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
Solution
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
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
Solution
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
