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- 1. Proteinconcentrationcalculation
- 2. Protein Assay by Absorbance Absorbance assays are fast andconvenient, since no additional reagentsor incubations are required. No protein standard need be prepared. Any non-protein component of thesolution that absorbs ultraviolet light willinterfere with the assay. These can benucleic acids, lipids.
- 3. Beer Lambert Law: A= l x c x αA= -log10 I/IoC= Analyte concentration ; α = absorption coefficient wavelength dependent
- 4. Principle of the test Proteins in solution absorb ultraviolet lightwith absorbance maxima at 280 and 200nm. Amino acids with aromatic rings are theprimary reason for the absorbance peakat 280 nm Peptide bonds are primarily responsiblefor the peak at 200 nm
- 5. The intensity of the absorbance is proportional to the number of aromaticAmino acids in the proteinBSA: bovineserum Albumin
- 6. AbsorbanceProtein Absorbance is Max at 280 nmDNA Absorbance is Max at 260 nmThe DNA/PROT at A280 is higher thanPROT only sample
- 7. How to calculate Concentration If nucleic acid are present in the protein samplethey will interfere with the absorbance at 280 nm(A280). Use the following formula to estimate proteinconcentration and remove nucleic Acidinterference at A260nm. Protein Concentration (mg/ml) = (1.55 x A280) -0.76 x A260). Example: if A280= 1.2 and A260 = 0.3the PC= (1.55 x 1.2) – (0.75 x 0.3) = 1.635 mg/ml
- 8. Serial Dilutions
- 9. IntroductionMany of the laboratory proceduresinvolve the use of dilutions.It is important to understand theconcept of dilutions, since they areused throughout all areas of theclinical laboratory.
- 10. Serial Dilutions A serial dilution is any dilution where theconcentration decreases by the samequantity in each successive step. Serial dilutions are mutiplicative.
- 11. What Does This Mean?? If a solution has a 1/10 dilution the numberrepresents 1 part of the patient sampleadded to 9 parts of diluent. So the volumes used would be 10-1= 9. This represents 1 part patient sampleadded to 9 parts of diluent.
- 12. Doubling Dilutions “Doubling dilutions” are very popular. This is a series of ½ dilutions. Eachsuccessive tube will ½ the amount of theoriginal concentrated solution. If this is done 6 times this is what you wouldend 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, eacha doubling dilution of the previous one
- 14. Dilution FactorThe dilution factor is the final usesthe formula volume/aliquot volume.EXAMPLE: What is the dilution factorif you add 0.1 mL aliquot of aspecimen to 9.9 mL of diluent? The final volume is equal to the aliquotvolume PLUS the diluent volume:0.1 mL + 9.9 mL = 10 mL The dilution factor is equal to the finalvolume divided by the aliquot volume:10 mL/0.1 mL = 1:100 dilution
- 15. Practice Problem: What is the dilution factor when0.2 mL is added to 3.8 mL diluent?
- 16. Set Up The Problem dilution factor = final volume/aliquotvolume 0.2 +3.8 = 4.0 total volume 4.0/0.2 = 1:20 dilution
- 17. Problem Continued Remember that serial dilutions are alwaysmade by taking a set quantity of the initialdilution and adding it successively totubes with the same volume. So each successive dilution would bemultiplied by the dilution factor.
- 18. Problem Continued So in the above problem all successivetubes would have 3.8 mLs of diluent. You would then transfer 0.2 of the initialdiluted sample into the next tube, mixtransfer 0.2, mix and so on. If you had 4 tubes what would be the finaldilution of tube 4?
- 19. Solving the Problem –*Calculate Dilution Factor of tube 1TubeTube 11 22 33 44AliquotAliquot 0.20.2 0.20.2 0.20.2 0.20.2DiluentDiluent 3.8 3.8 3.8 3.8MathMath *4/0.2*4/0.2 1/20x1/201/20x1/20 1/400x1/201/400x1/20 1/8000x1/201/8000x1/20DilutionDilution 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 thedilution of the final tube you could justmultiply them together:1/20 x 1/20 x 1/20 x 1/20 = 1:160,000
- 21. To Measure Immune reactionWe use the Titers TITERS are reported out as the reciprocalof the last tube giving a positive allergicreaction (ALR). So if tube 2 gave the lowest ALR, thedilution is 1:800 the titer is reported out as800/1= 800.

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