Dilutions AV

Dilutions

Dilutions for the laboratory

Dilution = making weaker solutions from stronger onesExample: Making orange juice from frozen concentrate. You mix one can of frozen orange juice with three (3) cans of water.

Introduction

Many of the laboratory procedures involve the use of dilutions.

It is important to understand the concept of dilutions, since they are a handy tool used throughout all areas of the clinical/research laboratory.

These dilutions have to be considered as they make a quantitative difference in what is going on.

Dilutions for the laboratory (cont’d)

Dilutions are expressed as the volume of the solution being diluted per the total final volume of the dilutionIn 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”.


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).


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”.


Notice that dilutions do NOT have units (cans, ml, or parts) but are expressed as one number to another numberExample: 1/10 or “one in ten”


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 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 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


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 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.


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 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/dl100 ml x 100 mg/dl = V2 x 500 mg/dlV2 = 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

Dilutions

If a 1/8 dilution of the stock solution is made followed by a 1/6 dilution what is the final dilution.

The final dilution is: 1/8 x 1/6 = 1/48 These type of dilutions are trickier and

not used very frequently in the lab.

Serial Dilutions

Dilutions can be made singly (as shown previously) or in series, in which case the original dilution is diluted further.

A serial dilution is any dilution where the concentration decreases by the same quantity in each successive step.

Serial dilutions are mutiplicative.

Serial Dilutions (cont’d)

Example of a serial dilution:


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.

Doubling dilutions

“Doubling dilutions” are very popular. This is a series of ½ dilutions. Each

successive tube will ½ the amount of the original concentrated solution.

If this is done 6 times this is what you would end up with:

Doubling dilutions 6 times

1st dilution = 1 /2 2nd dilution = 1 /2 x 1 /2 = 1/4 3rd dilution = 1/4 x 1 /2 = 1/8 4th dilution = 1/8 x 1 /2 = 1/16 5th dilution = 1/16 x 1 /2 - 1/32 6th dilution = 1/32 x 1 /2 = 1/64 This results in a series of dilutions, each

a doubling dilution of the previous one


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 = 1/10Tube #2 = 1/100Tube #3 = 1/1000Tube #4 = 1/10,000Tube #5 = 1/100,000Reactions occur in tubes 1 through 3, but NOT in tubes 4 or 5. The titer = 1000.

Dilution Factor The dilution factor uses the formula

volume/aliquot volume. EXAMPLE: What is the dilution factor if

you add 0.1 mL aliquot of a specimen to 9.9 mL of diluent? The final volume is equal to the aliquot

volume PLUS the diluent volume: 0.1 mL + 9.9 mL = 10 mL

The dilution factor is equal to the final volume divided by the aliquot volume: 10 mL/0.1 mL = 1:100 dilution

Practice

Problem: What is the dilution factor when 0.2 mL is added to 3.8 mL diluent?

Set Up The Problem

dilution factor = final volume/aliquot volume

0.2 +3.8 = 4.0 total volume 4.0/0.2 = 1:20 dilution

Problem Continued

Remember that serial dilutions are always made by taking a set quantity of the initial dilution and adding it successively to tubes with the same volume.

So each successive dilution would be multiplied by the dilution factor.

Problem Continued

So in the above problem all successive tubes would have 3.8 mLs of diluent.

You would then transfer 0.2 of the initial diluted sample into the next tube, mix transfer 0.2, mix and so on.

If you had 4 tubes what would be the final dilution of tube 4?

Solving the Problem - *Calculate DF of tube 1

TubeTube 11 22 33 44

AliquotAliquot 0.20.2 0.20.2 0.20.2 0.20.2

DiluenDiluentt

3.8 3.8 3.8 3.8

MathMath *4/0.*4/0.22

1/20x1/201/20x1/20 1/400x1/21/400x1/200

1/8000x1/201/8000x1/20

DilutioDilutionn

1:201:20 1:4001:400 1:80001:8000 1:160,0001:160,000

Solving the Problem

Or if you simply wanted to know the dilution of the final tube you could just multiply them together:

1/20 x 1/20 x 1/20 x 1/20 = 1:160,000

Applications Biology:

To determine concentrations of microscopic organisms or cells

Comparison of two samples Dilute reagents for protocols/storage

Medicine to determine microbial overload in a sample Test blood values of various components

Homeopathy Core foundational practice

Applications

Calibration of instruments Standard curves

Documents

Dilutions AV