Instructor: Cecile Sanders, M.Ed., MT(ASCP), CLS (NCA) Dilutions Instructor: Cecile Sanders, M.Ed., MT(ASCP), CLS (NCA)

Dilutions for the Clinical Laboratory Dilution = making weaker solutions from stronger ones Example: Making orange juice from frozen concentrate. You mix one can of frozen orange juice with three (3) cans of water.

Dilutions for the Clinical Laboratory (cont’d) Dilutions are expressed as the volume of the solution being diluted per the total final volume of the dilution In the orange juice example on the previous slide, the dilution would be expressed as 1/4, for one can of O.J. to a TOTAL of four cans of diluted O.J. When saying the dilution, you would say, in the O.J. example: “one in four”.

Dilutions for the Clinical Laboratory (cont’d) Another example: If you dilute 1 ml of serum with 9 ml of saline, the dilution would be written 1/10 or said “one in ten”, because you express the volume of the solution being diluted (1 ml of serum) per the TOTAL final volume of the dilution (10 ml total).

Dilutions for the Clinical Laboratory (cont’d) Another example: One (1) part of concentrated acid is diluted with 100 parts of water. The total solution volume is 101 parts (1 part acid + 100 parts water). The dilution is written as 1/101 or said “one in one hundred and one”.

Dilutions for the Clinical Laboratory (cont’d) Notice that dilutions do NOT have units (cans, ml, or parts) but are expressed as one number to another number Example: 1/10 or “one in ten”

Dilutions for the Clinical Laboratory (cont’d) Dilutions are always expressed with the original substance diluted as one (1). If more than one part of original substance is initially used, it is necessary to convert the original substance part to one (1) when the dilution is expressed.

Dilutions for the Clinical Laboratory (cont’d) Example: Two (2) parts of dye are diluted with eight (8) parts of diluent (the term often used for the diluting solution). The total solution volume is 10 parts (2 parts dye + 8 parts diluent). The dilution is initially expressed as 2/10, but the original substance must be expressed as one (1). To get the original volume to one (1), use a ratio and proportion equation, remembering that dilutions are stated in terms of 1 to something: ______2 parts dye = ___1.0___ 10 parts total volume x 2 x = 10 x = 5 The dilution is expressed as 1/5.

Dilutions for the Clinical Laboratory (cont’d) The dilution does not always end up in whole numbers. Example: Two parts (2) parts of whole blood are diluted with five (5) parts of saline. The total solution volume is seven (7) parts (2 parts of whole blood + 5 parts saline). The dilution would be 2/7, or, more correctly, 1/3.5. Again, this is calculated by using the ratio and proportion equation, remembering that dilutions are stated in terms of 1 to something: __2 parts blood_____ = ___1.0___ 7 parts total volume x 2 x = 7 x = 3.5 The dilution is expressed as 1/3.5

Dilutions for the Clinical Laboratory (cont’d) Dilution Factor – used to correct for having used a diluted sample in a lab test rather than the undiluted sample. The result (answer) using the diluted sample must be multiplied by the RECIPROCAL of the dilution made. The RECIPROCAL of a 1/5 dilution is 5.

Dilutions for the Clinical Laboratory (cont’d) Correction for using a diluted sample Example: A technician performed a laboratory analysis of patient’s serum for a serum glucose (blood sugar) determination. The patient’s serum glucose was too high to read on the glucose instrument. The technician diluted the patient’s serum 1/2 and reran the diluted specimen, obtaining a result of 210 g/dl. To correct for the dilution, it is necessary to multiply the result by the dilution factor (in this case x 2). The final result is 210 g/dl x 2 = 420 g/dl.

Dilutions for the Clinical Laboratory (cont’d) Sometimes it is necessary to make a dilution of an existing solution to make it weaker. Example: A 100 mg/dl solution of substrate is needed for a laboratory procedure. All that is available is a 500 mg/dl solution of substrate. A dilution of the stronger solution of substrate is needed.

Dilutions for the Clinical Laboratory (cont’d) To make a weaker solution from a stronger one, use this formula: V1 x C1 = V2 x C2 Example: To make 100 ml of the 100 mg/dl solution from the 500 mg/dl solution needed in the previous example: V1 = 100 ml V2 = V2 (unknown) C1 = 100 mg/dl C2 = 500 mg/dl 100 ml x 100 mg/dl = V2 x 500 mg/dl V2 = 20 ml Dilute 20 ml of 500 mg/dl solution up to 100 ml with water to obtain 100 ml of 100 mg/dl substrate solution

Serial Dilutions Dilutions can be made singly (as shown previously) or in series, in which case the original dilution is diluted further. A general rule for calculating the dilution of solutions obtained by diluting in a series is to MULTIPLY the original dilution by subsequent dilutions.

Serial Dilutions (cont’d) Example of a serial dilution:

Serial Dilutions (cont’d) In the serial dilution on the previous slide, 1 ml of stock solution is mixed with 9 ml of diluent, for a 1/10 dilution. Then 1 ml of the 1/10 dilution is mixed with another 9 ml of diluent. The second tube also has a 1/10 dilution, but the concentration of stock in the second tube is 1/10 x 1/10 for a 1/100 dilution.

Serial Dilutions (cont’d) Continuing with the serial dilution, in the third tube, you mix 1 ml of the 1/100 dilution from the second tube with 9 ml of diluent in the third tube. Again you have a 1/10 dilution in the third tube, but the concentration of stock in the third tube is 1/10 x 1/10 x 1/10 for a 1/1000 dilution. This dilution could be carried out over many subsequent tubes.

Serial Dilutions (cont’d) Serial dilutions are most often used in serological procedures, where technicians need to make dilutions of patient’s serum to determine the weakest concentration that still exhibits a reaction of some type. The RECIPROCAL of the weakest concentration exhibiting a reaction is called a “titer”.

Serial Dilutions (cont’d) Example of determining a titer: A technician makes a serial dilution using patient serum: Tube #1 = 1/10 Tube #2 = 1/100 Tube #3 = 1/1000 Tube #4 = 1/10,000 Tube #5 = 1/100,000 Reactions occur in tubes 1 through 3, but NOT in tubes 4 or 5. The titer = 1000.

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