ml of undiluted sample) and gotten possibly one or more positive tubes.

13.  You dilute a sample of vanilla pudding and plate it out onto Vogel-Johnson Agar. After incubation, you find 280 black colonies on a plate inoculated with 0.5 ml of a 10–2 dilution. You then pick 40 black colonies at random, and you find that 38 of them test out to be gram-positive, catalase-positive cocci arranged in clusters. Of these 38 isolates, you run the coagulase test, and you find that 30 of them are positive.

How many Staphylococcus aureus CFUs were there per gram of the pudding ?

As 30 out of 40 (=75%) black colonies were Staphylococcus aureus,
then 75% (=210) of the 280 total black colonies would likely be S. aureus.
Plated dilution of plate = 0.5 X 10–2 = 5 X 10–3.
Dilution factor of plate = 1/5 X 103 = 0.2 X 103 = 2.0 X 102 = 200.
Then, 200 X 210 = 4.2 X 104 S. aureus CFUs per gram.

Page last modified on 1/5/02 at 5:30 PM, CST.
John Lindquist, Department of Bacteriology
University of Wisconsin – Madison

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