Question

I had to make a calibration curve for the bradford dye-binding assay. My professor wanted us to fit the data to three different mathematical functions:

a straight line in the form of y=mx+b,

a second-order polynomial: y=a.x²+b*x+c

a rectangular hyperbola on the form:y=(a*x)/(b+x)

my results (I used the SciDavis program):

y=0.0526X+0.0353

y=0.00987+0.063X-0.000537x²

y=(5.07X)/(76.3+x)

I am supposed to use the graph i made to estimate the concentration of protein in my bacterial extract. So we set up 3 tubes of our E.coli extract as follows, adding 200 uL of dye to each tube after the bacteria and water were added. A₅₉₅ is the absorbance we obtained.

tube	uL E. coli extract	uL H20	A595
1	50	750	.1
2	100	700	.202
3	200	600	.340

I am not sure which absorbance I am supposed to use. I am thinking that it probably doesn't matter which one. I think I might have to multiply by a dilution factor? and i'm not sure which slope/formula to use. My data is actually pretty linear, and the R² (already .995.) didn't improve by much using the curved lines.

I belive that "y" is the absorbance value, and that I am solving for "x".

Answer 1

Lets start with the equation of straight line. y=0.0526X+0.0353
Y is the absorbance value. For test tube 1, it is 0.1
Hence, 0.1 =0.0526X+0.0353
X = 1.23
Here X can be micrograms of protein present per 100 microlitre of (or any other quantity) solution. It can be obtained from the data used for making the caliberation curve which I do not have.
Test tube 1 is prepared by diluting 50 microlitre of E.coli extract to total volume of 1000 microlitre. For undiluted E. coli extract, the value of X will be

Similarly we can calculate X value for diluted and undiluted E. coli extract for test tube 2 and test tube 3.

Now we can take average of the values of X for undiluted E.coli extract for test tubes 1, 2 and 3. This is the required answer.

Now sampe process should be repreated for a second-order polynomial and a rectangular hyperbola.