MLAB 2401:
Clinical Chemistry
Basic Principles and Practice of Clinical
Chemistry
Part One
1
UNITS OF MEASURE

Measurement requires a numerical value and a unit
 Laboratory results almost always have units of measurement associated
with them
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SI units:
 length ( meter )
 mass ( gram )
 quantity ( mole )
 Volume ( liter )
 Time ( second )
Basic units describe unrelated physical quantities
2
Unit of Measure: Prefixes

Common prefixes and abbreviations that are added to units of measure:

deci (d)
10-1

centi (c)
10-2

milli (m)
10-3
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micro ( μ)
10-6
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nano (n)
10-9
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pico (p)
10-12

femto (f)
10-15
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Example: A common unit of liquid measurement is a deciliter( dl ), or one –
tenth of a liter
Combine a prefix with a basic unit results in a statement of a specific length,
weight or volume

Reporting clinical chemistry results may be in units such as :

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mg / dL
g / dL
mEq / L
3
Scientific Notation

True scientific notation format:

1.22 X 104

BUT in hemo, for example a hemoglobin result
would look like = 12.2 X 103
4
Water Specifications

Tap water is unsuitable for lab use (too many impurities)
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Types of water purification techniques
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Reagent Grades of water
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Distillation – removes most organic matter
Reverse osmosis-removes organic, ionic, microbial, and viral
contaminants
Ultrafiltration – removes particulate matter, bacteria, emulsified solids
Deionization – ions removed
Type I
Type II
Type III
Purest – Required for sensitive tests
Acceptable for most uses
OK for washing glassware
CAP - QC of water : pH, electrical resistance, bacterial culture
5
Water filtration system for
Automated chemistry analyzer.
6
Solutions
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The clinical lab almost always uses solutions. A solution means that
something has been dissolved in a liquid. In the clinical laboratory the
solvent we measure most of the time is human plasma. The solute is
whatever the substance is we want to measure.
Mixtures of substances – the substances in a solution are not in
chemical combination with one another.
Dispersed phase - the substance is dissolved (the solute)
The substance in which the solute is dissolved is the solvent.
Solute + Solvent = Solution
7
Concentration

Amount of one substance relative to the amounts of the other substances in
the solution.

Concentration can be measured in many different units

% Solutions: w/w, v/v , w/v (parts of solute / 100 totals parts )
Note: liquids + liquids and solids + solids alters the total parts, but
solutes + solvents does not

Molarity: Moles / Liter

Molality: Moles / 1000 grams solvent

Normality: equivalent weight/ liter
8
Expressing Concentration:
Percent Solution (parts/100)

% w/w – percentage weight per weight

Most accurate method of expressing concentration, but can be
cumbersome (especially with liquids), not often used in clinical
labs.

% w/w = gram of solute OR gram of solute per 100.0 g of solution
100.0 g of solution

How many grams of NaOH are needed to make a 25.0% w/w
solution using deionized water as the solvent?
25.0% w/w = X g of solute in 100 g of solution
X= 25.0 g NaOH
9
Expressing Concentration:
Percent Solution (parts/100)

% w/v – percentage weight per volume

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Easiest & most commonly used, very accurate if temperature controlled.
%w/v= g of solute OR
100 mL of solution
g of solute per 100.0 mL of solution
What is the %w/v of a solution that has 15.0 g of NaCl dissolved into a total
volume of 100 mL deionized water?
X% w/v = 15.0 g NaCl
100 mL of solution
X= 15.0 %
10
Expressing Concentration:
Percent Solution (parts/100)

% v/v –percentage volume per volume
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Least accurate, but used when both substances are liquids
Note: volumes of liquids are not necessarily additive
%v/v= mL of solute
OR milliliter of solute per 100 mL of solution
100 mL of solution
How many milliliters of ethanol are needed to make a 75.0% v/v
solution using deionized water as the solvent?
75.0% v/v EtOH = X mL EtOH in 100 mL of solution
= 75.0 mL EtOH
11
Expressing Concentration: Molarity
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Three components of Molarity
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Gram weight of solute
Solute’s gram molecular weight
Solvent quantity
Number of moles per one liter of solution
Mole = 6.022 X 1023 number of atoms or
molecules OR
Mole= Molecular weight in grams
12
Determinig Molarity: First step
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Molecular Weight
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Sum of the atomic weights of each element in the compound
What is the molecular weight of Na3PO4?
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Step 1: Sodium has an atomic weight of 22.99, but there are 3 molecules so
22.99*3= 68.97
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Step 2: Phosphorus has an atomic weight of 30.97, and only 1 molecule, so
30.97 *1= 30.97
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Step 3: Oxygen has an atomic weight of 16, but there are 4 molecules ,so
16*4= 64.00
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Step 4: Add 68.97+ 30.97+ 64.00= 163.94 gram molecular weight
13
Determinig Molarity: Next Step


How many grams are contained in one mole of Na3PO4?
Use the formula for mole calculations
Number grams of solute
Gram molecular weight of solute
1 mole Na3PO4 = X g Na3PO4
gram molecular weight(gmw)
X= 163.94 g Na3PO4
So, 163.94 grams of trisodium phosphate are contained in 1 mole of
trisodium phosphate or 6.022 X 1023 trisodium phosphate molecules
weigh 163.94 grams
14
Determinig Molarity: Final Step
Molarity (M) = 1 mole of solute
1L of solution
We are asked to make a 1.00 L volume of a 0.100 molar solution of trisodium phosphate.
How many grams would we need?
M= grams
gmw
1.00 L of solution

0.100 molar= X grams of Na3PO4
163.94 gmw of Na3PO4
1.00 L of solution
(0.100M)(1.ooL) = X g
163.94 gmw
0.100= X
163.94
(0.100)(163.94)= X
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Expressing Concentration:
Molality
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Amount of solute per one kg of solvent
Expressed in terms of weight per weight or
moles per 1000 grams of solvent
Used to measure the physical properties of
solutions
Molality = 1 mole of solute
1 kg of solvent
16
Expressing Concentration:
Normality- First Step
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Equivalents Weights / Liter
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Equivalent weight is equal to the gram molecular weight of
a substance divided by its valence
Valence = the electrical charge of an ion, or the number
of moles that react with 1 Mole H+

Example

The MW of calcium = 40 grams

Calcium ions carry a +2 electrical charge ( valence = 2 )

Equivalent Weight of calcium = 40 / 2 = 20 gram equivalent
weight
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Normality:
N= number of grams of solute
Gram equivalent weight of solute
1.00 L of solution
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Normality (N)
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N = Molarity (M) x valence
Molarity = N / valence
M is always < N
18
Solution Properties
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Titration – Method of measuring concentration of one solution by comparing
it with a measured volume of a solution whose concentration is known
General formula: when you have a volume and concentration of one,
and either the volume or the concentration of the other: V1 C1 = V2 C2

For Example:
How many mls of 1.0 N HCl is required to prepare 25 mls of 0.5 N HCl ?
( 1.0 N ) ( ? mls ) = ( 0.5 N ) ( 25 mls)
? mls = 12.5 mls
You would need to add 12.5 mls of 1.0 N HCl to 12.5 mls of deionized water
( a total volume of 25 mls) to prepare 25 mls of 0.5 N HCl
19
Solution Properties
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Density – An expression in terms (usually) of
a mass per unit of volume
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Many examples - including specific gravity,
osmolality
20
pH and Buffers
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Buffers resist change in acidity
Buffers are usually weak acids ( or bases) and their salts
pH is the unit used to measure acidity ( Hydrogen ion concentration )
“p” = “negative log” of the concentration of a substance in solution.
Example: pH = - log [H+]
The Hydrogen ion concentration of deionized H2O is 1 x 10-7 M
The negative log of 10-7 = 7. The pH of H2O is 7.0
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The pH scale ranges from 0 - 14
pH 7 = neutral
pH > 7 = alkaline (basic)
pH < 7 = acid
21
Temperature

Measurement of temperature is an important component of
the clinical lab. Instruments, refrigerators and incubators are
required to operate within specific temperatures that must be
maintained and monitored daily.
 Examples
 Heat blocks, water baths, and incubators shall be
maintained at least +/- 1 degree C. of the desired
temperature
 Refrigerators shall be maintained at 2 -8 degrees C.

Each laboratory must have a NIST calibrated thermometer in
order to ensure the accuracy of other thermometers in the
laboratory
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Out-of-range temperatures should be addressed asap
22
Temperature

Scientific measurement of temperature is always expressed in the
Celsius ( C) scale , not Fahrenheit ( F )

Celsius scale: 0 degrees = freezing point of water
100 degrees = boiling point of water
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Conversion: Temperature

Conversion of Celsius to Fahrenheit and Fahrenheit to
Celsius

F° = ( C ° x 1.8 ) + 32

C° = ( F ° - 32 )
1.8
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For example:
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Your refrigerator at home is probably around 40 ° F. What is that in
Celsius?
 Celsius= 40-32 = 4.4
1.8
Water boils at 100 ° C. What is that expressed in Fahrenheit?
 (1.8)(100) +32 = 212
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Conversions
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Most conversions within the metric system occur in units of TEN
where changing a unit of measure to a higher or lower designation
requires moving the decimal one place either to the left or to the
right.
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When converting measures in either the high end of the scale
(example kilo to mega) or the low end of the scale (examples milli to
micro, micro to nano, etc.) the decimal must be moved three places
right or left as the prefix designations are assigned only to every
third unit in the extreme ends.
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Example of a conversion
How many mls are there in 2.5 liters?
The question you have to ask yourself is, what is the relationship between
liters and mls? The answer : 1 liter = 1000 ml
But now what?
We want to get rid of the “liters’ units and end up with “mls” … Right ?
 1000 mls 
2.5 Liter 
  2500 mls
 1 Liter 
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1.25 liters = _____ mls ?
Remember, write a fraction that does two things:
1. Equals 1
2. Gets rid of unwanted units and / or adds needed units
 1000 mls 
  1250 mls
. Liters 
125
 1 Liter 
100 mg =
_________ ug ?
 1000 ug 
100 mg  1 mg   10,0000 ug
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Dilutions
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A ratio of the concentrate to the total (final) volume.
 A 1:4 dilution has a 1 volume of sample and 3 volumes of diluent
mixed together.
Any volume can be used to create this dilution, but it must be the
same unit of volume
Keep in mind the sample size when making your dilution
 For example: a 2:3 dilution could contain:
 2 mL serum: 1 mL pure water
 20 µL of serum: 10 µL of pure water
 0.2 mL of serum: 0.1 mL of pure water
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Dilutions for the Clinical Laboratory
Example:
A technician performed a laboratory analysis of
patient’s serum for a serum glucose 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.
Examples of dilutions and dilution factors
Parts
Specimen
Parts
Diluent
Total
Volume
Dilution
Dilution
Factor
1.0
1.0
2.0
1:2
2
1.0
2.0
3.0
1:3
3
1.0
3.0
4.0
1:4
4
1.0
9.0
10.0
1 : 10
10
0.5
4.5
5.0
1 : 10
10
0.2
1.8
2.0
1 : 10
10
0.2
9.8
10.0
1 : 50
50
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Serial Dilutions
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In these types of questions, you are given a series of tubes.
Each tube having a measured amount of a diluent.
You are instructed to add a specified amount of specimen into the first
tube, mix well and transfer a specified amount of the mixture to the next
tube, etc.
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Serial Dilutions
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Example:
 6 tubes, each with 0.5 mL DI water
 Add 0.2 mL serum to first tube and serially dilute
 Find the dilution in tube # 6
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Find the dilution factor (will be the same in each of these tubes)
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1/dil factor x 1/dil factor x 1/dil factor (etc. 6 times)
Result multiplying the numerator 1x1x1x1x1x1x1x = 1
Multiplying the denominators
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Will give the result as 1 / 1838
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Resources
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Serial dilution
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http://tinyurl.com/cw7e3ok
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References
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Bishop, M., Fody, E., & Schoeff, l. (2010). Clinical Chemistry:
Techniques, principles, Correlations. Baltimore: Wolters
Kluwer Lippincott Williams & Wilkins.
Doucette, L. (2011). Mathematics for the Clinical Laboratory
(2nd ed.). Maryland Heights, MO: Saunders.
Sunheimer, R., & Graves, L. (2010). Clinical Laboratory
Chemistry. Upper Saddle River: Pearson .
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