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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 !!!

concentration The sample with unknown concentration OD The serial dilution of the standard According to OD: it could be anyone ? You should also dilute the unknown sample This region could indicate the concentration

OD Estimating the concentration with a „ruler” conc. of the standard (µg/ml) The OD are proportional with the concentrations in this range 2X4X8X16X32X64X128X256X Dilutions of the unknown sample points with identical OD The concentrations are equal in the tubes The 1.9μg/ml diluted standard corresponds to the… … 128-fold diluted unknown sample So, the concentration of the (undiluted) unknown sample: 1.9x128 = 243.2μg/ml

OD conc. of the standard (µg/ml) (two-fold dilution (log scale!)) 2X4X8X16X32X64X128X256X Dilutions of the unknown sample 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 y std =mx+b Y sample =mx+b

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! Dilution curve The fitted line with its equation(formula) y=mx+b  OD=m(concentration)+b OD concentration 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. This OD range results false concentrations also The range of suitable OD values

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

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

Dilution: dilution 2 x : =

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

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. 1x2x4x8x16x32x64x128x OD standard (1x: 100  g/ml) dilution Dilutions of the unknown dilution1x10x100x OD Value around the sensitivity threshold 2.1 Approx. 5 (in order to have the accurate value we could use the equation of the line between the two points) 100/5 = 20  g/ml The unknown: 20x10 = 200  g/ml

In this example the OD value of the undiluted (1x) sample should not be used. 1x2x4x8x16x32x64x128x OD standard (1x: 100mg/ml) dilution Dilutions of the unknown dilution1x10x100x OD The OD 4.0 is the value around the most concentrated standard dilution value It is possible that the dilution curve has other shape 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 Should we use the remaining OD value?

Presentation ELISA plate with serially diluted IFN γ 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

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

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

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

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

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

Do the same procedure with the unknown sample data also

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

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

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

Linear trendline Display equation on chart (Options tab)

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

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

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)