Agar plates can also be used to enumerate bacteria. You decide to enumerate the

Question

Agar plates can also be used to enumerate bacteria. You decide to enumerate the bacteria present in a sample of hamburger from Wildberries. You measure out 10 g of hamburger and disperse in 40 mL saline solution. You call this dilution A. You then prepare a dilution series, adding 1 mL of dilution A to 4 mL sterile saline to produce dilution B, and then take 1 ml of dilution B and add it to 4 ml sterile saline to produce dilution C, and so on. You take 0.2 mL from each dilution, spread it on plate count agar and incubate 36 h at 30°C. You count the colonies that develop on the plates, with the following results What was the total dilution in the tube labeled 'A'? (3) What was the total dilution in the tube labeled 'C'? (4) What was the concentration of cells in the original hamburger, in units of cell/g? (4) Dilution # Count >500 >500 200 2 0

Explanation / Answer

Before answering this question, let us summarize the given data

a. 10 g of hamburger was dispersed in 40 ml of saline. It is named as dilution A

b. 1 ml of the dilution A was transferred to 4 ml saline. It is named as dilution B

c. Again 1 ml of the B dilution was transferred to 4 ml saline. It is named as dilution C

Now let us solve the questions:

1. What was the total dilution in tube A?

Generally, in serial dilution, we take 1 gram of sample and dissolve in 9 ml of saline. This gives 1:10 dilution or 10-1 dilution. Similarly if we take 1 gram of sample and dissolve in 5 ml of saline, this gives 1:5 dilution or 5-1

In the given problem, they have taken 10 grams of sample and dispersed in 40 ml saline. So, from the above explanation, the dilution in the tube labeled A is 1:5 or 5-1 or 0.2 = 2 x 10-1

2. What was the total dilution in the tube labeled C?

To find the dilution of C, we must calculate the dilution of B.

For this we have to use the formula V1C1 = V2C2

Where V1 = Intial volume, C1 = Initial concentration, V2 = Final volume, C2 = Final concentration.

In the given problem,

V1 = 1 ml (that is, 1 ml from tube A was transferred to tube B)

C1 = 0.2 (Concentration of tube A, from previous answer)

V2 = 5 ml (1 ml from tube A + 4 ml of saline)

C2 = X (we have to calculate)

Now let us substitute the values

V1C1 = V2C2

1 x 0.2 = 5 x (X)

X = (1 x 0.2) / 5

X = 0.04 = 4 x 10-2

So, the total dilution in the tube B was 4 x 10-2

Now let us calculate the dilution in tube C by considering the dilution of tube B

Here again we have to use the formula described earlier V1C1 = V2C2

V1 = 1 ml ( volume added from tube B to C)

C1 = 0.04 (Dilution of tube B)

V2 = 5 ml (1 ml from tube B + 4 ml saline)

C2 = X (we have to calculate)

Let us substitute the values:

V1C1 = V2C2

1 x 0.04 = 5 x X

X = (1 x 0.04) / 5

X = 0.008 = 8 x 10-3

From the above calculation, the total dilution in the tube labeled C is 8 x 10-3

What was the concentration of cells in the original hamburger?

To calculate this we have to consider the dilution of the tube C and the volume of sample inoculated on the agar plate. The reason for using tube C is, it is in the range of 30-300 and it gives accurate results.

The formula for calculating the concentration of cells in the original hamburger is

number of cells present in 10 grams of hamburger = [(CFU/volume of inoculum) / dilution]

number of cells present in 10 grams of hamburger = [(200/0.2) / (8 x 10-3)] = [1000 / (8 x 10-3)] = 125 x 103

For 10 g = 125 x 103

For 1 g = ?

By solving we get 12.5 x 103 cells

So, from the above calculation, The number of cells present in the hamburger was 12.5 x 103 cells/ gram

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