1 answer

In a serial dilution, 0.1 mL of serum is added to 1.9 mL of saline in the first tube. The

Question:

In a serial dilution, 0.1 mL of serum is added to 1.9 mL of saline in the first tube. The next tubes contain 1.5 mL of saline, and 0.5 mL from the previous tube is added to the next tube. How do we calculate dilution in the next tubes?

Answers

Here's how you do it.

Explanation:

You can see the general method of calculating dilution factors in this Socratic answer.

The general method for calculating serial dilutions is in this answer

The general formula is for calculating a single dilution factor is

#"DF" = V_f/V_i#

Here you have two dilution factors, one for the first dilution and another for the other dilutions.

First dilution :

#V_f = "aliquot volume + diluent volume" = "(0.1 + 1.9) mL" = "2.0 mL"#

#"DF"_1 = V_f/V_i = (2.0 cancel("mL"))/(0.1cancel("mL")) = 20#

You have diluted the sample by a factor of 20.

The dilution factor is often used as the denominator of a fraction.

Thus, a #"DF"# of 20 means a 1:20 dilution.

The other dilutions:

#V_f = "(0.5 + 1.5) mL" = "2.0 mL"#

#"DF"_2 = V_f/V_i = (2.0 cancel("mL"))/(0.5cancel("mL")) = 4#

For each step, the dilution factor is 4 or 1:4.

The Serial Dilutions

The formula for serial dilutions is

#"DF" = "DF"_1 × "DF"_2 ×"DF"_3 × …#

We can generalize this formula as

#"DF"_n = "DF"_1× ("DF"_2)^(n-1) = 20 ×4^(n-1)#, where #n# is the number of dilutions.

Thus, #"DF"_1 = 20 × 4^(1-1) = 20 × 4^0 = 20 × 1 = 20# or #1:20#.

#"DF"_2 = 20 × 4^(2-1) = 20 × 4^1 = 20 × 4 = 80# or #1:80#.

#"DF"_3 = 20 × 4^(3-1) = 20 × 4^2 = 20 × 16 = 320# or #1:320#, etc.

.

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