LAB 1: SAMPLING AND SAMPLE HANDLING

QUESTION

1. One hundred grams of cabbage is blended with 250 ml water blank. Consecutive dilutions of 1:100 and 1:100 are made. From the last dilution, samples of 1.0 ml and 0.1 ml were plated in duplicate. The plates containing the 0.1 ml sample had colony counts of 60 and 70 respectively. What is the count per gram of the cabbage?

Dilution factor = 100 g

100g+250ml × 1

100 ×1

100

Amount plated = 0.1ml (another 1.0ml don’t have colony)

Average of colony = 60+70

2 = 65 colonies

Total Count = Averagenumber of colonies¿duplicate plates ¿

Dilution factor × Amount plated

= 75

[ 100 g100 g+250ml

× 1100

× 1100 ][0.1]

= 2.28 × 107 cfu/g= 2.3× 107 cfu/g

2. One gram of food is blended with 9 ml water blank. Consecutive dilutions of 1:10 and 1:10 are then made. Duplicate one ml portions from the last dilution are then plated. The plates are incubated, and the colony counts are determined to be 32 and 35. what is the count per gram of the food sample?

Dilution factor = 1 g

1g+9ml × 1

10 ×1

10

Amount plated = 1ml

Average of colony = 32+35

2 = 33.5 colonies

Total Count = Averagenumber of colonies¿duplicate plates ¿

Dilution factor × Amount plated

= 33.5

[ 1 g1 g+9ml

× 110× 1

10 ][1]

= 3.35 × 104 cfu/g= 3.4 × 104 cfu/g

3. Two grams of food are blended with 8 ml water blank. Consecutive dilutions of 1:100, 1:100, and 1:10 are then made. From the last dilution duplicate 0.1 ml

1

portions are plated, and after incubation it is determined that the plate count average is 245. What is the count per gram of the food sample?

Dilution factor = 2 g

2g+8ml × 1

100 ×1

100 ×1

10Amount plated = 0.1mlAverage of colony = 245 colonies

Total Count = Averagenumber of colonies¿duplicate plates ¿

Dilution factor × Amount plated

= 245

[ 2 g2g+8ml

× 1100

× 1100

× 110 ] [0.1]

= 1.225 × 109 cfu/g= 1.2 × 109 cfu/g

4. Sanitarians investigating a food poisoning outbreak could get only 4 grams of the suspected food. The sample was blended with 99 ml of water and five 1/10 additional dilutions were made. Duplicate one-tenth ml of each dilution was surface plated on Baird-Parker agar. After incubation the plates from the fifth dilution (don’t forget the initial dilution) showed 240 and 242 typical Staphylococcus colonies. Express the result as the number of staphylococci per gram of food.

Dilution factor = 4 g

4 g+99ml × 1

10 ×1

10 ×1

10× 1

10× 1

10Amount plated = 1ml

Average of colony = 240+242

2 = 241 colonies

Total Count = Averagenumber of colonies¿duplicate plates ¿

Dilution factor × Amount plated

= 241

[ 4 g4 g+99ml

× 110× 1

10× 1

10× 1

10× 1

10 ][1]

= 6.025 X 10 cfu/g=6.205 X 10

5. Forty-six grams of a food sample were blended with 54g of water. Two successive dilutions of 1:100 were made, and 1.0 ml portions of the final dilution were plated. After incubation, 93 and 99 colonies were counted on the duplicate plates. State the count in bacteria per gram of sample.

2

Dilution factor = 46 g

46 g+54 g × 1

100 ×1

100

Amount plated = 1.0ml

Average of colony = 93+99

2 = 96 colonies

Total Count = Averagenumber of colonies¿duplicate plates ¿

Dilution factor × Amount plated

= 96

[ 46 g46 g+54 g

× 1100

× 1100 ] [1.0 ]

= 2.087 × 106 cfu/g= 2.1× 106 cfu/g

6. A sample containing 1.6g pepper and 10 grams of water was heat shocked (heated at 80C for 10 min) to determine the number of aerobic spore forming bacteria. After rapidly cooling the sample, one ml was diluted to 100 ml. Another dilution consisting of 1:100 was made, and 0.1 ml portions were dispensed to two Petri dishes. Twenty ml of plate count agar was added to each plate and, after incubation, an average of 117 colonies was reported. What is the count of spores per gram of pepper?

Dilution factor = 1.6g

1.6g+10g × 1

100 ×1

100

Amount plated = 0.1ml Average of colony = 117 coloniesTotal Count = Averagenumber of colonies¿duplicate plates ¿

Dilution factor × Amount plated

= 117

[ 1.6 g1.6 g+10 g

× 1100

× 1100 ] [0.1]

= 8.4825 × 107 cfu/g= 8.5 × 107 cfu/g

7. Fifty-eight ml of whey were mixed with 100 ml of water, and three successive 1:10 dilutions were made. Two sets of duplicate pour plates were made containing 0.1 and 1.0 ml of final dilution. The plates containing 0.1 ml of the sample have 50 and 54 colonies after incubation, while the plates containing 1.0 ml contained 550 colonies on each. What was the total count of whey sample (bacteria per ml)?

3

Dilution factor = 58 g

58g+100ml × 1

10 ×1

10 ×1

10Amount plated = 0.1ml

Average of colony = 50+54

2 = 52 colonies

Total Count = Averagenumber of colonies¿duplicate plates ¿

Dilution factor × Amount plated

= 52

[ 58 g58 g+100 g

× 110× 1

10× 1

10 ][0.1]

= 1.4165 × 106 cfu/ml= 1.4 × 106 cfu/ml

8. Thirty-two g of a food sample were blended with 68g of water. Two successive dilutions of 1:10 were made. The technician prepared only one plate and accidentally added 1.1 ml instead of 1.0 ml to the plate. After incubation, 78 colonies were counted. State the count in bacteria per gram of sample.

Dilution factor = 32g

68g+32g × 1

10 ×110

Amount plated = 1.1ml Average of colony = 78 coloniesTotal Count = Averagenumber of colonies¿duplicate plates ¿

Dilution factor × Amount plated

= 78

[ 3268g+32 g

× 110× 1

10 ] [1.1]

= 22159 cfu/g= 2.2 × 104 cfu/g

9. How many grams of cheese must be added to 99 ml water blank to give a one-to-two dilution?

Let grams of cheese = YY

Y+99ml = 12Y = 99g

10. In your own words describe what makes a “representative” sample.Representative sample means a subset of statistical population which reflect accurately of the whole population.

4