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# Protein det

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### Protein det

1. 1. Proteinconcentrationcalculation
2. 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. 3. Beer Lambert Law: A= l x c x αA= -log10 I/IoC= Analyte concentration ; α = absorption coefficient wavelength dependent
4. 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. 5. The intensity of the absorbance is proportional to the number of aromaticAmino acids in the proteinBSA: bovineserum Albumin
6. 6. AbsorbanceProtein Absorbance is Max at 280 nmDNA Absorbance is Max at 260 nmThe DNA/PROT at A280 is higher thanPROT only sample
7. 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. 8. Serial Dilutions
9. 9. IntroductionMany 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. 10. Serial Dilutions A serial dilution is any dilution where theconcentration decreases by the samequantity in each successive step. Serial dilutions are mutiplicative.
11. 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. 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. 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. 14. Dilution FactorThe 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. 15. Practice Problem: What is the dilution factor when0.2 mL is added to 3.8 mL diluent?
16. 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. 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. 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. 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. 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. 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.