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Protein
concentration
calculation
Protein Assay by Absorbance
 Absorbance assays are fast and
convenient, since no additional reagents
or incubations are required.
 No protein standard need be prepared. 
 Any non-protein component of the
solution that absorbs ultraviolet light will
interfere with the assay. These can be
nucleic acids, lipids.
Beer Lambert Law: A= l x c x α
A= -log10 I/Io
C= Analyte concentration ; α = absorption coefficient wavelength dependent
Principle of the test
 Proteins in solution absorb ultraviolet light
with absorbance maxima at 280 and 200
nm.
 Amino acids with aromatic rings are the
primary reason for the absorbance peak
at 280 nm
 Peptide bonds are primarily responsible
for the peak at 200 nm
The intensity of the absorbance is proportional to the number of aromatic
Amino acids in the protein
BSA: bovine
serum Albumin
Absorbance
Protein Absorbance is Max at 280 nm
DNA Absorbance is Max at 260 nm
The DNA/PROT at A280 is higher than
PROT only sample
How to calculate Concentration
 If nucleic acid are present in the protein sample
they will interfere with the absorbance at 280 nm
(A280).
 Use the following formula to estimate protein
concentration and remove nucleic Acid
interference at A260nm.
 Protein Concentration (mg/ml) = (1.55 x A280) -
0.76 x A260).
 Example: if A280= 1.2 and A260 = 0.3
the PC= (1.55 x 1.2) – (0.75 x 0.3) = 1.635 mg/ml
Serial Dilutions
Introduction
Many of the laboratory procedures
involve the use of dilutions.
It is important to understand the
concept of dilutions, since they are
used throughout all areas of the
clinical laboratory.
Serial Dilutions
 A serial dilution is any dilution where the
concentration decreases by the same
quantity in each successive step.
 Serial dilutions are mutiplicative.
What Does This Mean??
 If a solution has a 1/10 dilution the number
represents 1 part of the patient sample
added to 9 parts of diluent.
 So the volumes used would be 10-1= 9.
 This represents 1 part patient sample
added to 9 parts of diluent.
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 Dilution 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
Dilution Factor
The dilution factor is the final 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 Dilution Factor of tube 1
TubeTube 11 22 33 44
AliquotAliquot 0.20.2 0.20.2 0.20.2 0.20.2
DiluentDiluent 3.8 3.8 3.8 3.8
MathMath *4/0.2*4/0.2 1/20x1/201/20x1/20 1/400x1/201/400x1/20 1/8000x1/201/8000x1/20
DilutionDilution 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
To Measure Immune reaction
We use the Titers
 TITERS are reported out as the reciprocal
of the last tube giving a positive allergic
reaction (ALR).
 So if tube 2 gave the lowest ALR, the
dilution is 1:800 the titer is reported out as
800/1= 800.

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Protein det

  • 2. Protein Assay by Absorbance  Absorbance assays are fast and convenient, since no additional reagents or incubations are required.  No protein standard need be prepared.   Any non-protein component of the solution that absorbs ultraviolet light will interfere with the assay. These can be nucleic acids, lipids.
  • 3. Beer Lambert Law: A= l x c x α A= -log10 I/Io C= Analyte concentration ; α = absorption coefficient wavelength dependent
  • 4. Principle of the test  Proteins in solution absorb ultraviolet light with absorbance maxima at 280 and 200 nm.  Amino acids with aromatic rings are the primary reason for the absorbance peak at 280 nm  Peptide bonds are primarily responsible for the peak at 200 nm
  • 5. The intensity of the absorbance is proportional to the number of aromatic Amino acids in the protein BSA: bovine serum Albumin
  • 6. Absorbance Protein Absorbance is Max at 280 nm DNA Absorbance is Max at 260 nm The DNA/PROT at A280 is higher than PROT only sample
  • 7. How to calculate Concentration  If nucleic acid are present in the protein sample they will interfere with the absorbance at 280 nm (A280).  Use the following formula to estimate protein concentration and remove nucleic Acid interference at A260nm.  Protein Concentration (mg/ml) = (1.55 x A280) - 0.76 x A260).  Example: if A280= 1.2 and A260 = 0.3 the PC= (1.55 x 1.2) – (0.75 x 0.3) = 1.635 mg/ml
  • 9. Introduction Many of the laboratory procedures involve the use of dilutions. It is important to understand the concept of dilutions, since they are used throughout all areas of the clinical laboratory.
  • 10. Serial Dilutions  A serial dilution is any dilution where the concentration decreases by the same quantity in each successive step.  Serial dilutions are mutiplicative.
  • 11. What Does This Mean??  If a solution has a 1/10 dilution the number represents 1 part of the patient sample added to 9 parts of diluent.  So the volumes used would be 10-1= 9.  This represents 1 part patient sample added to 9 parts of diluent.
  • 12. 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:
  • 13. Doubling Dilution 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
  • 14. Dilution Factor The dilution factor is the final 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
  • 15. Practice  Problem: What is the dilution factor when 0.2 mL is added to 3.8 mL diluent?
  • 16. 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
  • 17. 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.
  • 18. 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?
  • 19. Solving the Problem – *Calculate Dilution Factor of tube 1 TubeTube 11 22 33 44 AliquotAliquot 0.20.2 0.20.2 0.20.2 0.20.2 DiluentDiluent 3.8 3.8 3.8 3.8 MathMath *4/0.2*4/0.2 1/20x1/201/20x1/20 1/400x1/201/400x1/20 1/8000x1/201/8000x1/20 DilutionDilution 1:201:20 1:4001:400 1:80001:8000 1:160,0001:160,000
  • 20. 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
  • 21. To Measure Immune reaction We use the Titers  TITERS are reported out as the reciprocal of the last tube giving a positive allergic reaction (ALR).  So if tube 2 gave the lowest ALR, the dilution is 1:800 the titer is reported out as 800/1= 800.