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Determination of the concentrationA 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 !!!PARTIAL TRUTH !!!

concentration

10

00

50

0

25

0

12

5

62

31

16

7.8

3.9

1.9

0.9

7

0.4

9

0.2

4

0.1

2

0.0

30

0.0

61

0.0

07

0.0

15

0.0

04

0

The sample with unknown concentration

ODThe 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

This region could indicate the concentration

ODEstimating the concentration with a „ruler”

conc. of the standard (µg/ml)

The

OD

are

pro

port

iona

l with

the

co

ncen

trat

ions

in t

his

ran

ge

2X 4X 8X 16X 32X 64X 128X 256XDilutions of the unknown sample

10

00

50

0

25

0

12

5

62

31

16

7.8

3.9

1.9

0.9

7

0.4

9

0.2

4

0.1

2

0.0

30

0.0

61

0.0

07

0.0

15

0.0

04

0.0

02

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!))

2X 4X 8X 16X 32X 64X 128X 256XDilutions of the unknown sample

10

00

50

0

25

0

12

5

62

31

16

7.8

3.9

1.9

0.9

7

0.4

9

0.2

4

0.1

2

0.0

30

0.0

61

0.0

07

0.0

15

0.0

04

0.0

02

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

ystd=mx+b

Ysample=mx+b

Serious errorsYou 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!

0

1

2

3

4

5

6

1 10 100 1000 10000hígítás

OD

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

0

1

2

3

4

5

6

0 500 1000 1500 2000 2500hígítás

OD

1 2 4 8 16 32 64 128 256 512 1024 2048

Dilution:

0 1 2 3 4 5 6 7 8 9 10 11

dilution 2x:

0

1

2

3

4

5

6

1 10 100 1000 10000hígítás

OD

0

1

2

3

4

5

6

0 2 4 6 8 10 12hígítás 2x

OD

=

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.

1x 2x 4x 8x 16x 32x 64x 128x

OD

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5 standard (1x: 100 g/ml)

dilution

Dilutions of the unknown

dilution 1x 10x 100x

OD 4.0 2.1 1.1

1.1Value 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.

1x 2x 4x 8x 16x 32x 64x 128x

OD

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5 standard (1x: 100mg/ml)

dilution

Dilutions of the unknown

dilution 1x 10x 100x

OD 4.0 2.1 1.1

4.0 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?

PresentationELISA 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

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

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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 = -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

You can calculate the dilutions of different OD solutions with the equations

The sample is 2113489/22856= 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)