Confused on how to do these dilution problems can someone help me out please! 1.
ID: 310146 • Letter: C
Question
Confused on how to do these dilution problems can someone help me out please! 1. What is the final dilution factor for a suspension made from 10 ul of a 10-3 dilution diluted to 1000 ul with broth? 2. If your orignal culture had 6.7 x 106 cells/ml, what is the concentration of the final dilution made from a 10-2 dilution of a 103 dilution 3. How many cells are in 50 ul of a 2.2 x 104 cells/ml culture? 4. Calculate the original culture concentrations from the following counts: 112 colonies 0.1 ml plated 10-5 dilution 5. Calculate the original culture concentrations from the following counts: 121 colonies on 112 of the plate when 0.1 ml of a 10-8 dilution was spread over the whole plate 6. How would you prepare a 10 4 dilution of a bacterial culture? Assume you want 1000 ul total of the dilutionExplanation / Answer
Ans. 1.
I. First dilution factor ‘a’ = 10-3 ; [of 10 uL aliquot]
II. 10 uL is diluted to 1000 uL
Second dilution factor, b = Initial volume / Final volume = 10 uL / 1000 uL = 1/100 = 10-2
Final dilution factor = multiplication of suitable dilution factors
= dilution factor a x dilution factor b
= 10-3 x 10-2 = 10-5
Thus, final dilution factor of the suspension = 10-5
Ans. 2. Total dilution factor of the suspension = 10-2 dilution of a 10-3 dilution
= 10-2 of 10-3 = 10-2 x 10-3 = 10-5
Final Concertation = Initial concertation x Total dilution factor
= (6.7 x 106 cells/ mL) x (10-5)
= 6.7 x 101 cells/mL
= 67 cells/ mL
Ans. 3. Given,
Concertation = 2.2 x 104 cells/ mL
= 2.2 x 104 cells/ 1000 uL ; [1 mL = 1000 uL]
= 22 cells/ uL
Now,
Number of cells = volume of aliquot x concentration of aliquot
= 50 uL x (22 cells/ uL)
= 1100 cells = 1.1 x 103 cells
Note: The unit of volume shall be the same. Calculation can also be done by converting 50 uL to mL.
That is, Number of cells = 0.050 mL x (2.2 x 104 cells/ mL) = 1100 cells
Ans. 4. Original culture concentration = Plate count x volume of inoculum x dilution factor
= 112 colonies x 0.1 mL x 10-5
= 1.12 x 10-4 colonies/ mL
Since each colony is assumed to arise from a single cell (also called- colony forming unit, CFU) in the inoculum, the original culture concentration = 1.12 x 10-4 cells/ mL
Ans. 5. Total number of colonies on the plate = 2 x (number of colonies on ½ plate)
= 2 x 121 colonies = 242 colonies
Original culture concentration = Plate count x volume of inoculum x dilution factor
= 242 colonies x 0.1 mL x 10-8
= 2.42 x 10-7 colonies/ mL = 2.42 x 10-7 cells/ mL
Ans. 6. A 10-4 dilution means that the original culture is diluted 104 (= 10000) times. For example, taking 1 uL of culture and diluting it to 10000 uL with broth gives a 10-4 dilution.
However, we have to make 1000 uL solution. Required volume of bacterial culture to make 1000 uL of 10-4 dilution = final volume x dilution factor = 1000 uL x 10-4 = 0.1 uL. Since, measuring 0.1 uL may be difficult most often, we take an alternate route of multiple serial dilution as follow-
1st serial dilution: Take 10 uL of bacterial culture in a sterile Eppendorf tube. Add 990 uL sterile broth or saline to it.
Now, final volume of the solution = 10 uL (culture) + 990 uL (broth) = 1000 uL
Dilution factor = Initial volume taken / final volume made upto
= 10 uL / 1000 uL = 10-2
Thus, dilution factor for 1st serial dilution = 102
2nd serial dilution: Take 10 uL of 1st dilution (10-2 dilution prepared above) in a sterile Eppendorf tube. Add 990 uL sterile broth or saline to it.
Dilution factor of 2nd serial dilution= Initial volume taken / final volume made upto
= 10 uL / 1000 uL = 10-2
Total dilution = dilution factor for 1st serial dilution x dilution factor for 2nd serial dilution
= 10-2 x 10-2 = 10-4
Therefore, the second serial dilution gives a solution of 10-4 dilution.
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