KEY TO THE SAMPLE PLATING PROBLEMS

to go along with Exercise 1 and Appendix F.1

in the 1999 edition of the lab manual

- (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 10
^{6}CFUs/ml – which is. (Does the amount of medium in the plate matter? No.)__better rounded off to 2.2 X 10__^{6}

**dilutions**

made**X****amount**

inoculated**=****plated**

dilution1/10 X 1/100 **X**1/10 **=**1/10,000 or 10 ^{–4}

**dilution**

factor**X****no. of**

colonies**=****no. of**

CFUs/ml10 ^{4}**X**219 **=**2.19 X 10 ^{6}

- 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**

dilution1/10 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/g10 ^{6}**X**74 **=**7.4 X 10 ^{7}

- 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.

- 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.)

- In the setup for the problem above, you obtain 120 colonies on the plate inoculated with
**0.1 ml of the 10**of juice. What would be the no. of CFUs^{–2}dilution**per ml**of the**undiluted**sauerkraut juice?

**dilutions**

made**X****amount**

inoculated**=****plated**

dilution10 ^{–2}**X**1/10 **=**1/1000 or 10 ^{–3}

**dilution**

factor**X****no. of**

colonies**=****no. of**

CFUs/ml10 ^{3}**X**120 **=**1.2 X 10 ^{5}

- 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.

- 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.

- The
**same dilution**can be obtained in each of the following situations:

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

- The addition of
__11__ml of the same sample to 99 ml of diluent.

- The addition of 10 ml of the same sample to
__90__ml of diluent.

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

dilution1/10 X 1/100 **X**1/10 **=**1/10,000 or 10 ^{–4}

**dilution**

factor**X****no. of**

colonies**=****no. of**

CFUs/g10 ^{4}**X**50 **=**5.0 X 10 ^{5}

- 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**

dilution1/100 X 1/100 **X**2/10 **=**2/10 ^{5}(or 2 X 10^{–5})

**dilution**

factor**X****no. of**

colonies**=****no. of**

CFUs/g10 ^{5}/2 (or 5 X 10^{4})**X**92 **=**(46 X 10 ^{5}=) 4.6 X 10^{6}

- 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**

dilution1/100 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/g10 ^{6}**X**50 **=**(46 X 10 ^{5}=) 5.0 X 10^{7}

- 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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