1) Protein concentration can be calculated using absorbance assays which measure how much ultraviolet light is absorbed by protein solutions. 2) The Beer-Lambert law states that absorbance is directly proportional to analyte concentration. Proteins absorb UV light with maxima at 280nm and 200nm due to aromatic amino acids and peptide bonds. 3) Serial and doubling dilutions are used to successively decrease the concentration of solutions. Serial dilutions decrease the concentration by the same amount each time while doubling dilutions halve the concentration at each step.

1. Protein
concentration
calculation

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

8. Serial Dilutions

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.