Question #269be

1 Answer
Feb 5, 2015

The dilution factor for the fourth tube will be #"1:160,000"#.

You know that the dilution factor is defined as the final volume of the sample divided by the initial volume of the sample. So, if you add #"0.2 mL"# to #"3.8 mL"#, you initial volume will be

#V_("initial") = "0.2 mL"#

and your final volume will be

#V_("final") = "0.2 + 3.8" = "4.0 mL"#

This means that the dilution factor will be

#"DF" = V_("final")/V_("initial") = "4.0 mL"/"0.2 mL" = "20"#, or #"1:20"#

When you're doing a serial dilution, each tube will dilute the sample by the same calculated dilution factor.

http://www.biologyexams4u.com/2013/12/serial-dilution-protocol-pdf.html

Notice how in the above example each solution will be diluted by a factor of 10. If you apply this method to your dilution, you'll get a dilution factor for tube 4 equal to

#DF_("final tube") = "1:20"^(4) = "1:160,000"#

The first tube will have a #"1:20"# dilution factor. Now you take #"0.2 mL"# of this solution and add it to another #"3.8 mL"#. The second tube will have a dilution factor equal to #"1:20"#, but the original solution has now been diluted twice by a dilution factor of #"1:20"#, which means that the solution in tube 2 will have a total dilution factor of

#DF_("tube 2") = "1:20" * "1:20" = "1:20"^(2) = 1:400#

Now you repeat the process for tubes 3 and 4 to get the final dilution factor of #"1:160,000"#.

You can read more on dilution factors and serial dilutions here:

http://socratic.org/questions/how-do-you-calculate-dilution-factor?source=search

http://socratic.org/questions/how-do-you-do-serial-dilution-calculations?source=search