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.