Animated presentation, we suggest to switch slideshow mode on (ie. by pressing F5)

1. Animated presentation, we suggest to switch slideshow mode on (ie. by pressing F5) [Changing slides: cursors, space/backspace, mouse scroll, PageUp/PageDown]

2. Determination of the concentration • A quantitative property of an indicator refers to the concentration: • color (absorbance, optical density) • fluorescence • cell number (e.g. in determination of growth factor concentration) Quantified concentration can be obtained by comparison with known concentration sample (standard) The principle of comparison: equal absorbances  equal concentrations PARTIAL TRUTH !!!

3. The serial dilution of the standard OD The sample with unknown concentration You should also dilute the unknown sample This region could indicate the concentration According to OD: it could be anyone ? 62 31 16 7.8 3.9 1.9 500 250 125 0 0.97 0.49 0.24 0.12 1000 0.061 0.030 0.015 0.007 0.004 concentration

4. Estimating the concentration with a „ruler” The OD are proportional with the concentrations in this range points with identical OD 62 31 16 7.8 3.9 1.9 500 250 125 0.97 0.49 0.24 0.12 1000 0.061 0.030 0.015 0.007 0.004 0.002 2X 4X 8X 16X 32X 64X 128X 256X Dilutions of the unknown sample So, the concentration of the (undiluted) unknown sample: 1.9x128 = 243.2μg/ml OD The concentrations are equal in the tubes The 1.9μg/ml diluted standard corresponds to the… conc. of the standard (µg/ml) … 128-fold diluted unknown sample

5. You can use linear regression (Least-squares analysis), and calculate the concentrations with the equations (formula) of the lines fitted on the linear parts of the dilution curves 62 31 16 7.8 3.9 1.9 500 250 125 0.97 0.49 0.24 0.12 1000 0.061 0.030 0.015 0.007 0.004 0.002 2X 4X 8X 16X 32X 64X 128X 256X Dilutions of the unknown sample OD Ysample=mx+b ystd=mx+b conc. of the standard (µg/ml) (two-fold dilution (log scale!))

6. This OD range results false concentrations also The range of suitable OD values The OD of any highly diluted solutions will be located on this range of the dilution curve.If you insert this OD value into the formula and calculate the concentration by multiplying it with the dilution, then you get enormous high FALSE concentration. Serious errors You must know the optical density range that you should use to calculate the concentration with the equation(formula) of the dilution line! OD The fitted line with its equation(formula) y=mx+b  OD=m(concentration)+b Dilution curve concentration

7. The range of suitable OD values Don’t force fitting the line where it is unnecessary OD dilution curve Incorrectly fitted line The line has to be fitted to these points too! concentration

8. Notice that the dilution curve is represented on logaritmic function! Two different representations of the same results: Normal (linear) dilution curve Logaritmic dilution curve The correct representation helps to find the proper points of the curve

9. Dilution: = dilution 2x:

10. Good representation helps the correct data analysis Try to find the proper points of the ‘sigmoid’ curve (even if it’s not represented in the function completely), and fit the line to these proper points

11. In practice, it rarely happens that we are able to work with good standard dilution curves. It is the same with the dilution curve of the unknown sample. Usually we only make 2-3 dilutions of the samples. 1x 2x 4x 8x 16x 32x 64x 128x OD 4.5 standard (1x: 100 mg/ml) 4.0 3.5 3.0 The unknown: 20x10 = 200 mg/ml 2.5 2.1 2.0 1.5 Value around the sensitivity threshold 1.1 1.0 dilution Approx. 5 (in order to have the accurate value we could use the equation of the line between the two points) 100/5 = 20 mg/ml

12. 1x 2x 4x 8x 16x 32x 64x 128x Should we use the remaining OD value? OD 4.5 standard (1x: 100mg/ml) 4.0 4.0 The OD 4.0 is the value around the most concentrated standard dilution value 3.5 3.0 It is possible that the dilution curve has other shape 2.5 2.0 1.5 1.0 dilution For example if the missing 2 fold dilution value should be here Than it is the starting point of the plateau of the dilution curve In this example the OD value of the undiluted (1x) sample should not be used.

13. Presentation ELISA plate with serially diluted IFNγ standard and Tcell culture supernatants Which is the concentrated sample? • Try to calculate the concentration of the given ELISA data at home! • calculate the mean of the 3 parallel data • use the logarithm of the dilution to draw the dilution curves • try to use a computer with spreadsheet program

14. Calculate the concentration of the unknown sample. The concentration of the standard is 100μg/ml.

15. Write these formulas: the log of the dilution,… the average of the parallel standards,… and the average of the parallel samples

16. Select the cells with the formulas, click on the tiny square on the right bottom corner of the selected square, and drag it (autofill) into the next lines. Click on the chart wizard

17. Choose the „XY Scatter” chart type Choose the series tab Add data

18. Write the name of the first data series Choose the dilution (log) values Click on the X values (dilution) =Sheet1!$C$17:$C$28 Click on the Y values (OD values) =Sheet1!$A$17:$A$28 Choose the standard OD values

19. Do the same procedure with the unknown sample data also

20. Straight line can be fitted on point 5 to 8 of the standard curve, and the on point 3 to point 6 of the sample titration curve

21. Data of the standard curve linear part ! Data of the sample curve linear part ! You can use this data to draw the linear parts

22. Right click on the line and choose the „Add Trendline” option

23. Linear trendline (Options tab) Display equation on chart

24. The equation of the standard line Do the same with the sample line

25. You can calculate the dilutions of different OD solutions with the equations sample: y = -0.3778x + 2,8471 standard: y = -0.3678x + 3.5265 e.g. OD 1.2  x=4.359 OD 1.2  x=6.325 The 104.359=22856 –fold dilution of the sample has equal OD than…. …the 106.325=2113489 –fold diluted standard The sample is 2113489/22856= app. 92x thinner than the 100µg/ml standard 1.08 μg/ml

26. Note This is a demonstrative tutorial example! You can get the result easier! (eg. You need only the equation of the standard trend line compared to an appropriately chosen dilution and OD value of the sample for get the correct result)